diff --git "a/OE_MM_maths_en_COMP.tsv" "b/OE_MM_maths_en_COMP.tsv" new file mode 100644--- /dev/null +++ "b/OE_MM_maths_en_COMP.tsv" @@ -0,0 +1,1034 @@ +id question solution final_answer context image modality difficulty is_multiple_answer unit answer_type error question_type subfield subject language +0 "Turbo the snail sits on a point on a circle with circumference 1. Given an infinite sequence of positive real numbers $c_{1}, c_{2}, c_{3}, \ldots$. Turbo successively crawls distances $c_{1}, c_{2}, c_{3}, \ldots$ around the circle, each time choosing to crawl either clockwise or counterclockwise. + +For example, if the sequence $c_{1}, c_{2}, c_{3}, \ldots$ is $0.4,0.6,0.3, \ldots$, then Turbo may start crawling as follows: + + +Determine the largest constant $C>0$ with the following property: for every sequence of positive real numbers $c_{1}, c_{2}, c_{3}, \ldots$ with $c_{i}\\frac{1}{2}$, we write $C=\\frac{1}{2}+a$ with $a>0$, and we choose the sequence\n\n$$\n\\frac{1}{2}, \\quad \\frac{1+a}{2}, \\quad \\frac{1}{2}, \\quad \\frac{1+a}{2}, \\quad \\frac{1}{2}, \\ldots\n$$\n\nIn other words, $c_{i}=\\frac{1}{2}$ if $i$ is odd and $c_{i}=\\frac{1+a}{2}1$ for all $i$. Therefore, we are left with the case that Turbo alternates crawling clockwise and crawling counterclockwise. If it, without loss of generality, starts by going clockwise, then it will always crawl a distance $\\frac{1}{2}$ clockwise followed by a distance $\\frac{1+a}{2}$ counterclockwise. The net effect is that it crawls a distance $\\frac{a}{2}$ counterclockwise. Because $\\frac{a}{2}$ is positive, there exists a positive integer $N$ such that $\\frac{a}{2} \\cdot N>1$. After $2 N$ crawls, Turbo will have crawled a distance $\\frac{a}{2}$ counterclockwise $N$ times, therefore having covered a total distance of $\\frac{a}{2} \\cdot N>1$, meaning that it must have crawled over all points on the circle.\n\nNote: Every sequence of the form $c_{i}=x$ if $i$ is odd, and $c_{i}=y$ if $i$ is even, where $0\n\nFigure 5: Chet and Turbo equivalence\n\nClaim: $M \\geqslant 2$.\n\nProof. Suppose not, so $M<2$. Say $M=2-2 \\varepsilon$ for some $\\varepsilon>0$ and let $[-1+\\varepsilon, 1-\\varepsilon]$ be a minimal interval, that Chet cannot be forced out of. Then we can force Chet arbitrarily close to $\\pm(1-\\varepsilon)$. In partiular, we can force Chet out of $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$ by minimality of $M$. This means that there exists a sequence $d_{1}, d_{2}, \\ldots$ for which Chet has to leave $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$, which means he ends up either in the interval $\\left[-1+\\varepsilon,-1+\\frac{4}{3} \\varepsilon\\right)$ or in the interval $\\left(1-\\frac{4}{3} \\varepsilon, 1-\\varepsilon\\right]$.\n\nNow consider the sequence,\n\n$$\nd_{1}, 1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon, d_{2}, 1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon, d_{3}, \\ldots\n$$\n\nobtained by adding the sequence $1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon$ in between every two steps. We claim that this sequence forces Chet to leave the larger interval $[-1+\\varepsilon, 1-\\varepsilon]$. Indeed no two consecutive elements in the sequence $1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon$ can have the same sign, because the sum of any two consecutive terms is larger than $2-2 \\varepsilon$ and Chet would leave the interval $[-1+\\varepsilon, 1-\\varepsilon]$. It follows that the $\\left(1-\\frac{7}{6} \\varepsilon\\right)$ 's and the $\\left(1-\\frac{2}{3} \\varepsilon\\right)$ 's cancel out, so the position after $d_{k}$ is the same as before $d_{k+1}$. Hence, the positions after each $d_{k}$ remain the same as in the original sequence. Thus, Chet is also forced to the boundary in the new sequence.\n\nIf Chet is outside the interval $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$, then Chet has to move $1-\\frac{7}{6} \\varepsilon$ towards 0 , and ends in $\\left[-\\frac{1}{6} \\varepsilon, \\frac{1}{6} \\varepsilon\\right]$. Chet then has to move by $1-\\frac{2}{3} \\varepsilon$, which means that he has to leave the interval $[-1+\\varepsilon, 1-\\varepsilon]$. Indeed the absolute value of the final position is at least $1-\\frac{5}{6} \\varepsilon$. This contradicts the assumption, that we cannot force Chet out of $[-1+\\varepsilon, 1-\\varepsilon]$. Hence $M \\geqslant 2$ as needed.""]" ['$\\frac{1}{2}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +1 "In the diagram, $\angle A B F=41^{\circ}, \angle C B F=59^{\circ}, D E$ is parallel to $B F$, and $E F=25$. If $A E=E C$, determine the length of $A E$, to 2 decimal places. + +" ['Let the length of $A E=E C$ be $x$.\n\nThen $A F=x-25$.\n\nIn, $\\triangle B C F, \\frac{x+25}{B F}=\\tan \\left(59^{\\circ}\\right)$.\n\nIn $\\triangle A B F, \\frac{x-25}{B F}=\\tan \\left(41^{\\circ}\\right)$.\n\nSolving for $B F$ in these two equations and equating,\n\n$$\nB F=\\frac{x+25}{\\tan 59^{\\circ}}=\\frac{x-25}{\\tan 41^{\\circ}}\n$$\n\nso $\\quad\\left(\\tan 41^{\\circ}\\right)(x+25)=\\left(\\tan 59^{\\circ}\\right)(x-25)$\n\n$$\n\\begin{aligned}\n25\\left(\\tan 59^{\\circ}+\\tan 41^{\\circ}\\right) & =x\\left(\\tan 59^{\\circ}-\\tan 41^{\\circ}\\right) \\\\\nx & =\\frac{25\\left(\\tan 59^{\\circ}+\\tan 41^{\\circ}\\right)}{\\tan 59^{\\circ}-\\tan 41^{\\circ}} \\\\\nx & \\doteq 79.67 .\n\\end{aligned}\n$$\n\n\n\nTherefore the length of $A E$ is 79.67 .'] ['79.67'] 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Multimodal Competition False Numerical 1e-1 Open-ended Geometry Math English +2 "In triangle $A B C, A B=B C=25$ and $A C=30$. The circle with diameter $B C$ intersects $A B$ at $X$ and $A C$ at $Y$. Determine the length of $X Y$. + +" "['Join $B Y$. Since $B C$ is a diameter, then $\\angle B Y C=90^{\\circ}$. Since $A B=B C, \\triangle A B C$ is isosceles and $B Y$ is an altitude in $\\triangle A B C$, then $A Y=Y C=15$.\n\nLet $\\angle B A C=\\theta$.\n\nSince $\\triangle A B C$ is isosceles, $\\angle B C A=\\theta$.\n\nSince $B C Y X$ is cyclic, $\\angle B X Y=180-\\theta$ and so $\\angle A X Y=\\theta$.\n\n\n\nThus $\\triangle A X Y$ is isosceles and so $X Y=A Y=15$.\n\nTherefore $X Y=15$.' + 'Join $B Y . \\angle B Y C=90^{\\circ}$ since it is inscribed in a semicircle.\n\n$\\triangle B A C$ is isosceles, so altitude $B Y$ bisects the base.\n\nTherefore $B Y=\\sqrt{25^{2}-15^{2}}=20$.\n\nJoin $C X . \\angle C X B=90^{\\circ}$ since it is also inscribed in a semicircle.\n\n\n\nThe area of $\\triangle A B C$ is\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(A C)(B Y) & =\\frac{1}{2}(A B)(C X) \\\\\n\\frac{1}{2}(30)(20) & =\\frac{1}{2}(25)(C X) \\\\\nC X & =\\frac{600}{25}=24 .\n\\end{aligned}\n$$\n\nFrom $\\triangle A B Y$ we conclude that $\\cos \\angle A B Y=\\frac{B Y}{A B}=\\frac{20}{25}=\\frac{4}{5}$.\n\nIn $\\Delta B X Y$, applying the Law of Cosines we get $(X Y)^{2}=(B X)^{2}+(B Y)^{2}-2(B X)(B Y) \\cos \\angle X B Y$.\n\nNow (by Pythagoras $\\triangle B X C$ ),\n\n\n\n$$\n\\begin{aligned}\nB X^{2} & =B C^{2}-C X^{2} \\\\\n& =25^{2}-24^{2} \\\\\n& =49 \\\\\nB X & =7 .\n\\end{aligned}\n$$\n\nTherefore $X Y^{2}=7^{2}+20^{2}-2(7)(20) \\frac{4}{5}$\n\n$$\n=49+400-224\n$$\n\n$$\n=225 \\text {. }\n$$\n\nTherefore $X Y=15$.']" ['15'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +3 "Points $P$ and $Q$ are located inside the square $A B C D$ such that $D P$ is parallel to $Q B$ and $D P=Q B=P Q$. Determine the minimum possible value of $\angle A D P$. + +" "['Placing the information on the coordinate axes, the diagram is indicated to the right.\n\nWe note that $P$ has coordinates $(a, b)$.\n\nBy symmetry (or congruency) we can label lengths $a$ and $b$ as shown. Thus $Q$ has coordinates $(2-a, 2-b)$.\n\nSince $P D=P Q, a^{2}+b^{2}=(2-2 a)^{2}+(2-2 b)^{2}$\n\n$$\n\\begin{aligned}\n& 3 a^{2}+3 b^{2}-8 a-8 b+8=0 \\\\\n& \\left(a-\\frac{4}{3}\\right)^{2}+\\left(b-\\frac{4}{3}\\right)^{2}=\\frac{8}{9}\n\\end{aligned}\n$$\n\n\n\n$P$ is on a circle with centre $O\\left(\\frac{4}{3}, \\frac{4}{3}\\right)$ with $r=\\frac{2}{3} \\sqrt{2}$.\n\nThe minimum angle for $\\theta$ occurs when $D P$ is tangent to the circle.\n\nSo we have the diagram noted to the right.\n\nSince $O D$ makes an angle of $45^{\\circ}$ with the $x$-axis then $\\angle P D O=45-\\theta$ and $O D=\\frac{4}{3} \\sqrt{2}$.\n\nTherefore $\\sin (45-\\theta)=\\frac{\\frac{2}{3} \\sqrt{2}}{\\frac{4}{3} \\sqrt{2}}=\\frac{1}{2}$ which means $45^{\\circ}-\\theta=30^{\\circ}$ or $\\theta=15^{\\circ}$.\n\nThus the minimum value for $\\theta$ is $15^{\\circ}$.\n\n' + 'Let $A B=B C=C D=D A=1$.\n\nJoin $D$ to $B$. Let $\\angle A D P=\\theta$. Therefore, $\\angle P D B=45-\\theta$.\n\nLet $P D=a$ and $P B=b$ and $P Q=\\frac{a}{2}$.\n\n\n\nWe now establish a relationship between $a$ and $b$.\n\nIn $\\triangle P D B, b^{2}=a^{2}+2-2(a)(\\sqrt{2}) \\cos (45-\\theta)$\n\n$$\n\\text { or, } \\quad \\cos (45-\\theta)=\\frac{a^{2}-b^{2}+2}{2 \\sqrt{2} a}\n$$\n\n\n\nIn $\\triangle P D R,\\left(\\frac{a}{2}\\right)^{2}=a^{2}+\\left(\\frac{\\sqrt{2}}{2}\\right)^{2}-2 a \\frac{\\sqrt{2}}{2} \\cos (45-\\theta)$\n\nor, $\\cos (45-\\theta)=\\frac{\\frac{3}{4} a^{2}+\\frac{1}{2}}{a \\sqrt{2}}$\n\nComparing (1) and (2) gives, $\\frac{a^{2}-b^{2}+2}{2 \\sqrt{2} a}=\\frac{\\frac{3}{4} a^{2}+\\frac{1}{2}}{a \\sqrt{2}}$.\n\nSimplifying this, $a^{2}+2 b^{2}=2$\n\n$$\n\\text { or, } \\quad b^{2}=\\frac{2-a^{2}}{2}\n$$\n\nNow $\\cos (45-\\theta)=\\frac{a^{2}+2-\\left(\\frac{2-a^{2}}{2}\\right)}{2 a \\sqrt{2}}=\\frac{1}{4 \\sqrt{2}}\\left(3 a+\\frac{2}{a}\\right)$.\n\nNow considering $3 a+\\frac{2}{a}$, we know $\\left(\\sqrt{3 a}-\\sqrt{\\frac{2}{a}}\\right)^{2} \\geq 0$\n\n$$\n\\text { or, } \\quad 3 a+\\frac{2}{a} \\geq 2 \\sqrt{6}\n$$\n\nThus, $\\cos (45-\\theta) \\geq \\frac{1}{4 \\sqrt{2}}(2 \\sqrt{6})=\\frac{\\sqrt{3}}{2}$\n\n$$\n\\cos (45-\\theta) \\geq \\frac{\\sqrt{3}}{2}\n$$\n\n$\\cos (45-\\theta)$ has a minimum value for $45^{\\circ}-\\theta=30^{\\circ}$ or $\\theta=15^{\\circ}$.' + 'Join $B D$. Let $B D$ meet $P Q$ at $M$. Let $\\angle A D P=\\theta$.\n\nBy interior alternate angles, $\\angle P=\\angle Q$ and $\\angle P D M=\\angle Q B M$.\n\nThus $\\triangle P D M \\cong \\triangle Q B M$ by A.S.A., so $P M=Q M$ and $D M=B M$.\n\nSo $M$ is the midpoint of $B D$ and the centre of the square.\n\n\n\nWithout loss of generality, let $P M=1$. Then $P D=2$.\n\nSince $\\theta+\\alpha=45^{\\circ}$ (see diagram), $\\theta$ will be minimized when $\\alpha$ is maximized.\n\n\n\nConsider $\\triangle P M D$.\n\nUsing the sine law, $\\frac{\\sin \\alpha}{1}=\\frac{\\sin (\\angle P M D)}{2}$.\n\nTo maximize $\\alpha$, we maximize $\\sin \\alpha$.\n\nBut $\\sin \\alpha=\\frac{\\sin (\\angle P M D)}{2}$, so it is maximized when $\\sin (\\angle P M D)=1$.\n\nIn this case, $\\sin \\alpha=\\frac{1}{2}$, so $\\alpha=30^{\\circ}$.\n\nTherefore, $\\theta=45^{\\circ}-30^{\\circ}=15^{\\circ}$, and so the minimum value of $\\theta$ is $15^{\\circ}$.' + 'We place the diagram on a coordinate grid, with $D(0,0)$, $C(1,0), B(0,1), A(1,1)$.\n\nLet $P D=P Q=Q B=a$, and $\\angle A D P=\\theta$.\n\nDrop a perpendicular from $P$ to $A D$, meeting $A D$ at $X$.\n\nThen $P X=a \\sin \\theta, D X=a \\cos \\theta$.\n\nTherefore the coordinates of $P$ are $(a \\sin \\theta, a \\cos \\theta)$.\n\nSince $P D \\| B Q$, then $\\angle Q B C=\\theta$.\n\nSo by a similar argument (or by using the fact that $P Q$ are symmetric through the centre of the square), the coordinates of $Q$ are $(1-a \\sin \\theta, 1+a \\cos \\theta)$.\n\n\n\nNow $(P Q)^{2}=a^{2}$, so $(1-2 a \\sin \\theta)^{2}+(1-2 a \\cos \\theta)^{2}=a^{2}$\n\n$$\n2+4 a^{2} \\sin ^{2} \\theta+4 a^{2} \\cos ^{2} \\theta-4 a(\\sin \\theta+\\cos \\theta)=a^{2}\n$$\n\n\n\n$$\n\\begin{aligned}\n2+4 a^{2}-a^{2} & =4 a(\\sin \\theta+\\cos \\theta) \\\\\n\\frac{2+3 a^{2}}{4 a} & =\\sin \\theta+\\cos \\theta \\\\\n\\frac{2+3 a^{2}}{4 \\sqrt{2} a} & =\\frac{1}{\\sqrt{2}} \\sin \\theta+\\frac{1}{\\sqrt{2}} \\cos \\theta=\\cos \\left(45^{\\circ}\\right) \\sin \\theta+\\sin \\left(45^{\\circ}\\right) \\cos \\theta \\\\\n\\frac{2+3 a^{2}}{4 \\sqrt{2} a} & =\\sin \\left(\\theta+45^{\\circ}\\right)\n\\end{aligned}\n$$\n\n$$\n\\text { Now } \\begin{aligned}\n\\left(a-\\sqrt{\\frac{2}{3}}\\right)^{2} & \\geq 0 \\\\\na^{2}-2 a \\sqrt{\\frac{2}{3}}+\\frac{2}{3} & \\geq 0 \\\\\n3 a^{2}-2 a \\sqrt{6}+2 & \\geq 0 \\\\\n3 a^{2}+2 & \\geq 2 a \\sqrt{6} \\\\\n\\frac{3 a^{2}+2}{4 \\sqrt{2} a} & \\geq \\frac{\\sqrt{3}}{2}\n\\end{aligned}\n$$\n\nand equality occurs when $a=\\sqrt{\\frac{2}{3}}$.\n\nSo $\\sin \\left(\\theta+45^{\\circ}\\right) \\geq \\frac{\\sqrt{3}}{2}$ and thus since $0^{\\circ} \\leq \\theta \\leq 90^{\\circ}$, then $\\theta+45^{\\circ} \\geq 60^{\\circ}$ or $\\theta \\geq 15^{\\circ}$.\n\nTherefore the minimum possible value of $\\angle A D P$ is $15^{\\circ}$.']" ['$15$'] 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Multimodal Competition False ^{\circ} Numerical Open-ended Geometry Math English +4 "In the diagram, $\angle E A D=90^{\circ}, \angle A C D=90^{\circ}$, and $\angle A B C=90^{\circ}$. Also, $E D=13, E A=12$, $D C=4$, and $C B=2$. Determine the length of $A B$. + +" ['By the Pythagorean Theorem in $\\triangle E A D$, we have $E A^{2}+A D^{2}=E D^{2}$ or $12^{2}+A D^{2}=13^{2}$, and so $A D=\\sqrt{169-144}=5$, since $A D>0$.\n\nBy the Pythagorean Theorem in $\\triangle A C D$, we have $A C^{2}+C D^{2}=A D^{2}$ or $A C^{2}+4^{2}=5^{2}$, and so $A C=\\sqrt{25-16}=3$, since $A C>0$.\n\n(We could also have determined the lengths of $A D$ and $A C$ by recognizing 3-4-5 and 5-12-13 right-angled triangles.)\n\nBy the Pythagorean Theorem in $\\triangle A B C$, we have $A B^{2}+B C^{2}=A C^{2}$ or $A B^{2}+2^{2}=3^{2}$, and so $A B=\\sqrt{9-4}=\\sqrt{5}$, since $A B>0$.'] ['$\\sqrt{5}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +5 "In the diagram, $A B C D$ is a quadrilateral with $A B=B C=C D=6, \angle A B C=90^{\circ}$, and $\angle B C D=60^{\circ}$. Determine the length of $A D$. + +" "['Join $B$ to $D$.\n\n\n\nConsider $\\triangle C B D$.\n\nSince $C B=C D$, then $\\angle C B D=\\angle C D B=\\frac{1}{2}\\left(180^{\\circ}-\\angle B C D\\right)=\\frac{1}{2}\\left(180^{\\circ}-60^{\\circ}\\right)=60^{\\circ}$.\n\nTherefore, $\\triangle B C D$ is equilateral, and so $B D=B C=C D=6$.\n\nConsider $\\triangle D B A$.\n\nNote that $\\angle D B A=90^{\\circ}-\\angle C B D=90^{\\circ}-60^{\\circ}=30^{\\circ}$.\n\nSince $B D=B A=6$, then $\\angle B D A=\\angle B A D=\\frac{1}{2}\\left(180^{\\circ}-\\angle D B A\\right)=\\frac{1}{2}\\left(180^{\\circ}-30^{\\circ}\\right)=75^{\\circ}$.\n\nWe calculate the length of $A D$.\n\nMethod 1\n\nBy the Sine Law in $\\triangle D B A$, we have $\\frac{A D}{\\sin (\\angle D B A)}=\\frac{B A}{\\sin (\\angle B D A)}$.\n\nTherefore, $A D=\\frac{6 \\sin \\left(30^{\\circ}\\right)}{\\sin \\left(75^{\\circ}\\right)}=\\frac{6 \\times \\frac{1}{2}}{\\sin \\left(75^{\\circ}\\right)}=\\frac{3}{\\sin \\left(75^{\\circ}\\right)}$.\n\nMethod 2\n\nIf we drop a perpendicular from $B$ to $P$ on $A D$, then $P$ is the midpoint of $A D$ since $\\triangle B D A$ is isosceles. Thus, $A D=2 A P$.\n\nAlso, $B P$ bisects $\\angle D B A$, so $\\angle A B P=15^{\\circ}$.\n\nNow, $A P=B A \\sin (\\angle A B P)=6 \\sin \\left(15^{\\circ}\\right)$.\n\nTherefore, $A D=2 A P=12 \\sin \\left(15^{\\circ}\\right)$.\n\nMethod 3\n\nBy the Cosine Law in $\\triangle D B A$,\n\n$$\n\\begin{aligned}\nA D^{2} & =A B^{2}+B D^{2}-2(A B)(B D) \\cos (\\angle A B D) \\\\\n& =6^{2}+6^{2}-2(6)(6) \\cos \\left(30^{\\circ}\\right) \\\\\n& =72-72\\left(\\frac{\\sqrt{3}}{2}\\right) \\\\\n& =72-36 \\sqrt{3}\n\\end{aligned}\n$$\n\nTherefore, $A D=\\sqrt{36(2-\\sqrt{3})}=6 \\sqrt{2-\\sqrt{3}}$ since $A D>0$.' + 'Drop perpendiculars from $D$ to $Q$ on $B C$ and from $D$ to $R$ on $B A$.\n\n\n\nThen $C Q=C D \\cos (\\angle D C Q)=6 \\cos \\left(60^{\\circ}\\right)=6 \\times \\frac{1}{2}=3$.\n\nAlso, $D Q=C D \\sin (\\angle D C Q)=6 \\sin \\left(60^{\\circ}\\right)=6 \\times \\frac{\\sqrt{3}}{2}=3 \\sqrt{3}$.\n\nSince $B C=6$, then $B Q=B C-C Q=6-3=3$.\n\nNow quadrilateral $B Q D R$ has three right angles, so it must have a fourth right angle and so must be a rectangle.\n\nThus, $R D=B Q=3$ and $R B=D Q=3 \\sqrt{3}$.\n\nSince $A B=6$, then $A R=A B-R B=6-3 \\sqrt{3}$.\n\nSince $\\triangle A R D$ is right-angled at $R$, then using the Pythagorean Theorem and the fact that $A D>0$, we obtain\n\n$$\nA D=\\sqrt{R D^{2}+A R^{2}}=\\sqrt{3^{2}+(6-3 \\sqrt{3})^{2}}=\\sqrt{9+36-36 \\sqrt{3}+27}=\\sqrt{72-36 \\sqrt{3}}\n$$\n\nwhich we can rewrite as $A D=\\sqrt{36(2-\\sqrt{3})}=6 \\sqrt{2-\\sqrt{3}}$.']" ['$6\\sqrt{2-\\sqrt{3}}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +6 "A triangle has vertices $A(0,3), B(4,0)$, $C(k, 5)$, where $0" "['We ""complete the rectangle"" by drawing a horizontal line through $C$ which meets the $y$-axis at $P$ and the vertical line through $B$ at $Q$.\n\n\n\n\n\nSince $C$ has $y$-coordinate 5 , then $P$ has $y$-coordinate 5 ; thus the coordinates of $P$ are $(0,5)$.\n\nSince $B$ has $x$-coordinate 4 , then $Q$ has $x$-coordinate 4 .\n\nSince $C$ has $y$-coordinate 5 , then $Q$ has $y$-coordinate 5 .\n\nTherefore, the coordinates of $Q$ are $(4,5)$, and so rectangle $O P Q B$ is 4 by 5 and so has area $4 \\times 5=20$.\n\nNow rectangle $O P Q B$ is made up of four smaller triangles, and so the sum of the areas of these triangles must be 20 .\n\nLet us examine each of these triangles:\n\n- $\\triangle A B C$ has area 8 (given information)\n- $\\triangle A O B$ is right-angled at $O$, has height $A O=3$ and base $O B=4$, and so has area $\\frac{1}{2} \\times 4 \\times 3=6$.\n- $\\triangle A P C$ is right-angled at $P$, has height $A P=5-3=2$ and base $P C=k-0=k$, and so has area $\\frac{1}{2} \\times k \\times 2=k$.\n- $\\triangle C Q B$ is right-angled at $Q$, has height $Q B=5-0=5$ and base $C Q=4-k$, and so has area $\\frac{1}{2} \\times(4-k) \\times 5=10-\\frac{5}{2} k$.\n\nSince the sum of the areas of these triangles is 20 , then $8+6+k+10-\\frac{5}{2} k=20$ or $4=\\frac{3}{2} k$ and so $k=\\frac{8}{3}$.']" ['$\\frac{8}{3}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +7 "A helicopter hovers at point $H$, directly above point $P$ on level ground. Lloyd sits on the ground at a point $L$ where $\angle H L P=60^{\circ}$. A ball is droppped from the helicopter. When the ball is at point $B, 400 \mathrm{~m}$ directly below the helicopter, $\angle B L P=30^{\circ}$. What is the distance between $L$ and $P$ ? + +" "['Since $\\angle H L P=60^{\\circ}$ and $\\angle B L P=30^{\\circ}$, then $\\angle H L B=\\angle H L P-\\angle B L P=30^{\\circ}$.\n\nAlso, since $\\angle H L P=60^{\\circ}$ and $\\angle H P L=90^{\\circ}$, then $\\angle L H P=180^{\\circ}-90^{\\circ}-60^{\\circ}=30^{\\circ}$.\n\n\n\nTherefore, $\\triangle H B L$ is isosceles and $B L=H B=400 \\mathrm{~m}$.\n\nIn $\\triangle B L P, B L=400 \\mathrm{~m}$ and $\\angle B L P=30^{\\circ}$, so $L P=B L \\cos \\left(30^{\\circ}\\right)=400\\left(\\frac{\\sqrt{3}}{2}\\right)=200 \\sqrt{3}$ m.\n\nTherefore, the distance between $L$ and $P$ is $200 \\sqrt{3} \\mathrm{~m}$.' + 'Since $\\angle H L P=60^{\\circ}$ and $\\angle B L P=30^{\\circ}$, then $\\angle H L B=\\angle H L P-\\angle B L P=30^{\\circ}$.\n\nAlso, since $\\angle H L P=60^{\\circ}$ and $\\angle H P L=90^{\\circ}$, then $\\angle L H P=180^{\\circ}-90^{\\circ}-60^{\\circ}=30^{\\circ}$. Also, $\\angle L B P=60^{\\circ}$.\n\nLet $L P=x$.\n\n\n\nSince $\\triangle B L P$ is $30^{\\circ}-60^{\\circ}-90^{\\circ}$, then $B P: L P=1: \\sqrt{3}$, so $B P=\\frac{1}{\\sqrt{3}} L P=\\frac{1}{\\sqrt{3}} x$.\n\n\n\nSince $\\triangle H L P$ is $30^{\\circ}-60^{\\circ}-90^{\\circ}$, then $H P: L P=\\sqrt{3}: 1$, so $H P=\\sqrt{3} L P=\\sqrt{3} x$.\n\nBut $H P=H B+B P$ so\n\n$$\n\\begin{aligned}\n\\sqrt{3} x & =400+\\frac{1}{\\sqrt{3}} x \\\\\n3 x & =400 \\sqrt{3}+x \\\\\n2 x & =400 \\sqrt{3} \\\\\nx & =200 \\sqrt{3}\n\\end{aligned}\n$$\n\nTherefore, the distance from $L$ to $P$ is $200 \\sqrt{3} \\mathrm{~m}$.']" ['$200 \\sqrt{3}$'] 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Multimodal Competition False m Numerical Open-ended Geometry Math English +8 "In the diagram, $A B C D$ is a quadrilateral in which $\angle A+\angle C=180^{\circ}$. What is the length of $C D$ ? + +" ['In order to determine $C D$, we must determine one of the angles (or at least some information about one of the angles) in $\\triangle B C D$.\n\nTo do this, we look at $\\angle A$ use the fact that $\\angle A+\\angle C=180^{\\circ}$.\n\n\n\n\n\nUsing the cosine law in $\\triangle A B D$, we obtain\n\n$$\n\\begin{aligned}\n7^{2} & =5^{2}+6^{2}-2(5)(6) \\cos (\\angle A) \\\\\n49 & =61-60 \\cos (\\angle A) \\\\\n\\cos (\\angle A) & =\\frac{1}{5}\n\\end{aligned}\n$$\n\nSince $\\cos (\\angle A)=\\frac{1}{5}$ and $\\angle A+\\angle C=180^{\\circ}$, then $\\cos (\\angle C)=-\\cos \\left(180^{\\circ}-\\angle A\\right)=-\\frac{1}{5}$.\n\n(We could have calculated the actual size of $\\angle A$ using $\\cos (\\angle A)=\\frac{1}{5}$ and then used this to calculate the size of $\\angle C$, but we would introduce the possibility of rounding error by doing this.)\n\nThen, using the cosine law in $\\triangle B C D$, we obtain\n\n$$\n\\begin{aligned}\n7^{2} & =4^{2}+C D^{2}-2(4)(C D) \\cos (\\angle C) \\\\\n49 & =16+C D^{2}-8(C D)\\left(-\\frac{1}{5}\\right) \\\\\n0 & =5 C D^{2}+8 C D-165 \\\\\n0 & =(5 C D+33)(C D-5)\n\\end{aligned}\n$$\n\nSo $C D=-\\frac{33}{5}$ or $C D=5$. (We could have also determined these roots using the quadratic formula.)\n\nSince $C D$ is a length, it must be positive, so $C D=5$.\n\n(We could have also proceeded by using the sine law in $\\triangle B C D$ to determine $\\angle B D C$ and then found the size of $\\angle D B C$, which would have allowed us to calculate $C D$ using the sine law. However, this would again introduce the potential of rounding error.)'] ['5'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +9 "In the diagram, the parabola + +$$ +y=-\frac{1}{4}(x-r)(x-s) +$$ + +intersects the axes at three points. The vertex of this parabola is the point $V$. Determine the value of $k$ and the coordinates of $V$. + +" ['From the diagram, the $x$-intercepts of the parabola are $x=-k$ and $x=3 k$.\n\n\n\n\n\nSince we are given that $y=-\\frac{1}{4}(x-r)(x-s)$, then the $x$-intercepts are $r$ and $s$, so $r$ and $s$ equal $-k$ and $3 k$ in some order.\n\nTherefore, we can rewrite the parabola as $y=-\\frac{1}{4}(x-(-k))(x-3 k)$.\n\nSince the point $(0,3 k)$ lies on the parabola, then $3 k=-\\frac{1}{4}(0+k)(0-3 k)$ or $12 k=3 k^{2}$ or $k^{2}-4 k=0$ or $k(k-4)=0$.\n\nThus, $k=0$ or $k=4$.\n\nSince the two roots are distinct, then we cannot have $k=0$ (otherwise both $x$-intercepts would be 0 ).\n\nThus, $k=4$.\n\nThis tells us that the equation of the parabola is $y=-\\frac{1}{4}(x+4)(x-12)$ or $y=-\\frac{1}{4} x^{2}+$ $2 x+12$.\n\nWe still have to determine the coordinates of the vertex, $V$.\n\nSince the $x$-intercepts of the parabola are -4 and 12 , then the $x$-coordinate of the vertex is the average of these intercepts, or 4.\n\n(We could have also used the fact that the $x$-coordinate is $-\\frac{b}{2 a}=-\\frac{2}{2\\left(-\\frac{1}{4}\\right)}$.)\n\nTherefore, the $y$-coordinate of the vertex is $y=-\\frac{1}{4}\\left(4^{2}\\right)+2(4)+12=16$.\n\nThus, the coordinates of the vertex are $(4,16)$.'] ['$4,(4,16)$'] 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Multimodal Competition True Numerical,Tuple Open-ended Geometry Math English +10 "A school has a row of $n$ open lockers, numbered 1 through $n$. After arriving at school one day, Josephine starts at the beginning of the row and closes every second locker until reaching the end of the row, as shown in the example below. Then on her way back, she closes every second locker that is still open. She continues in this manner along the row, until only one locker remains open. Define $f(n)$ to be the number of the last open locker. For example, if there are 15 lockers, then $f(15)=11$ as shown below: + + +Determine $f(50)$." ['We proceed directly.\n\nOn the first pass from left to right, Josephine closes all of the even numbered lockers, leaving the odd ones open.\n\nThe second pass proceeds from right to left. Before the pass, the lockers which are open are $1,3, \\ldots, 47,49$.\n\nOn the second pass, she shuts lockers 47, 43, 39, .., 3 .\n\nThe third pass proceeds from left to right. Before the pass, the lockers which are open are $1,5, \\ldots, 45,49$.\n\nOn the third pass, she shuts lockers $5,13, \\ldots, 45$.\n\nThis leaves lockers 1, 9, 17, 25, 33, 41, 49 open.\n\nOn the fourth pass, from right to left, lockers 41, 25 and 9 are shut, leaving 1, 17, 33, 49.\n\nOn the fifth pass, from left to right, lockers 17 and 49 are shut, leaving 1 and 33 open.\n\nOn the sixth pass, from right to left, locker 1 is shut, leaving 33 open.\n\nThus, $f(50)=33$.'] ['33'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +11 "In the diagram, $P Q R S$ is a quadrilateral. What is its perimeter? + +" ['The length of $P Q$ is equal to $\\sqrt{(0-5)^{2}+(12-0)^{2}}=\\sqrt{(-5)^{2}+12^{2}}=13$.\n\nIn a similar way, we can see that $Q R=R S=S P=13$.\n\nTherefore, the perimeter of $P Q R S$ is $4 \\cdot 13=52$.\n\n(We can also see that if $O$ is the origin, then $\\triangle P O Q, \\triangle P O S, \\triangle R O Q$, and $\\triangle R O S$ are congruent because $O Q=O S$ and $O P=O R$, which means that $P Q=Q R=R S=S P$.)'] ['52'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +12 "In the diagram, $A$ has coordinates $(0,8)$. Also, the midpoint of $A B$ is $M(3,9)$ and the midpoint of $B C$ is $N(7,6)$. What is the slope of $A C$ ? + +" "['Suppose that $B$ has coordinates $(r, s)$ and $C$ has coordinates $(t, u)$.\n\nSince $M(3,9)$ is the midpoint of $A(0,8)$ and $B(r, s)$, then 3 is the average of 0 and $r$ (which gives $r=6)$ and 9 is the average of 8 and $s$ (which gives $s=10$ ).\n\nSince $N(7,6)$ is the midpoint of $B(6,10)$ and $C(t, u)$, then 7 is the average of 6 and $t$ (which gives $t=8$ ) and 6 is the average of 10 and $u$ (which gives $u=2$ ).\n\nThe slope of the line segment joining $A(0,8)$ and $C(8,2)$ is $\\frac{8-2}{0-8}$ which equals $-\\frac{3}{4}$.' + 'Since $M$ is the midpoint of $A B$ and $N$ is the midpoint of $B C$, then $M N$ is parallel to $A C$. Therefore, the slope of $A C$ equals the slope of the line segment joining $M(3,9)$ to $N(7,6)$, which is $\\frac{9-6}{3-7}$ or $-\\frac{3}{4}$.']" ['$-\\frac{3}{4}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +13 "In the diagram, $A B D E$ is a rectangle, $\triangle B C D$ is equilateral, and $A D$ is parallel to $B C$. Also, $A E=2 x$ for some real number $x$. + + +Determine the length of $A B$ in terms of $x$." ['We begin by determining the length of $A B$ in terms of $x$.\n\nSince $A B D E$ is a rectangle, $B D=A E=2 x$.\n\nSince $\\triangle B C D$ is equilateral, $\\angle D B C=60^{\\circ}$.\n\nJoin $A$ to $D$.\n\n\n\nSince $A D$ and $B C$ are parallel, $\\angle A D B=\\angle D B C=60^{\\circ}$.\n\nConsider $\\triangle A D B$. This is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle since $\\angle A B D$ is a right angle.\n\nUsing ratios of side lengths, $\\frac{A B}{B D}=\\frac{\\sqrt{3}}{1}$ and so $A B=\\sqrt{3} B D=2 \\sqrt{3} x$'] ['$2 \\sqrt{3} x$'] 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Multimodal Competition False Expression Open-ended Geometry Math English +14 "In the diagram, $A B D E$ is a rectangle, $\triangle B C D$ is equilateral, and $A D$ is parallel to $B C$. Also, $A E=2 x$ for some real number $x$. + + +Determine positive integers $r$ and $s$ for which + +$$ +\frac{A C}{A D}=\sqrt{\frac{r}{s}} +$$" ['We begin by determining the length of $A B$ in terms of $x$.\n\nSince $A B D E$ is a rectangle, $B D=A E=2 x$.\n\nSince $\\triangle B C D$ is equilateral, $\\angle D B C=60^{\\circ}$.\n\nJoin $A$ to $D$.\n\n\n\nSince $A D$ and $B C$ are parallel, $\\angle A D B=\\angle D B C=60^{\\circ}$.\n\nConsider $\\triangle A D B$. This is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle since $\\angle A B D$ is a right angle.\n\nUsing ratios of side lengths, $\\frac{A B}{B D}=\\frac{\\sqrt{3}}{1}$ and so $A B=\\sqrt{3} B D=2 \\sqrt{3} x$\n\nNext, we determine $\\frac{A C}{A D}$.\n\nNow, $\\frac{A D}{B D}=\\frac{2}{1}$ and so $A D=2 B D=4 x$.\n\nSuppose that $M$ is the midpoint of $A E$ and $N$ is the midpoint of $B D$.\n\nSince $A E=B D=2 x$, then $A M=M E=B N=N D=x$.\n\nJoin $M$ to $N$ and $N$ to $C$ and $A$ to $C$.\n\n\n\nSince $A B D E$ is a rectangle, then $M N$ is parallel to $A B$ and so $M N$ is perpendicular to both $A E$ and $B D$.\n\nAlso, $M N=A B=2 \\sqrt{3} x$.\n\nSince $\\triangle B C D$ is equilateral, its median $C N$ is perpendicular to $B D$.\n\nSince $M N$ and $N C$ are perpendicular to $B D, M N C$ is actually a straight line segment and so $M C=M N+N C$.\n\nNow $\\triangle B N C$ is also a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, and so $N C=\\sqrt{3} B N=\\sqrt{3} x$.\n\nThis means that $M C=2 \\sqrt{3} x+\\sqrt{3} x=3 \\sqrt{3} x$.\n\n\n\nFinally, $\\triangle A M C$ is right-angled at $M$ and so\n\n$$\nA C=\\sqrt{A M^{2}+M C^{2}}=\\sqrt{x^{2}+(3 \\sqrt{3} x)^{2}}=\\sqrt{x^{2}+27 x^{2}}=\\sqrt{28 x^{2}}=2 \\sqrt{7} x\n$$\n\nsince $x>0$.\n\nThis means that $\\frac{A C}{A D}=\\frac{2 \\sqrt{7} x}{4 x}=\\frac{\\sqrt{7}}{2}=\\sqrt{\\frac{7}{4}}$, which means that the integers $r=7$ and $s=4$ satisfy the conditions.'] ['7,4'] 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Multimodal Competition True Numerical Open-ended Geometry Math English +15 "Five distinct integers are to be chosen from the set $\{1,2,3,4,5,6,7,8\}$ and placed in some order in the top row of boxes in the diagram. Each box that is not in the top row then contains the product of the integers in the two boxes connected to it in the row directly above. Determine the number of ways in which the integers can be chosen and placed in the top row so that the integer in the bottom box is 9953280000 . + +" ['Suppose that the integers in the first row are, in order, $a, b, c, d, e$.\n\nUsing these, we calculate the integer in each of the boxes below the top row in terms of these variables, using the rule that each integer is the product of the integers in the two boxes above:\n\n$a$\n\n| $b$ | $c$ | $c$ | | $d$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $a b^{2} c$ | $b c$ | $c d$ | $c$ | |\n| | $a b^{3} c^{3} d$ | $b c^{2} d$ | $c d^{2} e$ | |$\\quad d e$\n\nTherefore, $a b^{4} c^{6} d^{4} e=9953280000$.\n\n\n\nNext, we determine the prime factorization of the integer 9953280000 :\n\n$$\n\\begin{aligned}\n9953280000 & =10^{4} \\cdot 995328 \\\\\n& =2^{4} \\cdot 5^{4} \\cdot 2^{3} \\cdot 124416 \\\\\n& =2^{7} \\cdot 5^{4} \\cdot 2^{3} \\cdot 15552 \\\\\n& =2^{10} \\cdot 5^{4} \\cdot 2^{3} \\cdot 1944 \\\\\n& =2^{13} \\cdot 5^{4} \\cdot 2^{3} \\cdot 243 \\\\\n& =2^{16} \\cdot 5^{4} \\cdot 3^{5} \\\\\n& =2^{16} \\cdot 3^{5} \\cdot 5^{4}\n\\end{aligned}\n$$\n\nThus, $a b^{4} c^{6} d^{4} e=2^{16} \\cdot 3^{5} \\cdot 5^{4}$.\n\nSince the right side is not divisible by 7 , none of $a, b, c, d$, $e$ can equal 7 .\n\nThus, $a, b, c, d, e$ are five distinct integers chosen from $\\{1,2,3,4,5,6,8\\}$.\n\nThe only one of these integers divisible by 5 is 5 itself.\n\nSince $2^{16} \\cdot 3^{5} \\cdot 5^{4}$ includes exactly 4 factors of 5 , then either $b=5$ or $d=5$. No other placement of the 5 can give exactly 4 factors of 5 .\n\nCase 1: $b=5$\n\nHere, $a c^{6} d^{4} e=2^{16} \\cdot 3^{5}$ and $a, c, d, e$ are four distinct integers chosen from $\\{1,2,3,4,6,8\\}$. Since $a c^{6} d^{4} e$ includes exactly 5 factors of 3 and the possible values of $a, c, d$, e that are divisible by 3 are 3 and 6 , then either $d=3$ and one of $a$ and $e$ is 6 , or $d=6$ and one of $a$ and $e$ is 3 . No other placements of the multiples of 3 can give exactly 5 factors of 3 .\n\nCase 1a: $b=5, d=3, a=6$\n\nHere, $a \\cdot c^{6} \\cdot d^{4} \\cdot e=6 \\cdot c^{6} \\cdot 3^{4} \\cdot e=2 \\cdot 3^{5} \\cdot c^{6} \\cdot e$.\n\nThis gives $c^{6} e=2^{15}$ and $c$ and $e$ are distinct integers from $\\{1,2,4,8\\}$.\n\nTrying the four possible values of $c$ shows that $c=4$ and $e=8$ is the only solution in this case. Here, $(a, b, c, d, e)=(6,5,4,3,8)$.\n\nCase 1b: $b=5, d=3, e=6$ We obtain $(a, b, c, d, e)=(8,5,4,3,6)$.\n\nCase 1c: $b=5, d=6, a=3$\n\nHere, $a \\cdot c^{6} \\cdot d^{4} \\cdot e=3 \\cdot c^{6} \\cdot 6^{4} \\cdot e=2^{4} \\cdot 3^{5} \\cdot c^{6} \\cdot e$.\n\nThis gives $c^{6} e=2^{12}$ and $c$ and $e$ are distinct integers from $\\{1,2,4,8\\}$.\n\nTrying the four possible values of $c$ shows that $c=4$ and $e=1$ is the only solution in this case. Here, $(a, b, c, d, e)=(3,5,4,6,1)$.\n\nCase 1d: $b=5, d=6, e=3$ We obtain $(a, b, c, d, e)=(1,5,4,6,3)$.\n\nCase 2: $d=5$ : A similar analysis leads to 4 further quintuples $(a, b, c, d, e)$.\n\nTherefore, there are 8 ways in which the integers can be chosen and placed in the top row to obtain the desired integer in the bottom box.'] ['8'] 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Multimodal Competition False Numerical Open-ended Number Theory Math English +16 "In the diagram, eleven circles of four different radius 1, each circle labelled $X$ has radius 2, the circle labelled $Y$ has radius 4 , and the circle labelled $Z$ has radius $r$. Each of the circles labelled $W$ or $X$ is tangent to three other circles. The circle labelled $Y$ is tangent to all ten of the other circles. The circle labelled $Z$ is tangent to three other circles. Determine positive integers $s$ and $t$ for which $r=\frac{s}{t}$. + + +" ['We label the centres of the outer circles, starting with the circle labelled $Z$ and proceeding clockwise, as $A, B, C, D, E, F, G, H, J$, and $K$, and the centre of the circle labelled $Y$ as $L$.\n\n\n\nJoin $L$ to each of $A, B, C, D, E, F, G, H, J$, and $K$. Join $A$ to $B, B$ to $C, C$ to $D, D$ to $E, E$ to $F, F$ to $G, G$ to $H, H$ to $J, J$ to $K$, and $K$ to $A$.\n\nWhen two circles are tangent, the distance between their centres equals the sum of their radii.\n\nThus,\n\n$$\n\\begin{array}{r}\nB C=C D=D E=E F=F G=G H=H J=J K=2+1=3 \\\\\nB L=D L=F L=H L=K L=2+4=6 \\\\\nC L=E L=G L=J L=1+4=5 \\\\\nA B=A K=r+2 \\\\\nA L=r+4\n\\end{array}\n$$\n\nBy side-side-side congruence, the following triangles are congruent:\n\n$$\n\\triangle B L C, \\triangle D L C, \\triangle D L E, \\triangle F L E, \\triangle F L G, \\triangle H L G, \\triangle H L J, \\triangle K L J\n$$\n\nSimilarly, $\\triangle A L B$ and $\\triangle A L K$ are congruent by side-side-side.\n\nLet $\\angle A L B=\\theta$ and let $\\angle B L C=\\alpha$.\n\n\n\nBy congruent triangles, $\\angle A L K=\\theta$ and\n\n$$\n\\angle B L C=\\angle D L C=\\angle D L E=\\angle F L E=\\angle F L G=\\angle H L G=\\angle H L J=\\angle K L J=\\alpha\n$$\n\nThe angles around $L$ add to $360^{\\circ}$ and so $2 \\theta+8 \\alpha=360^{\\circ}$ which gives $\\theta+4 \\alpha=180^{\\circ}$ and so $\\theta=180^{\\circ}-4 \\alpha$.\n\nSince $\\theta=180^{\\circ}-4 \\alpha$, then $\\cos \\theta=\\cos \\left(180^{\\circ}-4 \\alpha\\right)=-\\cos 4 \\alpha$.\n\nConsider $\\triangle A L B$ and $\\triangle B L C$.\n\n\n\nBy the cosine law in $\\triangle A L B$,\n\n$$\n\\begin{aligned}\nA B^{2} & =A L^{2}+B L^{2}-2 \\cdot A L \\cdot B L \\cdot \\cos \\theta \\\\\n(r+2)^{2} & =(r+4)^{2}+6^{2}-2(r+4)(6) \\cos \\theta \\\\\n12(r+4) \\cos \\theta & =r^{2}+8 r+16+36-r^{2}-4 r-4 \\\\\n\\cos \\theta & =\\frac{4 r+48}{12(r+4)} \\\\\n\\cos \\theta & =\\frac{r+12}{3 r+12}\n\\end{aligned}\n$$\n\nBy the cosine law in $\\triangle B L C$,\n\n$$\n\\begin{aligned}\nB C^{2} & =B L^{2}+C L^{2}-2 \\cdot B L \\cdot C L \\cdot \\cos \\alpha \\\\\n3^{2} & =6^{2}+5^{2}-2(6)(5) \\cos \\alpha \\\\\n60 \\cos \\alpha & =36+25-9 \\\\\n\\cos \\alpha & =\\frac{52}{60} \\\\\n\\cos \\alpha & =\\frac{13}{15}\n\\end{aligned}\n$$\n\nSince $\\cos \\alpha=\\frac{13}{15}$, then\n\n$$\n\\begin{aligned}\n\\cos 2 \\alpha & =2 \\cos ^{2} \\alpha-1 \\\\\n& =2 \\cdot \\frac{169}{225}-1 \\\\\n& =\\frac{338}{225}-\\frac{225}{225} \\\\\n& =\\frac{113}{225}\n\\end{aligned}\n$$\n\n\n\nand\n\n$$\n\\begin{aligned}\n\\cos 4 \\alpha & =2 \\cos ^{2} 2 \\alpha-1 \\\\\n& =2 \\cdot \\frac{113^{2}}{225^{2}}-1 \\\\\n& =\\frac{25538}{50625}-\\frac{50625}{50625} \\\\\n& =-\\frac{25087}{50625}\n\\end{aligned}\n$$\n\nFinally,\n\n$$\n\\begin{aligned}\n\\cos \\theta & =-\\cos 4 \\alpha \\\\\n\\frac{r+12}{3 r+12} & =\\frac{25087}{50625} \\\\\n\\frac{r+12}{r+4} & =\\frac{25087}{16875} \\\\\n\\frac{(r+4)+8}{r+4} & =\\frac{25087}{16875} \\\\\n1+\\frac{8}{r+4} & =\\frac{25087}{16875} \\\\\n\\frac{8}{r+4} & =\\frac{8212}{16875} \\\\\n\\frac{2}{r+4} & =\\frac{2053}{16875} \\\\\n\\frac{r+4}{2} & =\\frac{16875}{2053} \\\\\nr+4 & =\\frac{33750}{2053} \\\\\nr & =\\frac{25538}{2053}\n\\end{aligned}\n$$\n\nTherefore, the positive integers $s=25538$ and $t=2053$ satisfy the required conditions.'] ['$25538$,$2053$'] 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Multimodal Competition True Numerical Open-ended Geometry Math English +17 "A circular disc is divided into 36 sectors. A number is written in each sector. When three consecutive sectors contain $a, b$ and $c$ in that order, then $b=a c$. If the number 2 is placed in one of the sectors and the number 3 is placed in one of the adjacent sectors, as shown, what is the sum of the 36 numbers on the disc? + +" ['We are told that when $a, b$ and $c$ are the numbers in consecutive sectors, then $b=a c$. This means that if $a$ and $b$ are the numbers in consecutive sectors, then the number in the next sector is $c=\\frac{b}{a}$. (That is, each number is equal to the previous number divided by the one before that.)