competition_id string | problem_id int64 | difficulty int64 | category string | problem_type string | problem string | solutions list | solutions_count int64 | source_file string | competition string |
|---|---|---|---|---|---|---|---|---|---|
1977_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | For how many values of the coefficient a do the equations \begin{align*}x^2+ax+1=0 \\ x^2-x-a=0\end{align*} have a common real solution?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \infty$
| [
"Subtracting the equations, we get $ax+x+1+a=0$, or $(x+1)(a+1)=0$, so $x=-1$ or $a=-1$. If $x=-1$, then $a=2$, which satisfies the condition. If $a=-1$, then $x$ is nonreal. This means that $a=-1$ is the only number that works, so our answer is $(B)$.\n\n\n~alexanderruan\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/21.json | AHSME |
1977_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | If $(3x-1)^7 = a_7x^7 + a_6x^6 + \cdots + a_0$, then $a_7 + a_6 + \cdots + a_0$ equals
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 64 \qquad \text{(D)}\ -64 \qquad \text{(E)}\ 128$
| [
"Solution by e_power_pi_times_i\n\n\nNotice that if $x=1$, then $a_7x^7 + a_6x^6 + \\cdots + a_0 = a_7 + a_6 + \\cdots + a_0$. Therefore the answer is $(3(1)-1)^7) = \\boxed{\\text{(D)}\\ 128}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/10.json | AHSME |
1977_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | [asy] for (int i=0; i<9; ++i) { draw(dir(10+40*i)--dir(50+40*i)); } draw(dir(50) -- dir(90)); label("$a$", dir(50) -- dir(90), N); draw(dir(10) -- dir(90)); label("$b$", dir(10) -- dir(90), SW); draw(dir(-70) -- dir(90)); label("$d$", dir(-70) -- dir(90), E); //Credit to MSTang for the diagram [/asy]
If $a,b$, and $d... | [
"By the law of cosines we can get the following expressions for $d^{2}$ and $b^{2}$:\n\n\n\\[d^{2}=2b^{2}(1-\\cos (100^\\circ))\\] \\[b^{2}=2a^{2}(1-\\cos (140^\\circ))\\]\n\n\nWe can substitute what we got for $b^2$ into the expression for $d^{2}$:\n\n\n\\[d^{2}=4a^{2}(1-\\cos (140^\\circ))(1-\\cos (100^\\circ))=4... | 2 | ./CreativeMath/AHSME/1977_AHSME_Problems/30.json | AHSME |
1977_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If $y = 2x$ and $z = 2y$, then $x + y + z$ equals
$\text{(A)}\ x \qquad \text{(B)}\ 3x \qquad \text{(C)}\ 5x \qquad \text{(D)}\ 7x \qquad \text{(E)}\ 9x$
| [
"Solution by e_power_pi_times_i\n\n\n$x+y+z = x+(2x)+(4x) = \\boxed{\\text{(D)}\\ 7x}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/1.json | AHSME |
1977_AHSME_Problems | 11 | 0 | Number Theory | Multiple Choice | For each real number $x$, let $\textbf{[}x\textbf{]}$ be the largest integer not exceeding $x$
(i.e., the integer $n$ such that $n\le x<n+1$). Which of the following statements is (are) true?
$\textbf{I. [}x+1\textbf{]}=\textbf{[}x\textbf{]}+1\text{ for all }x \\ \textbf{II. [}x+y\textbf{]}=\textbf{[}x\textbf{]}+\te... | [
"Solution by e_power_pi_times_i\n\n\nNotice that $\\textbf{[}x\\textbf{]}$ is just $\\left \\lfloor x \\right \\rfloor$. We see that $\\textbf{I}$ is true, as adding by one does not change the fraction part of the number. Similarly, $\\textbf{II}$ is false, because $\\textbf{[}x+y\\textbf{]}$ does not always equal ... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/11.json | AHSME |
1977_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | Which one of the following statements is false? All equilateral triangles are
$\textbf{(A)}\ \text{ equiangular}\qquad \textbf{(B)}\ \text{isosceles}\qquad \textbf{(C)}\ \text{regular polygons }\qquad\\ \textbf{(D)}\ \text{congruent to each other}\qquad \textbf{(E)}\ \text{similar to each other}$
| [
"Solution by e_power_pi_times_i\n\n\nEquilateral triangles can be in different sizes, therefore they are not $\\boxed{\\textbf{(D)}\\ \\text{congruent to each other}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/2.json | AHSME |
1977_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | Let $g(x)=x^5+x^4+x^3+x^2+x+1$. What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$?
