id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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7,101 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_10 | 4 | Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall th... | Given that Mary's score is $30+4c-w$ , two other ways to get that score are $30+4(c+1)-(w+4)$ and $30+4(c-1)-(w-4)$ .
Since it is clear that $c>1$ , we must have $w<4$ . In order to minimize the score, assume that $w=3$ .
The number of problems left blank must be less than $5$ because of the $30+4(c+1)-(w+4)$ case. In ... | 119 |
7,102 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11 | 1 | A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$ | First notice that there is no difference between the maple trees and the oak trees; we have only two types, birch trees and "non-birch" trees. (If you don't believe this reasoning, think about it. You could also differentiate the tall oak trees from the short oak trees, and the maple trees with many branches as opposed... | 106 |
7,103 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11 | 2 | A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$ | Let $b$ $n$ denote birch tree and not-birch tree, respectively. Notice that we only need $4$ $n$ s to separate the $5$ $b$ s. Specifically, \[b,n,b,n,b,n,b,n,b\] Since we have $7$ $n$ s, we are placing the extra $3$ $n$ s into the $6$ intervals beside the $b$ s.
Now doing simple casework.
If all $3$ $n$ s are in the sa... | 106 |
7,104 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11 | 3 | A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$ | Note that the requested probability is computed by dividing the number of configurations with no adjacent Birch trees by the total number of configurations. We can compute the number of configurations with no adjacent Birch trees using complementary counting and then the Principle of Inclusion-Exclusion.
The number of ... | 106 |
7,105 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11 | 4 | A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$ | Here is a solution leaving out nothing. This solution is dedicated to those that are in self study and wish to learn the most they can. I will make it as elementary as possible and intuition based.
Arrange first the $3$ maple and $4$ oaks as $MMMOOOO$ . We then notice that for none of the $5$ birch trees to be adjacent... | 106 |
7,106 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12 | 1 | function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$ | If $f(2+x)=f(2-x)$ , then substituting $t=2+x$ gives $f(t)=f(4-t)$ . Similarly, $f(t)=f(14-t)$ . In particular, \[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\]
Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\pmod{10}$ are also roots. To see that these may be the only integer roots, observe that ... | 401 |
7,107 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12 | 2 | function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$ | We notice that the function has reflectional symmetry across both $x=2$ and $x=7$ . We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\pm 10$ as roots. Continuing this shows that the roots are $0 \mod 10$ or $4 \mod ... | 401 |
7,108 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12 | 3 | function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$ | Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only coun... | 401 |
7,109 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12 | 4 | function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$ | Let $z$ be an arbitrary zero. If $z=2-x$ , then $x=2-z$ and $2+x=4-z$ . Repeat with other equation to find if $z$ is a zero then so are $4-z$ and $14-z$ . From $0$ , we get $4$ and $14$ . Now note that applying either of these twice will return $z$ , so we must apply them in an alternating fashion for distinct roots. D... | 401 |
7,110 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_14 | 3 | What is the largest even integer that cannot be written as the sum of two odd composite numbers? | The easiest method is to notice that any odd number that ends in a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digi... | 38 |
7,111 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_15 | 2 | Determine $x^2+y^2+z^2+w^2$ if | As in Solution 1, we have \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\] where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$ , for $k=1,3,5,7$
Now the coefficient of $t^3$ on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the $t^4$ and $t^3$ terms ar... | 36 |
7,112 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_1 | 6 | Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$ | Converting all of the logarithms to exponentials gives $x^{24} = w, y^{40} =w,$ and $x^{12}y^{12}z^{12}=w.$ Thus, we have $y^{40} = x^{24} \Rightarrow z^3=y^2.$ We are looking for $\log_z w,$ which by substitution, is $\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.$ | 60 |
7,113 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_2 | 2 | Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$ | Let $p$ be equal to $15 - \varepsilon$ , where $\varepsilon$ is an almost neglectable value. Because of the small value $\varepsilon$ , the domain of $f(x)$ is basically the set ${15}$ . plugging in $15$ gives $\varepsilon + 0 + 15 - \varepsilon$ , or $15$ , so the answer is $\boxed{15}$ | 15 |
7,114 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_3 | 4 | What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$ | We are given the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] Squaring both sides yields \[(x^2+18x+30)^2=4(x^2+18x+45)\] \[(x^2+18x+30)^2=4(x^2+18x+30+15)\] \[(x^2+18x+30)^2=4(x^2+18x+30)+60\] \[(x^2+18x+30)^2-4(x^2+18x+30)-60=0\] Substituting $y=x^2+18x+30$ yields \[y^2-4y-60=0\] \[(y+6)(y-10)=0\] Thus $y=x^2+18x+30=-6... | 20 |