\n\nStarting with the given 2 and 3 and proceeding clockwise, we obtain\n\n$$\n2,3, \\quad \\frac{3}{2}, \\frac{3 / 2}{3}=\\frac{1}{2}, \\frac{1 / 2}{3 / 2}=\\frac{1}{3}, \\frac{1 / 3}{1 / 2}=\\frac{2}{3}, \\frac{2 / 3}{1 / 3}=2, \\frac{2}{2 / 3}=3, \\quad \\frac{3}{2}, \\ldots\n$$\n\nAfter the first 6 terms, the first 2 terms ( 2 and 3) reappear, and so the first 6 terms will repeat again. (This is because each term comes from the previous two terms, so when two consecutive terms reappear, then the following terms are the same as when these two consecutive terms appeared earlier.)\n\nSince there are 36 terms in total, then the 6 terms repeat exactly $\\frac{36}{6}=6$ times.\n\nTherefore, the sum of the 36 numbers is $6\\left(2+3+\\frac{3}{2}+\\frac{1}{2}+\\frac{1}{3}+\\frac{2}{3}\\right)=6(2+3+2+1)=48$.'] ['48'] 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Multimodal Competition False Numerical Open-ended Algebra Math English +18 "In the diagram, $A C D F$ is a rectangle with $A C=200$ and $C D=50$. Also, $\triangle F B D$ and $\triangle A E C$ are congruent triangles which are right-angled at $B$ and $E$, respectively. What is the area of the shaded region? + +" ['Join $B E$.\n\n\n\nSince $\\triangle F B D$ is congruent to $\\triangle A E C$, then $F B=A E$.\n\nSince $\\triangle F A B$ and $\\triangle A F E$ are each right-angled, share a common side $A F$ and have equal hypotenuses $(F B=A E)$, then these triangles are congruent, and so $A B=F E$.\n\nNow $B A F E$ has two right angles at $A$ and $F$ (so $A B$ and $F E$ are parallel) and has equal sides $A B=F E$ so must be a rectangle.\n\nThis means that $B C D E$ is also a rectangle.\n\nNow the diagonals of a rectangle partition it into four triangles of equal area. (Diagonal $A E$ of the rectangle splits the rectangle into two congruent triangles, which have equal area. The diagonals bisect each other, so the four smaller triangles all have equal area.) Since $\\frac{1}{4}$ of rectangle $A B E F$ is shaded and $\\frac{1}{4}$ of rectangle $B C D E$ is shaded, then $\\frac{1}{4}$ of the total area is shaded. (If the area of $A B E F$ is $x$ and the area of $B C D E$ is $y$, then the total shaded area is $\\frac{1}{4} x+\\frac{1}{4} y$, which is $\\frac{1}{4}$ of the total area $x+y$.)\n\nSince $A C=200$ and $C D=50$, then the area of rectangle $A C D F$ is $200(50)=10000$, so the total shaded area is $\\frac{1}{4}(10000)=2500$.'] ['2500'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +19 "In the diagram, $\triangle X Y Z$ is isosceles with $X Y=X Z=a$ and $Y Z=b$ where $b<2 a$. A larger circle of radius $R$ is inscribed in the triangle (that is, the circle is drawn so that it touches all three sides of the triangle). A smaller circle of radius $r$ is drawn so that it touches $X Y, X Z$ and the larger circle. Determine an expression for $\frac{R}{r}$ in terms of $a$ and $b$. + +" ['Suppose that $M$ is the midpoint of $Y Z$.\n\nSuppose that the centre of the smaller circle is $O$ and the centre of the larger circle is $P$. Suppose that the smaller circle touches $X Y$ at $C$ and $X Z$ at $D$, and that the larger circle touches $X Y$ at $E$ and $X Z$ at $F$.\n\nJoin $O C, O D$ and $P E$.\n\nSince $O C$ and $P E$ are radii that join the centres of circles to points of tangency, then $O C$ and $P E$ are perpendicular to $X Y$.\n\nJoin $X M$. Since $\\triangle X Y Z$ is isosceles, then $X M$ (which is a median by construction) is an altitude (that is, $X M$ is perpendicular to $Y Z)$ and an angle bisector (that is, $\\angle M X Y=\\angle M X Z$ ).\n\nNow $X M$ passes through $O$ and $P$. (Since $X C$ and $X D$ are tangents from $X$ to the same circle, then $X C=X D$. This means that $\\triangle X C O$ is congruent to $\\triangle X D O$ by side-side-side. This means that $\\angle O X C=\\angle O X D$ and so $O$ lies on the angle bisector of $\\angle C X D$, and so $O$ lies on $X M$. Using a similar argument, $P$ lies on $X M$.)\n\n\nDraw a perpendicular from $O$ to $T$ on $P E$. Note that $O T$ is parallel to $X Y$ (since each is perpendicular to $P E$ ) and that $O C E T$ is a rectangle (since it has three right angles).\n\nConsider $\\triangle X M Y$ and $\\triangle O T P$.\n\nEach triangle is right-angled (at $M$ and at $T$ ).\n\nAlso, $\\angle Y X M=\\angle P O T$. (This is because $O T$ is parallel to $X Y$, since both are perpendicular to $P E$.)\n\nTherefore, $\\triangle X M Y$ is similar to $\\triangle O T P$.\n\nThus, $\\frac{X Y}{Y M}=\\frac{O P}{P T}$.\n\nNow $X Y=a$ and $Y M=\\frac{1}{2} b$.\n\nAlso, $O P$ is the line segment joining the centres of two tangent circles, so $O P=r+R$.\n\nLastly, $P T=P E-E T=R-r$, since $P E=R, E T=O C=r$, and $O C E T$ is a rectangle. Therefore,\n\n$$\n\\begin{aligned}\n\\frac{a}{b / 2} & =\\frac{R+r}{R-r} \\\\\n\\frac{2 a}{b} & =\\frac{R+r}{R-r} \\\\\n2 a(R-r) & =b(R+r) \\\\\n2 a R-b R & =2 a r+b r \\\\\nR(2 a-b) & =r(2 a+b) \\\\\n\\frac{R}{r} & =\\frac{2 a+b}{2 a-b} \\quad(\\text { since } 2 a>b \\text { so } 2 a-b \\neq 0, \\text { and } r>0)\n\\end{aligned}\n$$\n\nTherefore, $\\frac{R}{r}=\\frac{2 a+b}{2 a-b}$.'] ['$\\frac{2 a+b}{2 a-b}$'] 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Multimodal Competition False Expression Open-ended Geometry Math English +20 "In the diagram, what is the area of figure $A B C D E F$ ? + +" ['Because all of the angles in the figure are right angles, then $B C=D E=4$.\n\nThus, we can break up the figure into a 4 by 8 rectangle and a 4 by 4 square, by extending $B C$ to hit $F E$. Therefore, the area of the figure is $(8)(4)+(4)(4)=48$.'] ['48'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +21 "In the diagram, $A B C D$ is a rectangle with $A E=15, E B=20$ and $D F=24$. What is the length of $C F$ ? + +" ['By the Pythagorean Theorem in triangle $A B E$, $A B^{2}=15^{2}+20^{2}=625$, so $A B=25$.\n\nSince $A B C D$ is a rectangle, $C D=A B=25$, so by the Pythagorean Theorem in triangle $C F D$, we have $625=25^{2}=24^{2}+C F^{2}$, so $C F^{2}=625-576=49$, or $C F=7$.'] ['7'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +22 "In the diagram, $A B C D$ is a square of side length 6. Points $E, F, G$, and $H$ are on $A B, B C, C D$, and $D A$, respectively, so that the ratios $A E: E B, B F: F C$, $C G: G D$, and $D H: H A$ are all equal to $1: 2$. + +What is the area of $E F G H$ ? + +" "['Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.\n\nThus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.\n\nThen the area of $E F G H$ is equal to the area of square $A B C D$ minus the combined area of the four triangles, or $6^{2}-4\\left[\\frac{1}{2}(2)(4)\\right]=36-16=20$ square units.' + 'Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.\n\nThus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.\n\nBy the Pythagorean Theorem,\n\n$E F=F G=G H=H E=\\sqrt{2^{2}+4^{2}}=\\sqrt{20}$.\n\nSince the two triangles $H A E$ and $E B F$ are congruent (we know the lengths of all three sides of each), then $\\angle A H E=\\angle B E F$. But $\\angle A H E+\\angle A E H=90^{\\circ}$, so $\\angle B E F+\\angle A E H=90^{\\circ}$, so $\\angle H E F=90^{\\circ}$.\n\nIn a similar way, we can show that each of the four angles of $E F G H$ is a right-angle, and so $E F G H$ is a square of side length $\\sqrt{20}$.\n\nTherefore, the area of $E F G H$ is $(\\sqrt{20})^{2}=20$ square units.']" ['20'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +23 "In the diagram, line $A$ has equation $y=2 x$. Line $B$ is obtained by reflecting line $A$ in the $y$-axis. Line $C$ is perpendicular to line $B$. What is the slope of line $C$ ? + +" ['When line $A$ with equation $y=2 x$ is reflected in the $y$-axis, the resulting line (line $B$ ) has equation $y=-2 x$. (Reflecting a line in the $y$-axis changes the sign of the slope.)\n\nSince the slope of line $B$ is -2 and line $C$ is perpendicular to line $B$, then the slope of line $C$ is $\\frac{1}{2}$ (the slopes of perpendicular lines are negative reciprocals).'] ['$\\frac{1}{2}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +24 "Three squares, each of side length 1 , are drawn side by side in the first quadrant, as shown. Lines are drawn from the origin to $P$ and $Q$. Determine, with explanation, the length of $A B$. + +" "['Consider the line through $O$ and $P$. To get from $O$ to $P$, we go right 2 and up 1. Since $B$ lies on this line and to get from $O$ to $B$ we go over 1, then we must go up $\\frac{1}{2}$, to keep the ratio constant.\n\nConsider the line through $O$ and $Q$. To get from $O$ to $Q$, we go right 3 and up 1. Since $A$ lies on this line and to get from $O$ to $A$ we go over 1, then we must go up $\\frac{1}{3}$, to keep the ratio constant.\n\nTherefore, since $A$ and $B$ lie on the same vertical line, then $A B=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$.' + 'Since the line through $P$ passes through the origin, then its equation is of the form $y=m x$. Since it passes through the point $(2,1)$, then $1=2 m$, so the line has equation $y=\\frac{1}{2} x$. Since $B$ has $x$-coordinate 1, then $y=\\frac{1}{2}(1)=\\frac{1}{2}$, so $B$ has coordinates $\\left(1, \\frac{1}{2}\\right)$. Similarly, we can determine that the equation of the line through $Q$ is $y=\\frac{1}{3} x$, and so $A$ has coordinates $\\left(1, \\frac{1}{3}\\right)$.\n\nTherefore, since $A$ and $B$ lie on the same vertical line, then $A B=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$.']" ['$\\frac{1}{6}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +25 "In the diagram, the parabola with equation $y=x^{2}+t x-2$ intersects the $x$-axis at points $P$ and $Q$. + +Also, the line with equation $y=3 x+3$ intersects the parabola at points $P$ and $R$. Determine the value of $t$ and the area of triangle $P Q R$. + +" ['Point $P$ is the point where the line $y=3 x+3$ crosses the $x$ axis, and so has coordinates $(-1,0)$.\n\nTherefore, one of the roots of the parabola $y=x^{2}+t x-2$ is $x=-1$, so\n\n$$\n\\begin{aligned}\n0 & =(-1)^{2}+t(-1)-2 \\\\\n0 & =1-t-2 \\\\\nt & =-1\n\\end{aligned}\n$$\n\nThe parabola now has equation $y=x^{2}-x-2=(x+1)(x-2)$ (we already knew one of the roots so this helped with the factoring) and so its two $x$-intercepts are -1 and 2 , ie. $P$ has coordinates $(-1,0)$ and $Q$ has coordinates $(2,0)$.\n\nWe now have to find the coordinates of the point $R$. We know that $R$ is one of the two points of intersection of the line and the parabola, so we equate their equations:\n\n$$\n\\begin{aligned}\n3 x+3 & =x^{2}-x-2 \\\\\n0 & =x^{2}-4 x-5 \\\\\n0 & =(x+1)(x-5)\n\\end{aligned}\n$$\n\n(Again, we already knew one of the solutions to this equation $(x=-1)$ so this made factoring easier.) Since $R$ does not have $x$-coordinate -1 , then $R$ has $x$-coordinate $x=5$. Since $R$ lies on the line, then $y=3(5)+3=18$, so $R$ has coordinates $(5,18)$.\n\nWe can now calculate the area of triangle $P Q R$. This triangle has base of length 3 (from $P$ to $Q$ ) and height of length 18 (from the $x$-axis to $R$ ), and so has area $\\frac{1}{2}(3)(18)=27$.\n\nThus, $t=-1$ and the area of triangle $P Q R$ is 27 .'] ['-1,27'] 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Multimodal Competition True Numerical Open-ended Geometry Math English +26 "In the diagram, $A C=B C, A D=7, D C=8$, and $\angle A D C=120^{\circ}$. What is the value of $x$ ? + +" ['We first calculate the length of $A C$ using the cosine law:\n\n$$\n\\begin{aligned}\nA C^{2} & =7^{2}+8^{2}-2(7)(8) \\cos \\left(120^{\\circ}\\right) \\\\\nA C^{2} & =49+64-112\\left(-\\frac{1}{2}\\right) \\\\\nA C^{2} & =169 \\\\\nA C & =13\n\\end{aligned}\n$$\n\nSince triangle $A B C$ is right-angled and isosceles, then $x=A B=\\sqrt{2}(A C)=13 \\sqrt{2}$.'] ['$13 \\sqrt{2}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +27 "Donna has a laser at $C$. She points the laser beam at the point $E$. The beam reflects off of $D F$ at $E$ and then off of $F H$ at $G$, as shown, arriving at point $B$ on $A D$. If $D E=E F=1 \mathrm{~m}$, what is the length of $B D$, in metres? + +" ['First, we note that a triangle with one right angle and one angle with measure $45^{\\circ}$ is isosceles.\n\nThis is because the measure of the third angle equals $180^{\\circ}-90^{\\circ}-45^{\\circ}=45^{\\circ}$ which means that the triangle has two equal angles.\n\nIn particular, $\\triangle C D E$ is isosceles with $C D=D E$ and $\\triangle E F G$ is isosceles with $E F=F G$. Since $D E=E F=1 \\mathrm{~m}$, then $C D=F G=1 \\mathrm{~m}$.\n\nJoin $C$ to $G$.\n\n\n\nConsider quadrilateral $C D F G$. Since the angles at $D$ and $F$ are right angles and since $C D=G F$, it must be the case that $C D F G$ is a rectangle.\n\nThis means that $C G=D F=2 \\mathrm{~m}$ and that the angles at $C$ and $G$ are right angles.\n\nSince $\\angle C G F=90^{\\circ}$ and $\\angle D C G=90^{\\circ}$, then $\\angle B G C=180^{\\circ}-90^{\\circ}-45^{\\circ}=45^{\\circ}$ and $\\angle B C G=90^{\\circ}$.\n\nThis means that $\\triangle B C G$ is also isosceles with $B C=C G=2 \\mathrm{~m}$.\n\nFinally, $B D=B C+C D=2 \\mathrm{~m}+1 \\mathrm{~m}=3 \\mathrm{~m}$.'] ['3'] 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rTPE0NoqSQ2lwk0q4G6QlFjyPXa0vPsK6w8gf1rzr4aeMfE/jyxfVbu20yx05JTEoSORnlIAJxlsAcgZ5716NtFADqRvunPSloIyMUAeE6hZ33wn+Kl54q+xz3XhrVt/2mSEbjAXYMdw7YYZHqDjrXpcPxL8GT2YuY/EmnbWHCGYCT6eX97Ptiup2Agg8gjBBqpDpOn28xmhsbaOU/xpCqt+YFAFDR9QvdYumvRDLaaWFKwRzxlZbg5/1hDDKL6A4Y5JIXAB3DyKTaM5paAPnrWfB9r44+PXi7SriZ7eWPTo57adOsUoS3AbHcYJGPfjB5rsfAfjrULHV/+EJ8bgQa5B8lrduflvV7fMerHHB/i6feBzT8Pc/tQeLP+wVH/wCg21dp478Bad450f7PcnyL6H5rS8Rfnhb+qnuP5HBoA6vJz0pa8r8B+OtQsdX/AOEJ8bgQa5B8lrduflvV7fMerHHB/i6feBz6nk56UALRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUHig8VheKvFul+D9Dl1TVJtka8Rxr9+V+yqO5/l3oAPFXi3S/CGhy6pqk2yNeI415eV+yqO5/l3rxy68N61440LXfHni9Xt4YdMupNI0xWIWJfKYq5/Q+rHk4AArd8K+FdV+IWuQ+M/G0OyzT5tL0lvuInUOwPUHg8/e6ngAH0Lx0uPh74lx/wBAq6/9FNQAeBOfh94aP/UKtf8A0UtTeLvD0PirwrqOiztsW6j2q+M7HBBVvwYA1F4E/wCSe+Gv+wVa/wDopa6AjIxQB458NPFK+CrE+C/GZGk3dnI/2Se5+WGeNmLfLIeOCT+BA7Gr/wAUfHGlXXgm+0fQNSttS1XUVFvHBYSid9rH58hM/wAO4fjXps1pBcoY7iJJoz1SRQwz64NMtdOsrEEWlpBb56+VEq/yFAGF8PdCbw54D0fS5IzHNHbh5lIIIkb5nB+jMR+FdPSAYpaACiiigAooooAKKKKAPH/Dv/J0Hiz/ALBUf/oNtXsB5rx/w7/ydB4s/wCwVH/6DbV7BQByfjvwFp3jnR/s9yfIvofmtLxF+eFv6qe4/kcGuX8B+OtQsdX/AOEJ8bgQa5B8lrduflvV7fMerHHB/i6feBz6oea5Px34C07xzo/2e5PkX0PzWl4i/PC39VPcfyODQB1eTnpS15X4D8dahY6v/wAIT43Ag1yD5LW7c/Ler2+Y9WOOD/F0+8Dn1PJz0oAWiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKDxQeKwvFXi3S/B+hy6pqk2yNeI41+/K/ZVHc/y70AHirxbpfg/Q5dU1SbZGvEca/flfsqjuf5d6858K+FdV+IWuQ+M/G0OyzT5tL0lvuInUOwPUHg8/e6ngAE8K+FdV+IWuQ+M/G0OyzT5tL0lvuInUOwPUHg8/e6ngAH2ILjpQABcdKwPHf/ACTzxL/2Crr/ANFNXQVz/jv/AJJ54l/7BV1/6KagA8Cf8k98Nf8AYKtf/RS10Fc/4E/5J74a/wCwVa/+ilroKACiiigAooooAKKKKACiiigAooooA8f8O/8AJ0Hiz/sFR/8AoNtXsFeP+Hf+ToPFn/YKj/8AQbavYKACg80UUAcn478Bad450f7PcnyL6H5rS8Rfnhb+qnuP5HBrl/AfjrULHV/+EJ8bgQa5B8lrduflvV7fMerHHB/i6feBz6oea5Px34C07xzo/wBnuT5F9D81peIvzwt/VT3H8jg0AdXk56UteV+A/HWoWOr/APCE+NwINcg+S1u3Py3q9vmPVjjg/wAXT7wOfU8nPSgBaKKKACiiigAooooAKKKKACiiigAooooAKDxQeKwvFXi3S/B+hy6pqk2yNeI41+/K/ZVHc/y70AHirxbpfg/Q5dU1SbZGvEca/flfsqjuf5d6858K+FdV+IWuQ+M/G0OyzT5tL0lvuInUOwPUHg8/e6ngAE8K+FdV+IWuQ+M/G0OyzT5tL0lvuInUOwPUHg8/e6ngAH2ILjpQABcdKWiigArn/Hf/ACTzxL/2Crr/ANFNXQVz/jv/AJJ54l/7BV1/6KagA8Cf8k98Nf8AYKtf/RS10Fc/4E/5J74a/wCwVa/+ilroKACiiigAooooAKKKKACiiigAooooA8f8O/8AJ0Hiz/sFR/8AoNtXsFeP+Hf+ToPFn/YKj/8AQbavYKACiiigAoPNFFAHJ+O/AWneOdH+z3J8i+h+a0vEX54W/qp7j+Rwa5fwH461Cx1f/hCfG4EGuQfJa3bn5b1e3zHqxxwf4un3gc+qHmuT8d+AtO8c6P8AZ7k+RfQ/NaXiL88Lf1U9x/I4NAHV5OelLXlfgPx1qFjq/wDwhPjcCDXIPktbtz8t6vb5j1Y44P8AF0+8Dn1PJz0oAWiiigAooooAKKKKACiiigAoPFB4rC8VeLdL8H6HLqmqTbI14jjX78r9lUdz/LvQAeKvFul+D9Dl1TVJtka8Rxr9+V+yqO5/l3rznwr4V1X4ha5D4z8bQ7LNPm0vSW+4idQ7A9QeDz97qeAATwr4V1X4ha5D4z8bQ7LNPm0vSW+4idQ7A9QeDz97qeAAfYguOlAAFx0paKKACiiigArn/Hf/ACTzxL/2Crr/ANFNXQVz/jv/AJJ54l/7BV1/6KagA8Cf8k98Nf8AYKtf/RS10Fc/4E/5J74a/wCwVa/+ilroKACiiigAooooAKKKKACiiigAooooA8f8O/8AJ0Hiz/sFR/8AoNtXsFeP+Hf+ToPFn/YKj/8AQbavYKACiiigAoopCcDNAC0HmkDZ6jFBOATxx60Acp478Bad450f7PcnyL6H5rS8Rfnhb+qnuP5HBrl/AfjrULHV/wDhCfG4EGuQfJa3bn5b1e3zHqxxwf4un3gc+pFvUVynjrwJp3jnRxb3P7i+hy1peIvzQv8A1U9xn8jg0AdZk56UteV+A/HWoWGsf8IR43Hka5DhbW7Y/JeoenzHqxxwf4uh+br6nk56UALRRRQAUUUUAFB4oPFYXinxZpfhDQ5dU1SbbGvEca8vK/ZVHc/y74oAPFXi3S/B+hy6pqk2yNeI41+/K/ZVHc/y715z4V8K6r8Qtch8Z+NodlmnzaXpLfcROodgeoPB5+91PAAKeFfCuq/ELWofGnjaLZZqN2l6SclETs7A9QeDz97gn5QBXsIIHTp7UAOC46UtNDZ7UFuOnP1oAdRSBsnFLQAUUUUAFc/47/5J54l/7BV1/wCimroK5/x3/wAk88S/9gq6/wDRTUAHgT/knvhr/sFWv/opa6Cuf8Cf8k98Nf8AYKtf/RS10FABRRRQAUUUUAFFFFABRRRQAUUUHgUAeP8Ah3/k6DxZ/wBgqP8A9Btq9gr581PxFqXhj9oXxPf6X4eu9cneyiha2td25FMcB3najcZUDoOtdIPi94vzz8J9cH4zf/GKAPX6K4T4d/EWbx1c6xb3GhPpM+mNGkkck/mNuYuCCCilSCn6+1d3QAGmSOqRszEKoGSxOABTj07/AIV53488W6JBq9r4X1XVI7Czmi8/UHfIMsOcCFcf3yDu/wBkEd+ADDHinVtV+M3hUxTy2+gXcd0bWBXKi5RYn/euvcMwyuewB6mu+8beJE8M+GprzzoIbiZlt7V5mAQSucKW/wBleWPspryzXPHPhWf4w+EdVtdXt202xtriOaZVbZETG4UYx6kCvWdV0PQPGel239pWkOoWbKJoC+QMMOGHTsaAOK0HwPr0l5ZakPihqGqaekquUiHyTqDkruEhGDjB4re1lJda8bRaBNqF3Z2MWni9KWkxhkuXMhXG8HcFUKDhSPvjJx14m78MzfCnxroupeHbic6Bq99HY3dhI5YIznCsvrjkjPIxjJ3cep6v4d03XjAb6FzLAS0M8MzwyxZ64dCGGe4zigDzzV/A9v4o1LV/ClzqFxcnT4IbzT9QmfzZ7KSQvmJn+86nYGwxzgjngETeA/HWoWOr/wDCE+NwINcg+S1u3Py3q9vmPVjjg/xdPvA59A0bw/pugWzwadB5SyP5krM7O8r/AN5nYlmP1NY/jvwFp3jnR/s9yfIvofmtLxF+eFv6qe4/kcGgDq8nPSlryvwH461Cx1f/AIQnxuBBrkHyWt25+W9Xt8x6sccH+Lp94HPqeTnpQAtB4oPFYXirxbpfg/Q5dU1SbZGvEca/flfsqjuf5d6ADxV4t0vwfocuqapNtjXiONfvyv2VR3P8u9ec+FfCuq/ELW4vGnjWHbZp82laS2diL1DsD1HQ+rdTwAKTwt4V1X4ha3F4z8bQ7bNPm0vSW+4idQzA9QeDz97qflAB674g+LbXw3Z2VhJqC6fNqjtGl4+cW8agGSQAdWAICj1IoA5L4i+K9Ul1vRItIuHt9Jh1uCzuJ4nKNczZyyAjqiDIbsWOP4a9S1nVbbQdFvNVvX221pE0rnucdh7noB6kV4l4+8WeDZdN8I2GgarbyW+m6tBJIqBvkiXOWYkc9ck9SSa9iH9g+OfDgbEOp6TdZxwdj7W/A8Mv6UAedaZ4W8SeLIP7XtvilOouT5kttYYkjt2b5jEGWToucDjPHSuy8USXJvNA0CDULmyt9QkkSe6jfEzLHHuCK55DNjqOcK2PbgPGngNfh2n/AAm3gqWayayZTeWW8tHLEWAOM5OOeQc+oxivWLvTdP8AE+jQx6ha+Zbyqk6KWKtG2MqysCCrDPUYNAHPWMMvhrxvp+kWmpXt5YajazSPbXk7TtbtHtw6uxLbTuwQSRnpXb1jaN4W0rQZ5riyila5mAWS4uJ5J5WUdF3uxOB6dK2aACiikJwM0ALXP+O/+SeeJf8AsFXX/opq85svjb4j1Kzju7D4Zapd20mdk0E0kiNgkHDLCQeQR+FZ3iv4teJLjwlq1rf/AA31XTrW6tJLZ7ud5AkXmKUDHMIHVhxkZ6UAer+BP+Se+Gv+wVa/+ilroK5/wJ/yT3w1/wBgq1/9FLXQUAFFFFABRRRQAUUUh6UAG7AzjijPtXD/ABW1G70TwNfaxY6pc2V1bbBD5JXDsXVcMCDnqaz/AAbo3ibWfBum6tqPjHWIb+7hE+yOO3Maq2SnytESfl29T6+tAHpAbJoboa8sn8a694B8VWOk+Lri31DR9QJS21aKIQvGwIBEiD5eCRnGODn1A9Szkf8A16APIfD2P+GnvFfr/Zaf+g21evnpzXimtaH8Q9H+Let+KfC2gWl9DeQR26PczxhSoSLcdvmKQd0fern/AAkPxy/6E3Q/+/y//JFAB8IePiF8TB/1FRk/9tbivYK+e/Del/F/wvrGuanY+E9Nkn1m4FxcLNcxFUbc7YQCYEDMh6k9BXSf8JF8cv8AoTdD/wC/y/8AyRQB6+eRVO50rT7uXzbqxtp5Mbd8kKscemSOnWvLP+Ei+OX/AEJuh/8Af5f/AJIo/wCEh+OX/Qm6H/3+X/5IoA19b8EPdfFTw1qttotqdFtYJ1vCFjChmjcLlDy3JHQGt7xLq3iPQ7rTxofhxNV07BS6SKZY5Ium3aDxjH+R1rif+Eh+OX/Qm6H/AN/k/wDkij/hIPjicf8AFGaHx/02T/5IoA7H+z9S8Vajpt1rGnf2bYWEwu4rWSRZJZpwCFZ9mVVVyTjJycHjHPXgYrx//hIPjj/0Juh/9/k/+SKX/hIvjl/0Juh/9/l/+SKAPYKQ815B/wAJF8cv+hN0P/v8v/yRR/wkXxy/6E3Q/wDv8v8A8kUAdp478B6d450f7PcnyL6H5rS8Rfnhb+qnuP5HBrl/AfjvULDVv+EJ8bgQa5B8trdOflvU7fMerHHB/i6H5gc0/wDhIfjl/wBCbof/AH+X/wCSK5rxho3xX8b2cMGqeCtISWBg0Nzb3EayxeoBM5GD6EEfjzQB7T4r8WaX4O0SXVNVm2Rp8sca/flfsqjuf5dTXnHhXwrqvxC1uLxn42h2WafNpekt9xE6h2B6g8Hn73U/KADxU3hD4uahrWnanrWgQ6w+noEghvruBouO7Ksq7j0JJPzYGcjiu3HiD44jp4N0P/v8v/yRQB6+FA6VXu9Osr7Z9rtILjZnb50QfbnGcZHHQflXlP8AwkXxy/6E3Q/+/wAv/wAkUf8ACRfHL/oTdD/7/L/8kUAa3xD8Evq//CO/2Jotq/2bVYp7ry1ij2wjrnJGR7DOa6XxLcazoWjQN4W0O2v5EmG+08xYR5ZyW29BnOPzPBrg/wDhIPjj/wBCbof/AH+X/wCSKP8AhIPjjn/kTtE/7/p/8kUAdVdxaz4405dMv9Fl0bTJWU3huJ0eaRAQ3loqE4DEYLMQcZ4rtUUIoUAADoAMAV5D/wAJD8cf+hN0P2/fL/8AJFH/AAkXxy/6E3Q/+/y//JFAHsFFeP8A/CRfHL/oTdD/AO/y/wDyRR/wkXxy/wChN0P/AL/L/wDJFAHsFIeleQf8JF8cv+hN0P8A7/L/APJFB8Q/HIjH/CG6H/3+X/5IoA3/AIKZ/wCFSaH/ANvH/o+Sl+Nn/JIdd/7d/wD0fHXD+F1+MvhLw7aaJYeEdKkt7Xfse4uI2c7nZzkicDqxHTpR4pX4zeLfDl3od/4R0qO1udm97e4jDja6uMEzkdVHagD1fwL/AMk98Nf9gq1/9FLXQE4rH8J2Vxpvg7Q7C7j8u5tbCCGVMg7XWNVIyODyDUfi52t/C2o3q6hcWLWtvJOJYGAOVUkZyCCOOlAG5mgtivIfhhD4o8ZeE/7Z1jxbrEDSzskAthAoZFwCTujP8W4f8BrQ8ZaB4t0zSEu9C8a6o8/nwxGK7ihYN5kioMFY1xgsD34B6UAenA5pajjTYqruLbQBk9T9akoAKD0pCcDNJu9KAPHPjnPJq134Y8HWrfvdSvRJIB1Vc7FJ9sux/wCA16/b28dtaxW0ChIokWNAOygYH9K8b0Yjxf8AtG6lqGd9poFuYIj1G8DZj/vppT+FezPIqKzsyqqgliTgD6mgDxz9owxf8IZpQOPOOoDZnrt8t8/hkrXqHhhrh/CejNdFjcNYwmUt13bFzn3zmvJtdtpPjF8QbG2sQX8K6K5+0XY+5PISCyof4sgKuRwOTzkZ9vACgAAADgAUAGOc5paKKACiiigAooooAKKKKACiiigAooooAQ8CjIpk8KXEDwyAlHUqwBI4PuK8K8D6Uviz4leLFkv9Tk0DT5WihhTUZ0G8uQpDBwSMI/fuKAPeM8ZoBz/+uvNvF3w3t38OX02gajrdrqcEDyQCPU55BKyjIUh3PXGOMfjjFd9plobDTLOzaV5jbwJEZZDuZyoAySepOKALlFFFABRRRQAUUUUAFFFFABRRRQAUhAIxS0UANI6+leY/HfWTpvw8ksYm/f6pOluqj7xUfM3/AKCB/wACr1A9K8V8bH/hK/jv4Y8PA7rbSk+13A7BvvkN6ghIx/wKgDUsvF0nw/8Ah9Z2zeDfEBXT7VVkkaKIRFx95mZXYqpYk5x3rvfDBvp/DGnzaq269njE8w7IznftHsucD2Aq7qFhBqmm3On3IZre5iaGRQcEqwwR+WathQoAAAA4AHagAA560tFFACEZGKqajZTX1sYodRubFyf9bbCMvj0+dWH6Zq5RQB55onwjsfDl1dXOleJvEVtNdf69xNAxk5Jyd0R5yT+daU3w50/UG/4nOr65q8PU293ekRH6pGEB/GuxooAr2dla6fax2tnbxW9vGMJFEgVVHsBViiigAooooAKKKKACiiigAooooAKKKKACg8CikPQ0AY3i7Wx4d8Iarq5YK1rbO6Z/v4wo/FiBXk3wkj8U6D8P3vNM8M2+of2hK9ys0moiKRx90DaUP90kc87q1Pj7qsg8M6Z4dtj/AKVq92Btz1RCOv8AwNk/I16hpOmw6RotlpkH+ptYEgUjuFUAGgDB+HV3qGq+E4ta1Q4vNUka7aNfuRqflRVB6DYifjk966wAg9a5ez8R2DeKbbwpovkyrY25e8KHK20ajYkYIP3ySvHYKeORXU0AFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAQ3UL3Fs8UdzLbO2MSxBSy89twI/MGuHsfhTZad4om8SW/iLXRqs+fNnZ4G3g4yCDFjHA6DjAxXfUUANVSAAWLY7nqadRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUhGRilooA4/VPhh4T1u8F5qljc3dyBhZZtQuGZf8Ad/eYUewwKkl+G/hqeIxSw6jJGRjY+rXZX8jLiusooA5/wz4I8PeDzcHQtPFp9p2+b+9eTdtzj77HHU9PxroKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKAP/9k='] Multimodal Competition False Numerical Open-ended Geometry Math English +28 "An L shape is made by adjoining three congruent squares. The L is subdivided into four smaller L shapes, as shown. Each of the resulting L's is subdivided in this same way. After the third round of subdivisions, how many L's of the smallest size are there? +" "[""After each round, each L shape is divided into 4 smaller $\\mathrm{L}$ shapes.\n\nThis means that the number of $\\mathrm{L}$ shapes increases by a factor of 4 after each round.\n\nAfter 1 round, there are $4 \\mathrm{~L}$ shapes.\n\nAfter 2 rounds, there are $4^{2}=16$ L's of the smallest size.\n\nAfter 3 rounds, there are $4^{3}=64$ L's of the smallest size.""]" ['64'] 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'] Multimodal Competition False Numerical Open-ended Number Theory Math English +29 "Jimmy is baking two large identical triangular cookies, $\triangle A B C$ and $\triangle D E F$. Each cookie is in the shape of an isosceles right-angled triangle. The length of the shorter sides of each of these triangles is $20 \mathrm{~cm}$. He puts the cookies on a rectangular baking tray so that $A, B, D$, and $E$ are at the vertices of the rectangle, as shown. If the distance between parallel sides $A C$ and $D F$ is $4 \mathrm{~cm}$, what is the width $B D$ of the tray? + +" ['We note that $B D=B C+C D$ and that $B C=20 \\mathrm{~cm}$, so we need to determine $C D$.\n\nWe draw a line from $C$ to $P$ on $F D$ so that $C P$ is perpendicular to $D F$.\n\nSince $A C$ and $D F$ are parallel, then $C P$ is also perpendicular to $A C$.\n\nThe distance between $A C$ and $D F$ is $4 \\mathrm{~cm}$, so $C P=4 \\mathrm{~cm}$.\n\nSince $\\triangle A B C$ is isosceles and right-angled, then $\\angle A C B=45^{\\circ}$.\n\n\n\nThus, $\\angle P C D=180^{\\circ}-\\angle A C B-\\angle P C A=180^{\\circ}-45^{\\circ}-90^{\\circ}=45^{\\circ}$.\n\nSince $\\triangle C P D$ is right-angled at $P$ and $\\angle P C D=45^{\\circ}$, then $\\triangle C P D$ is also an isosceles right-angled triangle.\n\nTherefore, $C D=\\sqrt{2} C P=4 \\sqrt{2} \\mathrm{~cm}$.\n\nFinally, $B D=B C+C D=(20+4 \\sqrt{2}) \\mathrm{cm}$.'] ['$(20+4 \\sqrt{2})$'] 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Multimodal Competition False cm Numerical Open-ended Geometry Math English +30 "In the diagram, $\angle A C B=\angle A D E=90^{\circ}$. If $A B=75, B C=21, A D=20$, and $C E=47$, determine the exact length of $B D$. + +" ['We use the cosine law in $\\triangle A B D$ to determine the length of $B D$ :\n\n$$\nB D^{2}=A B^{2}+A D^{2}-2(A B)(A D) \\cos (\\angle B A D)\n$$\n\nWe are given that $A B=75$ and $A D=20$, so we need to determine $\\cos (\\angle B A D)$.\n\nNow\n\n$$\n\\begin{aligned}\n\\cos (\\angle B A D) & =\\cos (\\angle B A C+\\angle E A D) \\\\\n& =\\cos (\\angle B A C) \\cos (\\angle E A D)-\\sin (\\angle B A C) \\sin (\\angle E A D) \\\\\n& =\\frac{A C}{A B} \\frac{A D}{A E}-\\frac{B C}{A B} \\frac{E D}{A E}\n\\end{aligned}\n$$\n\nsince $\\triangle A B C$ and $\\triangle A D E$ are right-angled.\n\nSince $A B=75$ and $B C=21$, then by the Pythagorean Theorem,\n\n$$\nA C=\\sqrt{A B^{2}-B C^{2}}=\\sqrt{75^{2}-21^{2}}=\\sqrt{5625-441}=\\sqrt{5184}=72\n$$\n\nsince $A C>0$.\n\nSince $A C=72$ and $C E=47$, then $A E=A C-C E=25$.\n\nSince $A E=25$ and $A D=20$, then by the Pythagorean Theorem,\n\n$$\nE D=\\sqrt{A E^{2}-A D^{2}}=\\sqrt{25^{2}-20^{2}}=\\sqrt{625-400}=\\sqrt{225}=15\n$$\n\nsince $E D>0$.\n\nTherefore,\n\n$$\n\\cos (\\angle B A D)=\\frac{A C}{A B} \\frac{A D}{A E}-\\frac{B C}{A B} \\frac{E D}{A E}=\\frac{72}{75} \\frac{20}{25}-\\frac{21}{75} \\frac{15}{25}=\\frac{1440-315}{75(25)}=\\frac{1125}{75(25)}=\\frac{45}{75}=\\frac{3}{5}\n$$\n\n\n\nFinally,\n\n$$\n\\begin{aligned}\nB D^{2} & =A B^{2}+A D^{2}-2(A B)(A D) \\cos (\\angle B A D) \\\\\n& =75^{2}+20^{2}-2(75)(20)\\left(\\frac{3}{5}\\right) \\\\\n& =5625+400-1800 \\\\\n& =4225\n\\end{aligned}\n$$\n\nSince $B D>0$, then $B D=\\sqrt{4225}=65$, as required.'] ['65'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +31 "A circle, with diameter $A B$ as shown, intersects the positive $y$-axis at point $D(0, d)$. Determine $d$. + +" "['The centre of the circle is $(3,0)$ and the circle has a radius of 5.\n\nThus $\\sqrt{d^{2}+3^{2}}=5$\n\n$$\n\\begin{aligned}\n& d^{2}=5^{2}-3^{2} \\\\\n& d^{2}=16\n\\end{aligned}\n$$\n\nTherefore $d=4$, since $d>0$.' + 'Since $A B$ is a diameter of the circle, $\\angle A D B=90^{\\circ}$ and $\\angle A O D=90^{\\circ}$.\n\n$\\triangle A D O \\sim \\triangle D B O$\n\nTherefore, $\\frac{O D}{A O}=\\frac{B O}{O D}$\n\nand $d^{2}=2(8)$\n\n$$\n\\begin{aligned}\nd^{2} & =16 \\\\\nd & =4, \\text { since } d>0 .\n\\end{aligned}\n$$' + '$\\angle A D B=\\angle A O D=\\angle B O D=90^{\\circ}$\n\nIn $\\triangle A O D, A D^{2}=4+d^{2}$.\n\nIn $\\triangle B O D, D B^{2}=64+d^{2}$.\n\nIn $\\Delta A D B,\\left(4+d^{2}\\right)+\\left(64+d^{2}\\right)=100$\n\n$$\n\\begin{aligned}\n2 d^{2} & =32 \\\\\nd & =4, d>0\n\\end{aligned}\n$$']" ['4'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +32 "A square $P Q R S$ with side of length $x$ is subdivided into four triangular regions as shown so that area (A) + area $(B)=\text{area}(C)$. If $P T=3$ and $R U=5$, determine the value of $x$. + +" ['Since the side length of the square is $x, T S=x-3$ and $V S=x-5$\n\nArea of triangle $A=\\frac{1}{2}(3)(x)$.\n\nArea of triangle $B=\\frac{1}{2}(5)(x)$\n\nArea of triangle $C=\\frac{1}{2}(x-5)(x-3)$.\n\nFrom the given information, $\\frac{1}{2}(3 x)+\\frac{1}{2}(5 x)=\\frac{1}{2}(x-5)(x-3)$. Labelled diagram\n\n$3 x+5 x=x^{2}-8 x+15$\n\n$x^{2}-16 x+15=0$\n\n\n\nThus $x=15$ or $x=1$.\n\nTherefore $x=15$ since $x=1$ is inadmissible.'] ['15'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +33 "In the diagram, $A D=D C, \sin \angle D B C=0.6$ and $\angle A C B=90^{\circ}$. What is the value of $\tan \angle A B C$ ? + +" ['Let $D B=10$.\n\nTherefore, $D C=A D=6$.\n\nBy the theorem of Pythagoras, $B C^{2}=10^{2}-6^{2}=64$.\n\nTherefore, $B C=8$.\n\n\n\nThus, $\\tan \\angle A B C=\\frac{12}{8}=\\frac{3}{2}$.'] ['$\\frac{3}{2}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +34 "On a cross-sectional diagram of the Earth, the $x$ and $y$-axes are placed so that $O(0,0)$ is the centre of the Earth and $C(6.40,0.00)$ is the location of Cape Canaveral. A space shuttle is forced to land on an island at $A(5.43,3.39)$, as shown. Each unit represents $1000 \mathrm{~km}$. + +Determine the distance from Cape Canaveral to the island, measured on the surface of the earth, to the nearest $10 \mathrm{~km}$. + +" ['$\\tan \\angle A O C=\\frac{3.39}{5.43}$\n\n$\\angle A O C=\\tan ^{-1}\\left(\\frac{3.39}{5.43}\\right)=31.97^{\\circ}$\n\nThe arc length $\\overparen{A C}=\\frac{31.97}{360^{\\circ}}[(2 \\pi)(6.40)]=3.57$ units\n\nThe distance is approximately $3570 \\mathrm{~km}$.'] ['3570'] 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Multimodal Competition False km Numerical Open-ended Geometry Math English +35 "The parabola $y=-x^{2}+4$ has vertex $P$ and intersects the $x$-axis at $A$ and $B$. The parabola is translated from its original position so that its vertex moves along the line $y=x+4$ to the point $Q$. In this position, the parabola intersects the $x$-axis at $B$ and $C$. Determine the coordinates of $C$. + +" "['The parabola $y=-x^{2}+4$ has vertex $P(0,4)$ and intersects the $x$-axis at $A(-2,0)$ and $B(2,0)$. The intercept $B(2,0)$ has its pre-image, $B^{\\prime}$ on the parabola $y=-x^{2}+4$. To find $B^{\\prime}$, we find the point of intersection of the line passing through $B(2,0)$, with slope 1 , and the parabola $y=-x^{2}+4$.\n\nThe equation of the line is $y=x-2$.\n\nIntersection points, $x-2=-x^{2}+4$\n\n$$\n\\begin{array}{r}\nx^{2}+x-6=0 \\\\\n(x+3)(x-2)=0 .\n\\end{array}\n$$\n\nTherefore, $x=-3$ or $x=2$.\n\nFor $x=-3, y=-3-2=-5$. Thus $B^{\\prime}$ has coordinates $(-3,-5)$.\n\nIf $(-3,-5) \\rightarrow(2,0)$ then the required general translation mapping $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $(x, y) \\rightarrow(x+5, y+5)$.\n\nPossibility 1\n\nUsing the general translation, we find the coordinates of $Q$ to be, $P(0,4) \\rightarrow Q(0+5,4+5)=Q(5,9)$.\n\nIf $C$ is the reflection of $B$ in the axis of symmetry of the parabola, i.e. $x=5, C$ has coordinates $(8,0)$.\n\nPossibility 2\n\nIf $B^{\\prime}$ has coordinates $(-3,-5)$ then $C^{\\prime}$ is the reflection of $B^{\\prime}$ in the $y$-axis. Thus $C^{\\prime}$ has coordinates $(3,-5)$.\n\nIf we apply the general translation then $C$ has coordinates $(3+5,-5+5)$ or $(8,0)$.\n\nThus $C$ has coordinates $(8,0)$.\n\nPossibility 3\n\nUsing the general translation, we find the coordinates of $Q$ to be, $P(0,4) \\rightarrow Q(0+5,4+5)=Q(5,9)$.\n\nThe equation of the image parabola is $y=-(x-5)^{2}+9$.\n\n\n\nTo find its intercepts, $-(x-5)^{2}+9=0$\n\n$$\n\\begin{aligned}\n(x-5)^{2} & =9 \\\\\nx-5 & = \\pm 3 .\n\\end{aligned}\n$$\n\nTherefore $x=8$ or $x=2$.\n\nThus $C$ has coordinates $(8,0)$.' + 'The translation moving the parabola with equation $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $T(t, t)$ because the slope of the line $y=x+4$ is 1 .\n\nThe pre-image of $B^{\\prime}$ is $(2-t,-t)$.\n\nSince $B^{\\prime}$ is on the parabola with vertex $P$, we have\n\n$$\n\\begin{aligned}\n-t & =-(2-t)^{2}+4 \\\\\n-t & =-4+4 t-t^{2}+4 \\\\\nt^{2}-5 t & =0 \\\\\nt(t-5) & =0\n\\end{aligned}\n$$\n\nTherefore, $t=0$ or $t=5$.\n\nThus $B^{\\prime}$ is $(-3,-5)$.\n\nLet $C$ have coordinates $(c, 0)$.\n\nThe pre-image of $C$ is $(c-5,-5)$.\n\nTherefore, $-5=-(c-5)^{2}+4$.\n\nOr, $(c-5)^{2}=9$.\n\nTherefore $c-5=3$ or $c-5=-3$.\n\n$$\nc=8 \\text { or } \\quad c=2\n$$\n\nThus $C$ has coordinates $(8,0)$.']" ['$(8,0)$'] 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Multimodal Competition False Tuple Open-ended Geometry Math English +36 "In the isosceles trapezoid $A B C D$, $A B=C D=x$. The area of the trapezoid is 80 and the circle with centre $O$ and radius 4 is tangent to the four sides of the trapezoid. Determine the value of $x$. + +" ['Using the tangent properties of a circle, the lengths of line segments are as shown on the diagram.\n\nArea of trapezoid $A B C D=\\frac{1}{2}(8)(B C+A D)$\n\n$$\n\\begin{aligned}\n& =4(2 b+2 x-2 b) \\\\\n& =8 x .\n\\end{aligned}\n$$\n\n\n\nThus, $8 x=80$.\n\nTherefore, $x=10$.'] ['10'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +37 "In the diagram, points $P(p, 4), B(10,0)$, and $O(0,0)$ are shown. If $\triangle O P B$ is right-angled at $P$, determine all possible values of $p$. + +" "['Since $\\angle O P B=90^{\\circ}$, then $O P$ and $P B$ are perpendicular, so the product of their slopes is -1 .\n\nThe slope of $O P$ is $\\frac{4-0}{p-0}=\\frac{4}{p}$ and the slope of $P B$ is $\\frac{4-0}{p-10}=\\frac{4}{p-10}$.\n\nTherefore, we need\n\n$$\n\\begin{aligned}\n\\frac{4}{p} \\cdot \\frac{4}{p-10} & =-1 \\\\\n16 & =-p(p-10) \\\\\np^{2}-10 p+16 & =0 \\\\\n(p-2)(p-8) & =0\n\\end{aligned}\n$$\n\nand so $p=2$ or $p=8$. Since each these steps is reversible, then $\\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.' + 'Since $\\triangle O P B$ is right-angled at $P$, then $O P^{2}+P B^{2}=O B^{2}$ by the Pythagorean Theorem. Note that $O B=10$ since $O$ has coordinates $(0,0)$ and $B$ has coordinates $(10,0)$.\n\nAlso, $O P^{2}=(p-0)^{2}+(4-0)^{2}=p^{2}+16$ and $P B^{2}=(10-p)^{2}+(4-0)^{2}=p^{2}-20 p+116$. Therefore,\n\n$$\n\\begin{aligned}\n\\left(p^{2}+16\\right)+\\left(p^{2}-20 p+116\\right) & =10^{2} \\\\\n2 p^{2}-20 p+32 & =0 \\\\\np^{2}-10 p+16 & =0\n\\end{aligned}\n$$\n\n\n\nand so $(p-2)(p-8)=0$, or $p=2$ or $p=8$. Since each these steps is reversible, then $\\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.']" ['2,8'] 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Multimodal Competition True Numerical Open-ended Geometry Math English +38 "A snail's shell is formed from six triangular sections, as shown. Each triangle has interior angles of $30^{\circ}, 60^{\circ}$ and $90^{\circ}$. If $A B$ has a length of $1 \mathrm{~cm}$, what is the length of $A H$, in $\mathrm{cm}$ ? + +" "['In a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, the ratio of the side opposite the $90^{\\circ}$ to the side opposite the $60^{\\circ}$ angle is $2: \\sqrt{3}$.\n\nNote that each of $\\triangle A B C, \\triangle A C D, \\triangle A D E, \\triangle A E F, \\triangle A F G$, and $\\triangle A G H$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nTherefore, $\\frac{A H}{A G}=\\frac{A G}{A F}=\\frac{A F}{A E}=\\frac{A E}{A D}=\\frac{A D}{A C}=\\frac{A C}{A B}=\\frac{2}{\\sqrt{3}}$.\n\nThus, $A H=\\frac{2}{\\sqrt{3}} A G=\\left(\\frac{2}{\\sqrt{3}}\\right)^{2} A F=\\left(\\frac{2}{\\sqrt{3}}\\right)^{3} A E=\\left(\\frac{2}{\\sqrt{3}}\\right)^{4} A D=\\left(\\frac{2}{\\sqrt{3}}\\right)^{5} A C=\\left(\\frac{2}{\\sqrt{3}}\\right)^{6} A B$.\n\n(In other words, to get from $A B=1$ to the length of $A H$, we multiply by the ""scaling factor"" $\\frac{2}{\\sqrt{3}}$ six times.)\n\nTherefore, $A H=\\left(\\frac{2}{\\sqrt{3}}\\right)^{6}=\\frac{64}{27}$.']" ['$\\frac{64}{27}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +39 "In rectangle $A B C D$, point $E$ is on side $D C$. Line segments $A E$ and $B D$ are perpendicular and intersect at $F$. If $A F=4$ and $D F=2$, determine the area of quadrilateral $B C E F$. + +" "['Since $\\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{A F^{2}+F D^{2}}=\\sqrt{4^{2}+2^{2}}=\\sqrt{20}=2 \\sqrt{5}\n$$\n\nsince $A D>0$.\n\nLet $\\angle F A D=\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B A F=90^{\\circ}-\\beta$.\n\nSince $\\triangle A F D$ is right-angled at $F$, then $\\angle A D F=90^{\\circ}-\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B D C=90^{\\circ}-\\left(90^{\\circ}-\\beta\\right)=\\beta$.\n\n\n\n\n\nTherefore, $\\triangle B F A, \\triangle A F D$, and $\\triangle D F E$ are all similar as each is right-angled and has either an angle of $\\beta$ or an angle of $90^{\\circ}-\\beta$ (and hence both of these angles).