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$
\subsubsection{Solution 1}
Let $r(x)$ be the remainde... | [
"Let $r(x)$ be the remainder when $g(x^{12})$ is divided by $g(x)$. Then $r(x)$ is the unique polynomial such that\n\\[g(x^{12}) - r(x) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 - r(x)\\]\nis divisible by $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1$, and $\\deg r(x) < 5$.\n\n\nNote that $(x - 1)(x^5 + x^4 + x^3 + ... | 3 | ./CreativeMath/AHSME/1977_AHSME_Problems/28.json | AHSME |
1977_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | Al's age is $16$ more than the sum of Bob's age and Carl's age, and the square of Al's age is $1632$ more than the square of the sum of
Bob's age and Carl's age. What is the sum of the ages of Al, Bob, and Carl?
$\text{(A)}\ 64 \qquad \text{(B)}\ 94 \qquad \text{(C)}\ 96 \qquad \text{(D)}\ 102 \qquad \text{... | [
"Solution by e_power_pi_times_i\n\n\nDenote Al's age, Bob's age, and Carl's age by $a$, $b$, and $c$, respectively. Then, $a = 16 + b + c$ and $a^2 = 1632 + b^2 + c^2$. Substituting the first equation into the second, $(16 + b + c)^2 = b^2 + c^2 + 2bc + 32b + 32c + 256 = b^2 + c^2 + 1632$. Thus, $bc + 16b + 16c = 6... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/12.json | AHSME |
1977_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | Find the sum $\frac{1}{1(3)}+\frac{1}{3(5)}+\dots+\frac{1}{(2n-1)(2n+1)}+\dots+\frac{1}{255(257)}$.
$\textbf{(A) }\frac{127}{255}\qquad \textbf{(B) }\frac{128}{255}\qquad \textbf{(C) }\frac{1}{2}\qquad \textbf{(D) }\frac{128}{257}\qquad \textbf{(E) }\frac{129}{257}$
| [
"Note that $\\frac{1}{(2n-1)(2n+1)} = \\frac{1/(2n-1)-1/(2n+1)}{2}$. Indeed, we find the series telescopes and is equal to $\\frac{1-\\frac{1}{257}}{2}$, which is evidently $\\boxed {\\frac{128}{257}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/24.json | AHSME |
1977_AHSME_Problems | 25 | 0 | Number Theory | Multiple Choice | Determine the largest positive integer $n$ such that $1005!$ is divisible by $10^n$.
$\textbf{(A) }102\qquad \textbf{(B) }112\qquad \textbf{(C) }249\qquad \textbf{(D) }502\qquad \textbf{(E) }\text{none of the above}\qquad$
\subsubsection{Solution}
We first observe that since there will be more 2s than 5s in $10... | [
"We first observe that since there will be more 2s than 5s in $1005!$, we are looking for the largest $n$ such that $5^n$ divides $1005!$. We will use the fact that:\n\n\n\\[n = \\left \\lfloor {\\frac{1005}{5^1}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^2}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/25.json | AHSME |
1977_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | If $a_1,a_2,a_3,\dots$ is a sequence of positive numbers such that $a_{n+2}=a_na_{n+1}$ for all positive integers $n$,
then the sequence $a_1,a_2,a_3,\dots$ is a geometric progression
$\textbf{(A) }\text{for all positive values of }a_1\text{ and }a_2\qquad\\ \textbf{(B) }\text{if and only if }a_1=a_2\qquad\\ \textbf... | [
"Solution by e_power_pi_times_i\n\n\nThe first few terms are $a_1,a_2,a_1a_2,a_1a_2^2,a_1^2a_2^3,\\dots$ . If this is a geometric progression, $\\dfrac{a_2}{a_1} = a_1 = a_2 = a_1a_2$. $a_1=0,1$, $a_2=0,1$. Since this is a sequence of positive integers, then the answer must be $\\boxed{\\textbf{(E) }\\text{if and o... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/13.json | AHSME |
1977_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | Find the smallest integer $n$ such that $(x^2+y^2+z^2)^2\le n(x^4+y^4+z^4)$ for all real numbers $x,y$, and $z$.
$\textbf{(A) }2\qquad \textbf{(B) }3\qquad \textbf{(C) }4\qquad \textbf{(D) }6\qquad \textbf{(E) }\text{There is no such integer n}$
Solution
| [
"We see squares and one number. And we see an inequality. This calls for Cauchy's inequality. EEEEWWW.\n\n\nAnyways, look at which side is which. The squared side is smaller-- so that's good. It's in the right format.\n\n\nCauchy's states that $(a_1b_1+a_2b_2+a_3b_3+......)^2 \\le (a_1^2+a_2^2+a_3^2+....)(b_1^2+b_2... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/29.json | AHSME |
1977_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | A man has $2.73 in pennies, nickels, dimes, quarters and half dollars. If he has an equal number of coins of each kind, then the total number of coins he has is
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 15$
| [
"Solution by e_power_pi_times_i\n\n\nDenote the number of pennies, nickels, dimes, quarters, and half-dollars by $x$. Then $x+5x+10x+25x+50x = 273$, and $x = 3$. The total amount of coins is $5x$, which is $\\textbf{\\text{(E)}}\\ 15$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/3.json | AHSME |
1977_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | For every triple $(a,b,c)$ of non-zero real numbers, form the number $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$.