7,115 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_5 | 4 | Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have? | $x^3 + y^3 = 10 = (x+y)(x^2-xy+y^2) = (x+y)(7-xy) \implies xy = 7 - \frac{10}{x+y}.$ Also, $(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 = 10 + 3xy(x+y).$ Substituting our above into this, we get $10 + 3(7-\frac{10}{x+y})(x+y) = 21x+21y-20 = (x+y)^3$ . Letting $p = x+y$ , we have that $p^3 - 21p + 20 = 0$ . Testing $p = 1$ , we f... | 4 |
7,116 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_8 | 2 | What is the largest $2$ -digit prime factor of the integer $n = {200\choose 100}$ | We know that \[{200\choose100}=\frac{200!}{100!100!}\] Since $p<100$ , there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\frac{200!}{100!100!}$ . (Note that here we assume the minimum because ... | 61 |
7,117 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10 | 1 | The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there? | Suppose that the two identical digits are both $1$ . Since the thousands digit must be $1$ , only one of the other three digits can be $1$ . This means the possible forms for the number are
Because the number must have exactly two identical digits, $x\neq y$ $x\neq1$ , and $y\neq1$ . Hence, there are $3\cdot9\cdot8=216... | 432 |
7,118 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10 | 2 | The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there? | Consider a sequence of $4$ digits instead of a $4$ -digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is $\frac{1}{10}$ . This means we can find all possible sequences with one digit repeated twice, and then divide by $10$
If ... | 432 |
7,119 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10 | 3 | The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there? | We'll use complementary counting. We will split up into $3$ cases: (1) no number is repeated, (2) $2$ numbers are repeated, and $2$ other numbers are repeated, (3) $3$ numbers are repeated, or (4) $4$ numbers are repeated.
Case 1:
There are $9$ choices for the hundreds digit (it cannot be $1$ ), $8$ choices for the ten... | 432 |
7,120 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10 | 4 | The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there? | Let us proceed by casework.
Case 1:
We will count the amount of numbers that have two identical digits that are not one. The thousands digit is fixed, and we are choosing two spots to hold two identical digits that are chosen from $0, 2-9$ , which is $9$ options. For the last digit, their are $8$ possibilities since it... | 432 |
7,121 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11 | 1 | The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid? | First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$ . The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$ , and one of the legs is $3\sqrt{2}$ . We apply the Pythagorean Theorem to deduce that the height is $6$
Next, we complete t he figure ... | 288 |
7,122 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11 | 2 | The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid? | Extend $EA$ and $FB$ to meet at $G$ , and $ED$ and $FC$ to meet at $H$ . Now, we have a regular tetrahedron $EFGH$ , which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\sqrt{2}$ . Using the formula for the volume of a regular tetrahedron, which is $V = \frac{\sqrt{2}S... | 288 |
7,123 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11 | 3 | The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid? | We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$ ; thus, we will integrate with respect to height from $0$ to $6$ , noting that each cross section of height $dh$ is a rectangle. The volume is then $\in... | 288 |
7,124 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11 | 4 | The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid? | Draw an altitude from a vertex of the square base to the top edge. By using $30,60, 90$ triangle ratios, we obtain that the altitude has a length of $3 \sqrt{6}$ , and that little portion that hangs out has a length of $3\sqrt2$ . This is a triangular pyramid with a base of $3\sqrt6, 3\sqrt6, 3\sqrt2$ , and a height of... | 288 |
7,125 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13 | 1 | For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | Let $S$ be a non- empty subset of $\{1,2,3,4,5,6\}$
Then the alternating sum of $S$ , plus the alternating sum of $S \cup \{7\}$ , is $7$ . This is because, since $7$ is the largest element, when we take an alternating sum, each number in $S$ ends up with the opposite sign of each corresponding element of $S\cup \{7\}$... | 448 |
7,126 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13 | 2 | For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | Consider a given subset $T$ of $S$ that contains $7$ ; then there is a subset $T'$ which contains all the elements of $T$ except for $7$ , and only those elements . Since each element of $T'$ has one fewer element preceding it than it does in $T$ , their signs are opposite. Thus the sum of the alternating sums of $T$ a... | 448 |
7,127 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13 | 3 | For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | Denote the desired total of all alternating sums of an $n$ -element set as $S_n$ . We are looking for $S_7$ . Notice that all alternating sums of an $n$ -element set are also alternating sums of an $n+1$ -element set. However, when we go from an $n$ to $n+1$ element set, for each subset with the new element, we are add... | 448 |
7,128 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13 | 4 | For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | We analyze all the numbers from 1 to 7 separately to see where the number contributes its positive or negative to the sum of the alternating sums. Whenever 7 appears, which it does 64 times, it contributes a positive because it is always first. This gives a net gain of $7 \cdot 64=448$
If we look at when 6 appears, whi... | 448 |