\n\nTherefore, $\\frac{A B}{A F}=\\frac{D A}{D F}$ and so $A B=\\frac{4(2 \\sqrt{5})}{2}=4 \\sqrt{5}$.\n\nAlso, $\\frac{F E}{F D}=\\frac{F D}{F A}$ and so $F E=\\frac{2(2)}{4}=1$.\n\nSince $A B C D$ is a rectangle, then $B C=A D=2 \\sqrt{5}$, and $D C=A B=4 \\sqrt{5}$.\n\nFinally, the area of quadrilateral $B C E F$ equals the area of $\\triangle D C B$ minus the area $\\triangle D F E$. Thus, the required area is\n\n$$\n\\frac{1}{2}(D C)(C B)-\\frac{1}{2}(D F)(F E)=\\frac{1}{2}(4 \\sqrt{5})(2 \\sqrt{5})-\\frac{1}{2}(2)(1)=20-1=19\n$$' + 'Since $\\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{A F^{2}+F D^{2}}=\\sqrt{4^{2}+2^{2}}=\\sqrt{20}=2 \\sqrt{5}\n$$\n\nsince $A D>0$.\n\nLet $\\angle F A D=\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B A F=90^{\\circ}-\\beta$. Since $\\triangle B A F$ is right-angled at $F$, then $\\angle A B F=\\beta$.\n\nSince $\\triangle A F D$ is right-angled at $F$, then $\\angle A D F=90^{\\circ}-\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B D C=90^{\\circ}-\\left(90^{\\circ}-\\beta\\right)=\\beta$.\n\n\n\nLooking at $\\triangle A F D$, we see that $\\sin \\beta=\\frac{F D}{A D}=\\frac{2}{2 \\sqrt{5}}=\\frac{1}{\\sqrt{5}}, \\cos \\beta=\\frac{A F}{A D}=\\frac{4}{2 \\sqrt{5}}=\\frac{2}{\\sqrt{5}}$, and $\\tan \\beta=\\frac{F D}{A F}=\\frac{2}{4}=\\frac{1}{2}$.\n\nSince $A F=4$ and $\\angle A B F=\\beta$, then $A B=\\frac{A F}{\\sin \\beta}=\\frac{4}{\\frac{1}{\\sqrt{5}}}=4 \\sqrt{5}$.\n\nSince $F D=2$ and $\\angle F D E=\\beta$, then $F E=F D \\tan \\beta=2 \\cdot \\frac{1}{2}=1$.\n\nSince $A B C D$ is a rectangle, then $B C=A D=2 \\sqrt{5}$, and $D C=A B=4 \\sqrt{5}$.\n\nFinally, the area of quadrilateral $E F B C$ equals the area of $\\triangle D C B$ minus the area $\\triangle D F E$. Thus, the required area is\n\n$$\n\\frac{1}{2}(D C)(C B)-\\frac{1}{2}(D F)(F E)=\\frac{1}{2}(4 \\sqrt{5})(2 \\sqrt{5})-\\frac{1}{2}(2)(1)=20-1=19\n$$']" ['19'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +40 "In the diagram, points $B, P, Q$, and $C$ lie on line segment $A D$. The semi-circle with diameter $A C$ has centre $P$ and the semi-circle with diameter $B D$ has centre $Q$. The two semi-circles intersect at $R$. If $\angle P R Q=40^{\circ}$, determine the measure of $\angle A R D$. + +" ['Suppose $\\angle P A R=x^{\\circ}$ and $\\angle Q D R=y^{\\circ}$.\n\n\n\nSince $P R$ and $P A$ are radii of the larger circle, then $\\triangle P A R$ is isosceles.\n\nThus, $\\angle P R A=\\angle P A R=x^{\\circ}$.\n\nSince $Q D$ and $Q R$ are radii of the smaller circle, then $\\triangle Q R D$ is isosceles.\n\nThus, $\\angle Q R D=\\angle Q D R=y^{\\circ}$.\n\nIn $\\triangle A R D$, the sum of the angles is $180^{\\circ}$, so $x^{\\circ}+\\left(x^{\\circ}+40^{\\circ}+y^{\\circ}\\right)+y^{\\circ}=180^{\\circ}$ or $2 x+2 y=140$ or $x+y=70$.\n\nTherefore, $\\angle C P D=x^{\\circ}+40^{\\circ}+y^{\\circ}=(x+y+40)^{\\circ}=110^{\\circ}$.'] ['$110$'] 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Multimodal Competition False ^{\circ} Numerical Open-ended Geometry Math English +41 "In the diagram, a line is drawn through points $P, Q$ and $R$. If $P Q=Q R$, what are the coordinates of $R$ ? + +" "['To get from $P$ to $Q$, we move 3 units right and 4 units up.\n\nSince $P Q=Q R$ and $R$ lies on the line through $Q$, then we must use the same motion to get from $Q$ to $R$.\n\nTherefore, to get from $Q(0,4)$ to $R$, we move 3 units right and 4 units up, so the coordinates of $R$ are $(3,8)$.' + 'The line through $P(-3,0)$ and $Q(0,4)$ has slope $\\frac{4-0}{0-(-3)}=\\frac{4}{3}$ and $y$-intercept 4 , so has equation $y=\\frac{4}{3} x+4$.\n\nThus, $R$ has coordinates $\\left(a, \\frac{4}{3} a+4\\right)$ for some $a>0$.\n\nSince $P Q=Q R$, then $P Q^{2}=Q R^{2}$, so\n\n$$\n\\begin{aligned}\n(-3)^{2}+4^{2} & =a^{2}+\\left(\\frac{4}{3} a+4-4\\right)^{2} \\\\\n25 & =a^{2}+\\frac{16}{9} a^{2} \\\\\n\\frac{25}{9} a^{2} & =25 \\\\\na^{2} & =9\n\\end{aligned}\n$$\n\nso $a=3$ since $a>0$.\n\nThus, $R$ has coordinates $\\left(3, \\frac{4}{3}(3)+4\\right)=(3,8)$.']" ['(3,8)'] 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Multimodal Competition False Tuple Open-ended Geometry Math English +42 "In the diagram, $O A=15, O P=9$ and $P B=4$. Determine the equation of the line through $A$ and $B$. Explain how you got your answer. + +" ['Since $O P=9$, then the coordinates of $P$ are $(9,0)$.\n\nSince $O P=9$ and $O A=15$, then by the Pythagorean Theorem,\n\n$$\nA P^{2}=O A^{2}-O P^{2}=15^{2}-9^{2}=144\n$$\n\nso $A P=12$.\n\nSince $P$ has coordinates $(9,0)$ and $A$ is 12 units directly above $P$, then $A$ has coordinates $(9,12)$.\n\nSince $P B=4$, then $B$ has coordinates $(13,0)$.\n\nThe line through $A(9,12)$ and $B(13,0)$ has slope $\\frac{12-0}{9-13}=-3$ so, using the point-slope form, has equation $y-0=-3(x-13)$ or $y=-3 x+39$.'] ['$y=-3 x+39$'] 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Multimodal Competition False Expression Open-ended Geometry Math English +43 "In the diagram, $\triangle A B C$ is right-angled at $B$ and $A B=10$. If $\cos (\angle B A C)=\frac{5}{13}$, what is the value of $\tan (\angle A C B)$ ? + +" ['Since $\\cos (\\angle B A C)=\\frac{A B}{A C}$ and $\\cos (\\angle B A C)=\\frac{5}{13}$ and $A B=10$, then $A C=\\frac{13}{5} A B=26$.\n\nSince $\\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem, $B C^{2}=A C^{2}-A B^{2}=26^{2}-10^{2}=576$ so $B C=24$ since $B C>0$.\n\nTherefore, $\\tan (\\angle A C B)=\\frac{A B}{B C}=\\frac{10}{24}=\\frac{5}{12}$.'] ['$\\frac{5}{12}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +44 "In the diagram, $A B=B C=2 \sqrt{2}, C D=D E$, $\angle C D E=60^{\circ}$, and $\angle E A B=75^{\circ}$. Determine the perimeter of figure $A B C D E$. Explain how you got your answer. + +" ['Since $\\triangle A B C$ is isosceles and right-angled, then $\\angle B A C=45^{\\circ}$.\n\nAlso, $A C=\\sqrt{2} A B=\\sqrt{2}(2 \\sqrt{2})=4$.\n\nSince $\\angle E A B=75^{\\circ}$ and $\\angle B A C=45^{\\circ}$, then $\\angle C A E=\\angle E A B-\\angle B A C=30^{\\circ}$.\n\nSince $\\triangle A E C$ is right-angled and has a $30^{\\circ}$ angle, then $\\triangle A E C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nThus, $E C=\\frac{1}{2} A C=2$ (since $E C$ is opposite the $30^{\\circ}$ angle) and $A E=\\frac{\\sqrt{3}}{2} A C=2 \\sqrt{3}$ (since $A E$ is opposite the $60^{\\circ}$ angle).\n\nIn $\\triangle C D E, E D=D C$ and $\\angle E D C=60^{\\circ}$, so $\\triangle C D E$ is equilateral.\n\nTherefore, $E D=C D=E C=2$.\n\nOverall, the perimeter of $A B C D E$ is\n\n$$\nA B+B C+C D+D E+E A=2 \\sqrt{2}+2 \\sqrt{2}+2+2+2 \\sqrt{3}=4+4 \\sqrt{2}+2 \\sqrt{3}\n$$'] ['$4+4 \\sqrt{2}+2 \\sqrt{3}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +45 "In the diagram, the parabola intersects the $x$-axis at $A(-3,0)$ and $B(3,0)$ and has its vertex at $C$ below the $x$-axis. The area of $\triangle A B C$ is 54 . Determine the equation of the parabola. Explain how you got your answer. + +" "['From the diagram, the parabola has $x$-intercepts $x=3$ and $x=-3$.\n\nTherefore, the equation of the parabola is of the form $y=a(x-3)(x+3)$ for some real number $a$.\n\nTriangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).\n\nSuppose $C$ has coordinates $(0,-c)$. Then $O C=c$.\n\nThus, the area of $\\triangle A B C$ is $\\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.\n\nSince the parabola passes through $C(0,-18)$, then this point must satisfy the equation of the parabola.\n\nTherefore, $-18=a(0-3)(0+3)$ or $-18=-9 a$ or $a=2$.\n\nThus, the equation of the parabola is $y=2(x-3)(x+3)=2 x^{2}-18$.' + 'Triangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).\n\nSuppose $C$ has coordinates $(0,-c)$. Then $O C=c$.\n\nThus, the area of $\\triangle A B C$ is $\\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.\n\nTherefore, the parabola has vertex $C(0,-18)$, so has equation $y=a(x-0)^{2}-18$.\n\n(The vertex of the parabola must lie on the $y$-axis since its roots are equally distant from the $y$-axis, so $C$ must be the vertex.)\n\nSince the parabola passes through $B(3,0)$, then these coordinates satisfy the equation, so $0=3^{2} a-18$ or $9 a=18$ or $a=2$.\n\nTherefore, the equation of the parabola is $y=2 x^{2}-18$.']" ['$y=2 x^{2}-18$'] 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Multimodal Competition False Expression Open-ended Geometry Math English +46 "In the diagram, $A(0, a)$ lies on the $y$-axis above $D$. If the triangles $A O B$ and $B C D$ have the same area, determine the value of $a$. Explain how you got your answer. + +" ['$\\triangle A O B$ is right-angled at $O$, so has area $\\frac{1}{2}(A O)(O B)=\\frac{1}{2} a(1)=\\frac{1}{2} a$.\n\nWe next need to calculate the area of $\\triangle B C D$.\n\nMethod 1: Completing the trapezoid\n\nDrop a perpendicular from $C$ to $P(3,0)$ on the $x$-axis.\n\n\n\nThen $D O P C$ is a trapezoid with parallel sides $D O$ of length 1 and $P C$ of length 2 and height $O P$ (which is indeed perpendicular to the parallel sides) of length 3.\n\nThe area of the trapezoid is thus $\\frac{1}{2}(D O+P C)(O P)=\\frac{1}{2}(1+2)(3)=\\frac{9}{2}$.\n\nBut the area of $\\triangle B C D$ equals the area of trapezoid $D O P C$ minus the areas of $\\triangle D O B$ and $\\triangle B P C$.\n\n$\\triangle D O B$ is right-angled at $O$, so has area $\\frac{1}{2}(D O)(O B)=\\frac{1}{2}(1)(1)=\\frac{1}{2}$.\n\n$\\triangle B P C$ is right-angled at $P$, so has area $\\frac{1}{2}(B P)(P C)=\\frac{1}{2}(2)(2)=2$.\n\nThus, the area of $\\triangle D B C$ is $\\frac{9}{2}-\\frac{1}{2}-2=2$.\n\n\n\n(A similar method for calculating the area of $\\triangle D B C$ would be to drop a perpendicular to $Q$ on the $y$-axis, creating a rectangle $Q O P C$.)\n\nMethod 2: $\\triangle D B C$ is right-angled\n\nThe slope of line segment $D B$ is $\\frac{1-0}{0-1}=-1$.\n\nThe slope of line segment $B C$ is $\\frac{2-0}{3-1}=1$.\n\nSince the product of these slopes is -1 (that is, their slopes are negative reciprocals), then $D B$ and $B C$ are perpendicular.\n\nTherefore, the area of $\\triangle D B C$ is $\\frac{1}{2}(D B)(B C)$.\n\nNow $D B=\\sqrt{(1-0)^{2}+(0-1)^{2}}=\\sqrt{2}$ and $B C=\\sqrt{(3-1)^{2}+(2-0)^{2}}=\\sqrt{8}$.\n\nThus, the area of $\\triangle D B C$ is $\\frac{1}{2} \\sqrt{2} \\sqrt{8}=2$.\n\nSince the area of $\\triangle A O B$ equals the area of $\\triangle D B C$, then $\\frac{1}{2} a=2$ or $a=4$.'] ['4'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +47 "The Little Prince lives on a spherical planet which has a radius of $24 \mathrm{~km}$ and centre $O$. He hovers in a helicopter $(H)$ at a height of $2 \mathrm{~km}$ above the surface of the planet. From his position in the helicopter, what is the distance, in kilometres, to the furthest point on the surface of the planet that he can see? + +" ['Suppose that $O$ is the centre of the planet, $H$ is the place where His Highness hovers in the helicopter, and $P$ is the furthest point on the surface of the planet that he can see.\n\n\n\nThen $H P$ must be a tangent to the surface of the planet (otherwise he could see further), so $O P$ (a radius) is perpendicular to $H P$ (a tangent).\n\nWe are told that $O P=24 \\mathrm{~km}$.\n\nSince the helicopter hovers at a height of $2 \\mathrm{~km}$, then $O H=24+2=26 \\mathrm{~km}$.\n\nTherefore, $H P^{2}=O H^{2}-O P^{2}=26^{2}-24^{2}=100$, so $H P=10 \\mathrm{~km}$.\n\nTherefore, the distance to the furthest point that he can see is $10 \\mathrm{~km}$.'] ['10'] 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Multimodal Competition False km Numerical Open-ended Geometry Math English +48 "In the diagram, points $A$ and $B$ are located on islands in a river full of rabid aquatic goats. Determine the distance from $A$ to $B$, to the nearest metre. (Luckily, someone has measured the angles shown in the diagram as well as the distances $C D$ and $D E$.) + +" ['Since we know the measure of $\\angle A D B$, then to find the distance $A B$, it is enough to find the distances $A D$ and $B D$ and then apply the cosine law.\n\nIn $\\triangle D B E$, we have $\\angle D B E=180^{\\circ}-20^{\\circ}-70^{\\circ}=90^{\\circ}$, so $\\triangle D B E$ is right-angled, giving $B D=100 \\cos \\left(20^{\\circ}\\right) \\approx 93.969$.\n\nIn $\\triangle D A C$, we have $\\angle D A C=180^{\\circ}-50^{\\circ}-45^{\\circ}=85^{\\circ}$.\n\nUsing the sine law, $\\frac{A D}{\\sin \\left(50^{\\circ}\\right)}=\\frac{C D}{\\sin \\left(85^{\\circ}\\right)}$, so $A D=\\frac{150 \\sin \\left(50^{\\circ}\\right)}{\\sin \\left(85^{\\circ}\\right)} \\approx 115.346$.\n\n\n\nFinally, using the cosine law in $\\triangle A B D$, we get\n\n$$\n\\begin{aligned}\nA B^{2} & =A D^{2}+B D^{2}-2(A D)(B D) \\cos (\\angle A D B) \\\\\nA B^{2} & \\approx(115.346)^{2}+(93.969)^{2}-2(115.346)(93.969) \\cos \\left(35^{\\circ}\\right) \\\\\nA B^{2} & \\approx 4377.379 \\\\\nA B & \\approx 66.16\n\\end{aligned}\n$$\n\nTherefore, the distance from $A$ to $B$ is approximately $66 \\mathrm{~m}$.'] ['66'] 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Multimodal Competition False m Numerical Open-ended Geometry Math English +49 "In the $4 \times 4$ grid shown, three coins are randomly placed in different squares. Determine the probability that no two coins lie in the same row or column. + +" ['We consider placing the three coins individually.\n\nPlace one coin randomly on the grid.\n\nWhen the second coin is placed (in any one of 15 squares), 6 of the 15 squares will leave two coins in the same row or column and 9 of the 15 squares will leave the two coins in different rows and different columns.\n\n\n\nTherefore, the probability that the two coins are in different rows and different columns is $\\frac{9}{15}=\\frac{3}{5}$.\n\nThere are 14 possible squares in which the third coin can be placed.\n\n\n\nOf these 14 squares, 6 lie in the same row or column as the first coin and an additional 4 lie the same row or column as the second coin. Therefore, the probability that the third coin is placed in a different row and a different column than each of the first two coins is $\\frac{4}{14}=\\frac{2}{7}$.\n\nTherefore, the probability that all three coins are placed in different rows and different columns is $\\frac{3}{5} \\times \\frac{2}{7}=\\frac{6}{35}$.'] ['$\\frac{6}{35}$'] 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'] Multimodal Competition False Numerical Open-ended Geometry Math English +50 "In the diagram, the area of $\triangle A B C$ is 1 . Trapezoid $D E F G$ is constructed so that $G$ is to the left of $F, D E$ is parallel to $B C$, $E F$ is parallel to $A B$ and $D G$ is parallel to $A C$. Determine the maximum possible area of trapezoid $D E F G$. + +" ['Suppose that $A B=c, A C=b$ and $B C=a$.\n\nSince $D G$ is parallel to $A C, \\angle B D G=\\angle B A C$ and $\\angle D G B=\\angle A C B$, so $\\triangle D G B$ is similar to $\\triangle A C B$.\n\n(Similarly, $\\triangle A E D$ and $\\triangle E C F$ are also both similar to $\\triangle A B C$.)\n\nSuppose next that $D B=k c$, with $0" ['Using the cosine law in $\\triangle P Q R$,\n\n$$\n\\begin{aligned}\nP R^{2} & =P Q^{2}+Q R^{2}-2 \\cdot P Q \\cdot Q R \\cdot \\cos (\\angle P Q R) \\\\\n21^{2} & =a^{2}+b^{2}-2 a b \\cos \\left(60^{\\circ}\\right) \\\\\n441 & =a^{2}+b^{2}-2 a b \\cdot \\frac{1}{2} \\\\\n441 & =a^{2}+b^{2}-a b\n\\end{aligned}\n$$\n\nUsing the sine law in $\\triangle S T U$, we obtain $\\frac{S T}{\\sin (\\angle T U S)}=\\frac{T U}{\\sin (\\angle T S U)}$ and so $\\frac{a}{4 / 5}=\\frac{b}{\\sin \\left(30^{\\circ}\\right)}$. Therefore, $\\frac{a}{4 / 5}=\\frac{b}{1 / 2}$ and so $a=\\frac{4}{5} \\cdot 2 b=\\frac{8}{5} b$.\n\nSubstituting into the previous equation,\n\n$$\n\\begin{aligned}\n& 441=\\left(\\frac{8}{5} b\\right)^{2}+b^{2}-\\left(\\frac{8}{5} b\\right) b \\\\\n& 441=\\frac{64}{25} b^{2}+b^{2}-\\frac{8}{5} b^{2} \\\\\n& 441=\\frac{64}{25} b^{2}+\\frac{25}{25} b^{2}-\\frac{40}{25} b^{2} \\\\\n& 441=\\frac{49}{25} b^{2} \\\\\n& 225=b^{2}\n\\end{aligned}\n$$\n\nSince $b>0$, then $b=15$ and so $a=\\frac{8}{5} b=\\frac{8}{5} \\cdot 15=24$.'] ['$24,15$'] 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Multimodal Competition True Numerical Open-ended Geometry Math English +52 "A triangle of area $770 \mathrm{~cm}^{2}$ is divided into 11 regions of equal height by 10 lines that are all parallel to the base of the triangle. Starting from the top of the triangle, every other region is shaded, as shown. What is the total area of the shaded regions? + +" "['We make two copies of the given triangle, labelling them $\\triangle A B C$ and $\\triangle D E F$, as shown:\n\n\nThe combined area of these two triangles is $2 \\cdot 770 \\mathrm{~cm}^{2}=1540 \\mathrm{~cm}^{2}$, and the shaded area in each triangle is the same.\n\nNext, we rotate $\\triangle D E F$ by $180^{\\circ}$ :\n\n\nand join the two triangles together:\n\n\n\nWe note that $B C$ and $A E$ (which was $F E$ ) are equal in length (since they were copies of each other) and parallel (since they are $180^{\\circ}$ rotations of each other). The same is true for $A B$ and $E C$.\n\nTherefore, $A B C E$ is a parallelogram.\n\nFurther, $A B C E$ is divided into 11 identical parallelograms (6 shaded and 5 unshaded) by the horizontal lines. (Since the sections of the two triangles are equal in height, the horizontal lines on both sides of $A C$ align.)\n\nThe total area of parallelogram $A B C E$ is $1540 \\mathrm{~cm}^{2}$.\n\nThus, the shaded area of $A B C E$ is $\\frac{6}{11} \\cdot 1540 \\mathrm{~cm}^{2}=840 \\mathrm{~cm}^{2}$.\n\nSince this shaded area is equally divided between the two halves of the parallelogram, then the combined area of the shaded regions of $\\triangle A B C$ is $\\frac{1}{2} \\cdot 840 \\mathrm{~cm}^{2}=420 \\mathrm{~cm}^{2}$.' + 'We label the points where the horizontal lines touch $A B$ and $A C$ as shown:\n\n\n\nWe use the notation $|\\triangle A B C|$ to represent the area of $\\triangle A B C$ and use similar notation for the area of other triangles and quadrilaterals.\n\nLet $\\mathcal{A}$ be equal to the total area of the shaded regions.\n\nThus,\n\n$$\n\\mathcal{A}=\\left|\\triangle A B_{1} C_{1}\\right|+\\left|B_{2} B_{3} C_{3} C_{2}\\right|+\\left|B_{4} B_{5} C_{5} C_{4}\\right|+\\left|B_{6} B_{7} C_{7} C_{6}\\right|+\\left|B_{8} B_{9} C_{9} C_{8}\\right|+\\left|B_{10} B C C_{10}\\right|\n$$\n\nThe area of each of these quadrilaterals is equal to the difference of the area of two triangles. For example,\n\n$$\n\\left|B_{2} B_{3} C_{3} C_{2}\\right|=\\left|\\triangle A B_{3} C_{3}\\right|-\\left|\\triangle A B_{2} C_{2}\\right|=-\\left|\\triangle A B_{2} C_{2}\\right|+\\left|\\triangle A B_{3} C_{3}\\right|\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\mathcal{A}=\\mid & \\triangle A B_{1} C_{1}|-| \\triangle A B_{2} C_{2}|+| \\triangle A B_{3} C_{3}|-| \\triangle A B_{4} C_{4}|+| \\triangle A B_{5} C_{5} \\mid \\\\\n& \\quad-\\left|\\triangle A B_{6} C_{6}\\right|+\\left|\\triangle A B_{7} C_{7}\\right|-\\left|\\triangle A B_{8} C_{8}\\right|+\\left|\\triangle A B_{9} C_{9}\\right|-\\left|\\triangle A B_{10} C_{10}\\right|+|\\triangle A B C|\n\\end{aligned}\n$$\n\nEach of $\\triangle A B_{1} C_{1}, \\triangle A B_{2} C_{2}, \\ldots, \\triangle A B_{10} C_{10}$ is similar to $\\triangle A B C$ because their two base angles are equal due.\n\nSuppose that the height of $\\triangle A B C$ from $A$ to $B C$ is $h$.\n\nSince the height of each of the 11 regions is equal in height, then the height of $\\triangle A B_{1} C_{1}$ is $\\frac{1}{11} h$, the height of $\\triangle A B_{2} C_{2}$ is $\\frac{2}{11} h$, and so on.\n\nWhen two triangles are similar, their heights are in the same ratio as their side lengths:\n\nTo see this, suppose that $\\triangle P Q R$ is similar to $\\triangle S T U$ and that altitudes are drawn from $P$ and $S$ to $V$ and $W$.\n\n\nSince $\\angle P Q R=\\angle S T U$, then $\\triangle P Q V$ is similar to $\\triangle S T W$ (equal angle; right angle), which means that $\\frac{P Q}{S T}=\\frac{P V}{S W}$. In other words, the ratio of sides is equal to the ratio of heights.\n\nSince the height of $\\triangle A B_{1} C_{1}$ is $\\frac{1}{11} h$, then $B_{1} C_{1}=\\frac{1}{11} B C$.\n\nTherefore, $\\left|\\triangle A B_{1} C_{1}\\right|=\\frac{1}{2} \\cdot B_{1} C_{1} \\cdot \\frac{1}{11} h=\\frac{1}{2} \\cdot \\frac{1}{11} B C \\cdot \\frac{1}{11} h=\\frac{1^{2}}{11^{2}} \\cdot \\frac{1}{2} \\cdot B C \\cdot h=\\frac{1^{2}}{11^{2}}|\\triangle A B C|$.\n\nSimilarly, since the height of $\\triangle A B_{2} C_{2}$ is $\\frac{2}{11} h$, then $B_{2} C_{2}=\\frac{2}{11} B C$.\n\n\n\nTherefore, $\\left|\\triangle A B_{2} C_{2}\\right|=\\frac{1}{2} \\cdot B_{2} C_{2} \\cdot \\frac{2}{11} h=\\frac{1}{2} \\cdot \\frac{2}{11} B C \\cdot \\frac{2}{11} h=\\frac{2^{2}}{11^{2}} \\cdot \\frac{1}{2} \\cdot B C \\cdot h=\\frac{2^{2}}{11^{2}}|\\triangle A B C|$.\n\nThis result continues for each of the triangles.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\mathcal{A} & =\\frac{1^{2}}{11^{2}}|\\triangle A B C|-\\frac{2^{2}}{11^{2}}|\\triangle A B C|+\\frac{3^{2}}{11^{2}}|\\triangle A B C|-\\frac{4^{2}}{11^{2}}|\\triangle A B C|+\\frac{5^{2}}{11^{2}}|\\triangle A B C| \\\\\n& \\quad-\\frac{6^{2}}{11^{2}}|\\triangle A B C|+\\frac{7^{2}}{11^{2}}|\\triangle A B C|-\\frac{8^{2}}{11^{2}}|\\triangle A B C|+\\frac{9^{2}}{11^{2}}|\\triangle A B C|-\\frac{10^{2}}{11^{2}}|\\triangle A B C|+\\frac{11^{2}}{11^{2}}|\\triangle A B C| \\\\\n& =\\frac{1}{11^{2}}|\\triangle A B C|\\left(11^{2}-10^{2}+9^{2}-8^{2}+7^{2}-6^{2}+5^{2}-4^{2}+3^{2}-2^{2}+1\\right) \\\\\n& =\\frac{1}{11^{2}}\\left(770 \\mathrm{~cm}^{2}\\right)((11+10)(11-10)+(9+8)(9-8)+\\cdots+(3+2)(3-2)+1) \\\\\n& =\\frac{1}{11^{2}}\\left(770 \\mathrm{~cm}^{2}\\right)(11+10+9+8+7+6+5+4+3+2+1) \\\\\n& =\\frac{1}{11}\\left(70 \\mathrm{~cm}^{2}\\right) \\cdot 66 \\\\\n& =420 \\mathrm{~cm}^{2}\n\\end{aligned}\n$$\n\nTherefore, the combined area of the shaded regions of $\\triangle A B C$ is $420 \\mathrm{~cm}^{2}$.']" ['$420$'] 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'] Multimodal Competition False Numerical Open-ended Geometry Math English +53 "A square lattice of 16 points is constructed such that the horizontal and vertical distances between adjacent points are all exactly 1 unit. Each of four pairs of points are connected by a line segment, as shown. The intersections of these line segments are the vertices of square $A B C D$. Determine the area of square $A B C D$. + +" "['We label five additional points in the diagram:\n\n\n\nSince $P Q=Q R=R S=1$, then $P S=3$ and $P R=2$.\n\nSince $\\angle P S T=90^{\\circ}$, then $P T=\\sqrt{P S^{2}+S T^{2}}=\\sqrt{3^{2}+1^{2}}=\\sqrt{10}$ by the Pythagorean Theorem.\n\nWe are told that $A B C D$ is a square.\n\nThus, $P T$ is perpendicular to $Q C$ and to $R B$.\n\nThus, $\\triangle P D Q$ is right-angled at $D$ and $\\triangle P A R$ is right-angled at $A$.\n\nSince $\\triangle P D Q, \\triangle P A R$ and $\\triangle P S T$ are all right-angled and all share an angle at $P$, then these three triangles are similar.\n\nThis tells us that $\\frac{P A}{P S}=\\frac{P R}{P T}$ and so $P A=\\frac{3 \\cdot 2}{\\sqrt{10}}$. Also, $\\frac{P D}{P S}=\\frac{P Q}{P T}$ and so $P D=\\frac{1 \\cdot 3}{\\sqrt{10}}$.\n\nTherefore,\n\n$$\nD A=P A-P D=\\frac{6}{\\sqrt{10}}-\\frac{3}{\\sqrt{10}}=\\frac{3}{\\sqrt{10}}\n$$\n\nThis means that the area of square $A B C D$ is equal to $D A^{2}=\\left(\\frac{3}{\\sqrt{10}}\\right)^{2}=\\frac{9}{10}$.' + 'We add coordinates to the diagram as shown:\n\n\n\nWe determine the side length of square $A B C D$ by determining the coordinates of $D$ and $A$ and then calculating the distance between these points.\n\nThe slope of the line through $(0,3)$ and $(3,2)$ is $\\frac{3-2}{0-3}=-\\frac{1}{3}$.\n\nThis equation of this line can be written as $y=-\\frac{1}{3} x+3$.\n\nThe slope of the line through $(0,0)$ and $(1,3)$ is 3.\n\nThe equation of this line can be written as $y=3 x$.\n\nThe slope of the line through $(1,0)$ and $(2,3)$ is also 3.\n\nThe equation of this line can be written as $y=3(x-1)=3 x-3$.\n\nPoint $D$ is the intersection point of the lines with equations $y=-\\frac{1}{3} x+3$ and $y=3 x$.\n\nEquating expressions for $y$, we obtain $-\\frac{1}{3} x+3=3 x$ and so $\\frac{10}{3} x=3$ which gives $x=\\frac{9}{10}$.\n\nSince $y=3 x$, we get $y=\\frac{27}{10}$ and so the coordinates of $D$ are $\\left(\\frac{9}{10}, \\frac{27}{10}\\right)$.\n\nPoint $A$ is the intersection point of the lines with equations $y=-\\frac{1}{3} x+3$ and $y=3 x-3$.\n\nEquating expressions for $y$, we obtain $-\\frac{1}{3} x+3=3 x-3$ and so $\\frac{10}{3} x=6$ which gives $x=\\frac{18}{10}$.\n\nSince $y=3 x-3$, we get $y=\\frac{24}{10}$ and so the coordinates of $A$ are $\\left(\\frac{18}{10}, \\frac{24}{10}\\right)$. (It is easier to not reduce these fractions.)\n\nTherefore,\n\n$$\nD A=\\sqrt{\\left(\\frac{9}{10}-\\frac{18}{10}\\right)^{2}+\\left(\\frac{27}{10}-\\frac{24}{10}\\right)^{2}}=\\sqrt{\\left(-\\frac{9}{10}\\right)^{2}+\\left(\\frac{3}{10}\\right)^{2}}=\\sqrt{\\frac{90}{100}}=\\sqrt{\\frac{9}{10}}\n$$\n\nThis means that the area of square $A B C D$ is equal to $D A^{2}=\\left(\\sqrt{\\frac{9}{10}}\\right)^{2}=\\frac{9}{10}$.']" ['$\\frac{9}{10}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +54 "At the Canadian Eatery with Multiple Configurations, there are round tables, around which chairs are placed. When a table has $n$ chairs around it for some integer $n \geq 3$, the chairs are labelled $1,2,3, \ldots, n-1, n$ in order around the table. A table is considered full if no more people can be seated without having two people sit in neighbouring chairs. For example, when $n=6$, full tables occur when people are seated in chairs labelled $\{1,4\}$ or $\{2,5\}$ or $\{3,6\}$ or $\{1,3,5\}$ or $\{2,4,6\}$. Thus, there are 5 different full tables when $n=6$. + + +A full table with $6 k+5$ chairs, for some positive integer $k$, has $t$ people seated in its chairs. Determine, in terms of $k$, the number of possible values of $t$." ['Suppose that $k$ is a positive integer.\n\nSuppose that $t$ people are seated at a table with $6 k+5$ chairs so that the table is full.\n\nWhen $t$ people are seated, there are $t$ gaps. Each gap consists of either 1 or 2 chairs. (A gap with 3 or more chairs can have an additional person seated in it, so the table is not full.)\n\nTherefore, there are between $t$ and $2 t$ empty chairs.\n\nThis means that the total number of chairs is between $t+t$ and $t+2 t$.\n\nIn other words, $2 t \\leq 6 k+5 \\leq 3 t$.\n\nSince $2 t \\leq 6 k+5$, then $t \\leq 3 k+\\frac{5}{2}$. Since $k$ and $t$ are integers, then $t \\leq 3 k+2$.\n\nWe note that it is possible to seat $3 k+2$ people around the table in seats\n\n$$\n\\{2,4,6, \\ldots, 6 k+2,6 k+4\\}\n$$\n\nThis table is full becase $3 k+1$ of the gaps consist of 1 chair and 1 gap consists of 2 chairs. Since $3 t \\geq 6 k+5$, then $t \\geq 2 k+\\frac{5}{3}$. Since $k$ and $t$ are integers, then $t \\geq 2 k+2$.\n\nWe note that it is possible to seat $2 k+2$ people around the table in seats\n\n$$\n\\{3,6,9, \\ldots, 6 k, 6 k+3,6 k+5\\}\n$$\n\nThis table is full becase $2 k+1$ of the gaps consist of 2 chairs and 1 gap consists of 1 chair.\n\nWe now know that, if there are $t$ people seated at a full table with $6 k+5$ chairs, then $2 k+2 \\leq t \\leq 3 k+2$.\n\nTo confirm that every such value of $t$ is possible, consider a table with $t$ people, $3 t-(6 k+5)$\n\n\n\ngaps of 1 chair, and $(6 k+5)-2 t$ gaps of 2 chairs.\n\nFrom the work above, we know that $3 t \\geq 6 k+5$ and so $3 t-(6 k+5) \\geq 0$, and that $2 t \\leq 6 k+5$ and so $(6 k+5)-2 t \\geq 0$.\n\nThe total number of gaps is $3 t-(6 k+5)+(6 k+5)-2 t=t$, since there are $t$ people seated.\n\nFinally, the total number of chairs is\n\n$$\nt+1 \\cdot(3 t-(6 k+5))+2 \\cdot((6 k+5)-2 t)=t+3 t-4 t-(6 k+5)+2(6 k+5)=6 k+5\n$$\n\nas expected.\n\nThis shows that every $t$ with $2 k+2 \\leq t \\leq 3 k+2$ can produce a full table.\n\nTherefore, the possible values of $t$ are those integers that satisfy $2 k+2 \\leq t \\leq 3 k+2$.\n\nThere are $(3 k+2)-(2 k+2)+1=k+1$ possible values of $t$.'] ['$k+1$'] 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Multimodal Competition False Expression Open-ended Combinatorics Math English +55 "At the Canadian Eatery with Multiple Configurations, there are round tables, around which chairs are placed. When a table has $n$ chairs around it for some integer $n \geq 3$, the chairs are labelled $1,2,3, \ldots, n-1, n$ in order around the table. A table is considered full if no more people can be seated without having two people sit in neighbouring chairs. For example, when $n=6$, full tables occur when people are seated in chairs labelled $\{1,4\}$ or $\{2,5\}$ or $\{3,6\}$ or $\{1,3,5\}$ or $\{2,4,6\}$. Thus, there are 5 different full tables when $n=6$. + + +Determine the number of different full tables when $n=19$." "['For each integer $n \\geq 3$, we define $f(n)$ to be the number of different full tables of size $n$. We can check that\n\n- $f(3)=3$ because the full tables when $n=3$ have people in chairs $\\{1\\},\\{2\\},\\{3\\}$,\n- $f(4)=2$ because the full tables when $n=4$ have people in chairs $\\{1,3\\},\\{2,4\\}$, and\n- $f(5)=5$ because the full tables when $n=4$ have people in chairs $\\{1,3\\},\\{2,4\\},\\{3,5\\}$, $\\{4,1\\},\\{5,2\\}$.\n\nIn the problem, we are told that $f(6)=5$ and in part (a), we determined that $f(8)=10$. This gives us the following table:\n\n| $n$ | $f(n)$ |\n| :---: | :---: |\n| 3 | 3 |\n| 4 | 2 |\n| 5 | 5 |\n| 6 | 5 |\n| 7 | $?$ |\n| 8 | 10 |\n\nBased on this information, we make the guess that for every integer $n \\geq 6$, we have $f(n)=f(n-2)+f(n-3)$.\n\nFor example, this would mean that $f(7)=f(5)+f(4)=5+2=7$ which we can verify is true.\n\nBased on this recurrence relation (which we have yet to prove), we deduce the values of $f(n)$ up to and including $n=19$ :\n\n| $n$ | $f(n)$ | $n$ | $f(n)$ |\n| :---: | :---: | :---: | :---: |\n| 3 | 3 | 11 | 22 |\n| 4 | 2 | 12 | 29 |\n| 5 | 5 | 13 | 39 |\n| 6 | 5 | 14 | 51 |\n| 7 | 7 | 15 | 68 |\n| 8 | 10 | 16 | 90 |\n| 9 | 12 | 17 | 119 |\n| 10 | 17 | 18 | 158 |\n| | | 19 | 209 |\n\nWe now need to prove that the equation $f(n)=f(n-2)+f(n-3)$ is true for all $n \\geq 6$.\n\n\n\nWe think about each full table as a string of 0 s and 1s, with 1 representing a chair that is occupied and 0 representing an empty chair.\n\nLet $a(n)$ be the number of full tables with someone in seat 1 (and thus nobody in seat 2). Let $b(n)$ be the number of full tables with someone in seat 2 (and thus nobody in seat 1). Let $c(n)$ be the number of full tables with nobody in seat 1 or in seat 2 .\n\nSince every full table must be in one of these categories, then $f(n)=a(n)+b(n)+c(n)$. A full table with $n$ seats $n \\geq 4$ must correspond to a string that starts with 10,01 or 00 . Since there cannot be more than two consecutive 0s, we can further specify this, namely to say that a full table with $n$ seats must correspond to a string that starts with 1010 or 1001 or 0100 or 0101 or 0010 . In each case, these are the first 4 characters of the string and correspond to full (1) and empty (0) chairs.\n\nConsider the full tables starting with 1010. Note that such strings end with 0 since the table is circular. Removing the 10 from positions 1 and 2 creates strings of length $n-2$ that begin 10. These strings will still correspond to a full table, and so there are $a(n-2)$ such strings. (We note that all possible strings starting 1010 of length $n$ will lead to all possible strings starting with 1010 of length $n-2$.)\n\nConsider the full tables starting with 1001. Note that such a string ends with 0 since the table is circular. Removing the 100 from positions 1, 2 and 3 creates strings of length $n-3$ that begin 10. (There must have been a 0 in position 5 after the 1 in position 4.) These strings will still correspond to full tables, and so there are $a(n-3)$ such strings. Consider the full tables starting with 0100 . Removing the 100 from positions 2,3 and 4 creates strings of length $n-3$ that begin 01 . (There must have been a 1 in position 5 after the 0 in position 4.) These strings will still correspond to full tables, and so there are $b(n-3)$ such strings.\n\nConsider the full tables starting with 0101. Removing the 01 from positions 3 and 4 creates strings of length $n-2$ that begin 01 . (The 1 in position 4 must have been followed by one or two 0s and so these strings maintains the desired properties.) These strings will still correspond to full tables, and so there are $b(n-2)$ such strings.\n\nConsider the full tables starting with 0010. These strings must begin with either 00100 or 00101.\n\nIf strings start 00100 , then they start 001001 and so we remove the 001 in positions 4,5 and 6 and obtain strings of length $n-3$ that start 001 (and thus start 00). There are $c(n-3)$ such strings.\n\nIf strings start 00101, we remove the 01 in positions 4 and 5 and obtain strings of length $n-2$ that start 001 (and thus start 00 ). There are $c(n-2)$ such strings.\n\nThese 6 cases and subcases count all strings counted by $f(n)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nf(n) & =a(n-2)+a(n-3)+b(n-3)+b(n-2)+c(n-3)+c(n-2) \\\\\n& =a(n-2)+b(n-2)+c(n-2)+a(n-3)+b(n-3)+c(n-3) \\\\\n& =f(n-2)+f(n-3)\n\\end{aligned}\n$$\n\nas required, which means that the number of different full tables when $n=19$ is 209 .' + 'Extending our approach from (b), the number of people seated at a full table with 19 chairs is at least $\\frac{19}{3}=6 \\frac{1}{3}$ and at most $\\frac{19}{2}=9 \\frac{1}{2}$.\n\nSince the number of people is an integer, there must be 7,8 or 9 people at the table, which means that the number of empty chairs is 12,11 or 10 , respectively.\n\n\n\nSuppose that there are 9 people and 9 gaps with a total of 10 empty chairs.\n\nIn this case, there is 1 gap with 2 empty chairs and 8 gaps with 1 empty chair.\n\nThere are 19 pairs of chairs in which we can put 2 people with a gap of 2 in between: $\\{1,4\\},\\{2,5\\}, \\ldots,\\{19,3\\}$.\n\nOnce we choose one of these pairs, the seat choice for the remaining 8 people is completely determined by placing people in every other chair.\n\nTherefore, there are 19 different full tables with 9 people.\n\nSuppose that there are 8 people and 8 gaps with a total of 11 empty chairs.\n\nIn this case, there are 3 gaps with 2 empty chairs and 5 gaps with 1 empty chair. There are 7 different circular orderings in which these 8 gaps can be arranged:\n\n$$\n\\begin{array}{lllllll}\n22211111 & 22121111 & 22112111 & 22111211 & 22111121 & 21212111 & 21211211\n\\end{array}\n$$\n\nWe note that ""22211111"" would be the same as, for example, ""11222111"" since these gaps are arranged around a circle.\n\nIf the three gaps of length 2 are consecutive, there is only one configuration (22211111). If there are exactly 2 consecutive gaps of length 2 , there are 4 relative places in which the third gap of length 2 can be placed.\n\nIf there are no consecutive gaps of length 2 , these gaps can either be separated by 1 gap each (21212111) with 3 gaps on the far side, or can be separated by 1 gap, 2 gaps, and 2 gaps (21211211). There is only one configuration for the gaps in this last situation.\n\nThere are 7 different circular orderings for these 8 gaps.\n\nEach of these 7 different orderings can be placed around the circle of 19 chairs in 19 different ways, because each can be started in 19 different places. Because 19 is prime, none of these orderings overlap.\n\nTherefore, there are $7 \\cdot 19=133$ different full tables with 8 people.\n\nSuppose that there are 7 people and 7 gaps with a total of 12 empty chairs.\n\nIn this case, there are 2 gaps with 1 empty chair and 5 gaps with 2 empty chairs.\n\nThe 2 gaps with 1 empty chair can be separated by 0 gaps with 2 empty chairs, 1 gap with 2 empty chairs, or 2 gaps with 2 empty chairs. Because the chairs are around a circle, if there were 3 , 4 or 5 gaps with 2 empty chairs between them, there would be 2,1 or 0 gaps going the other way around the circle.\n\nThis means that there are 3 different configurations for the gaps.\n\nEach of these configurations can be placed in 19 different ways around the circle of chairs. Therefore, there are $3 \\cdot 19=57$ full tables with 7 people.\n\nIn total, there are $19+133+57=209$ full tables with 19 chairs.' + 'As in Solution 2, there must be 7, 8 or 9 people in chairs, and so there are 7,8 or 9 gaps. If there are 7 gaps, there are 2 gaps of 1 chair and 5 gaps of 2 chairs.