The set of all numbers formed is
$\textbf{(A)}\ {0} \qquad \textbf{(B)}\ \{-4,0,4\} \qquad \textbf{(C)}\ \{-4,-2,0,2,4\} \qquad \textbf{(D)}\ \{-4,-2,2,4\}\qquad \textbf{(E)}\ \text{non... | [
"Solution by e_power_pi_times_i\n\n\n$\\dfrac{x}{|x|} = 1$ or $-1$ depending whether $x$ is positive or negative. If $a$, $b$, and $c$ are positive, then the entire thing amounts to $4$. If one of the three is negative and the other two positive, the answer is $0$. If two of the three is negative and one is positiv... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/8.json | AHSME |
1977_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | If $f(x)$ is a real valued function of the real variable $x$, and $f(x)$ is not identically zero,
and for all $a$ and $b$ $f(a+b)+f(a-b)=2f(a)+2f(b)$, then for all $x$ and $y$
$\textbf{(A) }f(0)=1\qquad \textbf{(B) }f(-x)=-f(x)\qquad \textbf{(C) }f(-x)=f(x)\qquad \\ \textbf{(D) }f(x+y)=f(x)+f(y) \qquad \\ \textbf{(E... | [
"We can start by finding the value of $f(0)$. \nLet $a = b = 0$\n\\[f(0) + f(0) = 2f(0) + 2f(0) \\Rrightarrow 2f(0) = 4f(0) \\Rrightarrow f(0) = 0\\]\nThus, $A$ is not true.\nTo check $B$, we let $a = 0$. We have\n\\[f(0+b) + f(0-b) = 2f(0) + 2f(b)\\]\n\\[f(b) + f(-b) = 0 + 2f(b)\\]\n\\[f(-b) = f(b)\\]\nThus, $B$ i... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/22.json | AHSME |
1977_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | If $y=(\log_23)(\log_34)\cdots(\log_n[n+1])\cdots(\log_{31}32)$ then
$\textbf{(A) }4<y<5\qquad \textbf{(B) }y=5\qquad \textbf{(C) }5<y<6\qquad \textbf{(D) }y=6\qquad \\ \textbf{(E) }6<y<7$
| [
"Solution by e_power_pi_times_i\n\n\n\n\nNote that $\\log_{a}b = \\dfrac{\\log{b}}{\\log{a}}$. Then $y=(\\dfrac{\\log3}{\\log2})(\\dfrac{\\log4}{\\log3})\\cdots(\\dfrac{\\log32}{\\log31}) = \\dfrac{\\log32}{\\log2} = \\log_232 = \\boxed{\\text{(B) }y=5}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/18.json | AHSME |
1977_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | [asy] size(130); pair A = (2, 2.4), C = (0, 0), B = (4.3, 0), E = 0.7*A, F = 0.57*A + 0.43*B, D = (2.4, 0); draw(A--B--C--cycle); draw(E--D--F); label("$A$", A, N); label("$B$", B, E); label("$C$", C, W); label("$D$", D, S); label("$E$", E, NW); label("$F$", F, NE); //Credit to MSTang for the diagram [/asy]
In trian... | [
"Solution by e_power_pi_times_i\n\n\nBecause $\\measuredangle A=80^\\circ$, $\\measuredangle B=\\measuredangle C=50^\\circ$. Then, because triangles $CDE$ and $BDF$ are isosceles, $\\measuredangle CDE=\\measuredangle BDF=65^\\circ$. $\\measuredangle EDF=180^\\circ - 2(65^\\circ)=\\textbf{(C) }50^\\circ$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/4.json | AHSME |
1977_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | How many pairs $(m,n)$ of integers satisfy the equation $m+n=mn$?
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }\text{more than }4$
| [
"Solution by e_power_pi_times_i\n\n\nIf $m+n=mn$, $mn-m-n = (m-1)(n-1)-1 = 0$. Then $(m-1)(n-1) = 1$, and $(m,n) = (2,2) or (0,0)$. The answer is $\\boxed{\\textbf{(B) }2}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/14.json | AHSME |
1977_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | Three circles are drawn, so that each circle is externally tangent to the other two circles. Each circle has a radius of $3.$ A triangle is then constructed, such that each side of the triangle is tangent to two circles, as shown below. Find the perimeter of the triangle.