7,129 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13 | 5 | For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | Let $\mathbb{N}_n := \{1, 2, 3, \dots n\}$ . Let the alternating sum of a certain subset of $S$ of $\mathbb{N}_n$ be $\xi(S),$ and let \[\mathcal{A}(\mathbb{N}_n) := \sum_{S \subseteq \mathbb{N}_n} \xi(S).\] We see that \[\mathcal{A}(\mathbb{N}_n) = \sum_{S \subseteq \mathbb{N}_n} \xi(S) = \sum_{n \in S, S \subseteq \m... | 448 |
7,130 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | 1 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | Firstly, notice that if we reflect $R$ over $P$ , we get $Q$ . Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$ , we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$ , and with radius $6$ ) that intersects circle $A$ at $Q$ . The rest is just finding lengths, as follows... | 130 |
7,131 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | 2 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | [asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label("$A$",a,S);... | 130 |
7,132 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | 3 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | Let $QP=PR=x$ . Angles $QPA$ $APB$ , and $BPR$ must add up to $180^{\circ}$ . By the Law of Cosines $\angle APB=\cos^{-1}\left(\frac{{-11}}{24}\right)$ . Also, angles $QPA$ and $BPR$ equal $\cos^{-1}\left(\frac{x}{16}\right)$ and $\cos^{-1}\left(\frac{x}{12}\right)$ . So we have
Taking the cosine of both sides, and sim... | 130 |
7,133 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | 4 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | Let $QP = PR = x$ . Extend the line containing the centers of the two circles to meet $R$ , and to meet the other side of the large circle at a point $S$
The part of this line from $R$ to the point nearest to $R$ where it intersects the larger circle has length $6+(12-8)=10$ .
The length of the diameter of the large... | 130 |
7,134 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | 5 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | [asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--a--m--n--cycle); draw(p--t); draw(q--m); draw(n--r); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); label("$A$",a,S); lab... | 130 |
7,135 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | 6 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | Looking at Drawing 2 (by the way, we don't need point $R$ ), we set $AM=a$ and $BN=b$ , and the desired length $QP=x=PR$ . We know that a radius perpendicular to a chord bisects the chord, so $MP=\frac{x}{2}$ and $PN=\frac{x}{2}$ . Draw line $AP$ and $PB$ , and we see that they are radii of Circles $A$ and $B$ , respec... | 130 |
7,136 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | 7 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | The centers are collinear, you can prove it (but that is already given in the later section [Proof that R,A, and B are collinear]). Drop a perpendicular from $P$ to the lines that the centers are on. You then have 2 separate segments, separated by the foot of the altitude of $P$ . Call them $a$ and $b$ respectively. Ca... | 130 |
7,137 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | 8 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | We use coordinate geometry to approach this problem. Let the center of larger circle be the origin $O_1$ , the smaller circle be $O_2$ , and the x-axis be $O_1O_2$ . Hence, we can get the two circle equations: $x^2+y^2 = 64$ and $(x-12)^2+y^2=36$
Let point $P$ be $(a, b)$ . Noting that it lies on both circles, we can p... | 130 |
7,138 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | 9 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | Let the center of the circle with radius $8$ be $A,$ and let the center of the one with radius $6$ be $B.$ Also, let $QP = PR = x.$ Using law of cosines on triangle $APB,$ we have that $\cos {APB} = \left(-\frac{{11}}{24}\right).$ Angle chasing gives that $\angle{QAR} = \angle{APB},$ so its cosines must be the same. Ap... | 130 |
7,139 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15 | 1 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | Let $M$ be the midpoint of the chord $BC$ . From right triangle $OMB$ , we have $OM = \sqrt{OB^2 - BM^2} =4$ . This gives $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$
Notice that the distance $OM$ equals $PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)$ , where $r$ is the radius of circle $P$
Hence \[\cos \angle AOM ... | 175 |
7,140 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15 | 2 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | This solution, while similar to Solution 2, is arguably more motivated and less contrived.
Firstly, we note the statement in the problem that " $AD$ is the only chord starting at $A$ and bisected by $BC$ " – what is its significance? What is the criterion for this statement to be true?
We consider the locus of midpoint... | 175 |
7,141 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15 | 3 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | Let the center of the circle be $O$ . Fix $B,C,$ and $A$ . Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$ , it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$
Now, ... | 175 |
7,142 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15 | 4 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | Let I be the intersection of AD and BC.
Lemma: $AI = ID$ if and only if $\angle AIO = 90$
Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If $\angle AIO = 90$ , We can get $\triangle AIO \cong \triangle OID$
Let be this the circle with diameter AO.
Thus, we get $\angle AIO = 90$ , imp... | 175 |
7,143 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15 | 5 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | Let $O$ be the center of the circle. The locus of midpoints of chords with $A$ as a endpoint is a circle with diameter $\overline{AO}$ . Additionally, this circle must be tangent to $\overline{BC}$ . Let the center of this circle be $P$ . Let $M$ be the midpoint of $BC$ $N$ be the foot of the perpendicular from $P$ to ... | 175 |
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