\n\nIf there are 8 gaps, there are 5 gaps of 1 chair and 3 gaps of 2 chairs.\n\nIf there are 9 gaps, there are 8 gaps of 1 chair and 1 gap of 2 chairs.\n\nWe consider three mutually exclusive cases: (i) there is a person in chair 1 and not in chair 2 , (ii) there is a person in chair 2 and not in chair 1 , and (iii) there is nobody in chair 1 or in chair 2. Every full table fits into exactly one of these three cases.\n\nCase (i): there is a person in chair 1 and not in chair 2\n\nWe use the person in chair 1 to ""anchor"" the arrangement, by starting at chair 1 and\n\n\n\narranging the gaps (and thus the full chairs) clockwise around the table from chair 1.\n\nIf there are 7 gaps, we need to choose 2 of them to be of length 1 , and so there are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIf there are 8 gaps, we need to choose 3 of them to be of length 2 , and so there are $\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIf there are 9 gaps, we need to choose 1 of them to be of length 2 , and so there are $\\left(\\begin{array}{l}9 \\\\ 1\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIn this case, there are a total of $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)+\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)+\\left(\\begin{array}{l}9 \\\\ 1\\end{array}\\right)=21+56+9=86$ full tables.\n\nCase (ii): there is a person in chair 2 and not in chair 1\n\nWe use the same reasoning starting with the person in chair 2 as the anchor.\n\nAgain, there are 86 full tables in this case.\n\nCase (iii): there is nobody in chair 1 or chair 2\n\nSince there is nobody in chair 1 or chair 2 , there must be a person in chair 3 and also in chair 19 , which fixes one gap of 2 chairs.\n\nHere, we use the person in chair 3 as the anchor.\n\nIf there are 7 gaps, there are 2 gaps of 1 chair and 4 gaps of 2 chairs left to place. There are $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)$ ways of doing this.\n\nIf there are 8 gaps, there are 5 gaps of 1 chair and 2 gaps of 2 chairs left to place. There are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)$ ways of doing this.\n\nIf there are 9 gaps, there are 8 gaps of 1 chair and 0 gaps of 2 chairs left to place. There is 1 way to do this.\n\nIn this case, there are a total of $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)+\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)+1=15+21+1=37$ full tables.\n\nIn total, there are $86+86+37=209$ full tables with 19 chairs.']" ['209'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +56 "In the diagram, $\triangle A B C$ is right-angled at $B$ and $\triangle A C D$ is right-angled at $A$. Also, $A B=3, B C=4$, and $C D=13$. What is the area of quadrilateral $A B C D$ ? + +" "['The area of quadrilateral $A B C D$ is the sum of the areas of $\\triangle A B C$ and $\\triangle A C D$.\n\nSince $\\triangle A B C$ is right-angled at $B$, its area equals $\\frac{1}{2}(A B)(B C)=\\frac{1}{2}(3)(4)=6$.\n\nSince $\\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem,\n\n$$\nA C=\\sqrt{A B^{2}+B C^{2}}=\\sqrt{3^{2}+4^{2}}=\\sqrt{25}=5\n$$\n\nbecause $A C>0$. (We could have also observed that $\\triangle A B C$ must be a ""3-4-5"" triangle.) Since $\\triangle A C D$ is right-angled at $A$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{C D^{2}-A C^{2}}=\\sqrt{13^{2}-5^{2}}=\\sqrt{144}=12\n$$\n\nbecause $A D>0$. (We could have also observed that $\\triangle A C D$ must be a "" $5-12-13$ "" triangle.) Thus, the area of $\\triangle A C D$ equals $\\frac{1}{2}(A C)(A D)=\\frac{1}{2}(5)(12)=30$.\n\nFinally, the area of quadrilateral $A B C D$ is thus $6+30=36$.']" ['36'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +57 "Three identical rectangles $P Q R S$, WTUV and $X W V Y$ are arranged, as shown, so that $R S$ lies along $T X$. The perimeter of each of the three rectangles is $21 \mathrm{~cm}$. What is the perimeter of the whole shape? +" ['Let the width of each of the identical rectangles be $a$.\n\nIn other words, $Q P=R S=T W=W X=U V=V Y=a$.\n\nLet the height of each of the identical rectangles be $b$.\n\nIn other words, $Q R=P S=T U=W V=X Y=b$.\n\nThe perimeter of the whole shape equals\n\n$$\nQ P+P S+S X+X Y+V Y+U V+T U+T R+Q R\n$$\n\nSubstituting for known lengths, we obtain\n\n$$\na+b+S X+b+a+a+b+T R+b\n$$\n\nor $3 a+4 b+(S X+T R)$.\n\nBut $S X+T R=(T R+R S+S X)-R S=(T W+W X)-R S=a+a-a=a$.\n\nTherefore, the perimeter of the whole shape equals $4 a+4 b$.\n\nThe perimeter of one rectangle is $2 a+2 b$, which we are told equals $21 \\mathrm{~cm}$.\n\nFinally, the perimeter of the whole shape is thus $2(2 a+2 b)$ which equals $42 \\mathrm{~cm}$.'] ['42'] 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'] Multimodal Competition False cm Numerical Open-ended Geometry Math English +58 "The diagram shows two hills that meet at $O$. One hill makes a $30^{\circ}$ angle with the horizontal and the other hill makes a $45^{\circ}$ angle with the horizontal. Points $A$ and $B$ are on the hills so that $O A=O B=20 \mathrm{~m}$. Vertical poles $B D$ and $A C$ are connected by a straight cable $C D$. If $A C=6 \mathrm{~m}$, what is the length of $B D$ for which $C D$ is as short as possible? + +" ['Extend $C A$ and $D B$ downwards until they meet the horizontal through $O$ at $P$ and $Q$, respectively.\n\n\n\nSince $C A$ and $D B$ are vertical, then $\\angle C P O=\\angle D Q O=90^{\\circ}$.\n\nSince $O A=20 \\mathrm{~m}$, then $A P=O A \\sin 30^{\\circ}=(20 \\mathrm{~m}) \\cdot \\frac{1}{2}=10 \\mathrm{~m}$.\n\nSince $O B=20 \\mathrm{~m}$, then $B Q=O B \\sin 45^{\\circ}=(20 \\mathrm{~m}) \\cdot \\frac{1}{\\sqrt{2}}=10 \\sqrt{2} \\mathrm{~m}$.\n\nSince $A C=6 \\mathrm{~m}$, then $C P=A C+A P=16 \\mathrm{~m}$.\n\nFor $C D$ to be as short as possible and given that $C$ is fixed, then it must be the case that $C D$ is horizontal:\n\nIf $C D$ were not horizontal, then suppose that $X$ is on $D Q$, possibly extended, so that $C X$ is horizontal.\n\n\n\nThen $\\angle C X D=90^{\\circ}$ and so $\\triangle C X D$ is right-angled with hypotenuse $C D$.\n\nIn this case, $C D$ is longer than $C X$ or $X D$.\n\nIn particular, $C D>C X$, which means that if $D$ were at $X$, then $C D$ would be shorter.\n\nIn other words, a horizontal $C D$ makes $C D$ as short as possible.\n\nWhen $C D$ is horizontal, $C D Q P$ is a rectangle, since it has two vertical and two horizontal sides. Thus, $D Q=C P=16 \\mathrm{~m}$.\n\nFinally, this means that $B D=D Q-B Q=(16-10 \\sqrt{2}) \\mathrm{m}$.'] ['$(16-10 \\sqrt{2})$'] 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Multimodal Competition False m Numerical Open-ended Geometry Math English +59 "In the diagram, line segments $A C$ and $D F$ are tangent to the circle at $B$ and $E$, respectively. Also, $A F$ intersects the circle at $P$ and $R$, and intersects $B E$ at $Q$, as shown. If $\angle C A F=35^{\circ}, \angle D F A=30^{\circ}$, and $\angle F P E=25^{\circ}$, determine the measure of $\angle P E Q$. + +" ['Let $\\angle P E Q=\\theta$.\n\nJoin $P$ to $B$.\n\nWe use the fact that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord. We prove this fact below.\n\nMore concretely, $\\angle D E P=\\angle P B E$ (using the chord $E P$ and the tangent through $E$ ) and $\\angle A B P=\\angle P E Q=\\theta$ (using the chord $B P$ and the tangent through $B$ ).\n\nNow $\\angle D E P$ is exterior to $\\triangle F E P$ and so $\\angle D E P=\\angle F P E+\\angle E F P=25^{\\circ}+30^{\\circ}$, and so $\\angle P B E=\\angle D E P=55^{\\circ}$.\n\nFurthermore, $\\angle A Q B$ is an exterior angle of $\\triangle P Q E$.\n\nThus, $\\angle A Q B=\\angle Q P E+\\angle P E Q=25^{\\circ}+\\theta$.\n\n\n\nIn $\\triangle A B Q$, we have $\\angle B A Q=35^{\\circ}, \\angle A B Q=\\theta+55^{\\circ}$, and $\\angle A Q B=25^{\\circ}+\\theta$.\n\nThus, $35^{\\circ}+\\left(\\theta+55^{\\circ}\\right)+\\left(25^{\\circ}+\\theta\\right)=180^{\\circ}$ or $115^{\\circ}+2 \\theta=180^{\\circ}$, and so $2 \\theta=65^{\\circ}$.\n\nTherefore $\\angle P E Q=\\theta=\\frac{1}{2}\\left(65^{\\circ}\\right)=32.5^{\\circ}$.\n\nAs an addendum, we prove that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord.\n\nConsider a circle with centre $O$ and a chord $X Y$, with tangent $Z X$ meeting the circle at $X$. We prove that if $Z X$ is tangent to the circle, then $\\angle Z X Y$ equals $\\angle X W Y$ whenever $W$ is a point on the circle on the opposite side of $X Y$ as $X Z$ (that is, the angle subtended by $X Y$ on the opposite side of the circle).\n\nWe prove this in the case that $\\angle Z X Y$ is acute. The cases where $\\angle Z X Y$ is a right angle or an obtuse angle are similar.\n\nDraw diameter $X O V$ and join $V Y$.\n\n\n\nSince $\\angle Z X Y$ is acute, points $V$ and $W$ are on the same arc of chord $X Y$.\n\nThis means that $\\angle X V Y=\\angle X W Y$, since they are angles subtended by the same chord.\n\nSince $O X$ is a radius and $X Z$ is a tangent, then $\\angle O X Z=90^{\\circ}$.\n\nThus, $\\angle O X Y+\\angle Z X Y=90^{\\circ}$.\n\nSince $X V$ is a diameter, then $\\angle X Y V=90^{\\circ}$.\n\nFrom $\\triangle X Y V$, we see that $\\angle X V Y+\\angle V X Y=90^{\\circ}$.\n\nBut $\\angle O X Y+\\angle Z X Y=90^{\\circ}$ and $\\angle X V Y+\\angle V X Y=90^{\\circ}$ and $\\angle O X Y=\\angle V X Y$ tells us that $\\angle Z X Y=\\angle X V Y$.\n\nThis gives us that $\\angle Z X Y=\\angle X W Y$, as required.'] ['$32.5$'] 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Multimodal Competition False ^{\circ} Numerical Open-ended Geometry Math English +60 "In the diagram, $A B C D$ and $P N C D$ are squares of side length 2, and $P N C D$ is perpendicular to $A B C D$. Point $M$ is chosen on the same side of $P N C D$ as $A B$ so that $\triangle P M N$ is parallel to $A B C D$, so that $\angle P M N=90^{\circ}$, and so that $P M=M N$. Determine the volume of the convex solid $A B C D P M N$. + +" "['Draw a line segment through $M$ in the plane of $\\triangle P M N$ parallel to $P N$ and extend this line until it reaches the plane through $P, A$ and $D$ at $Q$ on one side and the plane through $N, B$ and $C$ at $R$ on the other side.\n\nJoin $Q$ to $P$ and $A$. Join $R$ to $N$ and $B$.\n\n\n\nSo the volume of solid $A B C D P M N$ equals the volume of solid $A B C D P Q R N$ minus the volumes of solids $P M Q A$ and $N M R B$.\n\nSolid $A B C D P Q R N$ is a trapezoidal prism. This is because $N R$ and $B C$ are parallel (since they lie in parallel planes), which makes $N R B C$ a trapezoid. Similarly, $P Q A D$ is a trapezoid. Also, $P N, Q R, D C$, and $A B$ are all perpendicular to the planes of these trapezoids and equal in length, since they equal the side lengths of the squares.\n\nSolids $P M Q A$ and $N M R B$ are triangular-based pyramids. We can think of their bases as being $\\triangle P M Q$ and $\\triangle N M R$. Their heights are each equal to 2 , the height of the original solid. (The volume of a triangular-based pyramid equals $\\frac{1}{3}$ times the area of its base times its height.)\n\nThe volume of $A B C D P Q R N$ equals the area of trapezoid $N R B C$ times the width of the prism, which is 2.\n\nThat is, this volume equals $\\frac{1}{2}(N R+B C)(N C)(N P)=\\frac{1}{2}(N R+2)(2)(2)=2 \\cdot N R+4$.\n\nSo we need to find the length of $N R$.\n\nConsider quadrilateral $P N R Q$. This quadrilateral is a rectangle since $P N$ and $Q R$ are perpendicular to the two side planes of the original solid.\n\nThus, $N R$ equals the height of $\\triangle P M N$.\n\nJoin $M$ to the midpoint $T$ of $P N$.\n\nSince $\\triangle P M N$ is isosceles, then $M T$ is perpendicular to $P N$.\n\n\n\nSince $N T=\\frac{1}{2} P N=1$ and $\\angle P M N=90^{\\circ}$ and $\\angle T N M=45^{\\circ}$, then $\\triangle M T N$ is also right-angled and isosceles with $M T=T N=1$.\n\nTherefore, $N R=M T=1$ and so the volume of $A B C D P Q R N$ is $2 \\cdot 1+4=6$.\n\nThe volumes of solids $P M Q A$ and $N M R B$ are equal. Each has height 2 and their bases $\\triangle P M Q$ and $\\triangle N M R$ are congruent, because each is right-angled (at $Q$ and at $R$ ) with $P Q=N R=1$ and $Q M=M R=1$.\n\nThus, using the formula above, the volume of each is $\\frac{1}{3}\\left(\\frac{1}{2}(1)(1)\\right) 2=\\frac{1}{3}$.\n\nFinally, the volume of the original solid equals $6-2 \\cdot \\frac{1}{3}=\\frac{16}{3}$.' + 'We determine the volume of $A B C D P M N$ by splitting it into two solids: $A B C D P N$ and $A B N P M$ by slicing along the plane of $A B N P$.\n\nSolid $A B C D P N$ is a triangular prism, since $\\triangle B C N$ and $\\triangle A D P$ are each right-angled (at $C$ and $D$ ), $B C=C N=A D=D P=2$, and segments $P N, D C$ and $A B$ are perpendicular to each of the triangular faces and equal in length.\n\nThus, the volume of $A B C D P N$ equals the area of $\\triangle B C N$ times the length of $D C$, or $\\frac{1}{2}(B C)(C N)(D C)=\\frac{1}{2}(2)(2)(2)=4$. (This solid can also be viewed as ""half"" of a cube.)\n\nSolid $A B N P M$ is a pyramid with rectangular base $A B N P$. (Note that $P N$ and $A B$ are perpendicular to the planes of both of the side triangular faces of the original solid, that $P N=A B=2$ and $B N=A P=\\sqrt{2^{2}+2^{2}}=2 \\sqrt{2}$, by the Pythagorean Theorem.)\n\nTherefore, the volume of $A B N P M$ equals $\\frac{1}{3}(A B)(B N) h=\\frac{4 \\sqrt{2}}{3} h$, where $h$ is the height of the pyramid (that is, the distance that $M$ is above plane $A B N P$ ).\n\nSo we need to calculate $h$.\n\nJoin $M$ to the midpoint, $T$, of $P N$ and to the midpoint, $S$, of $A B$. Join $S$ and $T$. By symmetry, $M$ lies directly above $S T$. Since $A B N P$ is a rectangle and $S$ and $T$ are the midpoints of opposite sides, then $S T=A P=2 \\sqrt{2}$.\n\nSince $\\triangle P M N$ is right-angled and isosceles, then $M T$ is perpendicular to $P N$. Since $N T=\\frac{1}{2} P N=1$ and $\\angle T N M=45^{\\circ}$, then $\\triangle M T N$ is also right-angled and isosceles with $M T=T N=1$.\n\n\n\nAlso, $M S$ is the hypotenuse of the triangle formed by dropping a perpendicular from $M$ to $U$ in the plane of $A B C D$ (a distance of 2) and joining $U$ to $S$. Since $M$ is 1 unit horizontally from $P N$, then $U S=1$.\n\nThus, $M S=\\sqrt{2^{2}+1^{2}}=\\sqrt{5}$ by the Pythagorean Theorem.\n\n\n\nWe can now consider $\\triangle S M T . h$ is the height of this triangle, from $M$ to base $S T$.\n\n\n\nNow $h=M T \\sin (\\angle M T S)=\\sin (\\angle M T S)$.\n\nBy the cosine law in $\\triangle S M T$, we have\n\n$$\nM S^{2}=S T^{2}+M T^{2}-2(S T)(M T) \\cos (\\angle M T S)\n$$\n\nTherefore, $5=8+1-4 \\sqrt{2} \\cos (\\angle M T S)$ or $4 \\sqrt{2} \\cos (\\angle M T S)=4$.\n\nThus, $\\cos (\\angle M T S)=\\frac{1}{\\sqrt{2}}$ and so $\\angle M T S=45^{\\circ}$ which gives $h=\\sin (\\angle M T S)=\\frac{1}{\\sqrt{2}}$.\n\n(Alternatively, we note that the plane of $A B C D$ is parallel to the plane of $P M N$, and so since the angle between plane $A B C D$ and plane $P N B A$ is $45^{\\circ}$, then the angle between plane $P N B A$ and plane $P M N$ is also $45^{\\circ}$, and so $\\angle M T S=45^{\\circ}$.)\n\nFinally, this means that the volume of $A B N P M$ is $\\frac{4 \\sqrt{2}}{3} \\cdot \\frac{1}{\\sqrt{2}}=\\frac{4}{3}$, and so the volume of solid $A B C D P M N$ is $4+\\frac{4}{3}=\\frac{16}{3}$.']" ['$\\frac{16}{3}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +61 "In the diagram, $\triangle A B C$ is right-angled at $B$ and $A C=20$. If $\sin C=\frac{3}{5}$, what is the length of side $B C$ ? + +" "['Since $\\sin C=\\frac{A B}{A C}$, then $A B=A C \\sin C=20\\left(\\frac{3}{5}\\right)=12$.\n\nBy Pythagoras, $B C^{2}=A C^{2}-A B^{2}=20^{2}-12^{2}=256$ or $B C=16$.' + 'Using the standard trigonometric ratios, $B C=A C \\cos C$.\n\nSince $\\sin C=\\frac{3}{5}$, then $\\cos ^{2} C=1-\\sin ^{2} C=1-\\frac{9}{25}=\\frac{16}{25}$ or $\\cos C=\\frac{4}{5}$. (Notice that $\\cos C$ is positive since angle $C$ is acute in triangle $A B C$.)\n\nTherefore, $B C=20\\left(\\frac{4}{5}\\right)=16$.']" ['16'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +62 "A helicopter is flying due west over level ground at a constant altitude of $222 \mathrm{~m}$ and at a constant speed. A lazy, stationary goat, which is due west of the helicopter, takes two measurements of the angle between the ground and the helicopter. The first measurement the goat makes is $6^{\circ}$ and the second measurement, which he makes 1 minute later, is $75^{\circ}$. If the helicopter has not yet passed over the goat, as shown, how fast is the helicopter travelling to the nearest kilometre per hour? + +" ['Let $G$ be the point where the goat is standing, $H$ the position of the helicopter when the goat first measures the angle, $P$ the point directly below the helicopter at this time, $J$ the position of the helicopter one minute later, and $Q$ the point directly below the helicopter at this time.\n\n\n\nUsing the initial position of the helicopter, $\\tan \\left(6^{\\circ}\\right)=\\frac{H P}{P G}$ or $P G=\\frac{222}{\\tan \\left(6^{\\circ}\\right)} \\approx 2112.19 \\mathrm{~m}$.\n\nUsing the second position of the helicopter, $\\tan \\left(75^{\\circ}\\right)=\\frac{J Q}{Q G}$ or $Q G=\\frac{222}{\\tan \\left(75^{\\circ}\\right)} \\approx 59.48 \\mathrm{~m}$.\n\nSo in the one minute that has elapsed, the helicopter has travelled\n\n$2112.19 \\mathrm{~m}-59.48 \\mathrm{~m}=2052.71 \\mathrm{~m}$ or $2.0527 \\mathrm{~km}$.\n\nTherefore, in one hour, the helicopter will travel $60(2.0527)=123.162 \\mathrm{~km}$.\n\nThus, the helicopter is travelling $123 \\mathrm{~km} / \\mathrm{h}$.'] ['123'] 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Multimodal Competition False km/h Numerical Open-ended Geometry Math English +63 "A regular hexagon is a six-sided figure which has all of its angles equal and all of its side lengths equal. In the diagram, $A B C D E F$ is a regular hexagon with an area of 36. The region common to the equilateral triangles $A C E$ and $B D F$ is a hexagon, which is shaded as shown. What is the area of the shaded hexagon? + +" "['We label the vertices of the shaded hexagon $U, V, W, X$, $Y$, and $Z$.\n\nBy symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles. In order to determine the area of the inner hexagon, we determine the ratio of the side lengths of the two hexagons.\n\n\n\nLet the side length of the inner hexagon be $x$. Then $A U=U F=x$. Then triangle $A U F$ has a $120^{\\circ}$ between the two sides of length $x$. If we draw a perpendicular from $U$ to point $P$ on side $A F$, then $U P$ divides\n\n$\\triangle A U F$ into two $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangles. Thus, $F P=P A=\\frac{\\sqrt{3}}{2} x$ and so $A F=\\sqrt{3} x$.\n\nSo the ratio of the side lengths of the hexagons is $\\sqrt{3}: 1$, and so the ratio of\n\n\ntheir areas is $(\\sqrt{3})^{2}: 1=3: 1$.\n\nSince the area of the larger hexagon is 36 , then the area of the inner hexagon is 12.' + 'We label the vertices of the hexagon $U, V, W, X, Y$, and $Z$. By symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles.\n\nWe also join the opposite vertices of the inner hexagon, ie. we join $U$ to $X, V$ to $Y$, and $W$ to $Z$. (These 3 line segments all meet at a single point, say $O$.) This divides the inner hexagon into 6 small equilateral triangles identical to the\n\n\nsix earlier mentioned equilateral triangles.\n\nLet the area of one of these triangles be $a$. Then we can label the 12 small equilateral triangles as all having area $a$.\n\nBut triangle $A U F$ also has area $a$, because if we consider triangle $A F V$, then $A U$ is a median (since $F U=A U=U V$ by symmetry) and so divides triangle $A F V$ into two triangles of equal area. Since the area of\n\n\ntriangle $A U V$ is $a$, then the area of triangle $A U F$ is also $a$.\n\nTherefore, hexagon $A B C D E F$ is divided into 18 equal areas. Thus, $a=2$ since the area of the large hexagon is 36.\n\nSince the area of $U V W X Y Z$ is $6 a$, then its area is 12 .']" ['12'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +64 "At the Big Top Circus, Herc the Human Cannonball is fired out of the cannon at ground level. (For the safety of the spectators, the cannon is partially buried in the sand floor.) Herc's trajectory is a parabola until he catches the vertical safety net, on his way down, at point $B$. Point $B$ is $64 \mathrm{~m}$ directly above point $C$ on the floor of the tent. If Herc reaches a maximum height of $100 \mathrm{~m}$, directly above a point $30 \mathrm{~m}$ from the cannon, determine the horizontal distance from the cannon to the net. + +" ['We assign coordinates to the diagram, with the mouth of the cannon at the point $(0,0)$, with the positive $x$-axis in the horizontal direction towards the safety net from the cannon, and the positive $y$ axis upwards from $(0,0)$.\n\nSince Herc reaches his maximum height when his horizontal distance is $30 \\mathrm{~m}$, then the axis of symmetry of the parabola is the line $x=30$. Since the parabola has a root at $x=0$, then the other root must be at $x=60$.\n\nTherefore, the parabola has the form $y=\\operatorname{ax}(x-60)$.\n\nIn order to determine the value of $a$, we note that Herc passes through the point $(30,100)$, and so\n\n\n\n\n\n$$\n\\begin{aligned}\n100 & =30 a(-30) \\\\\na & =-\\frac{1}{9}\n\\end{aligned}\n$$\n\nThus, the equation of the parabola is $y=-\\frac{1}{9} x(x-60)$.\n\n(Alternatively, we could say that since the parabola has its maximum point at $(30,100)$, then it must be of the form $y=a(x-30)^{2}+100$.\n\nSince the parabola passes through $(0,0)$, then we have\n\n$$\n\\begin{aligned}\n& 0=a(0-30)^{2}+100 \\\\\n& 0=900 a+100 \\\\\n& a=-\\frac{1}{9}\n\\end{aligned}\n$$\n\nThus, the parabola has the equation $y=-\\frac{1}{9}(x-30)^{2}+100$.)\n\nWe would like to find the points on the parabola which have $y$-coordinate 64 , so we solve\n\n$$\n\\begin{aligned}\n64 & =-\\frac{1}{9} x(x-60) \\\\\n0 & =x^{2}-60 x+576 \\\\\n0 & =(x-12)(x-48)\n\\end{aligned}\n$$\n\nSince we want a point after Herc has passed his highest point, then $x=48$, ie. the horizontal distance from the cannon to the safety net is $48 \\mathrm{~m}$.'] ['48'] 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Multimodal Competition False m Numerical Open-ended Geometry Math English +65 "In the diagram, $V$ is the vertex of the parabola with equation $y=-x^{2}+4 x+1$. Also, $A$ and $B$ are the points of intersection of the parabola and the line with equation $y=-x+1$. Determine the value of $A V^{2}+B V^{2}-A B^{2}$. + +" ['First, we find the coordinates of $V$.\n\nTo do this, we use the given equation for the parabola and complete the square:\n\n$y=-x^{2}+4 x+1=-\\left(x^{2}-4 x-1\\right)=-\\left(x^{2}-4 x+2^{2}-2^{2}-1\\right)=-\\left((x-2)^{2}-5\\right)=-(x-2)^{2}+5$\n\nTherefore, the coordinates of the vertex $V$ are $(2,5)$.\n\nNext, we find the coordinates of $A$ and $B$.\n\nNote that $A$ and $B$ are the points of intersection of the line with equation $y=-x+1$ and the parabola with equation $y=-x^{2}+4 x+1$.\n\nWe equate $y$-values to obtain $-x+1=-x^{2}+4 x+1$ or $x^{2}-5 x=0$ or $x(x-5)=0$.\n\nTherefore, $x=0$ or $x=5$.\n\nIf $x=0$, then $y=-x+1=1$, and so $A$ (which is on the $y$-axis) has coordinates $(0,1)$.\n\nIf $x=5$, then $y=-x+1=-4$, and so $B$ has coordinates $(5,-4)$.\n\n\n\nWe now have the points $V(2,5), A(0,1), B(5,-4)$.\n\nThis gives\n\n$$\n\\begin{aligned}\nA V^{2} & =(0-2)^{2}+(1-5)^{2}=20 \\\\\nB V^{2} & =(5-2)^{2}+(-4-5)^{2}=90 \\\\\nA B^{2} & =(0-5)^{2}+(1-(-4))^{2}=50\n\\end{aligned}\n$$\n\nand so $A V^{2}+B V^{2}-A B^{2}=20+90-50=60$.'] ['60'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +66 "In the diagram, $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \mathrm{~cm}$. The piece of pizza is placed on a circular pan with $A, B$ and $C$ touching the circumference of the pan, as shown. What fraction of the pan is covered by the piece of pizza? + +" ['Since $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \\mathrm{~cm}$, then $A C=A B=20 \\mathrm{~cm}$.\n\nWe are also told that $\\angle C A B=90^{\\circ}$ (one-quarter of $360^{\\circ}$ ).\n\nSince $\\angle C A B=90^{\\circ}$ and $A, B$ and $C$ are all on the circumference of the circle, then $C B$ is a diameter of the pan. (This is a property of circles: if $X, Y$ and $Z$ are three points on a circle with $\\angle Z X Y=90^{\\circ}$, then $Y Z$ must be a diameter of the circle.)\n\nSince $\\triangle C A B$ is right-angled and isosceles, then $C B=\\sqrt{2} A C=20 \\sqrt{2} \\mathrm{~cm}$.\n\nTherefore, the radius of the circular plate is $\\frac{1}{2} C B$ or $10 \\sqrt{2} \\mathrm{~cm}$.\n\nThus, the area of the circular pan is $\\pi(10 \\sqrt{2} \\mathrm{~cm})^{2}=200 \\pi \\mathrm{cm}^{2}$.\n\nThe area of the slice of pizza is one-quarter of the area of a circle with radius $20 \\mathrm{~cm}$, or $\\frac{1}{4} \\pi(20 \\mathrm{~cm})^{2}=100 \\pi \\mathrm{cm}^{2}$.\n\nFinally, the fraction of the pan that is covered is the area of the slice of pizza divided by the area of the pan, or $\\frac{100 \\pi \\mathrm{cm}^{2}}{200 \\pi \\mathrm{cm}^{2}}=\\frac{1}{2}$.'] ['$\\frac{1}{2}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +67 "The deck $A B$ of a sailboat is $8 \mathrm{~m}$ long. Rope extends at an angle of $60^{\circ}$ from $A$ to the top $(M)$ of the mast of the boat. More rope extends at an angle of $\theta$ from $B$ to a point $P$ that is $2 \mathrm{~m}$ below $M$, as shown. Determine the height $M F$ of the mast, in terms of $\theta$. + +" ['Suppose that the length of $A F$ is $x \\mathrm{~m}$.\n\nSince the length of $A B$ is $8 \\mathrm{~m}$, then the length of $F B$ is $(8-x) \\mathrm{m}$.\n\nSince $\\triangle M A F$ is right-angled and has an angle of $60^{\\circ}$, then it is $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nTherefore, $M F=\\sqrt{3} A F$, since $M F$ is opposite the $60^{\\circ}$ angle and $A F$ is opposite the $30^{\\circ}$ angle.\n\nThus, $M F=\\sqrt{3} x \\mathrm{~m}$.\n\nSince $M P=2 \\mathrm{~m}$, then $P F=M F-M P=(\\sqrt{3} x-2) \\mathrm{m}$.\n\nWe can now look at $\\triangle B F P$ which is right-angled at $F$.\n\nWe have\n\n$$\n\\tan \\theta=\\frac{P F}{F B}=\\frac{(\\sqrt{3} x-2) \\mathrm{m}}{(8-x) \\mathrm{m}}=\\frac{\\sqrt{3} x-2}{8-x}\n$$\n\nTherefore, $(8-x) \\tan \\theta=\\sqrt{3} x-2$ or $8 \\tan \\theta+2=\\sqrt{3} x+(\\tan \\theta) x$.\n\nThis gives $8 \\tan \\theta+2=x(\\sqrt{3}+\\tan \\theta)$ or $x=\\frac{8 \\tan \\theta+2}{\\tan \\theta+\\sqrt{3}}$.\n\nFinally, $M F=\\sqrt{3} x=\\frac{8 \\sqrt{3} \\tan \\theta+2 \\sqrt{3}}{\\tan \\theta+\\sqrt{3}} \\mathrm{~m}$.\n\n'] ['$\\frac{8 \\sqrt{3} \\tan \\theta+2 \\sqrt{3}}{\\tan \\theta+\\sqrt{3}}$'] 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Multimodal Competition False \mathrm{~m} Expression Open-ended Geometry Math English +68 "In the diagram, triangle ABC is right-angled at B. MT is the perpendicular bisector of $B C$ with $M$ on $B C$ and $T$ on $A C$. If $A T=A B$, what is the size of $\angle A C B$ ? + +" "['Since $M T$ is the perpendicular bisector of $B C$, then\n\n$B M=M C$, and $T M$ is perpendicular to $B C$.\n\nTherefore, $\\triangle C M T$ is similar to $\\triangle C B A$, since they share a common angle and each have a right angle.\n\n\n\nBut $\\frac{C M}{C B}=\\frac{1}{2}$ so $\\frac{C T}{C A}=\\frac{C M}{C B}=\\frac{1}{2}$, and thus $C T=A T=A B$, ie. $\\frac{A B}{A C}=\\frac{1}{2}$ or $\\sin (\\angle A C B)=\\frac{1}{2}$.\n\nTherefore, $\\angle A C B=30^{\\circ}$.' + 'Since $T M \\| A B$, and $C M=M B$, then $C T=T A=A B$.\n\nJoin $T$ to $B$.\n\nSince $\\angle A B C=90^{\\circ}$, then $A C$ is the diameter of a circle passing through $A, C$ and $B$, with $T$ as its centre.\n\n\n\nThus, $T A=A B=B T$ (all radii), and so $\\triangle A B T$ is equilateral. Therefore, $\\angle B A C=60^{\\circ}$, and so $\\angle A C B=30^{\\circ}$.']" ['$30^{\\circ}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +69 "In the diagram, $A B C D E F$ is a regular hexagon with a side +length of 10 . If $X, Y$ and $Z$ are the midpoints of $A B, C D$ and $E F$, respectively, what is the length of $X Z$ ? + +" "['Extend $X A$ and $Z F$ to meet at point $T$.\n\nBy symmetry, $\\angle A X Z=\\angle F Z X=60^{\\circ}$ and $\\angle T A F=\\angle T F A=60^{\\circ}$, and so $\\triangle T A F$ and $\\triangle T X Z$ are both equilateral triangles.\n\nSince $A F=10$, then $T A=10$, which means\n\n$T X=10+5=15$, and so $X Z=T X=15$.\n\n' + 'We look at the quadrilateral $A X Z F$.\n\nSince $A B C D E F$ is a regular hexagon, then $\\angle F A X=\\angle A F Z=120^{\\circ}$.\n\nNote that $A F=10$, and also $A X=F Z=5$ since $X$ and $Z$ are midpoints of their respective sides.\n\n\n\nBy symmetry, $\\angle A X Z=\\angle F Z X=60^{\\circ}$, and so $A X Z F$ is a trapezoid.\n\nDrop perpendiculars from $A$ and $F$ to $P$ and $Q$, respectively, on $X Z$.\n\nBy symmetry again, $P X=Q Z$. Now, $P X=A X \\cos 60^{\\circ}=5\\left(\\frac{1}{2}\\right)=\\frac{5}{2}$.\n\nSince $A P Q F$ is a rectangle, then $P Q=10$.\n\nTherefore, $X Z=X P+P Q+Q Z=\\frac{5}{2}+10+\\frac{5}{2}=15$.']" ['15'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +70 "In the diagram, $A C=2 x, B C=2 x+1$ and $\angle A C B=30^{\circ}$. If the area of $\triangle A B C$ is 18 , what is the value of $x$ ? + +" "['Using a known formula for the area of a triangle, $A=\\frac{1}{2} a b \\sin C$,\n\n$$\n\\begin{aligned}\n18 & =\\frac{1}{2}(2 x+1)(2 x) \\sin 30^{\\circ} \\\\\n36 & =(2 x+1)(2 x)\\left(\\frac{1}{2}\\right) \\\\\n0 & =2 x^{2}+x-36 \\\\\n0 & =(2 x+9)(x-4)\n\\end{aligned}\n$$\n\nand so $x=4$ or $x=-\\frac{9}{2}$. Since $x$ is positive, then $x=4$.' + 'Draw a perpendicular from $A$ to $P$ on $B C$.\n\n\n\nUsing $\\triangle A P C, A P=A C \\sin 30^{\\circ}=2 x\\left(\\frac{1}{2}\\right)=x$.\n\nNow $A P$ is the height of $\\triangle A B C$, so Area $=\\frac{1}{2}(B C)(A P)$.\n\nThen\n\n$$\n\\begin{aligned}\n18 & =\\frac{1}{2}(2 x+1)(x) \\\\\n0 & =2 x^{2}+x-36 \\\\\n0 & =(2 x+9)(x-4)\n\\end{aligned}\n$$\n\nand so $x=4$ or $x=-\\frac{9}{2}$.\n\nSince $x$ is positive, then $x=4$.']" ['4'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +71 "A ladder, $A B$, is positioned so that its bottom sits on horizontal ground and its top rests against a vertical wall, as shown. In this initial position, the ladder makes an angle of $70^{\circ}$ with the horizontal. The bottom of the ladder is then pushed $0.5 \mathrm{~m}$ away from the wall, moving the ladder to position $A^{\prime} B^{\prime}$. In this new position, the ladder makes an angle of $55^{\circ}$ with the horizontal. Calculate, to the nearest centimetre, the distance that the ladder slides down the wall (that is, the length of $B B^{\prime}$ ). + +" ['Let the length of the ladder be $L$.\n\nThen $A C=L \\cos 70^{\\circ}$ and $B C=L \\sin 70^{\\circ}$. Also, $A^{\\prime} C=L \\cos 55^{\\circ}$ and $B^{\\prime} C=L \\sin 55^{\\circ}$.\n\nSince $A^{\\prime} A=0.5$, then\n\n$$\n0.5=L \\cos 55^{\\circ}-L \\cos 70^{\\circ}\n$$\n\n$$\nL=\\frac{0.5}{\\cos 55^{\\circ}-\\cos 70^{\\circ}}\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\nB B^{\\prime} & =B C-B^{\\prime} C \\\\\n& =L \\sin 70^{\\circ}-L \\sin 55^{\\circ} \\\\\n& =L\\left(\\sin 70^{\\circ}-\\sin 55^{\\circ}\\right) \\\\\n& =\\frac{(0.5)\\left(\\sin 70^{\\circ}-\\sin 55^{\\circ}\\right)}{\\left(\\cos 55^{\\circ}-\\cos 70^{\\circ}\\right)} \\quad(\\text { from }(*)) \\\\\n& \\approx 0.2603 \\mathrm{~m}\n\\end{aligned}\n$$\n\n\n\nTherefore, to the nearest centimetre, the distance that the ladder slides down the wall is $26 \\mathrm{~cm}$.'] ['26'] 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Multimodal Competition False cm Numerical Open-ended Geometry Math English +72 "In the diagram, $P Q R S$ is an isosceles trapezoid with $P Q=7, P S=Q R=8$, and $S R=15$. Determine the length of the diagonal $P R$. + +" ['Draw perpendiculars from $P$ and $Q$ to $X$ and $Y$, respectively, on $S R$.\n\n\n\nSince $P Q$ is parallel to $S R$ (because $P Q R S$ is a trapezoid) and $P X$ and $Q Y$ are perpendicular to $S R$, then $P Q Y X$ is a rectangle.\n\nThus, $X Y=P Q=7$ and $P X=Q Y$.\n\nSince $\\triangle P X S$ and $\\triangle Q Y R$ are right-angled with $P S=Q R$ and $P X=Q Y$, then these triangles are congruent, and so $S X=Y R$.\n\nSince $X Y=7$ and $S R=15$, then $S X+7+Y R=15$ or $2 \\times S X=8$ and so $S X=4$.\n\nBy the Pythagorean Theorem in $\\triangle P X S$,\n\n$$\nP X^{2}=P S^{2}-S X^{2}=8^{2}-4^{2}=64-16=48\n$$\n\nNow $P R$ is the hypotenuse of right-angled $\\triangle P X R$.\n\nSince $P R>0$, then by the Pythagorean Theorem,\n\n$$\nP R=\\sqrt{P X^{2}+X R^{2}}=\\sqrt{48+(7+4)^{2}}=\\sqrt{48+11^{2}}=\\sqrt{48+121}=\\sqrt{169}=13\n$$\n\nTherefore, $P R=13$.'] ['13'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +73 "In the diagram, $\triangle A B C$ has $A B=A C$ and $\angle B A C<60^{\circ}$. Point $D$ is on $A C$ with $B C=B D$. Point $E$ is on $A B$ with $B E=E D$. If $\angle B A C=\theta$, determine $\angle B E D$ in terms of $\theta$. + +" ['Since $A B=A C$, then $\\triangle A B C$ is isosceles and $\\angle A B C=\\angle A C B$. Note that $\\angle B A C=\\theta$.\n\n\n\nThe angles in $\\triangle A B C$ add to $180^{\\circ}$, so $\\angle A B C+\\angle A C B+\\angle B A C=180^{\\circ}$.\n\nThus, $2 \\angle A C B+\\theta=180^{\\circ}$ or $\\angle A B C=\\angle A C B=\\frac{1}{2}\\left(180^{\\circ}-\\theta\\right)=90^{\\circ}-\\frac{1}{2} \\theta$.\n\nNow $\\triangle B C D$ is isosceles as well with $B C=B D$ and so $\\angle C D B=\\angle D C B=90^{\\circ}-\\frac{1}{2} \\theta$.\n\nSince the angles in $\\triangle B C D$ add to $180^{\\circ}$, then\n\n$$\n\\angle C B D=180^{\\circ}-\\angle D C B-\\angle C D B=180^{\\circ}-\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)-\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)=\\theta\n$$\n\nNow $\\angle E B D=\\angle A B C-\\angle D B C=\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)-\\theta=90^{\\circ}-\\frac{3}{2} \\theta$.\n\nSince $B E=E D$, then $\\angle E D B=\\angle E B D=90^{\\circ}-\\frac{3}{2} \\theta$.\n\nTherefore, $\\angle B E D=180^{\\circ}-\\angle E B D-\\angle E D B=180^{\\circ}-\\left(90^{\\circ}-\\frac{3}{2} \\theta\\right)-\\left(90^{\\circ}-\\frac{3}{2} \\theta\\right)=3 \\theta$.'] ['$3 \\theta$'] 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Multimodal Competition False Expression Open-ended Geometry Math English +74 "In the diagram, the ferris wheel has a diameter of $18 \mathrm{~m}$ and rotates at a constant rate. When Kolapo rides the ferris wheel and is at its lowest point, he is $1 \mathrm{~m}$ above the ground. When Kolapo is at point $P$ that is $16 \mathrm{~m}$ above the ground and is rising, it takes him 4 seconds to reach the highest point, $T$. He continues to travel for another 8 seconds reaching point $Q$. Determine Kolapo's height above the ground when he reaches point $Q$. + +" "[""Let $O$ be the centre of the ferris wheel and $B$ the lowest point on the wheel.\n\nSince the radius of the ferris wheel is $9 \\mathrm{~m}$ (half of the diameter of $18 \\mathrm{~m}$ ) and $B$ is $1 \\mathrm{~m}$ above the ground, then $O$ is $9+1=10 \\mathrm{~m}$ above the ground.\n\nLet $\\angle T O P=\\theta$.\n\n\n\nSince the ferris wheel rotates at a constant speed, then in 8 seconds, the angle through which the wheel rotates is twice the angle through which it rotates in 4 seconds. In other words, $\\angle T O Q=2 \\theta$.\n\nDraw a perpendicular from $P$ to $R$ on $T B$ and from $Q$ to $G$ on $T B$.\n\nSince $P$ is $16 \\mathrm{~m}$ above the ground and $O$ is $10 \\mathrm{~m}$ above the ground, then $O R=6 \\mathrm{~m}$.\n\nSince $O P$ is a radius of the circle, then $O P=9 \\mathrm{~m}$.\n\nLooking at right-angled $\\triangle O R P$, we see that $\\cos \\theta=\\frac{O R}{O P}=\\frac{6}{9}=\\frac{2}{3}$.\n\nSince $\\cos \\theta=\\frac{2}{3}<\\frac{1}{\\sqrt{2}}=\\cos \\left(45^{\\circ}\\right)$, then $\\theta>45^{\\circ}$.\n\nThis means that $2 \\theta>90^{\\circ}$, which means that $Q$ is below the horizontal diameter through $O$ and so $G$ is below $O$.\n\nSince $\\angle T O Q=2 \\theta$, then $\\angle Q O G=180^{\\circ}-2 \\theta$.\n\nKolapo's height above the ground at $Q$ equals $1 \\mathrm{~m}$ plus the length of $B G$.\n\nNow $B G=O B-O G$. We know that $O B=9 \\mathrm{~m}$.\n\nAlso, considering right-angled $\\triangle Q O G$, we have\n\n$$\nO G=O Q \\cos (\\angle Q O G)=9 \\cos \\left(180^{\\circ}-2 \\theta\\right)=-9 \\cos (2 \\theta)=-9\\left(2 \\cos ^{2} \\theta-1\\right)\n$$\n\nSince $\\cos \\theta=\\frac{2}{3}$, then $O G=-9\\left(2\\left(\\frac{2}{3}\\right)^{2}-1\\right)=-9\\left(\\frac{8}{9}-1\\right)=1 \\mathrm{~m}$.