\subsection{Diagram}
[asy] unitsize(1 cm);... | [
"Solution by e_power_pi_times_i\n\n\nDraw perpendicular lines from the radii of the circles to the sides of the triangle, and lines from the radii of the circles to the vertices of the triangle. Because the triangle is equilateral, the lines divide the big triangle into a small triangle, three rectangles, and six s... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/15.json | AHSME |
1977_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | The set of all points $P$ such that the sum of the (undirected) distances from $P$ to two fixed points $A$ and $B$ equals the distance between $A$ and $B$ is
$\textbf{(A) }\text{the line segment from }A\text{ to }B\qquad \textbf{(B) }\text{the line passing through }A\text{ and }B\qquad\\ \textbf{(C) }\text{the perpen... | [
"Solution by e_power_pi_times_i\n\n\nThe answer is $\\textbf{(A)}$ because $P$ has to be on the line segment $AB$ in order to satisfy $PA+PB=AB$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/5.json | AHSME |
1977_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | Let $E$ be the point of intersection of the diagonals of convex quadrilateral $ABCD$, and let $P,Q,R$, and $S$ be the centers of the circles
circumscribing triangles $ABE, BCE, CDE$, and $ADE$, respectively. Then
$\textbf{(A) }PQRS\text{ is a parallelogram}\\ \textbf{(B) }PQRS\text{ is a parallelogram if an only if ... | [
"Let $L$, $M$, $N$, and $O$ be the intersections of $AC$ and $PS$, $BD$ and $PQ$, $CA$ and $QR$, and $DB$ and $RS$ respectively. Since the circumcenter of a triangle is determined by the intersection of its perpendicular bisectors, each of the pairs mentioned earlier are perpendicular to each other. Let $\\angle SP... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/19.json | AHSME |
1977_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | [asy] size(120); path c = Circle((0, 0), 1); pair A = dir(20), B = dir(130), C = dir(240), D = dir(330); draw(c); pair F = 3(A-B) + B; pair G = 3(D-C) + C; pair E = intersectionpoints(B--F, C--G)[0]; draw(B--E--C--A); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, E)... | [
"Solution by e_power_pi_times_i\n\n\nIf arcs $AB$, $BC$, and $CD$ are congruent, then $\\measuredangle ACB = \\measuredangle BDC = \\measuredangle CBD = \\theta$. Because $ABCD$ is cyclic, $\\measuredangle CAD = \\measuredangle CBD = \\theta$, and $\\measuredangle ADB = \\measuredangle ACB = \\theta$. Then, $\\meas... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/9.json | AHSME |
1962_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 36$
| [
"If the angles are in an arithmetic progression, they can be expressed as\n$a$, $a+n$, $a+2n$, $a+3n$, and $a+4n$ for some real numbers $a$ and $n$.\nNow we know that the sum of the degree measures of the angles of a pentagon is $180(5-2)=540$.\nAdding our expressions for the five angles together, we get $5a+10n=54... | 2 | ./CreativeMath/AHSME/1962_AHSME_Problems/20.json | AHSME |
1962_AHSME_Problems | 36 | 0 | Number Theory | Multiple Choice | If both $x$ and $y$ are both integers, how many pairs of solutions are there of the equation $(x-8)(x-10) = 2^y$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \text{more than 3}$
| [
"The equality implies $x-8$ and $x-10$ are both powers of two; since they differ by two, it must be the case that $(x-8,x-10) = (4,2)$ or $(x-8,x-10) = (-2,-4)$. (Note that $(1,-1)$ is not allowed because then the product is negative.) These yield $(x,y) = (12,3)$ or $(x,y) = (6,3)$, for a total of $\\boxed{2\\tex... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/36.json | AHSME |
1962_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | Given rectangle $R_1$ with one side $2$ inches and area $12$ square inches. Rectangle $R_2$ with diagonal $15$ inches is similar to $R_1$. Expressed in square inches the area of $R_2$ is:
$\textbf{(A)}\ \frac{9}2\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 9\sqrt{10}\qquad\textbf{(E)}\ ... | [
"Clearly, the other side of $R_1$ has length $\\frac{12}2=6$.\nNow call the sides of $R_2$ a and b, with $b>a$. We know that $\\frac{b}a=\\frac31=3$, because the two rectangles are similar. We also know that $a^2+b^2=15^2=225$. But $b=3a$, so substituting gives \n\\[a^2+(3a)^2=225\\]\n\\[10a^2=225\\]\n\\[a^2=\\frac... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/16.json | AHSME |
1962_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | A square and an equilateral triangle have equal perimeters. The area of the triangle is $9 \sqrt{3}$ square inches. Expressed in inches the diagonal of the square is:
$\textbf{(A)}\ \frac{9}{2}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ \frac{9\sqrt{2}}{2}\qquad\textbf{(E)}\ \text{n... | [
"To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is $\\dfrac{x^2\\sqrt{3}}{4}$. This has to be equal to $9 \\sqrt{3}$, which means that $x=6$, or the side length of the triangle is $6$. Thus, the triangle (and the square) have a perimeter of $18$. It f... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/6.json | AHSME |
1962_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | Let the bisectors of the exterior angles at $B$ and $C$ of triangle $ABC$ meet at D$.$ Then, if all measurements are in degrees, angle $BDC$ equals:
$\textbf{(A)}\ \frac{1}{2}(90-A)\qquad\textbf{(B)}\ 90-A\qquad\textbf{(C)}\ \frac{1}{2}(180-A)\qquad$
$\textbf{(D)}\ 180-A\qquad\textbf{(E)}\ 180-2A$
| [
"Calculating for angles $\\angle DBC$ and $\\angle DCB$, we get\n\n\n$\\angle DBC$ = $90 - \\frac{B}{2}$ and\n$\\angle DCB$ = $90 - \\frac{C}{2}$.\n\n\nIn triangle $BCD$, we have\n\n\n$\\angle BDC$ = $180 - (90 - \\frac{B}{2}) - (90 - \\frac{C}{2})$ = $\\frac{B+C}{2}$ = $\\frac{1}{2}\\cdot(180 - A)$, so the answer ... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/7.json | AHSME |
1962_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | If $a = \log_8 225$ and $b = \log_2 15$, then $a$, in terms of $b$, is:
$\textbf{(A)}\ \frac{b}{2}\qquad\textbf{(B)}\ \frac{2b}{3}\qquad\textbf{(C)}\ b\qquad\textbf{(D)}\ \frac{3b}{2}\qquad\textbf{(E)}\ 2b$
| [
"Using the change-of-base rule: $a = \\frac{\\log 225}{\\log 8}$ and $b = \\frac{\\log 15}{\\log 2}$.\n\\[\\frac{a}{b} = \\frac{\\log 225 \\log 2}{\\log 8 \\log 15}\\]\n\\[a = b \\cdot \\frac{\\log 225}{\\log 15} \\cdot \\frac{\\log 2}{\\log 8}\\]\n\\[a = b \\log_{15} 225 \\log_8 2\\]\n\\[a = \\boxed{\\frac{2b}3 \\... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/17.json | AHSME |
1962_AHSME_Problems | 40 | 0 | Algebra | Multiple Choice | The limiting sum of the infinite series, $\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots$ whose $n$th term is $\frac{n}{10^n}$ is:
$\textbf{(A)}\ \frac{1}9\qquad\textbf{(B)}\ \frac{10}{81}\qquad\textbf{(C)}\ \frac{1}8\qquad\textbf{(D)}\ \frac{17}{72}\qquad\textbf{(E)}\ \text{larger than any finite quantity}$... | [
"The series can be written as the following:\n\n\n$\\frac{1}{10} + \\frac{1}{10^2} + \\frac{1}{10^3} + ...$\n\n\n$+ \\frac{1}{10^2} + \\frac{1}{10^3} + \\frac{1}{10^4} + ...$\n\n\n$+ \\frac{1}{10^3} + \\frac{1}{10^4} + \\frac{1}{10^5} + ...$\n\n\nand so on.\n\n\nby using the formula for infinite geometric series $(... | 3 | ./CreativeMath/AHSME/1962_AHSME_Problems/40.json | AHSME |
1962_AHSME_Problems | 37 | 0 | Geometry | Multiple Choice | $ABCD$ is a square with side of unit length. Points $E$ and $F$ are taken respectively on sides $AB$ and $AD$ so that $AE = AF$ and the quadrilateral $CDFE$ has maximum area. In square units this maximum area is:
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ \frac{9}{16}\qquad\textbf{(C)}\ \frac{19}{32}\qquad\textbf{(... | [
"Let $AE=AF=x$\n$[CDFE]=[ABCD]-[AEF]-[EBC]=1-\\frac{x^2}{2}-\\frac{1-x}{2}$\nOr\n$[CDFE]=\\frac{\\frac{5}{4}-(x-\\frac{1}{2})^2}{2}\\le \\frac{5}{8}$\nAs $(x-\\frac{1}{2})^2\\ge 0$\nSo $[CDFE]\\le \\frac{5}{8}$\nEquality occurs when $AE=AF=x=\\frac{1}{2}$\nSo the maximum value is $\\frac{5}{8}$\n\n\n",
"Let us fi... | 4 | ./CreativeMath/AHSME/1962_AHSME_Problems/37.json | AHSME |
1962_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | It is given that one root of $2x^2 + rx + s = 0$, with $r$ and $s$ real numbers, is $3+2i (i = \sqrt{-1})$. The value of $s$ is:
$\textbf{(A)}\ \text{undetermined}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ -13\qquad\textbf{(E)}\ 26$
| [