\n\nTherefore, $B G=9-1=8 \\mathrm{~m}$ and so $Q$ is $1+8=9 \\mathrm{~m}$ above the ground.""]" ['9'] 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Multimodal Competition False m Numerical Open-ended Geometry Math English +75 "On Saturday, Jimmy started painting his toy helicopter between 9:00 a.m. and 10:00 a.m. When he finished between 10:00 a.m. and 11:00 a.m. on the same morning, the hour hand was exactly where the minute hand had been when he started, and the minute hand was exactly where the hour hand had been when he started. Jimmy spent $t$ hours painting. Determine the value of $t$. + +" "['The hour hand and minute hand both turn at constant rates. Since the hour hand moves $\\frac{1}{12}$ of the way around the clock in 1 hour and the minute hand moves all of the way around the clock in 1 hour, then the minute hand turns 12 times as quickly as the hour hand.\n\n\nSuppose also that the hour hand moves through an angle of $x^{\\circ}$ between Before and After. Therefore, the minute hand moves through an angle of $\\left(360^{\\circ}-x^{\\circ}\\right)$ between Before and After, since these two angles add to $360^{\\circ}$.\n\n\n\nSince the minute hand moves 12 times as quickly as the hour hand, then $\\frac{360^{\\circ}-x^{\\circ}}{x^{\\circ}}=12$ or $360-x=12 x$ and so $13 x=360$, or $x=\\frac{360}{13}$.\n\nIn one hour, the hour hand moves through $\\frac{1}{12} \\times 360^{\\circ}=30^{\\circ}$.\n\nSince the hour hand is moving for $t$ hours, then we have $30^{\\circ} t=\\left(\\frac{360}{13}\\right)^{\\circ}$ and so $t=\\frac{360}{30(13)}=\\frac{12}{13}$.' + ""Suppose that Jimmy starts painting $x$ hours after 9:00 a.m. and finishes painting $y$ hours after 10:00 a.m., where $0" "['Since $\\angle A O B=90^{\\circ}, A B$ is a diameter of the circle.\n\nJoin $A B$.\n\n\n\nSince $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.\n\nTherefore, the area of the part of the circle inside the first quadrant is equal to the area of $\\triangle A O B$ plus the area of the semi-circle above $A B$.\n\nThe radius of the circle is equal to the distance from $C$ to $B$, or $\\sqrt{(1-2)^{2}+(1-0)^{2}}=\\sqrt{2}$, so the area of the semi-circle is $\\frac{1}{2} \\pi(\\sqrt{2})^{2}=\\pi$.\n\nThe area of $\\triangle A O B$ is $\\frac{1}{2}(O B)(A O)=\\frac{1}{2}(2)(2)=2$.\n\nThus, the area of the part of the circle inside the first quadrant is $\\pi+2$.' + 'Since $\\angle A O B=90^{\\circ}, A B$ is a diameter of the circle.\n\nJoin $A B$.\n\n\n\nSince $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.\n\nThus, $A O=B O$.\n\nWe ""complete the square"" by adding point $D(2,2)$, which is on the circle, by symmetry.\n\n\n\n\n\nThe area of the square is 4 .\n\nThe radius of the circle is equal to the distance from $C$ to $B$, or $\\sqrt{(1-2)^{2}+(1-0)^{2}}=\\sqrt{2}$, so the area of the circle is $\\pi(\\sqrt{2})^{2}=2 \\pi$.\n\nThe area of the portion of the circle outside the square is thus $2 \\pi-4$. This area is divided into four equal sections (each of area $\\frac{1}{4}(2 \\pi-4)=\\frac{1}{2} \\pi-1$ ), two of which are the only portions of the circle outside the first quadrant.\n\nTherefore, the area of the part of the circle inside the the first quadrant is $2 \\pi-2\\left(\\frac{1}{2} \\pi-1\\right)=$ $\\pi+2$.\n\nTwo additional ways to find the coordinates of $A$ :\n\n$*$ The length of $O C$ is $\\sqrt{1^{2}+1^{2}}=\\sqrt{2}$.\n\nSince $C$ is the centre of the circle and $O$ lies on the circle, then the circle has radius $\\sqrt{2}$.\n\nSince the circle has centre $(1,1)$ and radius $\\sqrt{2}$, its equation is $(x-1)^{2}+(y-1)^{2}=2$. To find the coordinates of $A$, we substitute $x=0$ to obtain $(0-1)^{2}+(y-1)^{2}=2$ or $(y-1)^{2}=1$, and so $y=0$ or $y=2$.\n\nSince $y=0$ gives us the point $O$, then $y=2$ gives us $A$, ie. $A$ has coordinates $(0,2)$.\n\n* Since $O$ and $A$ are both on the circle and each has a horizontal distance of 1 from $C$, then their vertical distances from $C$ must be same, ie. must each be 1.\n\nThus, $A$ has coordinates $(0,2)$.']" ['$(0,2),\\pi+2$'] 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Multimodal Competition True Tuple,Numerical Open-ended Geometry Math English +77 "Survivors on a desert island find a piece of plywood $(A B C)$ in the shape of an equilateral triangle with sides of length $2 \mathrm{~m}$. To shelter their goat from the sun, they place edge $B C$ on the ground, lift corner $A$, and put in a vertical post $P A$ which is $h \mathrm{~m}$ long above ground. When the sun is directly overhead, the shaded region $(\triangle P B C)$ on the ground directly underneath the plywood is an isosceles triangle with largest angle $(\angle B P C)$ equal to $120^{\circ}$. Determine the value of $h$, to the nearest centimetre. + +" ['From the given information, $P C=P B$.\n\nIf we can calculate the length of $P C$, we can calculate the value of $h$, since we already know the length of $A C$.\n\nNow $\\triangle C P B$ is isosceles with $P C=P B, B C=2$ and $\\angle B P C=120^{\\circ}$.\n\nSince $\\triangle C P B$ is isosceles, $\\angle P C B=\\angle P B C=30^{\\circ}$.\n\n\n\nJoin $P$ to the midpoint, $M$, of $B C$.\n\nThen $P M$ is perpendicular to $B C$, since $\\triangle P C B$ is isosceles.\n\n\n\nTherefore, $\\triangle P M C$ is right-angled, has $\\angle P C M=30^{\\circ}$ and has $C M=1$.\n\nThus, $P C=\\frac{2}{\\sqrt{3}}$.\n\n(There are many other techniques that we can use to calculate the length of $P C$.)\n\nReturning to $\\triangle A P C$, we see $A P^{2}=A C^{2}-P C^{2}$ or $h^{2}=2^{2}-\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}=4-\\frac{4}{3}=\\frac{8}{3}$, and so $h=\\sqrt{\\frac{8}{3}}=2 \\sqrt{\\frac{2}{3}}=\\frac{2 \\sqrt{6}}{3} \\approx 1.630$.\n\nTherefore, the height is approximately $1.63 \\mathrm{~m}$ or $163 \\mathrm{~cm}$.'] ['163'] 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Multimodal Competition False cm Numerical Open-ended Geometry Math English +78 "Points $A_{1}, A_{2}, \ldots, A_{N}$ are equally spaced around the circumference of a circle and $N \geq 3$. Three of these points are selected at random and a triangle is formed using these points as its vertices. + +Through this solution, we will use the following facts: + +When an acute triangle is inscribed in a circle: + +- each of the three angles of the triangle is the angle inscribed in the major arc defined by the side of the triangle by which it is subtended, +- each of the three arcs into which the circle is divided by the vertices of the triangles is less than half of the circumference of the circle, and +- it contains the centre of the circle. + +Why are these facts true? + +- Consider a chord of a circle which is not a diameter. + +Then the angle subtended in the major arc of this circle is an acute angle and the angle subtended in the minor arc is an obtuse angle. + +Now consider an acute triangle inscribed in a circle. + +Since each angle of the triangle is acute, then each of the three angles is inscribed in the major arc defined by the side of the triangle by which it is subtended. + +- It follows that each arc of the circle that is outside the triangle must be a minor arc, thus less than the circumference of the circle. +- Lastly, if the centre was outside the triangle, then we would be able to draw a diameter of the circle with the triangle entirely on one side of the diameter. + + + +In this case, one of the arcs of the circle cut off by one of the sides of the triangle would have to be a major arc, which cannot happen, because of the above. + +Therefore, the centre is contained inside the triangle. +If $N=7$, what is the probability that the triangle is acute? (A triangle is acute if each of its three interior angles is less than $90^{\circ}$.)" "[""Since there are $N=7$ points from which the triangle's vertices can be chosen, there are $\\left(\\begin{array}{l}7 \\\\ 3\\end{array}\\right)=35$ triangles in total.\n\nWe compute the number of acute triangles.\n\nFix one of the vertices of such a triangle at $A_{1}$.\n\nWe construct the triangle by choosing the other two vertices in ascending subscript order. We choose the vertices by considering the arc length from the previous vertex - each of\n\n\n\nthese arc lengths must be smaller than half the total circumference of the circle.\n\nSince there are 7 equally spaced points on the circle, we assume the circumference is 7 , so the arc length formed by each side must be at most 3 .\n\nSince the first arc length is at most 3 , the second point can be only $A_{2}, A_{3}$ or $A_{4}$.\n\nIf the second point is $A_{2}$, then since the second and third arc lengths are each at most 3 , then the third point must be $A_{5}$. (Since the second arc length is at most 3, then the third point cannot be any further along than $A_{5}$. However, the arc length from $A_{5}$ around to $A_{1}$ is 3 , so it cannot be any closer than $A_{5}$.)\n\n\n\nIf the second point is $A_{3}$, the third point must be $A_{5}$ or $A_{6}$. If the second point is $A_{4}$, the third point must be $A_{5}$ or $A_{6}$ or $A_{7}$. Therefore, there are 6 acute triangles which include $A_{1}$ as one of its vertices.\n\nHow many acute triangles are there in total?\n\nWe can repeat the above process for each of the 6 other points, giving $7 \\times 6=42$ acute triangles.\n\nBut each triangle is counted three times here, as it has been counted once for each of its vertices.\n\nThus, there are $\\frac{7 \\times 6}{3}=14$ acute triangles.\n\nTherefore, the probability that a randomly chosen triangle is acute if $\\frac{14}{35}=\\frac{2}{5}$.""]" ['$\\frac{2}{5}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +79 "In the diagram, $\triangle P Q S$ is right-angled at $P$ and $\triangle Q R S$ is right-angled at $Q$. Also, $P Q=x, Q R=8, R S=x+8$, and $S P=x+3$ for some real number $x$. Determine all possible values of the perimeter of quadrilateral $P Q R S$. + +" ['Since $\\triangle P Q S$ is right-angled at $P$, then by the Pythagorean Theorem,\n\n$$\nS Q^{2}=S P^{2}+P Q^{2}=(x+3)^{2}+x^{2}\n$$\n\nSince $\\triangle Q R S$ is right-angled at $Q$, then by the Pythagorean Theorem, we obtain\n\n$$\n\\begin{aligned}\nR S^{2} & =S Q^{2}+Q R^{2} \\\\\n(x+8)^{2} & =\\left((x+3)^{2}+x^{2}\\right)+8^{2} \\\\\nx^{2}+16 x+64 & =x^{2}+6 x+9+x^{2}+64 \\\\\n0 & =x^{2}-10 x+9 \\\\\n0 & =(x-1)(x-9)\n\\end{aligned}\n$$\n\nand so $x=1$ or $x=9$.\n\n(We can check that if $x=1, \\triangle P Q S$ has sides of lengths 4,1 and $\\sqrt{17}$ and $\\triangle Q R S$ has sides of lengths $\\sqrt{17}, 8$ and 9 , both of which are right-angled, and if $x=9, \\triangle P Q S$ has sides of lengths 12,9 and 15 and $\\triangle Q R S$ has sides of lengths 15,8 and 17 , both of which are right-angled.)\n\nIn terms of $x$, the perimeter of $P Q R S$ is $x+8+(x+8)+(x+3)=3 x+19$.\n\nThus, the possible perimeters of $P Q R S$ are 22 (when $x=1$ ) and 46 (when $x=9$ ).'] ['22,46'] 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Multimodal Competition True Numerical Open-ended Geometry Math English +80 "In the diagram, $\triangle A B D$ has $C$ on $B D$. Also, $B C=2, C D=1, \frac{A C}{A D}=\frac{3}{4}$, and $\cos (\angle A C D)=-\frac{3}{5}$. Determine the length of $A B$. + +" ['Since $\\frac{A C}{A D}=\\frac{3}{4}$, then we let $A C=3 t$ and $A D=4 t$ for some real number $t>0$.\n\n\n\nUsing the cosine law in $\\triangle A C D$, the following equations are equivalent:\n\n$$\n\\begin{aligned}\nA D^{2} & =A C^{2}+C D^{2}-2 \\cdot A C \\cdot C D \\cdot \\cos (\\angle A C D) \\\\\n(4 t)^{2} & =(3 t)^{2}+1^{2}-2(3 t)(1)\\left(-\\frac{3}{5}\\right) \\\\\n16 t^{2} & =9 t^{2}+1+\\frac{18}{5} t \\\\\n80 t^{2} & =45 t^{2}+5+18 t \\\\\n35 t^{2}-18 t-5 & =0 \\\\\n(7 t-5)(5 t+1) & =0\n\\end{aligned}\n$$\n\nSince $t>0$, then $t=\\frac{5}{7}$.\n\nThus, $A C=3 t=\\frac{15}{7}$.\n\nUsing the cosine law in $\\triangle A C B$ and noting that\n\n$$\n\\cos (\\angle A C B)=\\cos \\left(180^{\\circ}-\\angle A C D\\right)=-\\cos (\\angle A C D)=\\frac{3}{5}\n$$\n\n\n\nthe following equations are equivalent:\n\n$$\n\\begin{aligned}\nA B^{2} & =A C^{2}+B C^{2}-2 \\cdot A C \\cdot B C \\cdot \\cos (\\angle A C B) \\\\\n& =\\left(\\frac{15}{7}\\right)^{2}+2^{2}-2\\left(\\frac{15}{7}\\right)(2)\\left(\\frac{3}{5}\\right) \\\\\n& =\\frac{225}{49}+4-\\frac{36}{7} \\\\\n& =\\frac{225}{49}+\\frac{196}{49}-\\frac{252}{49} \\\\\n& =\\frac{169}{49}\n\\end{aligned}\n$$\n\nSince $A B>0$, then $A B=\\frac{13}{7}$.'] ['$\\frac{13}{7}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +81 "Suppose that $a>\frac{1}{2}$ and that the parabola with equation $y=a x^{2}+2$ has vertex $V$. The parabola intersects the line with equation $y=-x+4 a$ at points $B$ and $C$, as shown. If the area of $\triangle V B C$ is $\frac{72}{5}$, determine the value of $a$. + +" ['The parabola with equation $y=a x^{2}+2$ is symmetric about the $y$-axis.\n\nThus, its vertex occurs when $x=0$ (which gives $y=a \\cdot 0^{2}+2=2$ ) and so $V$ has coordinates $(0,2)$.\n\nTo find the coordinates of $B$ and $C$, we use the equations of the parabola and line to obtain\n\n$$\n\\begin{aligned}\na x^{2}+2 & =-x+4 a \\\\\na x^{2}+x+(2-4 a) & =0\n\\end{aligned}\n$$\n\nUsing the quadratic formula,\n\n$$\nx=\\frac{-1 \\pm \\sqrt{1^{2}-4 a(2-4 a)}}{2 a}=\\frac{-1 \\pm \\sqrt{1-8 a+16 a^{2}}}{2 a}\n$$\n\nSince $1-8 a+16 a^{2}=(4 a-1)^{2}$ and $4 a-1>0\\left(\\right.$ since $\\left.a>\\frac{1}{2}\\right)$, then $\\sqrt{1-8 a+16 a^{2}}=4 a-1$ and so\n\n$$\nx=\\frac{-1 \\pm(4 a-1)}{2 a}\n$$\n\nwhich means that $x=\\frac{4 a-2}{2 a}=\\frac{2 a-1}{a}=2-\\frac{1}{a}$ or $x=\\frac{-4 a}{2 a}=-2$.\n\nWe can use the equation of the line to obtain the $y$-coordinates of $B$ and $C$.\n\nWhen $x=-2$ (corresponding to point $B$ ), we obtain $y=-(-2)+4 a=4 a+2$.\n\nWhen $x=2-\\frac{1}{a}$ (corresponding to point $C$ ), we obtain $y=-\\left(2-\\frac{1}{a}\\right)+4 a=4 a-2+\\frac{1}{a}$.\n\nLet $P$ and $Q$ be the points on the horizontal line through $V$ so that $B P$ and $C Q$ are perpendicular to $P Q$.\n\n\n\n\n\nThen the area of $\\triangle V B C$ is equal to the area of trapezoid $P B C Q$ minus the areas of right-angled $\\triangle B P V$ and right-angled $\\triangle C Q V$.\n\nSince $B$ has coordinates $(-2,4 a+2), P$ has coordinates $(-2,2), V$ has coordiantes $(0,2)$, $Q$ has coordinates $\\left(2-\\frac{1}{a}, 2\\right)$, and $C$ has coordinates $\\left(2-\\frac{1}{a}, 4 a-2+\\frac{1}{a}\\right)$, then\n\n$$\n\\begin{aligned}\nB P & =(4 a+2)-2=4 a \\\\\nC Q & =\\left(4 a-2+\\frac{1}{a}\\right)-2=4 a-4+\\frac{1}{a} \\\\\nP V & =0-(-2)=2 \\\\\nQ V & =2-\\frac{1}{a}-0=2-\\frac{1}{a} \\\\\nP Q & =P V+Q V=2+2-\\frac{1}{a}=4-\\frac{1}{a}\n\\end{aligned}\n$$\n\nTherefore, the area of trapezoid $P B C Q$ is\n\n$$\n\\frac{1}{2}(B P+C Q)(P Q)=\\frac{1}{2}\\left(4 a+4 a-4+\\frac{1}{a}\\right)\\left(4-\\frac{1}{a}\\right)=\\left(4 a-2+\\frac{1}{2 a}\\right)\\left(4-\\frac{1}{a}\\right)\n$$\n\nAlso, the area of $\\triangle B P V$ is $\\frac{1}{2} \\cdot B P \\cdot P V=\\frac{1}{2}(4 a)(2)=4 a$.\n\nFurthermore, the area of $\\triangle C Q V$ is\n\n$$\n\\frac{1}{2} \\cdot C Q \\cdot Q V=\\frac{1}{2}\\left(4 a-4+\\frac{1}{a}\\right)\\left(2-\\frac{1}{a}\\right)=\\left(2 a-2+\\frac{1}{2 a}\\right)\\left(2-\\frac{1}{a}\\right)\n$$\n\nFrom the given information,\n\n$$\n\\left(4 a-2+\\frac{1}{2 a}\\right)\\left(4-\\frac{1}{a}\\right)-4 a-\\left(2 a-2+\\frac{1}{2 a}\\right)\\left(2-\\frac{1}{a}\\right)=\\frac{72}{5}\n$$\n\nMultiplying both sides by $2 a^{2}$, which we distribute through the factors on the left side as $2 a \\cdot a$, we obtain\n\n$$\n\\left(8 a^{2}-4 a+1\\right)(4 a-1)-8 a^{3}-\\left(4 a^{2}-4 a+1\\right)(2 a-1)=\\frac{144}{5} a^{2}\n$$\n\nMultiplying both sides by 5 , we obtain\n\n$$\n5\\left(8 a^{2}-4 a+1\\right)(4 a-1)-40 a^{3}-5\\left(4 a^{2}-4 a+1\\right)(2 a-1)=144 a^{2}\n$$\n\nExpanding and simplifying, we obtain\n\n$$\n\\begin{aligned}\n\\left(160 a^{3}-120 a^{2}+40 a-5\\right)-40 a^{3}-\\left(40 a^{3}-60 a^{2}+30 a-5\\right) & =144 a^{2} \\\\\n80 a^{3}-204 a^{2}+10 a & =0 \\\\\n2 a\\left(40 a^{2}-102 a+5\\right) & =0 \\\\\n2 a(20 a-1)(2 a-5) & =0\n\\end{aligned}\n$$\n\nand so $a=0$ or $a=\\frac{1}{20}$ or $a=\\frac{5}{2}$. Since $a>\\frac{1}{2}$, then $a=\\frac{5}{2}$.'] ['$\\frac{5}{2}$'] 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Multimodal Competition False Numerical Open-ended Algebra Math English +82 "Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is + + + +The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$. +Determine the sum of the terms in the $(4,2)$-sawtooth sequence." ['The $(4,2)$-sawtooth sequence consists of the terms\n\n$$\n1, \\quad 2,3,4,3,2,1, \\quad 2,3,4,3,2,1\n$$\n\nwhose sum is 31 .'] ['31'] 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Multimodal Competition False Numerical Open-ended Number Theory Math English +83 "Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is + + + +The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$. +For each positive integer $m \geq 2$, determine a simplified expression for the sum of the terms in the $(m, 3)$-sawtooth sequence." "['Suppose that $m \\geq 2$.\n\nThe $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .\n\nConsider one of these teeth whose terms are\n\n$$\n2,3,4, \\ldots, m-1, m, m-1, m-2, m-3, \\ldots, 2,1\n$$\n\nWhen we write the ascending portion directly above the descending portion, we obtain\n\n$$\n\\begin{aligned}\n& 2, \\quad 3, \\quad 4, \\quad \\ldots, \\quad m-1, \\quad m \\\\\n& m-1, \\quad m-2, \\quad m-3, \\quad \\ldots, \\quad 2, \\quad 1\n\\end{aligned}\n$$\n\nFrom this presentation, we can see $m-1$ pairs of terms, the sum of each of which is $m+1$. $($ Note that $2+(m-1)=3+(m-2)=4+(m-3)=\\cdots=(m-1)+2=m+1$ and as we move from left to right, the terms on the top increase by 1 at each step and the terms on the bottom decrease by 1 at each step, so their sum is indeed constant.) Therefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$. This means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\\left(m^{2}-1\\right)$, which equals $3 m^{2}-2$.' + 'Suppose that $m \\geq 2$.\n\nThe $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .\n\nConsider one of these teeth whose terms are\n\n$$\n2,3,4, \\ldots, m-1, m, m-1, m-2, m-3, \\ldots, 2,1\n$$\n\nThis tooth includes one 1 , two $2 \\mathrm{~s}$, two $3 \\mathrm{~s}$, and so on, until we reach two $(m-1) \\mathrm{s}$, and one $m$.\n\nThe sum of these numbers is\n\n$$\n1(1)+2(2)+2(3)+\\cdots+2(m-1)+m\n$$\n\nwhich can be rewritten as\n\n$$\n2(1+2+3+\\cdots+(m-1)+m)-1-m=2 \\cdot \\frac{1}{2} m(m+1)-m-1=m^{2}+m-m-1=m^{2}-1\n$$\n\nTherefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$.\n\nThis means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\\left(m^{2}-1\\right)$, which equals $3 m^{2}-2$.']" ['$3 m^{2}-2$'] 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Multimodal Competition False Expression Open-ended Number Theory Math English +84 "At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected + + +uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping. +For a 2-topping pizza, determine the probability that at least $\frac{1}{4}$ of the pizza is covered by both toppings." ['Assume that the first topping is placed on the top half of the pizza. (We can rotate the pizza so that this is the case.)\n\nAssume that the second topping is placed on the half of the pizza that is above the horizontal diameter that makes an angle of $\\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\\theta$ to $\\theta+180^{\\circ}$.\n\n\nWe may assume that $0^{\\circ} \\leq \\theta \\leq 360^{\\circ}$.\n\nWhen $0^{\\circ} \\leq \\theta \\leq 90^{\\circ}$, the angle of the sector covered by both toppings is at least $90^{\\circ}$ (and so is at least a quarter of the circle).\n\nWhen $90^{\\circ}<\\theta \\leq 180^{\\circ}$, the angle of the sector covered by both toppings is less than $90^{\\circ}$ (and so is less than a quarter of the circle).\n\nWhen $\\theta$ moves past $180^{\\circ}$, the left-hand portion of the upper half circle starts to be covered with both toppings again. When $180^{\\circ} \\leq \\theta<270^{\\circ}$, the angle of the sector covered by both toppings is less than $90^{\\circ}$ (and so is less than a quarter of the circle).\n\nWhen $270^{\\circ} \\leq \\theta \\leq 360^{\\circ}$, the angle of the sector covered by both toppings at least $90^{\\circ}$ (and so is at least a quarter of the circle).\n\nTherefore, if $\\theta$ is chosen randomly between $0^{\\circ}$ and $360^{\\circ}$, the combined length of the intervals in which at least $\\frac{1}{4}$ of the pizza is covered with both toppings is $180^{\\circ}$.\n\nTherefore, the probability is $\\frac{180^{\\circ}}{360^{\\circ}}$, or $\\frac{1}{2}$.'] ['$\\frac{1}{2}$'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +85 "At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected + + +uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping. +For a 3-topping pizza, determine the probability that some region of the pizza with non-zero area is covered by all 3 toppings. (The diagram above shows an example where no region is covered by all 3 toppings.)" ['Suppose that the first topping is placed on the top half of the pizza. (Again, we can rotate the pizza so that this is the case.)\n\nAssume that the second topping is placed on the half of the pizza that is above the diameter that makes an angle of $\\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\\theta$ to $\\theta+180^{\\circ}$.\n\nWe may assume that $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$. If $180^{\\circ} \\leq \\theta \\leq 360^{\\circ}$, the resulting pizza can be seen as a reflection of the one shown.\n\n\nConsider the third diameter added, shown dotted in the diagram above. Suppose that its angle with the horizontal is $\\alpha$. (In the diagram, $\\alpha<90^{\\circ}$.) We assume that the topping is added on the half pizza clockwise beginning at the angle of $\\alpha$, and that this topping stays in the same relative position as the diameter sweeps around the circle.\n\nFor what angles $\\alpha$ will there be a portion of the pizza covered with all three toppings?\n\nIf $0^{\\circ} \\leq \\alpha<180^{\\circ}$, there will be a portion covered with three toppings; this portion is above the right half of the horizontal diameter.\n\nIf $180^{\\circ} \\leq \\alpha<180^{\\circ}+\\theta$, the third diameter will pass through the two regions with angle $\\theta$ and the third topping will be below this diameter, so there will not be a region covered\n\n\n\nwith three toppings.\n\nIf $180^{\\circ}+\\theta \\leq \\alpha \\leq 360^{\\circ}$, the third topping starts to cover the leftmost part of the region currently covered with two toppings, and so a region is covered with three toppings.\n\nTherefore, for an angle $\\theta$ with $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$, a region of the pizza is covered with three toppings when $0^{\\circ} \\leq \\alpha<180^{\\circ}$ and when $180^{\\circ}+\\theta \\leq \\alpha \\leq 360^{\\circ}$.\n\nTo determine the desired probability, we graph points $(\\theta, \\alpha)$. A particular choice of diameters corresponds to a choice of angles $\\theta$ and $\\alpha$ with $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$ and $0^{\\circ} \\leq \\alpha \\leq 360^{\\circ}$, which corresponds to a point on the graph below.\n\nThe probability that we are looking for then equals the area of the region of this graph where three toppings are in a portion of the pizza divided by the total allowable area of the graph.\n\nThe shaded region of the graph corresponds to instances where a portion of the pizza will be covered by three toppings.\n\n\n\nThis shaded region consists of the entire portion of the graph where $0^{\\circ} \\leq \\alpha \\leq 180^{\\circ}$ (regardless of $\\theta$ ) as well as the region above the line with equation $\\alpha=\\theta+180^{\\circ}$ (that is, the region with $\\theta+180^{\\circ} \\leq \\alpha \\leq 360^{\\circ}$ ).\n\nSince the slope of the line is 1 , it divides the upper half of the region, which is a square, into two pieces of equal area.\n\nTherefore, $\\frac{3}{4}$ of the graph is shaded, which means that the probability that a region of the pizza is covered by all three toppings is $\\frac{3}{4}$.'] ['$\\frac{3}{4}$'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +86 "At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected + + +uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping. +Suppose that $N$ is a positive integer. For an $N$-topping pizza, determine the probability, in terms of $N$, that some region of the pizza with non-zero area is covered by all $N$ toppings." "['The main idea of this solution is that the toppings all overlap exactly when there is one topping with the property that all other toppings ""begin"" somewhere in that toppings semi-circle. In the rest of this solution, we determine the probability using this fact and then justify this fact.\n\nSuppose that, for $1 \\leq j \\leq N$, topping $j$ is put on the semi-circle that starts at an angle of $\\theta_{j}$ clockwise from the horizontal left-hand radius and continues to an angle of $\\theta_{j}+180^{\\circ}$, where $0^{\\circ} \\leq \\theta_{j}<360^{\\circ}$. By establishing these variables and this convention, we are fixing both the angle of the diameter and the semi-circle defined by this diameter on which the topping is placed.\n\nSuppose that there is some region of the pizza with non-zero area that is covered by all $N$ toppings.\n\nThis region will be a sector with two bounding radii, each of which must be half of a diameter that defines one of the toppings.\n\nSuppose that the radius at the clockwise ""end"" of the sector is the end of the semi-circle where topping $X$ is placed, and that the radius at the counter-clockwise ""beginning"" of the sector is the start of the semi-circle where topping $Y$ is placed.\n\n\n\n\n\nThis means that each of the other $N-2$ toppings begins between (in the clockwise sense) the points where topping $X$ begins and where topping $Y$ begins.\n\nConsider the beginning angle for topping $X, \\theta_{X}$.\n\nTo say that the other $N-1$ toppings begin at some point before topping $X$ ends is the same as saying that each $\\theta_{j}$ with $j \\neq X$ is between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$.\n\nHere, we can allow for the possibility that $\\theta_{X}+180^{\\circ}$ is greater than $360^{\\circ}$ by saying that an angle equivalent to $\\theta_{j}$ (which is either $\\theta_{j}$ or $\\theta_{j}+360^{\\circ}$ ) is between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$. For each $j \\neq X$, the angle $\\theta_{j}$ is randomly, uniformly and independently chosen on the circle, so there is a probability of $\\frac{1}{2}$ that this angle (or its equivalent) will be in the semicircle between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$.\n\nSince there are $N-1$ such angles, the probability that all are between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$ is $\\frac{1}{2^{N-1}}$.\n\nSince there are $N$ possible selections for the first topping that can end the common sector, then the desired probability will be $\\frac{N}{2^{N-1}}$ as long as we can show that no set of angles can give two different sectors that are both covered with all toppings.\n\nTo show this last fact, we suppose without loss of generality that\n\n$$\n0^{\\circ}=\\theta_{1}<\\theta_{2}<\\theta_{3}<\\cdots<\\theta_{N-1}<\\theta_{N}<180^{\\circ}\n$$\n\n(We can relabel the toppings if necessary to obtain this order and rotate the pizza so that topping 1 begins at $0^{\\circ}$.)\n\nWe need to show that it is not possible to have a $Z$ for which $\\theta_{Z}, \\theta_{Z+1}, \\ldots, \\theta_{N}, \\theta_{1}, \\theta_{2}, \\ldots, \\theta_{Z-1}$ all lie in a semi-circle starting with $\\theta_{Z}$.\n\nSince $\\theta_{Z}<180^{\\circ}$ and $\\theta_{1}$ can be thought of as $360^{\\circ}$, then this is not possible as $\\theta_{1}$ and the angles after it are all not within $180^{\\circ}$ of $\\theta_{Z}$.\n\nTherefore, it is not possible to have two such regions with the same set of angles, and so the desired probability is $\\frac{N}{2^{N-1}}$.']" ['$\\frac{N}{2^{N-1}}$'] 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Multimodal Competition False Expression Open-ended Combinatorics Math English +87 "In rectangle $A B C D, F$ is on diagonal $B D$ so that $A F$ is perpendicular to $B D$. Also, $B C=30, C D=40$ and $A F=x$. Determine the value of $x$. + +" "['Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nWe calculate the area of $\\triangle A D B$ is two different ways.\n\nFirst, using $A B$ as base and $A D$ as height, we obtain an area of $\\frac{1}{2}(40)(30)=600$.\n\nNext, using $D B$ as base and $A F$ as height, we obtain an area of $\\frac{1}{2}(50) x=25 x$.\n\nWe must have $25 x=600$ and so $x=\\frac{600}{25}=24$.' + 'Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nSince $\\triangle D A B$ is right-angled at $A$, then $\\sin (\\angle A D B)=\\frac{A B}{B D}=\\frac{40}{50}=\\frac{4}{5}$.\n\nBut $\\triangle A D F$ is right-angled at $F$ and $\\angle A D F=\\angle A D B$.\n\nTherefore, $\\sin (\\angle A D F)=\\frac{A F}{A D}=\\frac{x}{30}$.\n\nThus, $\\frac{x}{30}=\\frac{4}{5}$ and so $x=\\frac{4}{5}(30)=24$.\n\nSolution 3\n\nSince $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nNote that $\\triangle B F A$ is similar to $\\triangle B A D$, since each is right-angled and they share a common angle at $B$.\n\nThus, $\\frac{A F}{A B}=\\frac{A D}{B D}$ and so $\\frac{x}{30}=\\frac{40}{50}$ which gives $x=\\frac{30(40)}{50}=24$.']" ['24'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +88 "In the diagram, $\triangle A B C$ is right-angled at $C$. Also, $2 \sin B=3 \tan A$. Determine the measure of angle $A$. + +" "['Since $\\triangle A B C$ is right-angled at $C$, then $\\sin B=\\cos A$.\n\nTherefore, $2 \\cos A=3 \\tan A=\\frac{3 \\sin A}{\\cos A}$ or $2 \\cos ^{2} A=3 \\sin A$.\n\nUsing the fact that $\\cos ^{2} A=1-\\sin ^{2} A$, this becomes $2-2 \\sin ^{2} A=3 \\sin A$\n\nor $2 \\sin ^{2} A+3 \\sin A-2=0$ or $(2 \\sin A-1)(\\sin A+2)=0$.\n\nSince $\\sin A$ is between -1 and 1 , then $\\sin A=\\frac{1}{2}$.\n\nSince $A$ is an acute angle, then $A=30^{\\circ}$.' + 'Since $\\triangle A B C$ is right-angled at $C$, then $\\sin B=\\frac{b}{c}$ and $\\tan A=\\frac{a}{b}$.\n\nThus, the given equation is $\\frac{2 b}{c}=\\frac{3 a}{b}$ or $2 b^{2}=3 a c$.\n\nUsing the Pythagorean Theorem, $b^{2}=c^{2}-a^{2}$ and so we obtain $2 c^{2}-2 a^{2}=3 a c$ or $2 c^{2}-3 a c-2 a^{2}=0$.\n\nFactoring, we obtain $(c-2 a)(2 c+a)=0$.\n\nSince $a$ and $c$ must both be positive, then $c=2 a$.\n\nSince $\\triangle A B C$ is right-angled, the relation $c=2 a$ means that $\\triangle A B C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, with $A=30^{\\circ}$.']" ['$30^{\\circ}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +89 "Alice drove from town $E$ to town $F$ at a constant speed of $60 \mathrm{~km} / \mathrm{h}$. Bob drove from $F$ to $E$ along the same road also at a constant speed. They started their journeys at the same time and passed each other at point $G$. + + + +Alice drove from $G$ to $F$ in 45 minutes. Bob drove from $G$ to $E$ in 20 minutes. Determine Bob's constant speed." "[""Since Alice drives at $60 \\mathrm{~km} / \\mathrm{h}$, then she drives $1 \\mathrm{~km}$ every minute.\n\nSince Alice drove from $G$ to $F$ in 45 minutes, then the distance from $G$ to $F$ is $45 \\mathrm{~km}$.\n\nLet the distance from $E$ to $G$ be $d \\mathrm{~km}$ and let Bob's speed be $B \\mathrm{~km} / \\mathrm{h}$.\n\nSince Bob drove from $G$ to $E$ in 20 minutes (or $\\frac{1}{3}$ of an hour), then $\\frac{d}{B}=\\frac{1}{3}$. Thus, $d=\\frac{1}{3} B$.\n\nThe time that it took Bob to drive from $F$ to $G$ was $\\frac{45}{B}$ hours.\n\nThe time that it took Alice to drive from $E$ to $G$ was $\\frac{d}{60}$ hours.\n\nSince the time that it took each of Alice and Bob to reach $G$ was the same, then $\\frac{d}{60}=\\frac{45}{B}$\n\nand so $B d=45(60)=2700$.\n\nThus, $B\\left(\\frac{1}{3} B\\right)=2700$ so $B^{2}=8100$ or $B=90$ since $B>0$.\n\nTherefore, Bob's speed was $90 \\mathrm{~km} / \\mathrm{h}$.""]" ['90'] 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Multimodal Competition False km/h Numerical Open-ended Geometry Math English +90 "In the diagram, $D$ is the vertex of a parabola. The parabola cuts the $x$-axis at $A$ and at $C(4,0)$. The parabola cuts the $y$-axis at $B(0,-4)$. The area of $\triangle A B C$ is 4. Determine the area of $\triangle D B C$. + +" "['First, we determine the coordinates of $A$.\n\nThe area of $\\triangle A B C$ is 4 . We can think of $A C$ as its base, and its height being the distance from $B$ to the $x$-axis.\n\nIf the coordinates of $A$ are $(a, 0)$, then the base has length $4-a$ and the height is 4 .\n\nThus, $\\frac{1}{2}(4-a)(4)=4$, so $4-a=2$ and so $a=2$.\n\nTherefore, the coordinates of $A$ are $(2,0)$.\n\nNext, we determine the equation of the parabola.\n\nThe parabola has $x$-intercepts 2 and 4 , so has equation $y=k(x-2)(x-4)$.\n\nSince the parabola passes through $(0,-4)$ as well, then $-4=k(-2)(-4)$ so $k=-\\frac{1}{2}$.\n\nTherefore, the parabola has equation $y=-\\frac{1}{2}(x-2)(x-4)$.\n\nNext, we determine the coordinates of $D$, the vertex of the parabola.\n\nSince the $x$-intercepts are 2 and 4 , then the $x$-coordinate of the vertex is the average of these, or 3.\n\n\n\nThe $y$-coordinate of $D$ can be obtained from the equation of the parabola; we obtain $y=-\\frac{1}{2}(3-2)(3-4)=-\\frac{1}{2}(1)(-1)=\\frac{1}{2}$.\n\nThus, the coordinates of $D$ are $\\left(3, \\frac{1}{2}\\right)$.\n\nLastly, we determine the area of $\\triangle B D C$, whose vertices have coordinates $B(0,-4)$, $D\\left(3, \\frac{1}{2}\\right)$, and $C(4,0)$.\n\nMethod 1 \n\nWe proceed be ""completing the rectangle"". That is, we draw the rectangle with horizontal sides along the lines $y=\\frac{1}{2}$ and $y=-4$ and vertical sides along the lines $x=0$ and $x=4$. We label this rectangle as $B P Q R$.\n\n\n\nThe area of $\\triangle B D C$ equals the area of the rectangle minus the areas of $\\triangle B P D, \\triangle D Q C$ and $\\triangle C R B$.\n\nRectangle $B P Q R$ has height $4+\\frac{1}{2}=\\frac{9}{2}$ and width 4 .\n\n$\\triangle B P D$ has height $\\frac{9}{2}$ and base 3 .\n\n$\\triangle D Q C$ has height $\\frac{1}{2}$ and base 1.\n\n$\\triangle C R B$ has height 4 and base 4.\n\nTherefore, the area of $\\triangle B D C$ is $4\\left(\\frac{9}{2}\\right)-\\frac{1}{2}\\left(\\frac{9}{2}\\right)(3)-\\frac{1}{2}\\left(\\frac{1}{2}\\right)(1)-\\frac{1}{2}(4)(4)=18-\\frac{27}{4}-\\frac{1}{4}-8=3$.\n\nMethod 2\n\nWe determine the coordinates of $E$, the point where $B D$ crosses the $x$-axis.\n\n\n\nOnce we have done this, then the area of $\\triangle B D C$ equals the sum of the areas of $\\triangle E C B$ and $\\triangle E C D$.\n\nSince $B$ has coordinates $(0,-4)$ and $D$ has coordinates $\\left(3, \\frac{1}{2}\\right)$, then the slope of $B D$ is $\\frac{\\frac{1}{2}-(-4)}{3-0}=\\frac{\\frac{9}{2}}{3}=\\frac{3}{2}$.\n\nSince $B$ is on the $y$-axis, then the equation of the line through $B$ and $D$ is $y=\\frac{3}{2} x-4$. To find the $x$-coordinate of $E$, we set $y=0$ to obtain $0=\\frac{3}{2} x-4$ or $\\frac{3}{2} x=4$ or $x=\\frac{8}{3}$.\n\nWe think of $E C$ as the base of each of the two smaller triangles. Note that $E C=4-\\frac{8}{3}=\\frac{4}{3}$. Thus, the area of $\\triangle E C D$ is $\\frac{1}{2}\\left(\\frac{4}{3}\\right)\\left(\\frac{1}{2}\\right)=\\frac{1}{3}$.\n\nAlso, the area of $\\triangle E C B$ is $\\frac{1}{2}\\left(\\frac{4}{3}\\right)(4)=\\frac{8}{3}$.\n\nTherefore, the area of $\\triangle B D C$ is $\\frac{1}{3}+\\frac{8}{3}=3$.']" ['3'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +91 "In the diagram, $P Q R S$ is a square with sides of length 4. Points $T$ and $U$ are on sides $Q R$ and $R S$ respectively such that $\angle U P T=45^{\circ}$. Determine the maximum possible perimeter of $\triangle R U T$. + +" "['Rotate a copy of $\\triangle P S U$ by $90^{\\circ}$ counterclockwise around $P$, forming a new triangle $P Q V$. Note that $V$ lies on the extension of $R Q$.\n\n\n\nThen $P V=P U$ by rotation.\n\nAlso, $\\angle V P T=\\angle V P Q+\\angle Q P T=\\angle U P S+\\angle Q P T=90^{\\circ}-\\angle U P T=90^{\\circ}-45^{\\circ}$.\n\nThis tells us that $\\triangle P T U$ is congruent to $\\triangle P T V$, by ""side-angle-side"".