"If a quadratic with real coefficients has two non-real roots, the two roots must be complex conjugates of one another.\nThis means the other root of the given quadratic is $\\overline{3+2i}=3-2i$. \nNow Vieta's formulas say that $s/2$ is equal to the product of the two roots, so\n$s = 2(3+2i)(3-2i) = \\boxed{26 \\... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/21.json | AHSME |
1962_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | A man drives $150$ miles to the seashore in $3$ hours and $20$ minutes. He returns from the shore to the starting point in $4$ hours and $10$ minutes. Let $r$ be the average rate for the entire trip. Then the average rate for the trip going exceeds $r$ in miles per hour, by:
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 4\fra... | [
"Since the equation for rate is $r=\\frac{d}{t}$, you only need to find $d$ and $t$. The distance from the seashore to the starting point is $150$ miles, and since he makes a round trip, $d=300$. Also, you know the times it took him to go both directions, so when you add them up ($3$ hours and $20$ minutes $+$ $4$ ... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/10.json | AHSME |
1962_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | For any real value of $x$ the maximum value of $8x - 3x^2$ is:
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{8}3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \frac{16}{3}$
| [
"Let $f(x) = 8x-3x^2$ Since $f(x)$ is a quadratic and the quadratic term is negative, the maximum will be $f\\bigg(- \\dfrac{b}{2a}\\bigg)$ when written in the form $ax^2+bx+c$. We see that $a=-3$, and so $- \\dfrac{b}{2a} = -\\bigg( \\dfrac{8}{-6}\\bigg) = \\dfrac{4}{3}$. Plugging in this value yields $f(\\dfrac{4... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/26.json | AHSME |
1962_AHSME_Problems | 30 | 0 | Other | Multiple Choice | Consider the statements:
$\textbf{(1)}\ \text{p and q are both true}\qquad\textbf{(2)}\ \text{p is true and q is false}\qquad\textbf{(3)}\ \text{p is false and q is true}\qquad\textbf{(4)}\ \text{p is false and q is false.}$
How many of these imply the negative of the statement "p and q are both true?"
$\textbf... | [
"By De Morgan's Law, the negation of $p$ and $q$ are both true is that at least one of them is false, with the exception of the statement 1, i.e.;\n\n\n$\\text{p and q are both true}.$\n\n\nThe other three statements state that at least one statement is false. So, 2, 3, and 4 work, yielding an answer of 3, or $\\te... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/30.json | AHSME |
1962_AHSME_Problems | 31 | 0 | Geometry | Multiple Choice | The ratio of the interior angles of two regular polygons with sides of unit length is $3: 2$. How many such pairs are there?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{infinitely many}$
| [
"The formula for the measure of the interior angle of a regular polygon with $n$-sides is $180 - \\frac{360}{n}$. Letting our two polygons have side length $r$ and $k$, we have that the ratio of the interior angles is $\\frac{180 - \\frac{360}{r}}{180 - \\frac{360}{k}} = \\frac{(r-2) \\cdot k}{(k-2) \\cdot r} = \\f... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/31.json | AHSME |
1962_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | Let $a @ b$ represent the operation on two numbers, $a$ and $b$, which selects the larger of the two numbers, with $a@a = a$. Let $a ! b$ represent the operator which selects the smaller of the two numbers, with $a ! a = a$. Which of the following three rules is (are) correct?
$\textbf{(1)}\ a@b = b@a\qquad\textbf{(... | [
"The first rule must be correct as both sides of the equation pick the larger out of a and b.\n\n\nThe second rule must also be correct as both sides would end up picking the largest out of a, b, and c.\n\n\nWLOG, lets assume b < c.\n\n\nThe third rule is a little more complex. To see if it works, let’s split the p... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/27.json | AHSME |
1962_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | The expression $\frac{1^{4y-1}}{5^{-1}+3^{-1}}$ is equal to:
$\textbf{(A)}\ \frac{4y-1}{8}\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{15}{2}\qquad\textbf{(D)}\ \frac{15}{8}\qquad\textbf{(E)}\ \frac{1}{8}$
| [
"We simplify the expression to yield:\n\n\n$\\dfrac{1^{4y-1}}{5^{-1}+3^{-1}}=\\dfrac{1}{5^{-1}+3^{-1}}=\\dfrac{1}{\\dfrac{1}{5}+\\dfrac{1}{3}}=\\dfrac{1}{\\dfrac{8}{15}}=\\dfrac{15}{8}$.\n\n\nThus our answer is $\\boxed{\\textbf{(D)}\\ \\frac{15}{8}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/1.json | AHSME |
1962_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | The difference between the larger root and the smaller root of $x^2 - px + (p^2 - 1)/4 = 0$ is:
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ p\qquad\textbf{(E)}\ p+1$
| [
"Call the two roots $r$ and $s$, with $r \\ge s$. \nBy Vieta's formulas, $p=r+s$ and $(p^2-1)/4=rs.$\n(Multiplying both sides of the second equation by 4 gives $p^2-1=4rs$.)\nThe value we need to find, then, is $r-s$.\nSince $p=r+s$, $p^2=r^2+2rs+s^2$.\nSubtracting $p^2-1=4rs$ from both sides gives $1=r^2-2rs+s^2$.... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/11.json | AHSME |
1962_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | The expression $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}}$ is equal to:
$\textbf{(A)}\ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{-\sqrt{3}}{6}\qquad\textbf{(C)}\ \frac{\sqrt{-3}}{6}\qquad\textbf{(D)}\ \frac{5\sqrt{3}}{6}\qquad\textbf{(E)}\ 1$
| [
"Simplifying $\\sqrt{\\dfrac{4}{3}}$ yields $\\dfrac{2}{\\sqrt{3}}=\\dfrac{2\\sqrt{3}}{3}$.\n\n\n\n\nSimplifying $\\sqrt{\\dfrac{3}{4}}$ yields $\\dfrac{\\sqrt{3}}{2}$.\n\n\n\n\n$\\dfrac{2\\sqrt{3}}{3}-\\dfrac{\\sqrt{3}}{2}=\\dfrac{4\\sqrt{3}}{6}-\\dfrac{3\\sqrt{3}}{6}=\\dfrac{\\sqrt{3}}{6}$.\n\n\n\n\nSince we cann... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/2.json | AHSME |
1962_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | The set of $x$-values satisfying the equation $x^{\log_{10} x} = \frac{x^3}{100}$ consists of:
$\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \text{10, only}\qquad\textbf{(C)}\ \text{100, only}\qquad\textbf{(D)}\ \text{10 or 100, only}\qquad\textbf{(E)}\ \text{more than two real numbers.}$
| [
"Taking the base-$x$ logarithm of both sides gives $\\log_{10}x=\\log_x\\frac{x^3}{100}$.\nThis simplifies to \n\\[\\log_{10}x=\\log_x{x^3} - \\log_x{100}\\]\n\\[\\log_{10}x+\\log_x{100}=3\\]\n\\[\\log_{10}x+2 \\log_x{10}=3\\]\nAt this point, we substitute $u=\\log_{10}x$.\n\\[u+\\frac2u=3\\]\n\\[u^2+2=3u\\]\n\\[u^... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/28.json | AHSME |
1962_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | When $\left ( 1 - \frac{1}{a} \right ) ^6$ is expanded the sum of the last three coefficients is:
$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ -10\qquad\textbf{(E)}\ -11$
| [
"This is equivalent to $\\frac{(a-1)^6}{a^6}.$\nIts expansion has 7 terms, whose coefficients are the same as those of $(a-1)^6$.\nBy the Binomial Theorem, the sum of the last three coefficients is \n$\\binom{6}{2}-\\binom{6}{1}+\\binom{6}{0}=15-6+1=\\boxed{10 \\textbf{ (C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/12.json | AHSME |
1962_AHSME_Problems | 32 | 0 | Algebra | Multiple Choice | If $x_{k+1} = x_k + \frac12$ for $k=1, 2, \dots, n-1$ and $x_1=1$, find $x_1 + x_2 + \dots + x_n$.
$\textbf{(A)}\ \frac{n+1}{2}\qquad\textbf{(B)}\ \frac{n+3}{2}\qquad\textbf{(C)}\ \frac{n^2-1}{2}\qquad\textbf{(D)}\ \frac{n^2+n}{4}\qquad\textbf{(E)}\ \frac{n^2+3n}{4}$
| [
"The sequence $x_1, x_2, \\dots, x_n$ is an arithmetic sequence since every term is $\\frac12$ more than the previous term. Letting $a=1$ and $r=\\frac12$, we can rewrite the sequence as $a, a+r, \\dots, a+(n-1)r$.\nRecall that the sum of the first $n$ terms of an arithmetic sequence is $na+\\binom{n}2r$.\nSubstitu... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/32.json | AHSME |
1962_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | Three machines $\text{P, Q, and R,}$ working together, can do a job in $x$ hours. When working alone, $\text{P}$ needs an additional $6$ hours to do the job; $\text{Q}$, one additional hour; and $R$, $x$ additional hours. The value of $x$ is:
$\textbf{(A)}\ \frac{2}3\qquad\textbf{(B)}\ \frac{11}{12}\qquad\textbf{(C)... | [
"Machine P takes $x+6$ hours, machine Q takes $x+1$ hours, and machine R takes $2x$ hours.\nWe also know that all three working together take $x$ hours. \nNow the time it takes for all three machines to complete the job is the harmonic mean of the times the three machines take individually; that is,\n\\[x=\\frac1{\... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/24.json | AHSME |