\n\nThus, the perimeter of $\\triangle R U T$ equals\n\n$$\n\\begin{aligned}\nU R+R T+U T & =U R+R T+T V \\\\\n& =U R+R T+T Q+Q V \\\\\n& =U R+R Q+S U \\\\\n& =S U+U R+R Q \\\\\n& =S R+R Q \\\\\n& =8\n\\end{aligned}\n$$\n\nThat is, the perimeter of $\\triangle R U T$ always equals 8 , so the maximum possible perimeter is 8 .' + 'Let $\\angle S P U=\\theta$. Note that $0^{\\circ} \\leq \\theta \\leq 45^{\\circ}$.\n\nThen $\\tan \\theta=\\frac{S U}{P S}$, so $S U=4 \\tan \\theta$.\n\nSince $S R=4$, then $U R=S R-S U=4-4 \\tan \\theta$.\n\nSince $\\angle U P T=45^{\\circ}$, then $\\angle Q P T=90^{\\circ}-45^{\\circ}-\\theta=45^{\\circ}-\\theta$.\n\nThus, $\\tan \\left(45^{\\circ}-\\theta\\right)=\\frac{Q T}{P Q}$ and so $Q T=4 \\tan \\left(45^{\\circ}-\\theta\\right)$.\n\nSince $Q R=4$, then $R T=4-4 \\tan \\left(45^{\\circ}-\\theta\\right)$.\n\nBut $\\tan (A-B)=\\frac{\\tan A-\\tan B}{1+\\tan A \\tan B}$, so $\\tan \\left(45^{\\circ}-\\theta\\right)=\\frac{\\tan \\left(45^{\\circ}\\right)-\\tan \\theta}{1+\\tan \\left(45^{\\circ}\\right) \\tan \\theta}=\\frac{1-\\tan \\theta}{1+\\tan \\theta}$, since $\\tan \\left(45^{\\circ}\\right)=1$.\n\nThis gives $R T=4-4\\left(\\frac{1-\\tan \\theta}{1+\\tan \\theta}\\right)=\\frac{4+4 \\tan \\theta}{1+\\tan \\theta}-\\frac{4-4 \\tan \\theta}{1+\\tan \\theta}=\\frac{8 \\tan \\theta}{1+\\tan \\theta}$.\n\nBy the Pythagorean Theorem in $\\triangle U R T$, we obtain\n\n$$\n\\begin{aligned}\nU T & =\\sqrt{U R^{2}+R T^{2}} \\\\\n& =\\sqrt{(4-4 \\tan \\theta)^{2}+\\left(\\frac{8 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =4 \\sqrt{(1-\\tan \\theta)^{2}+\\left(\\frac{2 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =4 \\sqrt{\\left(\\frac{1-\\tan ^{2} \\theta}{1+\\tan \\theta}\\right)^{2}+\\left(\\frac{2 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =\\sqrt[4]{\\frac{1-2 \\tan ^{2} \\theta+\\tan ^{4} \\theta+4 \\tan ^{2} \\theta}{\\left(1+\\tan ^{2}\\right)^{2}}} \\\\\n& =\\sqrt[4]{\\frac{1+2 \\tan ^{2} \\theta+\\tan ^{4} \\theta}{\\left(1+\\tan ^{2}\\right.}} \\\\\n& =4 \\sqrt{\\frac{\\left(1+\\tan ^{2} \\theta\\right)^{2}}{\\left(1+\\tan ^{2}\\right.}} \\\\\n& =4\\left(\\frac{1+\\tan ^{2} \\theta}{1+\\tan ^{2}}\\right)\n\\end{aligned}\n$$\n\nTherefore, the perimeter of $\\triangle U R T$ is\n\n$$\n\\begin{aligned}\nU R+R T+U T & =4-4 \\tan \\theta+\\frac{8 \\tan \\theta}{1+\\tan \\theta}+4\\left(\\frac{1+\\tan ^{2} \\theta}{1+\\tan \\theta}\\right) \\\\\n& =4\\left(\\frac{1-\\tan ^{2} \\theta}{1+\\tan \\theta}+\\frac{2 \\tan \\theta}{1+\\tan \\theta}+\\frac{1+\\tan ^{2} \\theta}{1+\\tan \\theta}\\right) \\\\\n& =4\\left(\\frac{2+2 \\tan \\theta}{1+\\tan \\theta}\\right) \\\\\n& =8\n\\end{aligned}\n$$\n\nThus, the perimeter is always 8 , regardless of the value of $\\theta$, so the maximum possible perimeter is 8 .']" ['8'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +92 "Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below. + + +Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also ""adjacent"" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$. +Determine the value of $f(7,3)$." ['Suppose that $p_{1}=1$.\n\nThen $p_{2}=p_{7}=0$, so the string is of the form $10 p_{3} p_{4} p_{5} p_{6} 0$.\n\nSince $k=3$, then 2 of $p_{3}, p_{4}, p_{5}, p_{6}$ equal 1 , but in such a way that no two adjacent digits are both 1 .\n\nThe possible strings in this case are 1010100, 1010010 and 1001010.\n\nSuppose that $p_{1}=0$. Then $p_{2}$ can equal 1 or 0 .\n\nIf $p_{2}=1$, then $p_{3}=0$ as well. This means that the string is of the form $010 p_{4} p_{5} p_{6} p_{7}$, which is the same as the general string in the first case, but shifted by 1 position around the circle, so there are again 3 possibilities.\n\nIf $p_{2}=0$, then the string is of the form $00 p_{3} p_{4} p_{5} p_{6} p_{7}$ and 3 of the digits $p_{3}, p_{4}, p_{5}, p_{6}, p_{7}$ equal 1 in such a way that no 2 adjacent digits equal 1.\n\nThere is only 1 way in which this can happen: 0010101.\n\nOverall, this gives 7 possible configurations, so $f(7,3)=7$.'] ['7'] 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Multimodal Competition False Numerical Open-ended Algebra Math English +93 "Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below. + + +Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also ""adjacent"" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$. +Determine the smallest possible value of $n+k$ among all possible ordered pairs of integers $(n, k)$ for which $f(n, k)$ is a positive multiple of 2009 , where $n \geq 3$ and $k \geq 2$." "[""We develop an explicit formula for $f(n, k)$ by building these strings.\n\nConsider the allowable strings of length $n$ that include $k$ 1's. Either $p_{n}=0$ or $p_{n}=1$.\n\nConsider first the case when $p_{n}=0$. (Here, $p_{1}$ can equal 0 or 1.)\n\nThese strings are all of the form $p_{1} p_{2} p_{3} \\cdots p_{n-1} 0$.\n\nIn this case, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nThus, these strings can all be built using $k 10$ blocks and $n-2 k 0$ blocks. This gives $k$ 1 's and $k+(n-2 k)=n-k 0$ 's. Note that any string built with these blocks will be allowable and will end with a 0 , and any such allowable string can be built in this way.\n\nThe number of ways of arranging $k$ blocks of one kind and $n-2 k$ blocks of another kind is $\\left(\\begin{array}{c}k+(n-2 k) \\\\ k\\end{array}\\right)$, which simplifies to $\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)$.\n\nConsider next the case when $p_{n}=1$.\n\nHere, we must have $p_{n-1}=p_{1}=0$, since these are the two digits adjacent to $p_{n}$.\n\nThus, these strings are all of the form $0 p_{2} p_{3} \\cdots 01$.\n\nConsider the strings formed by removing the first and last digits.\n\nThese strings are allowable, are of length $n-2$, include $k-11$ 's, end with 0 , and can begin with 0 or 1 .\n\nAgain, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nTranslating our method of counting from the first case, there are $\\left(\\begin{array}{c}(n-2)-(k-1) \\\\ k-1\\end{array}\\right)$ or\n\n\n\n$\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nThus, $f(n, k)=\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)+\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nIn order to look at divisibility, we need to first simplify the formula:\n\n$$\n\\begin{aligned}\nf(n, k) & =\\left(\\begin{array}{c}\nn-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right) \\\\\n& =\\frac{(n-k) !}{k !(n-k-k) !}+\\frac{(n-k-1) !}{(k-1) !((n-k-1)-(k-1)) !} \\\\\n& =\\frac{(n-k) !}{k !(n-2 k) !}+\\frac{(n-k-1) !}{(k-1) !(n-2 k) !} \\\\\n& =\\frac{(n-k-1) !(n-k)}{k !(n-2 k) !}+\\frac{(n-k-1) ! k}{k !(n-2 k) !} \\\\\n& =\\frac{(n-k-1) !(n-k+k)}{k !(n-2 k) !} \\\\\n& =\\frac{n(n-k-1) !}{k !(n-2 k) !} \\\\\n& =\\frac{n(n-k-1)(n-k-2) \\cdots(n-2 k+2)(n-2 k+1)}{k !}\n\\end{aligned}\n$$\n\nNow that we have written $f(n, k)$ as a product, it is significantly easier to look at divisibility.\n\nNote that $2009=41 \\times 49=7^{2} \\times 41$, so we need $f(n, k)$ to be divisible by 41 and by 7 twice. For this to be the case, the numerator of $f(n, k)$ must have at least one more factor of 41 and at least two more factors of 7 than the denominator.\n\nAlso, we want to minimize $n+k$, so we work to keep $n$ and $k$ as small as possible.\n\nIf $n=49$ and $k=5$, then\n\n$$\nf(49,5)=\\frac{49(43)(42)(41)(40)}{5 !}=\\frac{49(43)(42)(41)(40)}{5(4)(3)(2)(1)}=49(43)(14)\n$$\n\nwhich is divisible by 2009 .\n\nWe show that this pair minimizes the value of $n+k$ with a value of 54 .\n\nWe consider the possible cases by looking separately at the factors of 41 and 7 that must occur. We focus on the factor of 41 first.\n\nFor the numerator to contain a factor of 41 , either $n$ is divisible by 41 or one of the terms in the product $(n-k-1)(n-k-2) \\cdots(n-2 k+1)$ is divisible by 41 .\n\nCase 1: $n$ is divisible by 41\n\nWe already know that $n=82$ is too large, so we consider $n=41$. From the original interpretation of $f(n, k)$, we see that $k \\leq 20$, as there can be no more than 20 gifts placed on 41 plates.\n\nHere, the numerator becomes 41 times the product of $k-1$ consecutive integers, the largest of which is $40-k$.\n\nNow the numerator must also contain at least two factors of 7 more than the denominator. But the denominator is the product of $k$ consecutive integers. Since the numerator contains the product of $k-1$ consecutive integers and the denominator contains the product of $k$ consecutive integers, then the denominator will always include at least as many multiples of 7 as the numerator (since there are more consecutive integers in the product in the denominator). Thus, it is impossible for the numerator to contain even one more\n\n\n\nadditional factor of 7 than the denominator.\n\nTherefore, if $n=41$, then $f(n, k)$ cannot be divisible by 2009 .\n\nCase 2: $n$ is not divisible by 41\n\nThis means that the factor of 41 in the numerator must occur in the product\n\n$$\n(n-k-1)(n-k-2) \\cdots(n-2 k+1)\n$$\n\nIn this case, the integer 41 must occur in this product, since an occurrence of 82 would make $n$ greater than 82 , which does not minimize $n+k$.\n\nSo we try to find values of $n$ and $k$ that include the integer 41 in this list.\n\nNote that $n-k-1$ is the largest factor in the product and $n-2 k+1$ is the smallest.\n\nSince 41 is contained somewhere in the product, then $n-2 k+1 \\leq 41$ (giving $n \\leq 40+2 k$ ) and $41 \\leq n-k-1$ (giving $n \\geq 42+k$ ).\n\nCombining these restrictions, we get $42+k \\leq n \\leq 40+2 k$.\n\nNow, we focus on the factors of 7 .\n\nEither $n$ is not divisible by 7 or $n$ is divisible by 7 .\n\n* If $n$ is not divisible by 7 , then at least two factors of 7 must be included in the product\n\n$$\n(n-k-1)(n-k-2) \\cdots(n-2 k+1)\n$$\n\nwhich means that either $k \\geq 8$ (to give two multiples of 7 in this list of $k-1$ consecutive integers) or one of the factors is divisible by 49 .\n\n- If $k \\geq 8$, then $n \\geq 42+k \\geq 50$ so $n+k \\geq 58$, which is not minimal.\n- If one of the factors is a multiple of 49 , then 49 must be included in the list so $n-2 k+1 \\leq 49$ (giving $n \\leq 48+2 k$ ) and $49 \\leq n-k-1$ (giving $n \\geq 50+k$ ). In this case, we already know that $42+k \\leq n \\leq 40+2 k$ and now we also have $50+k \\leq n \\leq 48+2 k$.\n\nFor these ranges to overlap, we need $50+k \\leq 40+2 k$ and so $k \\geq 10$, which means that $n \\geq 50+k \\geq 60$, and so $n+k \\geq 70$, which is not minimal.\n\n* Next, we consider the case where $n$ is a multiple of 7 .\n\nHere, $42+k \\leq n \\leq 40+2 k$ (to include 41 in the product) and $n$ is a multiple of 7 .\n\nSince $k$ is at least 2 by definition, then $n \\geq 42+k \\geq 44$, so $n$ is at least 49 .\n\nIf $n$ was 56 or more, we do not get a minimal value for $n+k$.\n\nThus, we need to have $n=49$. In this case, we do not need to look for another factor of 7 in the list.\n\nTo complete this case, we need to find the smallest value of $k$ for which 49 is in the range from $42+k$ to $40+2 k$ because we need to have $42+k \\leq n \\leq 40+2 k$.\n\nThis value of $k$ is $k=5$, which gives $n+k=49+5=54$.\n\nSince $f(49,5)$ is divisible by 2009 , as determined above, then this is the case that minimizes $n+k$, giving a value of 54 .""]" ['54'] 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Multimodal Competition False Numerical Open-ended Algebra Math English +94 "A regular pentagon covers part of another regular polygon, as shown. This regular polygon has $n$ sides, five of which are completely or partially visible. In the diagram, the sum of the measures of the angles marked $a^{\circ}$ and $b^{\circ}$ is $88^{\circ}$. Determine the value of $n$. + +(The side lengths of a regular polygon are all equal, as are the measures of its interior angles.) +" "['The angles in a polygon with $n$ sides have a sum of $(n-2) \\cdot 180^{\\circ}$.\n\nThis means that the angles in a pentagon have a sum of $3 \\cdot 180^{\\circ}$ or $540^{\\circ}$, which means that each interior angle in a regular pentagon equals $\\frac{1}{5} \\cdot 540^{\\circ}$ or $108^{\\circ}$.\n\nAlso, each interior angle in a regular polygon with $n$ sides equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$. (This is the general version of the statement in the previous sentence.)\n\nConsider the portion of the regular polygon with $n$ sides that lies outside the pentagon and join the points from which the angles that measure $a^{\\circ}$ and $b^{\\circ}$ emanate to form a hexagon.\n\n\n\nThis polygon has 6 sides, and so the sum of its 6 angles is $4 \\cdot 180^{\\circ}$.\n\nFour of its angles are the original angles from the $n$-sided polygon, so each equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$.\n\nThe remaining two angles have measures $a^{\\circ}+c^{\\circ}$ and $b^{\\circ}+d^{\\circ}$.\n\nWe are told that $a^{\\circ}+b^{\\circ}=88^{\\circ}$.\n\nAlso, the angles that measure $c^{\\circ}$ and $d^{\\circ}$ are two angles in a triangle whose third angle is $108^{\\circ}$.\n\nThus, $c^{\\circ}+d^{\\circ}=180^{\\circ}-108^{\\circ}=72^{\\circ}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n4 \\cdot \\frac{n-2}{n} \\cdot 180^{\\circ}+88^{\\circ}+72^{\\circ} & =4 \\cdot 180^{\\circ} \\\\\n160^{\\circ} & =\\left(4-\\frac{4(n-2)}{n}\\right) \\cdot 180^{\\circ} \\\\\n160^{\\circ} & =\\frac{4 n-(4 n-8)}{n} \\cdot 180^{\\circ} \\\\\n\\frac{160^{\\circ}}{180^{\\circ}} & =\\frac{8}{n} \\\\\n\\frac{8}{9} & =\\frac{8}{n}\n\\end{aligned}\n$$\n\nand so the value of $n$ is 9 .' + ""The angles in a polygon with $n$ sides have a sum of $(n-2) \\cdot 180^{\\circ}$.\n\nThis means that the angles in a pentagon have a sum of $3 \\cdot 180^{\\circ}$ or $540^{\\circ}$, which means that each interior angle in a regular pentagon equals $\\frac{1}{5} \\cdot 540^{\\circ}$ or $108^{\\circ}$.\n\nAlso, each interior angle in a regular polygon with $n$ sides equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$. (This is the general version of the statement in the previous sentence.)\n\nConsider the portion of the regular polygon with $n$ sides that lies outside the pentagon.\n\n\n\nThis polygon has 7 sides, and so the sum of its 7 angles is $5 \\cdot 180^{\\circ}$.\n\nFour of its angles are the original angles from the $n$-sided polygon, so each equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$.\n\nTwo of its angles are the angles equal to $a^{\\circ}$ and $b^{\\circ}$, whose sum is $88^{\\circ}$.\n\nIts seventh angle is the reflex angle corresponding to the pentagon's angle of $108^{\\circ}$, which equals $360^{\\circ}-108^{\\circ}$ or $252^{\\circ}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n4 \\cdot \\frac{n-2}{n} \\cdot 180^{\\circ}+88^{\\circ}+252^{\\circ} & =5 \\cdot 180^{\\circ} \\\\\n340^{\\circ} & =\\left(5-\\frac{4(n-2)}{n}\\right) \\cdot 180^{\\circ} \\\\\n340^{\\circ} & =\\frac{5 n-(4 n-8)}{n} \\cdot 180^{\\circ} \\\\\n\\frac{340^{\\circ}}{180^{\\circ}} & =\\frac{n+8}{n} \\\\\n\\frac{17}{9} & =\\frac{n+8}{n} \\\\\n17 n & =9(n+8) \\\\\n17 n & =9 n+72 \\\\\n8 n & =72\n\\end{aligned}\n$$\n\nand so the value of $n$ is 9 .""]" ['9'] 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0Web7f58rlp/Li+cwu/VlDKOTyQygk459arMl0e2l1+31qTebq3t5LeMZ+Xa5VifXPyDHPc1p0AFcX4qtPGlx4m0OTw7eW0Okxyj+0I5QuWXcM9RkjbkDBHP4Y7SvNPH+ueJvDOv6DLZanbHTNR1KG0a2+yDfGCRn5yTnOG7DHvQB6Tu6+1AbPauV8W+I30W80XTopobWXVbryPtM4ykICkk4yMseFUE4ye+MGDxRB4k0fw9c3uka3cXVxFtZoru3hbKbhuKbEXDAEnncOOlAHZ0UUUAJuo3e1cD8U9V8SeG/DN14g0XVLeCG1WMSW0loHLlpAu7eTx95eMHp71e0tfFGu6Fa6hJq0OmSzwLJHDFbLLjIyDIW6k9SFC4zjJ60Adhmk3VxXgvxpdatq2p+G9ct4bXX9MbMiwk+XPEcYlQHkDkce49eGS+J7vVPiPeeEYJrzTltLVbj7TBbo5lJwT8zhlAG5R905O7kY5AO5z7UZrgrbxJqmh/ES28J6xOt/a6lA01jemJY5Qy5LRyBQFPCnkAdRxzw34p6r4k8N+GrnxBouq20ENqsYktpLQOXLSBdwcnj7y8YPT3oA77d7Ub+cY5rzy51Txxf6Vo2taBb280V1NGz2kgRB9mIz5jFiDk8EBcbc4wa0PGfi2bQ9b8PaNCk0bazcNAbqKESGLBUDAJxnLDJOcAE7TQB2m4Vl65osetaYbR72/tADuEtlctBIDgjqvUc9DxwK5DxXr+r/AA/k07U7nUpNV0W5uVtrpbiKNZoS2SHRo1UEAA8EenPNehP9w/SgDzf4K3N1q3wyilv7y5uJpLiZWlklZpMZx94nNUfDENzpfx01fRv7V1O7sU0cTRx3t282xmeLONxPqfzq18BP+SW23/X1N/6FTdM/5OQ1n/sAp/6HFQB6eWx2o3DvXFeJvFd5F4o0zwnoJh/ta9UzTTyrvWzgGcuV4yxwcAnH51X8VarrfgOwg1yXU5tW0uOVI7+G4iiR0R2C+ZGY0XoSODnOeooAofGiyntvBN7r9nq2qWd3ZiJFS2vZI4mDShTuQHBOHPPXgV3+jsz6JYOzFma2jJJ6klRXE/GiVJ/g7rEsbBkcW7Kw6EGeMg12uif8gHTv+vWL/wBBFAF+ml8dRTqoara6jdWDxaZqEdjctwJ5Lfzto9l3KM9OuR7UAV9H1pdWvdZhjVNmnXv2TcrZ3ERRu2R2wXK9/umtbPtXinwcsvEVzoWuNZ+IIYcaxMsrXFj57yyBEy5bzF68cY/HmvR/F3iq38GeGX1O9U3M+Vhhhj+UzzHoo64zgnvgA0AdJn86N1czbaV4kuLBZ73xC1rfuu429rbRG3jbGduGUu2O53jPtUPg/wAXSa5c6npGowR2+t6TL5VzHESY3U/dkTPO0jt1FAHW0UUUAFUdX1WDRtKuNQuQxjhXIVRlpG6KqjuWJAA9SKu5rlv+Ri8Vd20vRZOnaW8x39RGD/323+xQBwnwm1XVLz4j+OYdbK/b2kjMiK25UCM67VPcAFR+Fey14t4UxYftKeK7Q8LPaeYAB1J8lx+hNe00AeV/GyeTQ/Cj6zZ6rqlnqMk0VtALe9kSPPJOUBwTtDVv6L4If+w7H+0tc8Qvfm3Q3D/2tOv7wqN2AGwOa5H4qD/hIPiX4J8Jr80Zn+2XMfUFAf8A4mOT869ix6UAeWeKrnXPhkLbXINWvtX0AzrFe2V84lliDdHjkxn8D6j149OtrmK8tYbmBg8MyLJGw/iUjIP5YrzT49X0dp8MbiByN95cwwp74bf/ACQ12ngu2mtPA+g21wGE0WnW6OG6giNcj8OlAG7RRRQAUUUUAFFFFABRRRQAUmaWsfxFrqaFpvniFri7mcQ2lqpw08x6KPTuSewBPagDK8QeNLfSfFWh+HLcJNqOpTfOpPEMIBLMfc4IA9j6c9bXgen6RPbftHaUt7P9q1BbJ7m+m52tK0cg+UHoqhkUD0UV75QAUUUUAFFFFABRRRQAUUUUAFFFFABRRTd3TjigB1FN3YFOoAKKKKACiiigAooooAKKKKACiiigArzf4jfEPVfAlzYLFpNpfxXztHAouGWQMNvUbSP4h3r0ivFvEx/4S39oTQtGHzWuhxC6mHZXH7wfmfJH40AdzLqvjq0gMreGNJvABnyrTVWVz9N8QBP41N4R8e6X4v8AtEFvHPaajaHbdWF0u2WIg4PHcZGM9u+DxXU7fevBvFczeH/2ldCuLH5W1GOBLkL0cOzRHI+gU/UCgD3qiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAyfEGpyaPo8moLam5hgYNcRIMv5WfnKjuVGWx3wRXmelfArwlLrB1pLyW90i4xPaWSnEe1hkZcHLrzkdO2c9/YCnviuHsz/AMIP4kXS5AF8O6rMTYt0WzuW5MJ9Ec5ZfQ5XGCKANXxP4J0rxN4XfQ5YI4IkUfZXhjA+ysBhSg6ADpgY4yKxPh14rv7573wv4iwniPRyEmOf+PmLjbMPqCuf95TxuwO+3e1eefEjw/fJLZ+NPD0Z/tzRfmaMDIurfnfGR1OATjHqR1xgA9FrhvGXizUrPXdN8LeHLe3m1zUUaUSXJPlWsS5/eMAOejY+nuAei8NeIrHxToFpq+nsTDcJkqesbDhkb3B4rg/Fol8IfFLTPGtwGOi3VqNMv5Rz9nJbKMw/u525PONp7kZAPMPE5174a/F3Ttf1a/TWLqdRNLNHCIvMTBjZNo4DBBx+FepPf2cXi7w3410Wf/iVeIyNPvjjAeQg+QxH98MpjJPTpx1rB/aK05bzwzo2swsri3uTFlcHKSLu3Z9P3Y/76rF8E6g998Fdbk2kzWGsxXdug6J+8hcKvPAzv/BqAPRtKu4PCWvQQQuT4X12UtZOchbO6YktEQfupIclfRsjHNegbunFedaNAmvHXPDGqbbjTZmunQKDlD9uuAcN2I+QD0KHGa2PCGsXiz3XhrXJPM1nTlDCY/8AL7bnhJx/6C3XDDrzQB19FFFABRRRQB5xqANv+0FpM78Lc6FJAhPdlkZyPyNej1wvxDtmsbnQfF0aM39h3ZNzt6i2lGyU4744P0Brt1lR0V0YMjAEMDkEHoaAPLPDWB+0N4zAPWxt8f8AfuGt/wCLNxDbfC7XmmYAPbiNc92ZgB+v5YrRk8D6W2u3+uxSXMGrXkkb/a42XfFsjEYVcrgqQMlWDAnnsMJceDIdVv7a617ULnVEtX8yC1kVEgV+zlFA3sP9okdeKAI/hvY3OnfDrQbW9VluEtVLK3VQTuAPuAQPwrE8WKbn4y+A4V6wR3sz+wMYAz+IxXoee2PpXCeH4f7f+JmteJNpNpp0K6PaMejsrFpm/BiFz3waAO+rgvjISfhNr3H8EX/o5K72uU8T+BYvFkFxaX+u6vHYTlS9nbvEsfy4I6xljyN3LHmgCH4Vn/i1/h4f9Oo/9CNc5qAB/aT0rjpoTY5/25a63w14MTwtDb2tnrmrzWMClY7S5eJkAPuIw3Hbmsx/hjBJ4ij19/E/iA6rHH5SXPmwZCc/Lt8rbjk9qAMf4+/8kyl9ruL+tenqflFcl4q+H1r4xt0tdU1rVhaLtP2eF4kQsARuP7snJz649q39J0qTS7M28uqXuoHdlZbwoXUYA25RVyOOpyeTzQBo01uUYD0p1FAHk3wMuVs/hNJcNgrBcTucnHQA/hXUeNrr7d8KNauvLMZn0p5ShOSuY84otfh5Zae2oW+n6je2uk6jKZbnTo9nlkt94KxXcisOCAenTFaXibwqnifTzp82q6hZ2boY5YbNo1EqnHDFkY9uxHWgCn8NwE+G/h1QMZsIj+a5/rWX408X6jZ+KNG8IaD5Eeq6rmRrqdd628K5ywQfebCvgHj5ffjc8M+EV8L28Vpba1ql1ZQxeVFbXbRMkYzkEFUDZ7ckjB6dMJ4j8F6f4ivbHUXmuLPVLBt1te2zAOn+yQwKsp7gj19TQBwnxd0ddO+Fd/NdatqV9dmSFQ9xcEKzGRSR5a7U6AkfLxivRdCv4Ws7PTxu+0RafbzsMcbGBUYP1Q/pVPUPBlvr1jNaeIL641KN4yiqypGsRIwXUKB8/uc4ycYyc2fD/huLQLcr9tu76Xyo4fPuipfyo8+WvyhRxubnGTk5JoA4f4JEDSfEw4P/ABPZ8gc/wp/9eqXx/Bv9D0HRrZTJqN5qam3jH3iArKcd+rpUfwf0r7TZ+I722vZ7O8GtTxmSMKwkTCkKysCCASSDwRk88nPoOn+CrK28QHX7+5uNU1cJ5cVxdbcQJ/djRQFUfmevPJoA4D4tWkkPjPwHqNzLKunpeC3mnRtvluzIQ2RjBIBP/ASa9Gfw3byo8bX+qsjDDD7dJg5696vaxomn6/pc+m6nbJcWk4w8b9PqPQjqCOQayLbwneWtqtivijVmsEAVUbyjKEH8Pm7N2Pf73vQBHpcfhbwr4P8ANsmSLRtLadlkkJkMTK7LIVLZOdxZeOueOtcn4F0268beJH+IWuQMlupMei2T9Io8/wCtI/vE559cnptNdb4n8A6Z4p0K10Se5vbLT7Zgyw2TogbAwA25WyB6evrTpfCF5JZfY18Xa7DBs8sLAtpEVX0UrACvtgigDjPixcz+JfhPqtzZRERWmoFHCEtvSGUxs3A6ZGe+AteieGrmK58L6TPCwMUllCykd/kFM8P+GbHw74bh0K3aW4tIg4zdFXZw7Fm3cAHJY9qz7PwW2jRPbaDrd9ptixLLaBI5Y4Sck+XvUleTnGSM9qAOLvonv/2mtNe0O5dO0om8K/w5EuAf+/kf6Ve8SH/i/wB4Lx0+x3Wfb93JXbaD4W07w8ly1qJZbq7fzLq8uH3zTt2LN7ZOAMAZ4ArBvfhpBf8AiK316bxLr39pWwZbeVZIB5StnKgeVjGGI5zQBzHjiN7X46+B9RnI+wyo1vGx6CXLjH1JkjxXrhcDJPAHJJ4wKytZ8Nad4h0YaXqyNdQjawkY7ZFdRw6suNrdeR6kdDiqq+GLmS1Flfa/qF5ZAbXjdY0eZf7sjqoJGOuME9zycgHGfAFgfAF2Qcj+05uc5/hSvVa4rwb8OovBW+Ox13U5bV3aRrSXy/KLEYz93cDwOjDpXa0AFZHiDWP7H0tp44/Ou5HWG1gzgyzMcKv07k9gCe1a9cRqXw1h1XWbfVbrxR4iN1ayGS2KzwhYSf7q+VgenTkcHNAE1h4AtYLctPqusNdzN511Jb6lPCksrfeYKjAKCew6AAdq868VaXD4b+OfgueK5vZkucRF7u5eZslmXAZyTj5xxXucaFI1UuzlQAWbqfeuE1b4U2Ouata6pqPiPX5ry0cPbSedCvkkEH5QIgByB27UAd5uHFeQ2f8AxUf7SF3OPmt9AsBGj9t7DBH1zLJ1/u16PLod1JpEViPEOqxyo25rxBB5zjn5TmIrjkdFB4HPXPN6R8LbbQr6/vtO8UeIYbq/ffdSmS3cytknJ3Qn+8fzoA7ssAMngD1rxDwpK3ib9oPWdf0Yk6RawmCe4X7sx2KgHocsu4eyZr0K+8Arq0Jt9V8Ta/e2jcSW7TxRJIPRvKjQkfjW/pGiadoOnx2GlWkVpax/djiXAz3J9SfU80AeVfHqxlQeF9eKF7PTb4i5AGcByhB+n7sj8RXr8U0c8KSxOrxuAyspyGB6EY7Go7/T7XU7GayvYI57WZdksUi5Vx6GuWsPBWp6DEtpoPii4g01D+7tL22W6EQ9EYlWA9AScUAdRqGo22l2Mt5eSCOCIZLHkkk4AA6kk4AA5JIAqroP299MSbUsi6ndpmiP/LEMxKx/VVwp9wfWq9l4aWO7ivdTv7jVLyI7onuAqxwnGMpGoCg4JGTlsEjODW2Fx/8AqoA5n4h6z/YPw+1vUQ211tWjiYHo7/Ip/Ngayvg5o/8AY3ww0lWTbLdK13Jx13nKn/vnbVvxT8PIPF8UtvqfiHW/sTyeYLSF4VjU9gP3W4j6k1NbeCrq0sobODxh4gS3hjWONF+ygKqjAGfIz0FAGh4o8Vad4T0WbUtQkACgiGEH555OyIOpJP5da474WeEdQ0+XUvFniCMLrusuZGiI5giLbtvtk447BQK6jTvA2kWOpLqc5udS1FOEu9QmMzoM5woPyr/wEA10e33oArXOp2NnPbwXV5bwTXDbII5ZQrSt6KD94/Srdef/ABK8GaV4itLfU7qS6j1Cx/d2JgcKHmkZVjDcHI37ehB969AoAKbu5x/WnVS1Kwlv7MwQ6hdWDEg+da7N49vnVh+lAHlmqY8S/tGaZZfettBsjPKvYSN8w/V4vyr14t+dcJp/wsttL1q81iz8T+IE1G9GLidnt3MgyD/FCQOg6elX7zwI2pwNbah4r8Q3Ns4w8PnQxBx6ExxKSPbNAHntpM3if9pH+0NFy9lpUBhvbmP7jHY649+WC/8AACegrQ/aF0q4vPBtlqEUZkjsLsNMoHARlxk+24KP+BV6Zonh3SfDmmpp+kWUVpbL/Ag5Y+rE8sfcmr1zaw3ltLbXMaSwSqUeNxkMp4IP1oAjsr231Cwt7y1kWS3uI1kidejKwyCPwp13e29haS3d3MkNvEu6SRzgKPU1ydj4EvPD/wC48N+I7mx0/cWWxuoFuo4snJCEkMoz2ya1rfwzvuobvWNQn1SeFt8KSqqQxN/eWNRgn0LFiOxFAE2gSX1zZS3t+Hje7mM0UDjBgiIARD6NtUMR2ZiK2Kbt96dQAV5f8YCfP8FkZ48QW/P4mvUK43xN8O7XxXfw3Woa5rCC3mE1tDBJEiQuMYK/uyc8dyaAHeNLDw/4ik03w1rtu0n9o+a9rIjBTG8ag5U9jgkjtxzmuE1XwV4j+G+lTa74X8V3lzY2A82XTL47o2jB+bGDjoOwB988V6FqXga01g6Q+oalqNxPpbSSQXBkVJGdsYZiigZXHAAA9c1Lc+FZdThFtrGtXl9ZAjfblI4hMAcgSFFBIz1AKg9wRxQBt2N39tsLe6EZTzolk2HquQDj9as0gUAYHAHYUtAHn/xsH/FotcP/AFw/9Hx11HhkqfCmjEcj7DBgj02KeK5b41n/AItFrvH/ADw/9Hx1c0Dw9cL4U02LTtbvrG3ls4mMaJG/l7kBJQupK8nOOQOwFAHI6fG2p/tL6jdWPMNhYhLp1zgsUUbSfXJ6f7B9K2bDxFqvjnxrrOk6bfPpejaLIILieFFNxcy7mXapYEIuUbkAnpzzx2Hh3wtpfhaxe102Jh5rmSeaVi8sznqzsep5/nWavgW2svEV7rmi6hdaZdX/ADeRxqjwzt/eKMDhuTyCOp9TQBxPiGxg0743+CIIri6lkMU7SG4uXlbG1gCNxwBwemOlbHxcvodR+DevT2+7YsqwksMYaO7VG/8AHkOK3NR+H1jqVzZ6hJqN+ms2kxmi1NGTzgSMbcFSmzH8O3HX1OW658O7LXdHXSJ9Y1aHTdoElvBJGBM+8uZHJQsWLHJwQOBxQBt6Aqr4b0sLgKLSLaPT5BXJeKfFmqSeOdN8E+H5I7a8uojcXN/LGJPs8Qz91TwXwp+9xyOOeOo0Dw++gW/2cazqV/AqLHGl60beWF44KopPHqTVPXPBNhrOtWeuRXNzp+sWa7Iry127inPyMrAqy8ngjvQB5z8bdIi0z4eRPLqWo3lw97GnmXV0xDcMT8gIj/h7LXr6ajBPfXVim4z26I0gxwA+7b/6CawtX8CWfiPTbi01+9utQMqbEchI/I5B3RhRgNkDk5PGOhIMyeEXTTJLVfEGrLczMDPfo0QmlUJsCnMZUADnhQc85ySSAcp8BTj4XWw6/wClTf8AoVN0zj9o/WT/ANQFOP8AgcVdH4W8AW3g+2e10zW9WNowbFvO8TorN/EP3YOR6Zx7VUi+GcEPiGXX08T+IBqk0flSXHmQZZOPlK+VtxwO3agDkbnTjH+0lJ9tluIYtS07NpJDKYySqLuUMP8Acfj3+ld7rPhHSL/SLq21e71GXT2TdOkt9Js2qd/PPbANX/EXhXTPE9vDHfpIstvJ5ttcwNsmgk/vI3Y8DjpwMg4FQDwvPcqkOr65eajaIQTbvHFGkuP+emxQWGecZCnuDQBy3xbS3T4IaglmSLURWoh3Z+55seOvPTHWu70U40LTuP8Al1i/9BFZHizwTb+MbZrO/wBW1OCxdVWS1tWiVJCrbgzEoWJzjjOPlHFaOg6I2hWC2f8Aal9fxoAsZvGQtGoGAoKquR9cn3oA1qKKKAPK/gadmgeIVPUa9cZHp8sdZPxdupdV8IeDvEvktHYi6hubiPO7yhIoYZ6dMEZ9SPWu8j8A29nqep3OmarfafBqr+Ze20Owq7nOWUspKE5OcHvxjjHQT6Pp91pJ0qe1ik08xCE27LldgAAX6DFAFsOrKGUgg8gjmvJfB0cl7+0B4z1OA5sordLaRl6ebiIY+o8uSuztvB11YWQ0/T/E+q29gq7I4iIpHiTsqSMhYADpnJHYitXQvD2neG9NFhpkPlRbjI7MxZ5XPV3Y8sx7k0AatBOKKayllIDEEjG4daAMTxJq01jaRWlgVOq38n2ezU8hWI+aQjuqAFj9MdSKzrP4f6ZY2iQRalrqKuSdmrToCSSWbCsACSSTjHJqqfhpE2vwa5J4q8RyajACqStPCQFPVQnlbQp7gCu0ljMkbIsjRlgQGTGVz3GcjP1FAHiUGnp4e/aW0+1jnupUvLBjvuZ3mc/u36s5JP8Aq/XtXuG4ZxXA3PwosrzxDDr9x4k8QSarBgQ3JmhzGBngDytuOT27munu9DurrTre0XxDqtu8Qw9xD5Akm/3sxkD/AICFoA808K48S/tBeJtYzvt9IhFnEeu1+EP6rL+dewNIqIzuQqqMkscYrhdD+Flt4ba7bSfE/iC3e7cSXDb7dzI3PJ3Qn+8fzrQn8A22o/JrOua3qtueWtri5WOJvqsSoCPY5FAHEapan4tfEC0SELJ4Q0KQmW4/5Z3k5PKIf4hwBkcYyc/MM+xgY6VDa2dvY2sdraQxQW8Q2xxRIFVR6ADgCp6ACiiigAooooAKKKKACiiigCC6u4rO0lurh1igiQySO5wEUDJJPYAV59b23i3XddTxTaRaXFatCY9NttQ80PDETzIVUcPIAD6hcD1BteMtF8ZeIb20trNNFj0SC5SaeCe6l33iqwYI+IiFU45Hzf0rsZmvUst9tbwPd4B8qSdkj3d/nCE4/wCA8+lAHjfgw39/+0Vr0uqfZjdWun7WNtnywQIlAXPI6mvca8o8K+CfGmg+PdX8TXY0K4GqbhJDHdzKYgWDDaTCc4xivSdW1SHR9HvtTuEdobOCS4kWPBYqiliBnAzgetAF6ivCbL9oq71K6S1sPBE11cP92KC+Lu30UQkmtf8A4W/4u/6JRrf/AH1L/wDGKAPX6K8g/wCFv+L/APok2ufnN/8AGKP+Fv8Ai/8A6JNrn5zf/GKAPX6K8g/4W/4v/wCiTa5+c3/xij/hb/i//ok2ufnN/wDGKAPX6K8g/wCFv+L/APok2ufnN/8AGKP+Fv8Ai/8A6JNrn5zf/GKAPX6K8g/4W/4v/wCiTa5+c3/xij/hb/i//ok2ufnN/wDGKAPX6878I/EhtT8R33hrxFp40fXYZX8mAsWWaPqu1j1bbj/e6juBif8AC3/F/wD0SfXPzm/+MVxHxF8T6r4n0tL7UPhzrOj3tgQ8GrF5FMA3dCfJXIz7jB5HXBAPU7z4jvoXxDbQPEWnjT9OulX+ztQ3ZSRu+89Bzgf7Jxng5r0HeD05+lfN0njvW/GvgOPTdZ+H+oa4mw7dUtN6fOuQJF2xMAw5B5wecgDil+HPxX8WWWhHTk8M33iWK0KpHLAX3QqeiMVRs9OM4/HAwAfSVFeQf8Lf8X/9Em1z85v/AIxR/wALf8X/APRJtc/Ob/4xQB6/RXkH/C3/ABf/ANEm1z85v/jFH/C3/F//AE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Multimodal Competition False Numerical Open-ended Geometry Math English +95 "A circle has centre $O$ and radius 1. Quadrilateral $A B C D$ has all 4 sides tangent to the circle at points $P, Q, S$, and $T$, as shown. Also, $\angle A O B=\angle B O C=\angle C O D=\angle D O A$. If $A O=3$, determine the length of $D S$. + +" ['Since $\\angle A O B=\\angle B O C=\\angle C O D=\\angle D O A$ and these angles form a complete circle around $O$, then $\\angle A O B=\\angle B O C=\\angle C O D=\\angle D O A=\\frac{1}{4} \\cdot 360^{\\circ}=90^{\\circ}$.\n\nJoin point $O$ to $P, B, Q, C, S, D, T$, and $A$.\n\n\n\nSince $P, Q, S$, and $T$ are points of tangency, then the radii meet the sides of $A B C D$ at right angles at these points.\n\nSince $A O=3$ and $O T=1$ and $\\angle O T A=90^{\\circ}$, then by the Pythagorean Theorem, $A T=\\sqrt{A O^{2}-O T^{2}}=\\sqrt{8}=2 \\sqrt{2}$.\n\nSince $\\triangle O T A$ is right-angled at $T$, then $\\angle T A O+\\angle A O T=90^{\\circ}$.\n\nSince $\\angle D O A=90^{\\circ}$, then $\\angle A O T+\\angle D O T=90^{\\circ}$.\n\nThus, $\\angle T A O=\\angle D O T$.\n\nThis means that $\\triangle A T O$ is similar to $\\triangle O T D$.\n\nThus, $\\frac{D T}{O T}=\\frac{O T}{A T}$ and so $D T=\\frac{O T^{2}}{A T}=\\frac{1}{2 \\sqrt{2}}$.\n\nSince $D S$ and $D T$ are tangents to the circle from the same point, then $D S=D T=\\frac{1}{2 \\sqrt{2}}$.'] ['$\\frac{1}{2 \\sqrt{2}}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +96 "In the diagram, $A B=21$ and $B C=16$. Also, $\angle A B C=60^{\circ}, \angle C A D=30^{\circ}$, and $\angle A C D=45^{\circ}$. Determine the length of $C D$, to the nearest tenth. + +" ['By the cosine law in $\\triangle C B A$,\n\n$$\n\\begin{aligned}\nC A^{2} & =C B^{2}+B A^{2}-2(C B)(B A) \\cos (\\angle C B A) \\\\\nC A^{2} & =16^{2}+21^{2}-2(16)(21) \\cos \\left(60^{\\circ}\\right) \\\\\nC A^{2} & =256+441-2(16)(21)\\left(\\frac{1}{2}\\right) \\\\\nC A^{2} & =256+441-(16)(21) \\\\\nC A^{2} & =361 \\\\\nC A & =\\sqrt{361}=19 \\quad(\\text { since } C A>0)\n\\end{aligned}\n$$\n\nIn $\\triangle C A D, \\angle C D A=180^{\\circ}-\\angle D C A-\\angle D A C=180^{\\circ}-45^{\\circ}-30^{\\circ}=105^{\\circ}$.\n\nBy the sine law in $\\triangle C D A$,\n\n$$\n\\begin{aligned}\n\\frac{C D}{\\sin (\\angle D A C)} & =\\frac{C A}{\\sin (\\angle C D A)} \\\\\nC D & =\\frac{19 \\sin \\left(30^{\\circ}\\right)}{\\sin \\left(105^{\\circ}\\right)} \\\\\nC D & =\\frac{19\\left(\\frac{1}{2}\\right)}{\\sin \\left(105^{\\circ}\\right)} \\\\\nC D & =\\frac{19}{2 \\sin \\left(105^{\\circ}\\right)} \\\\\nC D & \\approx 9.835\n\\end{aligned}\n$$\n\nso, to the nearest tenth, $C D$ equals 9.8 .\n\n(Note that we could have used\n\n$$\n\\begin{aligned}\n\\sin \\left(105^{\\circ}\\right) & =\\sin \\left(60^{\\circ}+45^{\\circ}\\right)=\\sin \\left(60^{\\circ}\\right) \\cos \\left(45^{\\circ}\\right)+\\cos \\left(60^{\\circ}\\right) \\sin \\left(45^{\\circ}\\right) \\\\\n& =\\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{\\sqrt{2}}+\\frac{1}{2} \\cdot \\frac{1}{\\sqrt{2}}=\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\n\\end{aligned}\n$$\n\nto say that $C D=\\frac{19}{2\\left(\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\\right)}=\\frac{19 \\sqrt{2}}{\\sqrt{3}+1}$ exactly, and then evaluated this expression.)'] ['$\\frac{19 \\sqrt{2}}{\\sqrt{3}+1}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +97 "In the diagram, the large circle has radius 9 and centre $C(15,0)$. The small circles have radius 4 and centres $A$ and $B$ on the horizontal line $y=12$. Each of the two small circles is tangent to the large circle. It takes a bug 5 seconds to walk at a constant speed from $A$ to $B$ along the line $y=12$. How far does the bug walk in 1 second? + +" ['Consider $P$ on $A B$ with $C P$ perpendicular to $A B$. Note that $C P=12$.\n\nSince the small circle with centre $A$ is tangent to the large circle with centre $C$, then $A C$ equals the sum of the radii of these circles, or $A C=4+9=13$. Similarly, $B C=13$.\n\nThis tells us that $\\triangle A P C$ is congruent to $\\triangle B P C$ (they have equal hypotenuses and each is right-angled and has a common side), so $B P=A P$.\n\nBy the Pythagorean Theorem in $\\triangle A P C$,\n\n$$\nA P^{2}=A C^{2}-P C^{2}=13^{2}-12^{2}=169-144=25\n$$\n\nso $A P=5$ (since $A P>0)$.\n\nTherefore, $B P=A P=5$ and so $A B=10$.\n\nSince it takes the bug 5 seconds to walk this distance, then in 1 second, the bug walks a distance of $\\frac{10}{5}=2$.'] ['2'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +98 "In the diagram, $A B C$ is a right-angled triangle with $P$ and $R$ on $A B$. Also, $Q$ is on $A C$, and $P Q$ is parallel to $B C$. If $R P=2$, $B R=3, B C=4$, and the area of $\triangle Q R C$ is 5 , determine the length of $A P$. + +" "['Let $A P=x$ and $Q P=h$.\n\nSince $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.\n\nConsider trapezoid $C B P Q$. We can think of this as having parallel bases of lengths 4 and $h$ and height 5 . Thus, its area is $\\frac{1}{2}(4+h)(5)$.\n\nHowever, we can also compute its area by adding the areas of $\\triangle C B R$ (which is $\\left.\\frac{1}{2}(4)(3)\\right)$, $\\triangle C R Q$ (which is given as 5), and $\\triangle R P Q$ (which is $\\left.\\frac{1}{2}(2)(h)\\right)$.\n\nThus,\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(4+h)(5) & =\\frac{1}{2}(4)(3)+5+\\frac{1}{2}(2)(h) \\\\\n20+5 h & =12+10+2 h \\\\\n3 h & =2 \\\\\nh & =\\frac{2}{3}\n\\end{aligned}\n$$\n\nNow, $\\triangle A P Q$ is similar to $\\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,\n\n$$\n\\begin{aligned}\n\\frac{A P}{P Q} & =\\frac{A B}{B C} \\\\\n(A P)(B C) & =(P Q)(A B) \\\\\n4 x & =\\frac{2}{3}(x+5) \\\\\n4 x & =\\frac{2}{3} x+\\frac{10}{3} \\\\\n\\frac{10}{3} x & =\\frac{10}{3} \\\\\nx & =1\n\\end{aligned}\n$$\n\n\n\nTherefore, $A P=x=1$.' + 'Let $A P=x$ and $Q P=h$.\n\nSince $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.\n\nSince $\\triangle A B C$ is right-angled at $B$, its area is $\\frac{1}{2}(4)(5+x)=10+2 x$.\n\nHowever, we can look at the area of the $\\triangle A B C$ in terms of its four triangular pieces:\n\n$\\triangle C B R$ (which has area $\\left.\\frac{1}{2}(4)(3)\\right), \\triangle C R Q$ (which has area 5), $\\triangle Q P R$ (which has area $\\left.\\frac{1}{2} h(2)\\right)$, and $\\triangle Q P A$ (which has area $\\frac{1}{2} x h$ ).\n\nTherefore, $10+2 x=6+5+h+\\frac{1}{2} x h$ so $x h-4 x+2 h+2=0$.\n\nNow, $\\triangle A P Q$ is similar to $\\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,\n\n$$\n\\begin{aligned}\n\\frac{A P}{P Q} & =\\frac{A B}{B C} \\\\\n(A P)(B C) & =(P Q)(A B) \\\\\nx(4) & =h(x+5) \\\\\n4 x & =h x+5 h \\\\\n-5 h & =h x-4 x\n\\end{aligned}\n$$\n\nSubstituting this into the equation above, $x h+2 h-4 x+2=0$ becomes $-5 h+2 h+2=0$ or $3 h=2$ or $h=\\frac{2}{3}$.\n\nLastly, we solve for $x$ by subsituting our value for $h$ : $-5\\left(\\frac{2}{3}\\right)=\\frac{2}{3} x-4 x$ or $-\\frac{10}{3}=-\\frac{10}{3} x$ and so $x=1$.\n\nTherefore, $A P=x=1$.']" ['1'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +99 "In the diagram, sector $A O B$ is $\frac{1}{6}$ of an entire circle with radius $A O=B O=18$. The sector is cut into two regions with a single straight cut through $A$ and point $P$ on $O B$. The areas of the two regions are equal. Determine the length of $O P$. + +" ['Since sector $A O B$ is $\\frac{1}{6}$ of a circle with radius 18 , its area is $\\frac{1}{6}\\left(\\pi \\cdot 18^{2}\\right)$ or $54 \\pi$.\n\nFor the line $A P$ to divide this sector into two pieces of equal area, each piece has area $\\frac{1}{2}(54 \\pi)$ or $27 \\pi$.\n\nWe determine the length of $O P$ so that the area of $\\triangle P O A$ is $27 \\pi$.\n\nSince sector $A O B$ is $\\frac{1}{6}$ of a circle, then $\\angle A O B=\\frac{1}{6}\\left(360^{\\circ}\\right)=60^{\\circ}$.\n\nDrop a perpendicular from $A$ to $T$ on $O B$.\n\n\n\nThe area of $\\triangle P O A$ is $\\frac{1}{2}(O P)(A T)$.\n\n$\\triangle A O T$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nSince $A O=18$, then $A T=\\frac{\\sqrt{3}}{2}(A O)=9 \\sqrt{3}$.\n\nFor the area of $\\triangle P O A$ to equal $27 \\pi$, we have $\\frac{1}{2}(O P)(9 \\sqrt{3})=27 \\pi$ which gives $O P=\\frac{54 \\pi}{9 \\sqrt{3}}=\\frac{6 \\pi}{\\sqrt{3}}=2 \\sqrt{3} \\pi$.\n\n(Alternatively, we could have used the fact that the area of $\\triangle P O A$ is $\\frac{1}{2}(O A)(O P) \\sin (\\angle P O A)$.)'] ['$2 \\sqrt{3} \\pi$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +100 "In the diagram, a straight, flat road joins $A$ to $B$. + + + +Karuna runs from $A$ to $B$, turns around instantly, and runs back to $A$. Karuna runs at $6 \mathrm{~m} / \mathrm{s}$. Starting at the same time as Karuna, Jorge runs from $B$ to $A$, turns around instantly, and runs back to $B$. Jorge runs from $B$ to $A$ at $5 \mathrm{~m} / \mathrm{s}$ and from $A$ to $B$ at $7.5 \mathrm{~m} / \mathrm{s}$. The distance from $A$ to $B$ is $297 \mathrm{~m}$ and each runner takes exactly $99 \mathrm{~s}$ to run their route. Determine the two values of $t$ for which Karuna and Jorge are at the same place on the road after running for $t$ seconds." ['Suppose that Karuna and Jorge meet for the first time after $t_{1}$ seconds and for the second time after $t_{2}$ seconds.\n\nWhen they meet for the first time, Karuna has run partway from $A$ to $B$ and Jorge has run partway from $B$ to $A$.\n\n\n\nAt this instant, the sum of the distances that they have run equals the total distance from $A$ to $B$.\n\nSince Karuna runs at $6 \\mathrm{~m} / \\mathrm{s}$ for these $t_{1}$ seconds, she has run $6 t_{1} \\mathrm{~m}$.\n\nSince Jorge runs at $5 \\mathrm{~m} / \\mathrm{s}$ for these $t_{1}$ seconds, he has run $5 t_{1} \\mathrm{~m}$.\n\nTherefore, $6 t_{1}+5 t_{1}=297$ and so $11 t_{1}=297$ or $t_{1}=27$.\n\nWhen they meet for the second time, Karuna has run from $A$ to $B$ and is running back to $A$ and Jorge has run from $B$ to $A$ and is running back to $B$. This is because Jorge gets to $A$ halfway through his run before Karuna gets back to $A$ at the end of her run.\n\n\n\nSince they each finish running after 99 seconds, then each has $99-t_{2}$ seconds left to run. At this instant, the sum of the distances that they have left to run equals the total distance from $A$ to $B$.\n\nSince Karuna runs at $6 \\mathrm{~m} / \\mathrm{s}$ for these $\\left(99-t_{2}\\right)$ seconds, she has to run $6\\left(99-t_{2}\\right) \\mathrm{m}$.\n\nSince Jorge runs at $7.5 \\mathrm{~m} / \\mathrm{s}$ for these $\\left(99-t_{2}\\right)$ seconds, he has to run $7.5\\left(99-t_{2}\\right) \\mathrm{m}$.\n\nTherefore, $6\\left(99-t_{2}\\right)+7.5\\left(99-t_{2}\\right)=297$ and so $13.5\\left(99-t_{2}\\right)=297$ or $99-t_{2}=22$ and so $t_{2}=77$.\n\nAlternatively, to calculate the value of $t_{2}$, we note that when Karuna and Jorge meet for the second time, they have each run the distance from $A$ to $B$ one full time and are on their return trips.\n\nThis means that they have each run the full distance from $A$ to $B$ once and the distances that they have run on their return trip add up to another full distance from $A$ to $B$, for a total distance of $3 \\cdot 297 \\mathrm{~m}=891 \\mathrm{~m}$.\n\nKaruna has run at $6 \\mathrm{~m} / \\mathrm{s}$ for $t_{2}$ seconds, for a total distance of $6 t_{2} \\mathrm{~m}$.\n\nJorge ran the first $297 \\mathrm{~m}$ at $5 \\mathrm{~m} / \\mathrm{s}$, which took $\\frac{297}{5} \\mathrm{~s}$ and ran the remaining $\\left(t_{2}-\\frac{297}{5}\\right)$ seconds at $7.5 \\mathrm{~m} / \\mathrm{s}$, for a total distance of $\\left(297+7.5\\left(t_{2}-\\frac{297}{5}\\right)\\right) \\mathrm{m}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n6 t_{2}+297+7.5\\left(t_{2}-\\frac{297}{5}\\right) & =891 \\\\\n13.5 t_{2} & =891-297+7.5 \\cdot \\frac{297}{5} \\\\\n13.5 t_{2} & =1039.5 \\\\\nt_{2} & =77\n\\end{aligned}\n$$\n\nTherefore, Karuna and Jorge meet after 27 seconds and after 77 seconds.'] ['27, 77'] 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Multimodal Competition True Numerical Open-ended Algebra Math English +101 "In the diagram, rectangle $P Q R S$ is placed inside rectangle $A B C D$ in two different ways: first, with $Q$ at $B$ and $R$ at $C$; second, with $P$ on $A B, Q$ on $B C, R$ on $C D$, and $S$ on $D A$. + + +If $A B=718$ and $P Q=250$, determine the length of $B C$." ['Let $B C=x, P B=b$, and $B Q=a$.\n\nSince $B C=x$, then $A D=P S=Q R=x$.\n\nSince $B C=x$ and $B Q=a$, then $Q C=x-a$.\n\nSince $A B=718$ and $P B=b$, then $A P=718-b$.\n\nNote that $P Q=S R=250$.\n\nLet $\\angle B Q P=\\theta$.\n\nSince $\\triangle P B Q$ is right-angled at $B$, then $\\angle B P Q=90^{\\circ}-\\theta$.\n\nSince $B Q C$ is a straight angle and $\\angle P Q R=90^{\\circ}$, then $\\angle R Q C=180^{\\circ}-90^{\\circ}-\\theta=90^{\\circ}-\\theta$.\n\nSince $A P B$ is a straight angle and $\\angle S P Q=90^{\\circ}$, then $\\angle A P S=180^{\\circ}-90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$.\n\nSince $\\triangle S A P$ and $\\triangle Q C R$ are each right-angled and have another angle in common with $\\triangle P B Q$, then these\n\n\nthree triangles are similar.\n\nContinuing in the same way, we can show that $\\triangle R D S$ is also similar to these three triangles.\n\nSince $R S=P Q$, then $\\triangle R D S$ is actually congruent to $\\triangle P B Q$ (angle-side-angle).\n\nSimilarly, $\\triangle S A P$ is congruent to $\\triangle Q C R$.\n\nIn particular, this means that $A S=x-a, S D=a, D R=b$, and $R C=718-b$.\n\nSince $\\triangle S A P$ and $\\triangle P B Q$ are similar, then $\\frac{S A}{P B}=\\frac{A P}{B Q}=\\frac{S P}{P Q}$.\n\nThus, $\\frac{x-a}{b}=\\frac{718-b}{a}=\\frac{x}{250}$.\n\nAlso, by the Pythagorean Theorem in $\\triangle P B Q$, we obtain $a^{2}+b^{2}=250^{2}$.\n\nBy the Pythagorean Theorem in $\\triangle S A P$,\n\n$$\n\\begin{aligned}\nx^{2} & =(x-a)^{2}+(718-b)^{2} \\\\\nx^{2} & =x^{2}-2 a x+a^{2}+(718-b)^{2} \\\\\n0 & =-2 a x+a^{2}+(718-b)^{2}\n\\end{aligned}\n$$\n\nSince $a^{2}+b^{2}=250^{2}$, then $a^{2}=250^{2}-b^{2}$.\n\nSince $\\frac{718-b}{a}=\\frac{x}{250}$, then $a x=250(718-b)$.\n\nTherefore, substituting into $(*)$, we obtain\n\n$$\n\\begin{aligned}\n0 & =-2(250)(718-b)+250^{2}-b^{2}+(718-b)^{2} \\\\\nb^{2} & =250^{2}-2(250)(718-b)+(718-b)^{2} \\\\\nb^{2} & =((718-b)-250)^{2} \\quad\\left(\\text { since } y^{2}-2 y z+z^{2}=(y-z)^{2}\\right) \\\\\nb^{2} & =(468-b)^{2} \\\\\nb & =468-b \\quad(\\text { since } b \\neq b-468) \\\\\n2 b & =468 \\\\\nb & =234\n\\end{aligned}\n$$\n\nTherefore, $a^{2}=250^{2}-b^{2}=250^{2}-234^{2}=(250+234)(250-234)=484 \\cdot 16=22^{2} \\cdot 4^{2}=88^{2}$ and so $a=88$.\n\nFinally, $x=\\frac{250(718-b)}{a}=\\frac{250 \\cdot 484}{88}=1375$. Therefore, $B C=1375$.'] ['1375'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +102 "How many equilateral triangles of side $1 \mathrm{~cm}$, placed as shown in the diagram, are needed to completely cover the interior of an equilateral triangle of side $10 \mathrm{~cm}$ ? + +" "['If we proceed by pattern recognition, we find after row 1 we have a total of 1 triangle, after two rows we have $2^{2}$ or 4 triangles. After ten rows we have $10^{2}$ or 100 triangles.\n\n' + 'This solution is based on the fact that the ratio of areas for similar triangles is the square of the ratio of corresponding sides. Thus the big triangle with side length ten times that of the smaller triangle has 100 times the area.']" ['100'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +103 "A rectangle PQRS has side PQ on the x-axis and touches the graph of $y=k \cos x$ at the points $S$ and $R$ as shown. If the length of $P Q$ is $\frac{\pi}{3}$ and the area of the rectangle is $\frac{5 \pi}{3}$, what is the value of $k ?$ + +" ['If $P Q=\\frac{\\pi}{3}$, then by symmetry the coordinates of $R$\n\nare $\\left(\\frac{\\pi}{6}, k \\cos \\frac{\\pi}{6}\\right)$.\n\nArea of rectangle $P Q R S=\\frac{\\pi}{3}\\left(k \\cos \\frac{\\pi}{6}\\right)=\\frac{\\pi}{3}(k)\\left(\\frac{\\sqrt{3}}{2}\\right)$\n\nBut $\\frac{\\sqrt{3} k \\pi}{6}=\\frac{5 \\pi}{3} \\quad \\therefore k=\\frac{10}{\\sqrt{3}}$ or $\\frac{10}{3} \\sqrt{3}$.\n\n'] ['$\\frac{10}{\\sqrt{3}}$,$\\frac{10}{3} \\sqrt{3}$'] 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Multimodal Competition True Numerical Open-ended Geometry Math English +104 "In determining the height, $M N$, of a tower on an island, two points $A$ and $B, 100 \mathrm{~m}$ apart, are chosen on the same horizontal plane as $N$. If $\angle N A B=108^{\circ}$, $\angle A B N=47^{\circ}$ and $\angle M B N=32^{\circ}$, determine the height of the tower to the nearest metre. + +" ['In $\\triangle B A N, \\angle B N A=25^{\\circ}$\n\nUsing the Sine Law in $\\triangle B A N$,\n\n$\\frac{N B}{\\sin 108^{\\circ}}=\\frac{100}{\\sin 25^{\\circ}}$\n\nTherefore $N B=\\frac{100 \\sin 108^{\\circ}}{\\sin 25^{\\circ}} \\approx 225.04$,\n\n\n\nNow in $\\triangle M N B, \\frac{M N}{N B}=\\tan 32^{\\circ}$\n\n$$\nM N=\\frac{100 \\sin 108^{\\circ}}{\\sin 25^{\\circ}} \\times \\tan 32^{\\circ} \\doteq 140.6\n$$\n\nThe tower is approximately $141 \\mathrm{~m}$ high.'] ['141'] 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Multimodal Competition False m Numerical Open-ended Geometry Math English +105 "The points $A, P$ and a third point $Q$ (not shown) are the vertices of a triangle which is similar to triangle $A B C$. What are the coordinates of all possible positions for $Q$ ? + +" ['$Q(4,0), Q(0,4)$\n\n$Q(2,0), Q(0,2)$\n\n$Q(-2,2), Q(2,-2)$\n\n'] ['$(4,0),(0,4),(2,0),(0,2),(-2,2),(2,-2)$'] 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Multimodal Competition True Tuple Open-ended Geometry Math English +106 "In triangle $A B C, B C=2$. Point $D$ is on $\overline{A C}$ such that $A D=1$ and $C D=2$. If $\mathrm{m} \angle B D C=2 \mathrm{~m} \angle A$, compute $\sin A$. + +" ['Let $[A B C]=K$. Then $[B C D]=\\frac{2}{3} \\cdot K$. Let $\\overline{D E}$ be the bisector of $\\angle B D C$, as shown below.\n\n\n\nNotice that $\\mathrm{m} \\angle D B A=\\mathrm{m} \\angle B D C-\\mathrm{m} \\angle A=\\mathrm{m} \\angle A$, so triangle $A D B$ is isosceles, and $B D=1$. (Alternately, notice that $\\overline{D E} \\| \\overline{A B}$, and by similar triangles, $[C D E]=\\frac{4}{9} \\cdot K$, which means $[B D E]=\\frac{2}{9} \\cdot K$. Because $[C D E]:[B D E]=2$ and $\\angle B D E \\cong \\angle C D E$, conclude that $\\frac{C D}{B D}=2$, thus $B D=1$.) Because $B C D$ is isosceles, it follows that $\\cos \\angle B D C=\\frac{1}{2} B D / C D=\\frac{1}{4}$. By the half-angle formula,\n\n$$\n\\sin A=\\sqrt{\\frac{1-\\cos \\angle B D C}{2}}=\\sqrt{\\frac{3}{8}}=\\frac{\\sqrt{6}}{\\mathbf{4}}\n$$'] ['$\\frac{\\sqrt{6}}{4}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +107 "Two equilateral triangles of side length 1 and six isosceles triangles with legs of length $x$ and base of length 1 are joined as shown below; the net is folded to make a solid. If the volume of the solid is 6 , compute $x$. + +" ['First consider a regular octahedron of side length 1. To compute its volume, divide it into two square-based pyramids with edges of length 1 . Such a pyramid has slant height $\\frac{\\sqrt{3}}{2}$ and height $\\sqrt{\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}}=\\sqrt{\\frac{1}{2}}=\\frac{\\sqrt{2}}{2}$, so its volume is $\\frac{1}{3} \\cdot 1^{2} \\cdot \\frac{\\sqrt{2}}{2}=\\frac{\\sqrt{2}}{6}$. Thus the octahedron has volume twice that, or $\\frac{\\sqrt{2}}{3}$. The result of folding the net shown is actually the image of a regular octahedron after being stretched along an axis perpendicular to one face by a factor of $r$. Because the octahedron is only being stretched in one dimension, the volume changes by the same factor $r$. So the problem reduces to computing the factor $r$ and the edge length of the resulting octahedron.\n\nFor convenience, imagine that one face of the octahedron rests on a plane. Seen from above the plane, the octahedron appears as shown below.\n\n\n\n\n\nLet $P$ be the projection of $A$ onto the plane on which the octahedron rests, and let $Q$ be the foot of the perpendicular from $P$ to $\\overline{B C}$. Then $P Q=R_{C}-R_{I}$, where $R_{C}$ is the circumradius and $R_{I}$ the inradius of the equilateral triangle. Thus $P Q=\\frac{2}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{1}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{\\sqrt{3}}{6}$. Then $B P^{2}=P Q^{2}+B Q^{2}=\\frac{3}{36}+\\frac{1}{4}=\\frac{1}{3}$, so $A P^{2}=A B^{2}-B P^{2}=\\frac{2}{3}$, and $A P=\\frac{\\sqrt{6}}{3}$.\n\nNow let a vertical stretch take place along an axis parallel to $\\overleftrightarrow{A P}$. If the scale factor is $r$, then $A P=\\frac{r \\sqrt{6}}{3}$, and because the stretch occurs on an axis perpendicular to $\\overline{B P}$, the length $B P$ is unchanged, as can be seen below.\n\n\n\nThus $A B^{2}=\\frac{6 r^{2}}{9}+\\frac{1}{3}=\\frac{6 r^{2}+3}{9}$. It remains to compute $r$. But $r$ is simply the ratio of the new volume to the old volume:\n\n$$\nr=\\frac{6}{\\frac{\\sqrt{2}}{3}}=\\frac{18}{\\sqrt{2}}=9 \\sqrt{2}\n$$\n\nThus $A B^{2}=\\frac{6(9 \\sqrt{2})^{2}+3}{9}=\\frac{975}{9}=\\frac{325}{3}$, and $A B=\\frac{5 \\sqrt{39}}{3}$.'] ['$\\frac{5 \\sqrt{39}}{3}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +108 "Let $T=5$. The diagram at right consists of $T$ congruent circles, each of radius 1 , whose centers are collinear, and each pair of adjacent circles are externally tangent to each other. Compute the length of the tangent segment $\overline{A B}$. + +" ['For each point of tangency of consecutive circles, drop a perpendicular from that point to $\\overline{A B}$. For each of the $T-2$ circles between the first and last circles, the distance between consecutive perpendiculars is $2 \\cdot 1=2$. Furthermore, the distance from $A$ to the first perpendicular equals 1 (i.e., the common radius of the circles), which also equals the distance from the last perpendicular to $B$. Thus $A B=1+(T-2) \\cdot 2+1=2(T-1)$. With $T=5$, it follows that $A B=2 \\cdot 4=8$.'] ['8'] 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'] Multimodal Competition False Numerical Open-ended Geometry Math English +109 "Square $A B C D$ has side length 22. Points $G$ and $H$ lie on $\overline{A B}$ so that $A H=B G=5$. Points $E$ and $F$ lie outside square $A B C D$ so that $E F G H$ is a square. Compute the area of hexagon $A E F B C D$. + +" ['Note that $G H=A B-A H-B G=22-5-5=12$. Thus\n\n$$\n\\begin{aligned}\n{[A E F B C D] } & =[A B C D]+[E F G H]+[A E H]+[B F G] \\\\\n& =22^{2}+12^{2}+\\frac{1}{2} \\cdot 5 \\cdot 12+\\frac{1}{2} \\cdot 5 \\cdot 12 \\\\\n& =484+144+30+30 \\\\\n& =\\mathbf{6 8 8} .\n\\end{aligned}\n$$'] ['688'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +110 "Suppose that Xena traces a path along the segments in the figure shown, starting and ending at point $A$. The path passes through each of the eleven vertices besides $A$ exactly once, and only visits $A$ at the beginning and end of the path. Compute the number of possible paths Xena could trace. + +" ['Count the number of complete paths that pass through all vertices exactly once (such a path is called a Hamiltonian path). The set of vertices can be split into two rings:\n\n$$\n\\mathcal{I}=\\left\\{A_{1}, A_{2}, \\ldots, A_{6}\\right\\} \\text { (i.e., the inner ring), } \\quad \\mathcal{O}=\\left\\{B_{1}, B_{2}, \\ldots, B_{6}\\right\\} \\text { (i.e., the outer ring). }\n$$\n\nwhere $A_{1}=A$. The two rings are connected by the edges $E=\\left\\{A_{1} B_{1}, A_{2} B_{2}, \\ldots, A_{6} B_{6}\\right\\}$. Each vertex in the figure has exactly three edges joining it with the neighboring vertices. Also note that any closed loop must use exactly two edges (out of three) for each vertex.\n\nFurther note that a loop must use at least one edge from $E$ to move from one ring to the other. Consider two cases: the loop uses all six edges from $E$, or it uses some but not all of them.\n\nIf all edges from $E$ are used, there are two possible undirected loops. It is not possible to use both edges $A_{1} A_{2}$ and $B_{1} B_{2}$, so either $A_{1} A_{2}$ or $B_{1} B_{2}$ will be used. This choice determines how the entire loop is constructed.\n\nIf not all edges from $E$ are used, then there must be some $i$ for which the loop uses the edge $A_{i} B_{i}$ and does not use $A_{i+1} B_{i+1}$ (where $A_{j} B_{j}$ represents $A_{j-6} B_{j-6}$ if $7 \\leq j \\leq 12$ ). Because $A_{i+1}$ is only connected to three other vertices, the loop must use $A_{i} A_{i+1}$ and $A_{i+1} A_{i+2}$, and similarly must use $B_{i} B_{i+1}$ and $B_{i+1} B_{i+2}$. This also precludes using $A_{i+2} B_{i+2}$, because doing so would close the loop before it visits all 12 vertices. Therefore the loop must also use $A_{i+2} A_{i+3}$ and $B_{i+2} B_{i+3}$, which now precludes using $A_{i+3} B_{i+3}$. This continues to force the structure of the loop until it closes by using $A_{i+5} B_{i+5}=A_{i-1} B_{i-1}$. Hence the loop must use exactly two edges from $E$, and they must be consecutive: $A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose those two consecutive edges, so there are 6 possible undirected loops in this case.\n\nThe forced path is a loop, and the only way the given conditions are satisfied if $A_{i+k}=A_{i-1}$ and $B_{i+k}=B_{i-1}$. Hence the loop must use (precisely) two consecutive edges from $E: A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose two consecutive edges, so there are 6 possible undirected loops in this case.\n\nEach undirected loop can be traced in two ways, and thus the number of ways for Xena to trace the path is $(6+2) \\cdot 2=16$.'] ['16'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +111 "Let $T$ be a rational number. Two coplanar squares $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ each have area $T$ and are arranged as shown to form a nonconvex octagon. The center of $\mathcal{S}_{1}$ is a vertex of $\mathcal{S}_{2}$, and the center of $\mathcal{S}_{2}$ is a vertex of $\mathcal{S}_{1}$. Compute $\frac{\text { area of the union of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}{\text { area of the intersection of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}$. + +" ['Let $2 x$ be the side length of the squares. Then the intersection of $\\mathcal{S}_{1}$ and $\\mathcal{S}_{2}$ is a square of side length $x$, so its area is $x^{2}$. The area of the union of $\\mathcal{S}_{1}$ and $\\mathcal{S}_{2}$ is $(2 x)^{2}+(2 x)^{2}-x^{2}=7 x^{2}$. Thus the desired ratio of areas is $\\frac{7 x^{2}}{x^{2}}=7$ (independent of $T$ ).'] ['7'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +112 "In acute triangle $I L K$, shown in the figure, point $G$ lies on $\overline{L K}$ so that $\overline{I G} \perp \overline{L K}$. Given that $I L=\sqrt{41}$ and $L G=I K=5$, compute $G K$. + +" ['Using the Pythagorean Theorem, $I G=\\sqrt{(I L)^{2}-(L G)^{2}}=\\sqrt{41-25}=4$, and $G K=\\sqrt{(I K)^{2}-(I G)^{2}}=$ $\\sqrt{25-16}=3$.'] ['3'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +113 "This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. + + + +Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . + +The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. + +Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. + +The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. +Find $C(M)$ for the map below. + +" ['First we prove that for all maps $M, C(M) + +Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . + +The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. + +Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. + +The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. + + +The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. + + +Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. + +Determine $W(M)$ for each of the maps in figure a and figure b. + +figure a + + +figure b" "['or the map $M$ from figure a, $W(M)=7$. The most efficient strategy is to use 7 Cops to blanket all the hideouts on the first day. Any strategy using fewer than 7 Cops would require 6 Cops on each of two consecutive days: given that any hideout can be reached from any other hideout, leaving more than one hideout unsearched on one day makes it is impossible to eliminate any hideouts the following day. So any other strategy would require a minimum of 12 Cop workdays.\n\nFor the map $M$ from figure b, $W(M)=8$. The strategy outlined in $2 \\mathrm{~b}$ used three Cops for a maximum of four workdays, yielding 12 Cop workdays. The most efficient strategy is to use 4 Cops, positioned at $\\{B, E, H, K\\}$ for 2 days each. The following argument demonstrates that 8 Cop workdays is in fact minimal. First, notice that there is no advantage to searching one of the hideouts between vertices of the square (for example, $C$ ) without searching the other hideout between the same vertices (for example, $D$ ). The Robber can reach $C$ on day $n$ if and only if he is at either $B$ or $E$ on day $n-1$, and in either case he could just as well go to $D$ instead of $C$. So there is no situation in which the Robber is certain to be caught at $C$ rather than at $D$. Additionally, the Robber\'s possible locations on day $n+1$ are the same whether he is at $C$ or $D$ on day $n$, so searching one rather than the other fails to rule out any locations for future days. So any successful strategy that involves searching $C$ should also involve searching $D$ on the same day, and similarly for $F$ and $G, I$ and $J$, and $L$ and $A$. On the other hand, if the Robber must be at one of $C$ and $D$ on day $n$, then he must be at either $B$ or $E$ on day $n+1$, because those are the only adjacent hideouts. So any strategy that involves searching both hideouts of one of the off-the-square pairs on day $n$ is equivalent to a strategy that searches the adjacent on-the-square hideouts on day $n+1$; the two strategies use the same number of Cops for the same number of workdays. Thus the optimal number of Cop workdays can be achieved using strategies that only search the ""corner"" hideouts $B, E, H, K$.\n\nRestricting the search to only those strategies searching corner hideouts $B, E, H, K$, a total of 8 workdays can be achieved by searching all four hideouts on two consecutive days: if the Robber is at one of the other eight hideouts the first day, he must move to one of the two adjacent corner hideouts the second day. But each of these corner hideouts is adjacent to two other corner hideouts. So if only one hideout is searched, for no matter how many consecutive days, the following day, the Robber could either be back at the previously-searched hideout or be at any other hideout: no possibilities are ruled out. If two adjacent corners are searched, the Cops do no better, as the following argument shows. Suppose that $B$ and $E$ are both searched for two consecutive days. Then the Cops can rule out $B, E, C$, and $D$ as possible locations, but if the Cops then switch to searching either $H$ or $K$ instead of $B$ or $E$, the Robber can go back to $C$ or $D$ within two days. So searching two adjacent corner hideouts for two days is fruitless and costs four Cop workdays. Searching diagonally opposite corner hideouts is even less fruitful, because doing so rules out none of the other hideouts as possible Robber locations. Using three Cops each day, it is easy to imagine scenarios in which the Robber evades capture for three days before being caught: for example, if $B, E, H$ are searched for two consecutive days, the Robber goes from $I$ or $J$ to $K$ to $L$. Therefore if three Cops are used, four days are required for a total of 12 Cop workdays.\n\nIf there are more than four Cops, the preceding arguments show that the number of Cops must be even to produce optimal results (because there is no advantage to searching one hideout between vertices of the square without searching the other). Using six Cops with four at corner hideouts yields no improvement, because the following day, the Robber could get to any of the four corner hideouts, requiring at least four Cops the second day, for ten Cop workdays. If two or fewer Cops are at corner hideouts, the situation is even worse, because if the Robber is not caught that day, he has at least nine possible hideouts the following day (depending on whether the unsearched corners are adjacent or diagonally opposite to each other). Using eight Cops (with four at corner vertices) could eliminate one corner vertex as a possible location for the second day (if the non-corner hideouts searched are on adjacent sides of the square), but eight Cop workdays have already been used on the first day. So 8 Cop workdays is minimal.']" ['12, 8'] 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Multimodal Competition True Numerical Open-ended Combinatorics Math English +115 "In $\triangle A B C, \mathrm{~m} \angle A=\mathrm{m} \angle B=45^{\circ}$ and $A B=16$. Mutually tangent circular arcs are drawn centered at all three vertices; the arcs centered at $A$ and $B$ intersect at the midpoint of $\overline{A B}$. Compute the area of the region inside the triangle and outside of the three arcs. + +" ['Because $A B=16, A C=B C=\\frac{16}{\\sqrt{2}}=8 \\sqrt{2}$. Then each of the large arcs has radius 8 , and the small arc has radius $8 \\sqrt{2}-8$. Each large arc has measure $45^{\\circ}$ and the small arc has measure $90^{\\circ}$. Therefore the area enclosed by each large arc is $\\frac{45}{360} \\cdot \\pi \\cdot 8^{2}=8 \\pi$, and the area enclosed by the small arc is $\\frac{90}{360} \\cdot \\pi \\cdot(8 \\sqrt{2}-8)^{2}=48 \\pi-32 \\pi \\sqrt{2}$. Thus the sum of the areas enclosed by the three arcs is $64 \\pi-32 \\pi \\sqrt{2}$. On the other hand, the area of the triangle is $\\frac{1}{2}(8 \\sqrt{2})^{2}=64$. So the area of the desired region is $64-64 \\pi+32 \\pi \\sqrt{2}$.'] ['$\\quad 64-64 \\pi+32 \\pi \\sqrt{2}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +116 "Given noncollinear points $A, B, C$, segment $\overline{A B}$ is trisected by points $D$ and $E$, and $F$ is the midpoint of segment $\overline{A C} . \overline{D F}$ and $\overline{B F}$ intersect $\overline{C E}$ at $G$ and $H$, respectively. If $[D E G]=18$, compute $[F G H]$. + +" "[""Compute the desired area as $[E G F B]-[E H B]$. To compute the area of concave quadrilateral $E G F B$, draw segment $\\overline{B G}$, which divides the quadrilateral into three triangles, $\\triangle D E G, \\triangle B D G$, and $\\triangle B G F$. Then $[B D G]=[D E G]=18$ because the triangles have equal bases and heights. Because $D, G$, and $F$ are collinear, to compute $[B G F]$ it suffices to find the ratio $D G / G F$. Use Menelaus's Theorem on $\\triangle A D F$ with Menelaus Line $\\overline{E C}$ to obtain\n\n$$\n\\frac{A E}{E D} \\cdot \\frac{D G}{G F} \\cdot \\frac{F C}{C A}=1\n$$\n\n\n\nBecause $E$ and $F$ are the midpoints of $\\overline{A D}$ and $\\overline{C A}$ respectively, $A E / E D=1$ and $F C / C A=$ $1 / 2$. Therefore $D G / G F=2 / 1$, and $[B G F]=\\frac{1}{2}[B D G]=9$. Thus $[E G F B]=18+18+9=45$.\n\nTo compute $[E H B]$, consider that its base $\\overline{E B}$ is twice the base of $\\triangle D E G$. The ratio of their heights equals the ratio $E H / E G$ because the altitudes from $H$ and $G$ to $\\overleftrightarrow{B E}$ are parallel to each other. Use Menelaus's Theorem twice more on $\\triangle A E C$ to find these values:\n\n$$\n\\begin{gathered}\n\\frac{A D}{D E} \\cdot \\frac{E G}{G C} \\cdot \\frac{C F}{F A}=1 \\Rightarrow \\frac{E G}{G C}=\\frac{1}{2} \\Rightarrow E G=\\frac{1}{3} E C, \\text { and } \\\\\n\\frac{A B}{B E} \\cdot \\frac{E H}{H C} \\cdot \\frac{C F}{F A}=1 \\Rightarrow \\frac{E H}{H C}=\\frac{2}{3} \\Rightarrow E H=\\frac{2}{5} E C .\n\\end{gathered}\n$$\n\nTherefore $\\frac{E H}{E G}=\\frac{2 / 5}{1 / 3}=\\frac{6}{5}$. Thus $[E H B]=\\frac{6}{5} \\cdot 2 \\cdot[D E G]=\\frac{216}{5}$. Thus $[F G H]=45-\\frac{216}{5}=\\frac{9}{5}$."" + ""The method of mass points leads to the same results as Menelaus's Theorem, but corresponds to the physical intuition that masses on opposite sides of a fulcrum balance if and only if the products of the masses and their distances from the fulcrum are equal (in physics-speak, the net torque is zero). If a mass of weight 1 is placed at vertex $B$ and masses of weight 2 are placed at vertices $A$ and $C$, then $\\triangle A B C$ balances on the line $\\overleftrightarrow{B F}$ and also on the line $\\overleftrightarrow{C E}$. Thus it balances on the point $H$ where these two lines intersect. Replacing the masses at $A$ and $C$ with a single mass of weight 4 at their center of mass $F$, the triangle still balances at $H$. Thus $B H / H F=4$.\n\nNext, consider $\\triangle B E F$. Placing masses of weight 1 at the vertices $B$ and $E$ and a mass of weight 4 at $F$, the triangle balances at $G$. A similar argument shows that $D G / G F=2$ and that $E G / G H=5$. Because $\\triangle D E G$ and $\\triangle F H G$ have congruent (vertical) angles at $G$, it follows that $[D E G] /[F H G]=(D G / F G) \\cdot(E G / H G)=2 \\cdot 5=10$. Thus $[F G H]=[D E G] / 10=$ $\\frac{18}{10}=\\frac{9}{5}$.""]" ['$\\frac{9}{5}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +117 "Let $T=6$. In the square $D E F G$ diagrammed at right, points $M$ and $N$ trisect $\overline{F G}$, points $A$ and $B$ are the midpoints of $\overline{E F}$ and $\overline{D G}$, respectively, and $\overline{E M} \cap \overline{A B}=S$ and $\overline{D N} \cap \overline{A B}=H$. If the side length of square $D E F G$ is $T$, compute $[D E S H]$. + +" ['Note that $D E S H$ is a trapezoid with height $\\frac{T}{2}$. Because $\\overline{A S}$ and $\\overline{B H}$ are midlines of triangles $E F M$ and $D G N$ respectively, it follows that $A S=B H=\\frac{T}{6}$. Thus $S H=T-2 \\cdot \\frac{T}{6}=\\frac{2 T}{3}$. Thus $[D E S H]=\\frac{1}{2}\\left(T+\\frac{2 T}{3}\\right) \\cdot \\frac{T}{2}=\\frac{5 T^{2}}{12}$. With $T=6$, the desired area is 15 .'] ['15'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +118 "Let $R$ be the larger number you will receive, and let $r$ be the smaller number you will receive. In the diagram at right (not drawn to scale), circle $D$ has radius $R$, circle $K$ has radius $r$, and circles $D$ and $K$ are tangent at $C$. Line $\overleftrightarrow{Y P}$ is tangent to circles $D$ and $K$. Compute $Y P$. + +" ['Note that $\\overline{D Y}$ and $\\overline{K P}$ are both perpendicular to line $\\overleftrightarrow{Y P}$. Let $J$ be the foot of the perpendicular from $K$ to $\\overline{D Y}$. Then $P K J Y$ is a rectangle and $Y P=J K=\\sqrt{D K^{2}-D J^{2}}=$ $\\sqrt{(R+r)^{2}-(R-r)^{2}}=2 \\sqrt{R r}$. With $R=450$ and $r=\\frac{1}{3}$, the answer is $2 \\sqrt{150}=\\mathbf{1 0} \\sqrt{\\mathbf{6}}$.'] ['$10 \\sqrt{6}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +119 "An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: + +$$ +\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 +$$ + +Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write + +$$ +S_{3}[263415]=(132,312,231,213) +$$ + +More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. + +In this power question, you will be asked to analyze some of the properties of labels and signatures. + + +We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : + + + + + +A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: + + +Consider the shape below: + + +Find the 2-signature that corresponds to this shape." ['The first pair indicates an increase; the next three are decreases, and the last pair is an increase. So the 2-signature is $(12,21,21,21,12)$.'] ['$(12,21,21,21,12)$'] 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Multimodal Competition False Tuple Open-ended Combinatorics Math English +120 "An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: + +$$ +\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 +$$ + +Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write + +$$ +S_{3}[263415]=(132,312,231,213) +$$ + +More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. + +In this power question, you will be asked to analyze some of the properties of labels and signatures. + + +We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : + + + + + +A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: + + +List all 5-labels with 2-signature $(12,12,21,21)$." ['12543,13542,14532,23541,24531,34521'] ['12543,13542,14532,23541,24531,34521'] 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Multimodal Competition True Numerical Open-ended Combinatorics Math English +121 "An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: + +$$ +\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 +$$ + +Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write + +$$ +S_{3}[263415]=(132,312,231,213) +$$ + +More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. + +In this power question, you will be asked to analyze some of the properties of labels and signatures. + + +We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : + + + + + +A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: + + +Find a formula for the number of $(2 n+1)$-labels with the 2 -signature + +$$ +(\underbrace{12,12, \ldots, 12}_{n}, \underbrace{21,21, \ldots, 21}_{n}) +$$" ['The answer is $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$. The shape of this signature is a wedge: $n$ up steps followed by $n$ down steps. The wedge for $n=3$ is illustrated below:\n\n\n\nThe largest number in the label, $2 n+1$, must be placed at the peak in the center. If we choose the numbers to put in the first $n$ spaces, then they must be placed in increasing order. Likewise, the remaining $n$ numbers must be placed in decreasing order on the downward sloping piece of the shape. Thus there are exactly $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$ such labels.\n\n'] ['$\\binom{2n}{n}$'] 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Multimodal Competition False Expression Open-ended Combinatorics Math English +122 "An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: + +$$ +\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 +$$ + +Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write + +$$ +S_{3}[263415]=(132,312,231,213) +$$ + +More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. + +In this power question, you will be asked to analyze some of the properties of labels and signatures. + + +We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : + + + + + +A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: + + +Compute the number of 5-labels with 2 -signature $(12,21,12,21)$." "[""The answer is 16 . We have a shape with two peaks and a valley in the middle. The 5 must go on one of the two peaks, so we place it on the first peak. By the shape's symmetry, we will double our answer at the end to account for the 5 -labels where the 5 is on the other peak.\n\n\n\nThe 4 can go to the left of the 5 or at the other peak. In the first case, shown below left, the 3 must go at the other peak and the 1 and 2 can go in either order. In the latter case, shown below right, the 1,2 , and 3 can go in any of 3 ! arrangements.\n\n\n\n\nSo there are $2 !+3 !=8$ possibilities. In all, there are 165 -labels (including the ones where the 5 is at the other peak).""]" ['16'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +123 "An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: + +$$ +\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 +$$ + +Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write + +$$ +S_{3}[263415]=(132,312,231,213) +$$ + +More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. + +In this power question, you will be asked to analyze some of the properties of labels and signatures. + + +We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : + + + + + +A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: + + +Determine the number of 9-labels with 2-signature + +$$ +(12,21,12,21,12,21,12,21) \text {. } +$$ + +Justify your answer." ['The answer is 7936. The shape of this 2-signature has four peaks and three intermediate valleys:\n\n\n\nWe will solve this problem by building up from smaller examples. Let $f_{n}$ equal the number of $(2 n+1)$-labels whose 2 -signature consists of $n$ peaks and $n-1$ intermediate valleys. In part (b) we showed that $f_{2}=16$. In the case where we have one peak, $f_{1}=2$. For the trivial case (no peaks), we get $f_{0}=1$. These cases are shown below.\n\n1\n\n\n\n\n\nSuppose we know the peak on which the largest number, $2 n+1$, is placed. Then that splits our picture into two shapes with fewer peaks. Once we choose which numbers from $1,2, \\ldots, 2 n$ to place each shape, we can compute the number of arrangements of the numbers on each shape, and then take the product. For example, if we place the 9 at the second peak, as shown below, we get a 1-peak shape on the left and a 2-peak shape on the right.\n\n\n\nFor the above shape, there are $\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)$ ways to pick the three numbers to place on the left-hand side, $f_{1}=2$ ways to place them, and $f_{2}=16$ ways to place the remaining five numbers on the right.\n\nThis argument works for any $n>1$, so we have shown the following:\n\n$$\nf_{n}=\\sum_{k=1}^{n}\\left(\\begin{array}{c}\n2 n \\\\\n2 k-1\n\\end{array}\\right) f_{k-1} f_{n-k}\n$$\n\n\n\nSo we have:\n\n$$\n\\begin{aligned}\n& f_{1}=\\left(\\begin{array}{l}\n2 \\\\\n1\n\\end{array}\\right) f_{0}^{2}=2 \\\\\n& f_{2}=\\left(\\begin{array}{l}\n4 \\\\\n1\n\\end{array}\\right) f_{0} f_{1}+\\left(\\begin{array}{l}\n4 \\\\\n3\n\\end{array}\\right) f_{1} f_{0}=16 \\\\\n& f_{3}=\\left(\\begin{array}{l}\n6 \\\\\n1\n\\end{array}\\right) f_{0} f_{2}+\\left(\\begin{array}{l}\n6 \\\\\n3\n\\end{array}\\right) f_{1}^{2}+\\left(\\begin{array}{l}\n6 \\\\\n5\n\\end{array}\\right) f_{2} f_{0}=272 \\\\\n& f_{4}=\\left(\\begin{array}{l}\n8 \\\\\n1\n\\end{array}\\right) f_{0} f_{3}+\\left(\\begin{array}{l}\n8 \\\\\n3\n\\end{array}\\right) f_{1} f_{2}+\\left(\\begin{array}{l}\n8 \\\\\n5\n\\end{array}\\right) f_{2} f_{1}+\\left(\\begin{array}{l}\n8 \\\\\n7\n\\end{array}\\right) f_{3} f_{0}=7936 .\n\\end{aligned}\n$$'] ['7936'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +124 "An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: + +$$ +\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 +$$ + +Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write + +$$ +S_{3}[263415]=(132,312,231,213) +$$ + +More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. + +In this power question, you will be asked to analyze some of the properties of labels and signatures. + + +We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : + + + + + +A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: + + +For a general $n$, determine the number of distinct possible $p$-signatures." "['The answer is $p ! \\cdot p^{n-p}$.\n\nCall two consecutive windows in a $p$-signature compatible if the last $p-1$ numbers in the first label and the first $p-1$ numbers in the second label (their ""overlap"") describe the same ordering. For example, in the $p$-signature $(. ., 2143,2431, \\ldots), 2143$ and 2431 are compatible. Notice that the last three digits of 2143 and the first three digits of 2431 can be described by the same 3-label, 132 .\n\nTheorem: A signature $\\sigma$ is possible if and only if every pair of consecutive windows is compatible.\n\nProof: $(\\Rightarrow)$ Consider a signature $\\sigma$ describing a $p$-label $L$. If some pair in $\\sigma$ is not compatible, then there is some string of $p-1$ numbers in our label $L$ that has two different $(p-1)$-signatures. This is impossible, since the $p$-signature is well-defined.\n\n$(\\Leftarrow)$ Now suppose $\\sigma$ is a $p$-signature such that that every pair of consecutive windows is compatible. We need to show that there is at least one label $L$ with $S_{p}[L]=\\sigma$. We do so by induction on the number of windows in $\\sigma$, using the results from $5(\\mathrm{~b})$.\n\nLet $\\sigma=\\left\\{\\omega_{1}, \\omega_{2}, \\ldots, \\omega_{k+1}\\right\\}$, and suppose $\\omega_{1}=a_{1}, a_{2}, \\ldots, a_{p}$. Set $L_{1}=\\omega_{1}$.\n\nSuppose that $L_{k}$ is a $(p+k-1)$-label such that $S_{p}\\left[L_{k}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k}\\right\\}$. We will construct $L_{k+1}$ for which $S_{p}\\left[L_{k+1}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k+1}\\right\\}$.\n\nAs in $5(\\mathrm{~b})$, denote by $L_{k}^{(j)}$ the label $L_{k}$ with a $j+0.5$ appended; we will eventually renumber the elements in the label to make them all integers. Appending $j+0.5$ does not affect any of the non-terminal windows of $S_{p}\\left[L_{k}\\right]$, and as $j$ varies from 0 to $p-k+1$ the final window of $S_{p}\\left[L_{k}^{(j)}\\right]$ varies over each of the $p$ windows compatible with $\\omega_{k}$. Since $\\omega_{k+1}$ is compatible with $\\omega_{k}$, there exists some $j$ for which $S_{p}\\left[L_{k}^{(j)}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k+1}\\right\\}$. Now we renumber as follows: set $L_{k+1}=S_{k+p}\\left[L_{k}^{(j)}\\right]$, which replaces $L_{k}^{(j)}$ with the integers 1 through $k+p$ and preserves the relative order of all integers in the label.\n\nBy continuing this process, we conclude that the $n$-label $L_{n-p+1}$ has $p$-signature $\\sigma$, so $\\sigma$ is possible.\n\nTo count the number of possible $p$-signatures, we choose the first window ( $p$ ! choices), then choose each of the remaining $n-p$ compatible windows ( $p$ choices each). In all, there are $p ! \\cdot p^{n-p}$ possible $p$-signatures.']" ['$p ! \\cdot p^{n-p}$'] 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Multimodal Competition False Expression Open-ended Combinatorics Math English +125 "An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: + +$$ +\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 +$$ + +Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write + +$$ +S_{3}[263415]=(132,312,231,213) +$$ + +More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. + +In this power question, you will be asked to analyze some of the properties of labels and signatures. + + +We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : + + + + + +A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: + + +If a randomly chosen $p$-signature is 575 times more likely of being impossible than possible, determine $p$ and $n$." ['The answer is $n=7, p=5$.\n\nLet $P$ denote the probability that a randomly chosen $p$-signature is possible. We are\n\n\n\ngiven that $1-P=575$, so $P=\\frac{1}{576}$. We want to find $p$ and $n$ for which\n\n$$\n\\begin{aligned}\n\\frac{p ! \\cdot p^{n-p}}{(p !)^{n-p+1}} & =\\frac{1}{576} \\\\\n\\frac{p^{n-p}}{(p !)^{n-p}} & =\\frac{1}{576} \\\\\n((p-1) !)^{n-p} & =576\n\\end{aligned}\n$$\n\nThe only factorial that has 576 as an integer power is $4 !=\\sqrt{576}$. Thus $p=5$ and $n-p=2 \\Rightarrow n=7$.'] ['7,5'] 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Multimodal Competition True Numerical Open-ended Combinatorics Math English +126 "An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: + +$$ +\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 +$$ + +Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write + +$$ +S_{3}[263415]=(132,312,231,213) +$$ + +More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. + +In this power question, you will be asked to analyze some of the properties of labels and signatures. + + +We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : + + + + + +A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: + + +Find two 5-labels with unique 2-signatures." ['12345 and 54321 are the only ones.'] ['12345, 54321'] 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Multimodal Competition True Numerical Open-ended Combinatorics Math English +127 "An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: + +$$ +\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 +$$ + +Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write + +$$ +S_{3}[263415]=(132,312,231,213) +$$ + +More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. + +In this power question, you will be asked to analyze some of the properties of labels and signatures. + + +We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : + + + + + +A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: + + +Determine the smallest $p$ for which the 20-label + + +$$ +L=3,11,8,4,17,7,15,19,6,2,14,1,10,16,5,12,20,9,13,18 +$$ + +has a unique $p$-signature." "['The answer is $p=16$. To show this fact we will need to extend the idea from part 8(b) about ""linking"" inequalities forced by the various windows:\n\nTheorem: A $p$-signature for an $n$-label $L$ is unique if and only if for every $k" ['Because $A$ is on diagonal $\\overline{N Q}$, rectangles $N X A B$ and $A C Q Y$ are similar. Thus $\\frac{A B}{A X}=\\frac{Q Y}{Q C}=$ $\\frac{A C}{A Y} \\Rightarrow A B \\cdot A Y=A C \\cdot A X$. Therefore, we have $2009=[\\mathrm{I}]+2[\\mathrm{II}]+[\\mathrm{III}]$.\n\nLet the common ratio of the geometric progression be $\\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers ( $q$ may equal 1 ). Then [I] must be some integer multiple of $q^{2}$, which we will call $a q^{2}$. This gives $[\\mathrm{II}]=a p q$ and [III $=a p^{2}$. By factoring, we get\n\n$$\n2009=a q^{2}+2 a p q+a p^{2} \\Rightarrow 7^{2} \\cdot 41=a(p+q)^{2}\n$$\n\nThus we must have $p+q=7$ and $a=41$. Since $[\\mathrm{I}]=a q^{2}$ and $p, q>0$, the area is maximized when $\\frac{p}{q}=\\frac{1}{6}$, giving $[\\mathrm{I}]=41 \\cdot 36=\\mathbf{1 4 7 6}$. The areas of the other regions are 246,246, and 41 .'] ['1476'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +129 "The numbers $1,2, \ldots, 8$ are placed in the $3 \times 3$ grid below, leaving exactly one blank square. Such a placement is called okay if in every pair of adjacent squares, either one square is blank or the difference between the two numbers is at most 2 (two squares are considered adjacent if they share a common side). If reflections, rotations, etc. of placements are considered distinct, compute the number of distinct okay placements. + +" "[""We say that two numbers are neighbors if they occupy adjacent squares, and that $a$ is a friend of $b$ if $0<|a-b| \\leq 2$. Using this vocabulary, the problem's condition is that every pair of neighbors must be friends of each other. Each of the numbers 1 and 8 has two friends, and each number has at most four friends.\n\nIf there is no number written in the center square, then we must have one of the cycles in the figures below. For each cycle, there are 8 rotations. Thus there are 16 possible configurations with no number written in the center square.\n\n| 2 | 1 | 3 |\n| :--- | :--- | :--- |\n| 4 | - | 5 |\n| 6 | 8 | 7 |\n\n\n| 3 | 1 | 2 |\n| :--- | :--- | :--- |\n| 5 | - | 4 |\n| 7 | 8 | 6 |\n\nNow assume that the center square contains the number $n$. Because $n$ has at least three neighbors, $n \\neq 1$ and $n \\neq 8$. First we show that 1 must be in a corner. If 1 is a neighbor of $n$, then one of the corners neighboring 1 must be empty, because 1 has only two friends ( 2 and $3)$. If $c$ is in the other corner neighboring 1 , then $\\{n, c\\}=\\{2,3\\}$. But then $n$ must have three\n\n\n\nmore friends $\\left(n_{1}, n_{2}, n_{3}\\right)$ other than 1 and $c$, for a total of five friends, which is impossible, as illustrated below. Therefore 1 must be in a corner.\n\n| - | 1 | $c$ |\n| :--- | :--- | :--- |\n| $n_{1}$ | $n$ | $n_{2}$ |\n| | $n_{3}$ | |\n\nNow we show that 1 can only have one neighbor, i.e., one of the squares adjacent to 1 is empty. If 1 has two neighbors, then we have, up to a reflection and a rotation, the configuration shown below. Because 2 has only one more friend, the corner next to 2 is empty and $n=4$. Consequently, $m_{1}=5$ (refer to the figure below). Then 4 has one friend (namely 6) left with two neighbors $m_{2}$ and $m_{3}$, which is impossible. Thus 1 must have exactly one neighbor. An analogous argument shows that 8 must also be at a corner with exactly one neighbor.\n\n| 1 | 2 | - |\n| :--- | :--- | :--- |\n| 3 | $n$ | $m_{3}$ |\n| $m_{1}$ | $m_{2}$ | |\n\nTherefore, 8 and 1 must be in non-opposite corners, with the blank square between them. Thus, up to reflections and rotations, the only possible configuration is the one shown at left below.\n\n| 1 | - | 8 |\n| :--- | :--- | :--- |\n| $m$ | | |\n| | | |\n\n\n| 1 | - | 8 |\n| :---: | :---: | :---: |\n| $2 / 3$ | $4 / 5$ | $6 / 7$ |\n| $3 / 2$ | $5 / 4$ | $7 / 6$ |\n\nThere are two possible values for $m$, namely 2 and 3 . For each of the cases $m=2$ and $m=3$, the rest of the configuration is uniquely determined, as illustrated in the figure above right. We summarize our process: there are four corner positions for 1; two (non-opposite) corner positions for 8 (after 1 is placed); and two choices for the number in the square neighboring 1 but not neighboring 8 . This leads to $4 \\cdot 2 \\cdot 2=16$ distinct configurations with a number written in the center square.\n\nTherefore, there are 16 configurations in which the center square is blank and 16 configurations with a number in the center square, for a total of $\\mathbf{3 2}$ distinct configurations.""]" ['32'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +130 "Let $T=80$. In circle $O$, diagrammed at right, minor arc $\widehat{A B}$ measures $\frac{T}{4}$ degrees. If $\mathrm{m} \angle O A C=10^{\circ}$ and $\mathrm{m} \angle O B D=5^{\circ}$, compute the degree measure of $\angle A E B$. Just pass the number without the units. + +" ['Note that $\\mathrm{m} \\angle A E B=\\frac{1}{2}(\\mathrm{~m} \\widehat{A B}-m \\widehat{C D})=\\frac{1}{2}(\\mathrm{~m} \\widehat{A B}-\\mathrm{m} \\angle C O D)$. Also note that $\\mathrm{m} \\angle C O D=$ $360^{\\circ}-(\\mathrm{m} \\angle A O C+\\mathrm{m} \\angle B O D+\\mathrm{m} \\angle A O B)=360^{\\circ}-\\left(180^{\\circ}-2 \\mathrm{~m} \\angle O A C\\right)-\\left(180^{\\circ}-2 \\mathrm{~m} \\angle O B D\\right)-$ $\\mathrm{m} \\widehat{A B}=2(\\mathrm{~m} \\angle O A C+\\mathrm{m} \\angle O B D)-\\mathrm{m} \\widehat{A B}$. Thus $\\mathrm{m} \\angle A E B=\\mathrm{m} \\widehat{A B}-\\mathrm{m} \\angle O A C-\\mathrm{m} \\angle O B D=$ $\\frac{T}{4}-10^{\\circ}-5^{\\circ}$, and with $T=80$, the answer is 5 .'] ['5'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +131 "Let $T=24$. A regular $n$-gon is inscribed in a circle; $P$ and $Q$ are consecutive vertices of the polygon, and $A$ is another vertex of the polygon as shown. If $\mathrm{m} \angle A P Q=\mathrm{m} \angle A Q P=T \cdot \mathrm{m} \angle Q A P$, compute the value of $n$. + +" ['Let $\\mathrm{m} \\angle A=x$. Then $\\mathrm{m} \\angle P=\\mathrm{m} \\angle Q=T x$, and $(2 T+1) x=180^{\\circ}$, so $x=\\frac{180^{\\circ}}{2 T+1}$. Let $O$ be the center of the circle, as shown below.\n\n\n\nThen $\\mathrm{m} \\angle P O Q=2 \\mathrm{~m} \\angle P A Q=2\\left(\\frac{180^{\\circ}}{2 T+1}\\right)=\\frac{360^{\\circ}}{2 T+1}$. Because $\\mathrm{m} \\angle P O Q=\\frac{360^{\\circ}}{n}$, the denominators must be equal: $n=2 T+1$. Substitute $T=24$ to find $n=\\mathbf{4 9}$.'] ['49'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +132 "A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. + +This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. + +Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. + + + +Circle $A$ has curvature 2; circle $B$ has curvature 1 . + + + +Circle $A$ has curvature 2; circle $B$ has curvature -1 . + +Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then + +$$ +(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), +$$ + +where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. +Without using Descartes' Circle Formula, Find the combined area of the six remaining curvilinear territories after day 1." "['The four ""removed"" circles have radii $\\frac{1}{2}, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{3}$ so the combined area of the six remaining curvilinear territories is:\n\n$$\n\\pi\\left(1^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}\\right)=\\frac{5 \\pi}{18}\n$$']" ['$\\frac{5 \\pi}{18}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +133 "A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. + +This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. + +Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. + + + +Circle $A$ has curvature 2; circle $B$ has curvature 1 . + + + +Circle $A$ has curvature 2; circle $B$ has curvature -1 . + +Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then + +$$ +(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), +$$ + +where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. +Determine the number of curvilinear territories remaining at the end of day 3." "[""At the beginning of day 2, there are six c-triangles, so six incircles are sold, dividing each of the six territories into three smaller curvilinear triangles. So a total of 18 curvilinear triangles exist at the start of day 3, each of which is itself divided into three pieces that day (by the sale of a total of 18 regions bounded by the territories' incircles). Therefore there are 54 regions at the end of day 3.""]" ['54'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +134 "A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. + +This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. + +Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. + + + +Circle $A$ has curvature 2; circle $B$ has curvature 1 . + + + +Circle $A$ has curvature 2; circle $B$ has curvature -1 . + +Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then + +$$ +(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), +$$ + +where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. +Determine the total number of plots sold up to and including day $n$." ['The total number of plots sold up to and including day $n$ is\n\n$$\n\\begin{aligned}\n2+\\sum_{k=1}^{n} X_{k} & =2+2 \\sum_{k=1}^{n} 3^{k-1} \\\\\n& =2+2 \\cdot\\left(1+3+3^{2}+\\ldots+3^{n-1}\\right) \\\\\n& =3^{n}+1\n\\end{aligned}\n$$\n\nAlternatively, proceed by induction: on day 0 , there are $2=3^{0}+1$ plots sold, and for $n \\geq 0$,\n\n$$\n\\begin{aligned}\n\\left(3^{n}+1\\right)+X_{n+1} & =\\left(3^{n}+1\\right)+2 \\cdot 3^{n} \\\\\n& =3 \\cdot 3^{n}+1 \\\\\n& =3^{n+1}+1 .\n\\end{aligned}\n$$'] ['$3^{n}+1$'] 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Multimodal Competition False Expression Open-ended Geometry Math English +135 "A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. + +This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. + +Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. + + + +Circle $A$ has curvature 2; circle $B$ has curvature 1 . + + + +Circle $A$ has curvature 2; circle $B$ has curvature -1 . + +Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then + +$$ +(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), +$$ + +where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. +Two unit circles and a circle of radius $\frac{2}{3}$ are mutually externally tangent. Compute all possible values of $r$ such that a circle of radius $r$ is tangent to all three circles." "[""Use Descartes' Circle Formula with $a=b=1$ and $c=\\frac{3}{2}$ to solve for $d$ :\n\n$$\n\\begin{aligned}\n2 \\cdot\\left(1^{2}+1^{2}+\\left(\\frac{3}{2}\\right)^{2}+d^{2}\\right) & =\\left(1+1+\\frac{3}{2}+d\\right)^{2} \\\\\n\\frac{17}{2}+2 d^{2} & =\\frac{49}{4}+7 d+d^{2} \\\\\nd^{2}-7 d-\\frac{15}{4} & =0\n\\end{aligned}\n$$\n\nfrom which $d=\\frac{15}{2}$ or $d=-\\frac{1}{2}$. These values correspond to radii of $\\frac{2}{15}$, a small circle nestled between the other three, or 2 , a large circle enclosing the other three.\n\nAlternatively, start by scaling the kingdom with the first four circles removed to match the situation given. Thus the three given circles are internally tangent to a circle of radius $r=2$ and curvature $d=-\\frac{1}{2}$. Descartes' Circle Formula gives a quadratic equation for $d$, and the sum of the roots is $2 \\cdot\\left(1+1+\\frac{3}{2}\\right)=7$, so the second root is $7+\\frac{1}{2}=\\frac{15}{2}$, corresponding to a circle of radius $r=\\frac{2}{15}$.""]" ['$2$, $\\frac{2}{15}$'] 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Multimodal Competition True Numerical Open-ended Geometry Math English +136 "A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. + +This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. + +Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. + + + +Circle $A$ has curvature 2; circle $B$ has curvature 1 . + + + +Circle $A$ has curvature 2; circle $B$ has curvature -1 . + +Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then + +$$ +(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), +$$ + +where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. +Find the areas of the circles removed on day 3." ['Day 3 begins with two circles of curvature 15 from the configuration $(2,2,3,15)$, and four circles of curvature 6 from the configuration $(-1,2,3,6)$. Consider the following two cases:\n\nCase 1: $(a, b, c, d)=(2,2,3,15), s=22$\n\n- $a=2: a^{\\prime}=2 s-3 a=\\mathbf{3 8}$\n- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{3 8}$\n- $c=3: c^{\\prime}=2 s-3 c=\\mathbf{3 5}$\n- $d=15: d^{\\prime}=2 s-3 d=-1$, which is the configuration from day 1 .\n\nCase 2: $(a, b, c, d)=(-1,2,3,6), s=10$\n\n- $a=-1: a^{\\prime}=2 s-3 a=\\mathbf{2 3}$\n- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{1 4}$\n- $c=3: c^{\\prime}=2 s-3 c=\\mathbf{1 1}$\n- $d=6: d^{\\prime}=2 s-3 d=2$, which is the configuration from day 1 .\n\n\n\nSo the areas of the plots removed on day 3 are:\n\n$$\n\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\text { and } \\frac{\\pi}{11^{2}}\n$$\n\nThere are two circles with area $\\frac{\\pi}{35^{2}}$, and four circles with each of the other areas, for a total of 18 plots.'] ['$\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\frac{\\pi}{11^{2}}$'] 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Vz/jTw2vinwne6ZnbcMnmWsgODHMvKMD25Az7ZpngTxC/ibwlZ6hcLsvQDDeRkYKTodrjHbkZA7BhQB0tFFFABRRRQAUUUUAFIeATS0HpQByHibxufDuvaLpX9k3M39qXkdqtyXVY1LFQcckkjd6Ae9daOteb/FL/AJD/AIB/7D8P8xXpVABRRRQAUUUUAFFFFAHM/EDRf+Eg8BazpyKWle2aSEA4PmJ86f8AjyirnhLWR4h8JaTq2QWurVJJMdA+MMPwYEfhWyfunjPtXB/CoCx0fWNAC7E0fV7m1hXP/LIt5iH8d5/KgDvaKKKACkPApaRiFUknAHJNAHh3w7tX8R/GnxTrlxseDS7qZIGA53ufKU59PLiI/wCBV7lXlvwHtHHgi51abBm1S/luGb2Hy4/NW/OvUqAEPSvO4Vlt/j40H2u6lgk8PNP5U0pZEczqp2r0HCjpXoteeyEf8NBxDPP/AAjB4/7eaAOm8U3M0WgyW1s5S7vmWzgYdUeQ7S4/3VJf/gNZuqeBdP12VIdUkuDpltEkVnYwXDxRoFH322kFm7DJIAAx1NXJ/wDiY+NLaAcw6Vb/AGlv+u0u6NPxCLL/AN9iovEniqy0qaPSl1OxtNSuVLK11KqJAnQyMCRn2Xqx9AGZQDE+Fy6jaDxDo91ezXtjpepPa2VxO259oAJQnvtyPzPoAPQq53wtd+H0g/sjQtRgvfsy+ZM8MwmJZySXkZeN7Nub1PPFdFQAUUUUAFFFFABRRRQAUUUUAIelcB4c/wCJD8V/EehgBbXVYU1i2VRgK/8Aq5vxLbWr0A9K4Lxp/wAS7x34I1rfsQ3kunSc43CaM7Af+BJQB3tFNGc4p1ABRRRQAUUUUAFIehpaKAPMfiYbm41/wf8AZdO1K5Fjq8dzctbWM0ixxgrkkqpB+gJr0iCQTRJIofDLkb0KH8VPI+h5qaigAooooAKKKKACiiigBD0rhPDyGw+L3i+1LYS9trO9jTHAwrRsfzAruz0rhJ4/s/x0s7jkC78Pyw+xKTq354NAHeUUUUAB6VzvjPxPpvhTw3c3+oXXkh0aOAAEmSQqSqrjvx16DvgV0J6V5b8bfBWoeLPDtpdaYwa40xpJWgZ9qyRlcnAPBYbRj6tQBY+B+r6XefDqx02yulkurDcLqLBDRl5HYZyOc54Iz068GvS6+dvhD8LrzUdHk1y513VdKgvFUW39kXnkySKGYMZPlPGQMD6+1ek/8Ks/6nzxx/4OP/sKAPQDWW3h3RDqP9pHR9P+3A7vtP2ZPNz67sZ/WuT/AOFWf9T544/8HH/2FH/CrP8AqfPHH/g4/wDsKAO4jtYIZ5po4lWWdg0rAcuQoUZ/AD8qzL3wn4c1K7ku7/QNKu7mQ/PNPZxu7Y4GWKknAAH4VzX/AAqz/qfPHH/g4/8AsKP+FWf9T544/wDBx/8AYUAdfpeh6VowkXS9MsrFZDmRbWBYgxHTO0DPWtA9K4D/AIVZ/wBT544/8HH/ANhR/wAKtx/zPnjj/wAHH/2FAG14u8caH4KtYp9Zumjafd5EUaF3lx1AHTjI5JA5q34Y8U6R4u0z+0dGuvtEAcxvlSrI2M7SCBg4I9vQmvHfiP8ABjWruKzudD1XVdcljVlli1W+EkijjHlkgDHXIz6Ve+H3wa1PT9HlfV9f1vR7ydwTb6RqAjUKBgbyFIZs7uhwBjBNAHt9Fef/APCrP+p88cf+Dj/7Cj/hVn/U+eOP/Bx/9hQB6BRXn/8Awqz/AKnzxx/4OP8A7Cj/AIVZ/wBT544/8HH/ANhQB6BRXn//AAqz/qfPHH/g4/8AsKP+FWf9T544/wDBx/8AYUAd+3CnnFeWfGPxLo2lQaJb3V0q6hbana6jHAEJby0chm46cbvrg1pn4W4/5nzxwf8AuMf/AGFeUfFf4W6ppl5a6pZ6ne6tbTGK1MmpXYkuBIzEKNxCjbyMehznigD6G0TW9P8AEOlwanpVytzZzg7JACMkEg5B5BBHQ1pV494M+Dl5pnh2GHUfFPiHTryRmklt9J1DyoVJOBxtOWwBk/gOma6H/hVn/U+eOP8Awcf/AGFAHoFFef8A/CrP+p88cf8Ag4/+wo/4VZ/1Pnjj/wAHH/2FAHoFFef/APCrP+p88cf+Dj/7Cj/hVn/U+eOP/Bx/9hQB6BRXn/8Awqz/AKnzxx/4OP8A7Cj/AIVZ/wBT544/8HH/ANhQB6BRXn//AAqz/qfPHH/g4/8AsKP+FWf9T544/wDBx/8AYUAegUV5/wD8Ks/6nzxx/wCDj/7Cj/hVn/U+eOP/AAcf/YUAegUV5/8A8Ks/6nzxx/4OP/sKP+FWf9T544/8HH/2FAHoFFef/wDCrP8AqfPHH/g4/wDsKP8AhVn/AFPnjj/wcf8A2FAHft9015B4p8f+GdO+LmhvNqeBp0N1bX7ojMsbOF2KcDnkHOM4OM4xW+fhbgZ/4TzxwfY6v/8AYV4j4p+EXiG18fR6Tayi8GqPLNa3NxMN7qvLGU/3xkZOPmzkd8AH1TFMssSSRsXjdQysvIIPQ0/efRvyrN8P6Y2ieHNM0rzjL9jtY7fzNv3tqhc/TitHLf3v/HDQBJUVzCtxaywuAVkQowPoRipaQ9KAPOvgbdrdfCrTIw2WtpJon9j5jMB+TCvRq8b+Ct+lj4j8Z+Fiyg2upSzxLnkjeY2/Lan517JQAUUUUAFFFFABRRRQAUUUjfdPOKAKmqalb6Tp0t7clvLjAwqLuZ2JwqqO7EkAD1Irjfhbr2peJrLXdV1IlHbVJIIoA+5YY0RAFH4k5Pc5Pelm1a+1LVxq6aFeX2jWQL2LxzQqsj8hpiHcEgDIXjGCWGcjFL4GRMPhpa3D8tdXM8xOep3lf/ZTQB6VRRRQAUUUUAB6VwXxIVru98HaZG3zz69BM68fNHEGdv6V3p6V5/dFdc+NlhAvzQ+H9NkndgeFmnwgU++wE0Ad8cYrz34v3t1ZeE7caZdXdvq13ew2tm1vcPGS7NkghSN2QCOQeteh9q8u8VW//CWfGDQ9BE8sdvo9pJqU7wttYSMQqc9mB2n6NQBF8QHvvh9oGnatpGtajNfC6jga2u7p7hLzIO4bXJIPHVcEV6lCxaNCy7GKjK+ntXkF/bL4Z+M+g217LPrkOpxnyH1FzLLYuM/NH/CBwM8Z6817Ep56/hQA6iiigAooooAKKKKACiiigAooooAKKKKAA9K4bU8XHxn8Pxct9l0q6nx027mRM+9dwehrhLJFvPjlqdypyLDQobVh6NJK0n8hQB3lFFFABQaKQnAoA8E1GVvB3xp1TXApSyiuoDdnput7mMKzn2SVB+de9j2rzPxnpVtN8SNJS9jLWPiDTbjR7hs8KR+9jP8AvZ3AfStP4bavdNYXPhnV3H9taEwtpuf9dFj91KPUFcDPXjnrQB3VFFFABRRRQAUUUUAB6Vz3ivRdV1/SnsLDWI9NhlG2dvshld17qDvXbkcH2PauhooAx9T06+vPD0mnWd7b2c0sBhM/2YsqgqQSibxjHbJOPeqHgXwvdeD/AA5b6JJqMN7b227ymW1MLjcxY7vnYHlu2K6eigAooooAKKKD0oAhvLmGysp7q5lWKCGNpJJGOAiqMkn6AZriPhdbTXenal4rvIil34gujdKGHzJbr8sKn2C8j/eqH4g3EviDUbDwHp7sr6hifUpUPMFmp5+hcjaPXkd67y2ghtLeK3t41ihiQIkaDCqoGAB6AAUASTBzC4iZVkI+VmXIB7ZGRn8xXB6d4J8Q6Z4l1jXYvEmnSXWqlPN83SXIjVBhVTE4OAMdc9BXf0UAcfongSOz8St4m1fU59X1ryzFFNJGI44EPGI0HTgkZyTz7knsKKKACiiigAooooAKKKKACiiigAooooAKKKKAEPSuC+H4W/8AFPjjWwD+/wBVFkCe4t0CZHtkmuw1nUY9H0S/1OYZjtLd52HqFUnH6Vzfws0+Ww+HWkm5Yvc3aNezOerNKxkyffDD8qAOyooooAKDRRQBw3xVs55PBp1W0TdeaJcxanCM4/1bZb/xwtVXxZY3Nwum+PvCqm41C2twzQIP+P8As2AYx8fxAHcvXnPXiu/uIIrq2lt50WSGVCjo3RlIwQfwrhfhdcS2elaj4UvHZrzw/dtagv8AeeBiXhc+xXp7LQB1Ph/XrDxLo9tq2mTebbXC5GeCh7qR2YHOR+vStWvNtZ0rU/A2t3Pijw3bvdaVdMZNX0pG6t3nhHZxzuHGf1XttE1vTvEWmwanpV0lxaSjKsv6gjqCOhBoA06KKKACiiigAooooAKKKKACiig9KAEPSsDxd4ptPCehPf3MbTzORFa2sfL3MzfdjUe/48Z61L4m8T6Z4U0p7/VJiqk7IokGZJ3PREXqT/Lqa5nwx4d1PW9bTxl4uiEd4EK6bph5WwiPOW9ZD3Pbp6BQDR8CeGrvSLa61bW3WbxDqzie+kHSMY+WJf8AZQcfXPtXYU0dRTqACiiigAooooAKKKKACiiigAooooAKKKKACiiigAoPSig9KAOB+K00t14fsfDdszC416+isgUPKRZDSP8AQKuD7NXdQxJDGkUahY0UKqjooHAArgrPPiX4xXV5jdYeGrX7LCexuphmQg+yYUj1Ir0GgAooooAKKKKAEb7przvxQT4T+IWj+KF+TTtSUaVqTDOEYnMEhA44OVJPQcd69FrL8SaFa+JfDl/o14P3F3EYy2MlD1Vh7ggEe4oA0TwCAM8dK4PV/BWoaTq0/iDwPcRWd/Md15p0xP2W+PqR/A/+0MZJ5xkk3Phzr11quhyadqxI1zR5DZXyk5LMvCye4ZRnPcg12ZoA4/w78QNN1i9OlahDLo+ux8Sade/KxPrG3Rx3BHbnGK68VjeIvC2ieKbL7LrOnRXSL9wtwyH/AGWHIrlR4f8AHHhVf+Kf1uPXbBOlhrJxMq/7M68k9AA3AFAHolFefL8UI9MBTxX4d1jQXX787QG4tgfQSxg5/Kt2w8feEtTCfZf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Multimodal Competition True Numerical Open-ended Geometry Math English +137 "Points $A$ and $L$ lie outside circle $\omega$, whose center is $O$, and $\overline{A L}$ contains diameter $\overline{R M}$, as shown below. Circle $\omega$ is tangent to $\overline{L K}$ at $K$. Also, $\overline{A K}$ intersects $\omega$ at $Y$, which is between $A$ and $K$. If $K L=3, M L=2$, and $\mathrm{m} \angle A K L-\mathrm{m} \angle Y M K=90^{\circ}$, compute $[A K M]$ (i.e., the area of $\triangle A K M$ ). + +" "['Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\nCompute $[A K M]$ as $\\frac{1}{2}$ base $\\cdot$ height, using base $\\overline{A M}$.\n\n\n\nBecause of (*), $\\triangle A Y O \\sim \\triangle A K L$. To compute $A M$, notice that in $\\triangle A Y O, A O=A M-r$, while in $\\triangle A K L$, the corresponding side $A L=A M+M L=A M+2$. Therefore:\n\n$$\n\\begin{aligned}\n\\frac{A O}{A L} & =\\frac{Y O}{K L} \\\\\n\\frac{A M-\\frac{5}{4}}{A M+2} & =\\frac{5 / 4}{3}\n\\end{aligned}\n$$\n\nfrom which $A M=\\frac{25}{7}$. Draw the altitude of $\\triangle A K M$ from vertex $K$, and let $h$ be its length. In right triangle $O K L, h$ is the altitude to the hypotenuse, so $\\frac{h}{3}=\\sin (\\angle K L O)=\\frac{r}{r+2}$. Hence $h=\\frac{15}{13}$. Therefore $[A K M]=\\frac{1}{2} \\cdot \\frac{25}{7} \\cdot \\frac{15}{13}=\\frac{375}{182}$.' + 'Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\nBy the Power of the Point Theorem, $L K^{2}=L M \\cdot L R$, so\n\n$$\n\\begin{aligned}\nL R & =\\frac{9}{2} \\\\\nR M & =L R-L M=\\frac{5}{2} \\\\\nO L & =r+M L=\\frac{13}{4}\n\\end{aligned}\n$$\n\nFrom (*), we know that $\\triangle A Y O \\sim \\triangle A K L$. Hence by $(\\dagger)$,\n\n$$\n\\frac{A L}{A O}=\\frac{A L}{A L-O L}=\\frac{K L}{Y O}=\\frac{3}{5 / 4}=\\frac{12}{5}, \\quad \\text { thus } \\quad A L=\\frac{12}{7} \\cdot O L=\\frac{12}{7} \\cdot \\frac{13}{4}=\\frac{39}{7}\n$$\n\nHence $A M=A L-2=\\frac{25}{7}$. The ratio between the areas of triangles $A K M$ and $R K M$ is equal to\n\n$$\n\\frac{[A K M]}{[R K M]}=\\frac{A M}{R M}=\\frac{25 / 7}{5 / 2}=\\frac{10}{7}\n$$\n\nThus $[A K M]=\\frac{10}{7} \\cdot[R K M]$.\n\nBecause $\\angle K R L$ and $\\angle M K L$ both subtend $\\widehat{K M}, \\triangle K R L \\sim \\triangle M K L$. Therefore $\\frac{K R}{M K}=\\frac{L K}{L M}=$ $\\frac{3}{2}$. Thus let $K R=3 x$ and $M K=2 x$ for some positive real number $x$. Because $R M$ is a diameter of $\\omega$ (see left diagram below), $\\mathrm{m} \\angle R K M=90^{\\circ}$. Thus triangle $R K M$ is a right triangle with hypotenuse $\\overline{R M}$. In particular, $13 x^{2}=K R^{2}+M K^{2}=R M^{2}=\\frac{25}{4}$, so $x^{2}=\\frac{25}{52}$ and $[R K M]=\\frac{R K \\cdot K M}{2}=3 x^{2}$. Therefore\n\n$$\n[A K M]=\\frac{10}{7} \\cdot[R K M]=\\frac{10}{7} \\cdot 3 \\cdot \\frac{25}{52}=\\frac{\\mathbf{3 7 5}}{\\mathbf{1 8 2}}\n$$\n\n' + 'Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\n\nLet $U$ and $V$ be the respective feet of the perpendiculars dropped from $A$ and $M$ to $\\overleftrightarrow{K L}$. From (*), $\\triangle A K L$ can be dissected into two infinite progressions of triangles: one progression of triangles similar to $\\triangle O K L$ and the other similar to $\\triangle Y O K$, as shown in the right diagram above. In both progressions, the corresponding sides of the triangles have common ratio equal to\n\n$$\n\\frac{Y O}{K L}=\\frac{5 / 4}{3}=\\frac{5}{12}\n$$\n\n\n\nThus\n\n$$\nA U=\\frac{5}{4}\\left(1+\\frac{5}{12}+\\left(\\frac{5}{12}\\right)^{2}+\\cdots\\right)=\\frac{5}{4} \\cdot \\frac{12}{7}=\\frac{15}{7}\n$$\n\nBecause $\\triangle L M V \\sim \\triangle L O K$, and because $L O=\\frac{13}{4}$ by $(\\dagger)$,\n\n$$\n\\frac{M V}{O K}=\\frac{L M}{L O}, \\quad \\text { thus } \\quad M V=\\frac{O K \\cdot L M}{L O}=\\frac{\\frac{5}{4} \\cdot 2}{\\frac{13}{4}}=\\frac{10}{13}\n$$\n\nFinally, note that $[A K M]=[A K L]-[K L M]$. Because $\\triangle A K L$ and $\\triangle K L M$ share base $\\overline{K L}$,\n\n$$\n[A K M]=\\frac{1}{2} \\cdot 3 \\cdot\\left(\\frac{15}{7}-\\frac{10}{13}\\right)=\\frac{\\mathbf{3 7 5}}{\\mathbf{1 8 2}}\n$$']" ['$\\frac{375}{182}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +138 "Let $T=8 \sqrt{2}$. In the diagram at right, the smaller circle is internally tangent to the larger circle at point $O$, and $\overline{O P}$ is a diameter of the larger circle. Point $Q$ lies on $\overline{O P}$ such that $P Q=T$, and $\overline{P Q}$ does not intersect the smaller circle. If the larger circle's radius is three times the smaller circle's radius, find the least possible integral radius of the larger circle. + +" ['Let $r$ be the radius of the smaller circle. Then the conditions defining $Q$ imply that $P Q=$ $T<4 r$. With $T=8 \\sqrt{2}$, note that $r>2 \\sqrt{2} \\rightarrow 3 r>6 \\sqrt{2}=\\sqrt{72}$. The least integer greater than $\\sqrt{72}$ is 9 .'] ['9'] 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Multimodal Competition False Numerical Open-ended Geometry Math English +139 "The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. + +Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. + +PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. + + + +As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. +For $n=1,2,3,4$, and $k=4$, find $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\operatorname{Pa}(n+k, n)$." ['$$\n\\begin{aligned}\n& \\mathrm{Pa}(1,1)+\\mathrm{Pa}(2,1)+\\mathrm{Pa}(3,1)+\\mathrm{Pa}(4,1)+\\mathrm{Pa}(5,1)=1+2+3+4+5=\\mathbf{1 5} \\\\\n& \\mathrm{Pa}(2,2)+\\mathrm{Pa}(3,2)+\\mathrm{Pa}(4,2)+\\mathrm{Pa}(5,2)+\\mathrm{Pa}(6,2)=1+3+6+10+15=\\mathbf{3 5} \\\\\n& \\mathrm{Pa}(3,3)+\\mathrm{Pa}(4,3)+\\mathrm{Pa}(5,3)+\\mathrm{Pa}(6,3)+\\mathrm{Pa}(7,3)=1+4+10+20+35=\\mathbf{7 0} \\\\\n& \\mathrm{Pa}(4,4)+\\mathrm{Pa}(5,4)+\\mathrm{Pa}(6,4)+\\mathrm{Pa}(7,4)+\\mathrm{Pa}(8,4)=1+5+15+35+70=\\mathbf{1 2 6}\n\\end{aligned}\n$$'] ['15, 35, 70, 126'] 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Multimodal Competition True Numerical Open-ended Combinatorics Math English +140 "The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. + +Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. + +PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. + + + +As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. +If $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\mathrm{Pa}(n+k, n)=\mathrm{Pa}(m, j)$, find and justify formulas for $m$ and $j$ in terms of $n$ and $k$." ['Notice that $\\mathrm{Pa}(n, n)+\\operatorname{Pa}(n+1, n)+\\cdots+\\operatorname{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, so $m=n+k+1$ and $j=n+1$. (By symmetry, $j=k$ is also correct.) The equation is true for all $n$ when $k=0$, because the sum is simply $\\mathrm{Pa}(n, n)$ and the right side is $\\mathrm{Pa}(n+1, n+1)$, both of which are 1 . Proceed by induction on $k$. If $\\mathrm{Pa}(n, n)+\\mathrm{Pa}(n+1, n)+\\cdots+\\mathrm{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, then adding $\\mathrm{Pa}(n+k+1, n)$ to both sides yields $\\mathrm{Pa}(n+k+1, n)+\\mathrm{Pa}(n+k+1, n+1)=$ $\\mathrm{Pa}(n+k+2, n+1)$ by the recursive rule for $\\mathrm{Pa}$.'] ['$m=n+k+1$, $j=n+1$'] 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Multimodal Competition True Expression Open-ended Combinatorics Math English +141 "The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. + +Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. + +PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. + + + +As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. + +Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: + +$$ +\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} +$$ + +The first four rows of Clark's Triangle are given below. + + +If $\mathrm{Cl}(n, 1)=a n^{2}+b n+c$, determine the values of $a, b$, and $c$." ['Using the given values yields the system of equations below.\n\n$$\n\\left\\{\\begin{array}{l}\n\\mathrm{Cl}(1,1)=1=a(1)^{2}+b(1)+c \\\\\n\\mathrm{Cl}(2,1)=7=a(2)^{2}+b(2)+c \\\\\n\\mathrm{Cl}(3,1)=19=a(3)^{2}+b(3)+c\n\\end{array}\\right.\n$$\n\nSolving this system, $a=3, b=-3, c=1$.'] ['$3,-3,1$'] 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Multimodal Competition True Numerical Open-ended Combinatorics Math English +142 "The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. + +Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. + +PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. + + + +As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. + +Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: + +$$ +\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} +$$ + +The first four rows of Clark's Triangle are given below. + + +Compute $\mathrm{Cl}(11,2)$." ['$\\mathrm{Cl}(11,2)=1000$.'] ['1000'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +143 "The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. + +Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. + +PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. + + + +As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. + +Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: + +$$ +\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} +$$ + +The first four rows of Clark's Triangle are given below. + + +Compute $\mathrm{Cl}(11,3)$." ['$\\mathrm{Cl}(11,3)=2025$.'] ['2025'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +144 "Leibniz's Harmonic Triangle: Consider the triangle formed by the rule + +$$ +\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} +$$ + +This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. + + + +For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. +Compute Le(17,1)." ['$\\operatorname{Le}(17,1)=\\operatorname{Le}(16,0)-\\operatorname{Le}(17,0)=\\frac{1}{17}-\\frac{1}{18}=\\frac{1}{306}$.'] ['$\\frac{1}{306}$'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +145 "Leibniz's Harmonic Triangle: Consider the triangle formed by the rule + +$$ +\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} +$$ + +This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. + + + +For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. +Compute $\operatorname{Le}(17,2)$." ['$\\operatorname{Le}(17,2)=\\operatorname{Le}(16,1)-\\operatorname{Le}(17,1)=\\operatorname{Le}(15,0)-\\operatorname{Le}(16,0)-\\operatorname{Le}(17,1)=\\frac{1}{2448}$.'] ['$\\frac{1}{2448}$'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +146 "Leibniz's Harmonic Triangle: Consider the triangle formed by the rule + +$$ +\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} +$$ + +This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. + + + +For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. +Compute $\sum_{n=1}^{2011} \operatorname{Le}(n, 1)$." ['Because $\\operatorname{Le}(n, 1)=\\frac{1}{n}-\\frac{1}{n+1}$,\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{2011} \\operatorname{Le}(i, 1) & =\\sum_{i=1}^{2011}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right) \\\\\n& =\\left(\\frac{1}{1}-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\cdots+\\left(\\frac{1}{2010}-\\frac{1}{2011}\\right)+\\left(\\frac{1}{2011}-\\frac{1}{2012}\\right) \\\\\n& =1-\\frac{1}{2012} \\\\\n& =\\frac{2011}{2012} .\n\\end{aligned}\n$$'] ['$\\frac{2011}{2012}$'] 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Multimodal Competition False Numerical Open-ended Combinatorics Math English +147 "Leibniz's Harmonic Triangle: Consider the triangle formed by the rule + +$$ +\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} +$$ + +This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. + + + +For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. +If $\sum_{i=1}^{\infty} \operatorname{Le}(i, 1)=\operatorname{Le}(n, k)$, determine the values of $n$ and $k$." ['Extending the result of $8 \\mathrm{~b}$ gives\n\n$$\n\\sum_{i=1}^{n} \\operatorname{Le}(i, 1)=\\frac{1}{1}-\\frac{1}{n}\n$$\n\nso as $n \\rightarrow \\infty, \\sum_{i=1}^{n} \\operatorname{Le}(i, 1) \\rightarrow 1$. This value appears as $\\operatorname{Le}(0,0)$, so $n=k=0$.'] ['$0,0$'] 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Multimodal Competition True Numerical Open-ended Combinatorics Math English +148 "Leibniz's Harmonic Triangle: Consider the triangle formed by the rule + +$$ +\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} +$$ + +This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. + + + +For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. +If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, compute expressions for $n$ and $k$ in terms of $m$." ['$n=k=m-1$.'] ['$m-1,m-1$'] 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Multimodal Competition True Expression Open-ended Combinatorics Math English +149 "$\quad$ Let $T=12$. As shown, three circles are mutually externally tangent. The large circle has a radius of $T$, and the smaller two circles each have radius $\frac{T}{2}$. Compute the area of the triangle whose vertices are the centers of the three circles. + +" ['The desired triangle is an isosceles triangle whose base vertices are the centers of the two smaller circles. The congruent sides of the triangle have length $T+\\frac{T}{2}$. Thus the altitude to the base has length $\\sqrt{\\left(\\frac{3 T}{2}\\right)^{2}-\\left(\\frac{T}{2}\\right)^{2}}=T \\sqrt{2}$. Thus the area of the triangle is $\\frac{1}{2} \\cdot\\left(\\frac{T}{2}+\\frac{T}{2}\\right) \\cdot T \\sqrt{2}=\\frac{T^{2} \\sqrt{2}}{2}$. With $T=12$, the area is $\\mathbf{7 2} \\sqrt{\\mathbf{2}}$.'] ['$72 \\sqrt{2}$'] 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Multimodal Competition False Numerical Open-ended Geometry Math English