1962_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | Given square $ABCD$ with side $8$ feet. A circle is drawn through vertices $A$ and $D$ and tangent to side $BC$. The radius of the circle, in feet, is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4\sqrt{2}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5\sqrt{2}\qquad\textbf{(E)}\ 6$
| [
"Let $O$ be the center of the circle and $E$ be the point of tangency of the circle and $BC$ and let $F$ be the point of intersection of lines $OE$ and $AD$ Because of the symmetry, $BE=EC=AF=FD=4$ feet. Let the length of $OF$ be $x$. The length of $OE$ is $EF-OF=-x+8$. By Pythagorean Theorem, $OA=OD=\\sqrt{x^2+4^2... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/25.json | AHSME |
1962_AHSME_Problems | 33 | 0 | Algebra | Multiple Choice | The set of $x$-values satisfying the inequality $2 \leq |x-1| \leq 5$ is:
$\textbf{(A)}\ -4\leq x\leq-1\text{ or }3\leq x\leq 6\qquad$
$\textbf{(B)}\ 3\leq x\leq 6\text{ or }-6\leq x\leq-3\qquad\textbf{(C)}\ x\leq-1\text{ or }x\geq 3\qquad$
$\textbf{(D)}\ -1\leq x\leq 3\qquad\textbf{(E)}\ -4\leq x\leq 6$
| [
"This inequality can be split into two inequalities:\n$2\\le x-1\\le5$ or $2\\le1-x\\le5$. Solving for x gives\n$3\\le x\\le6$ or $-4\\le x\\le-1$. $\\boxed{\\textbf{(A)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/33.json | AHSME |
1962_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | $R$ varies directly as $S$ and inversely as $T$. When $R = \frac{4}{3}$ and $T = \frac {9}{14}$, $S = \frac37$. Find $S$ when $R = \sqrt {48}$ and $T = \sqrt {75}$.
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60$
| [
"\\[R=c\\cdot\\frac{S}T\\]\n\n\nfor some constant $c$. \n\n\nYou know that\n\n\n\\[\\frac43=c\\cdot\\frac{3/7}{9/14}=c\\cdot\\frac37\\cdot\\frac{14}9=c\\cdot\\frac23\\,,\\]\n\n\nso\n\n\n\\[c=\\frac{4/3}{2/3}=2\\,.\\]\n\n\nWhen $R=\\sqrt{48}$ and $T=\\sqrt{75}$ we have\n\n\n\\[\\sqrt{48}=\\frac{2S}{\\sqrt{75}}\\,,\\... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/13.json | AHSME |
1962_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | Which of the following sets of $x$-values satisfy the inequality $2x^2 + x < 6$?
$\textbf{(A)}\ -2 < x <\frac{3}{2}\qquad\textbf{(B)}\ x >\frac{3}2\text{ or }x <-2\qquad\textbf{(C)}\ x <\frac{3}2\qquad$
$\textbf{(D)}\ \frac{3}2 < x < 2\qquad\textbf{(E)}\ x <-2$
| [
"First, subtract 6 from both sides of the inequality, \n$2x^2 + x - 6 < 0$.\nThis is a parabola that opens upward when graphed, it has a positive leading coefficient. So any negative x values must be between its x-axis intersections, namely $x = -2, 1.5$. The answer is A.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/29.json | AHSME |
1962_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$, in the order shown. The value of $x$ is:
$\textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined}$
| [
"Let $y$ represent the common difference between the terms. We have $(x+1)-y=(x-1)\\implies y=2$.\n\n\nSubstituting gives us $(2x+3)-2=(x+1)\\implies 2x+1=x+1\\implies x=0$.\n\n\nTherefore, our answer is $\\boxed{\\textbf{(B)}\\ 0}$;\n\n\n"
] | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/3.json | AHSME |
1962_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | For what real values of $K$ does $x = K^2 (x-1)(x-2)$ have real roots?
$\textbf{(A)}\ \text{none}\qquad\textbf{(B)}\ -2<K<1\qquad\textbf{(C)}\ -2\sqrt{2}< K < 2\sqrt{2}\qquad$
$\textbf{(D)}\ K>1\text{ or }K<-2\qquad\textbf{(E)}\ \text{all}$
| [
"Factoring, we get\n\\[x = K^2x^2 - 3K^2x + 2K^2\\]\n\\[K^2x^2 - (3K^2+1)x + 2K^2 = 0\\]\nAt this point, we could use the quadratic formula, but we're only interested in whether the roots are real, which occurs if and only if the discriminant ($b^2-4ac$) is non-negative.\n\\[(-3K^2-1)^2-8K^4\\ge0\\]\n\\[9K^4+6K^2+1... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/34.json | AHSME |
1962_AHSME_Problems | 8 | 0 | Arithmetic | Multiple Choice | Given the set of $n$ numbers; $n > 1$, of which one is $1 - \frac {1}{n}$ and all the others are $1$. The arithmetic mean of the $n$ numbers is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ n-\frac{1}{n}\qquad\textbf{(C)}\ n-\frac{1}{n^2}\qquad\textbf{(D)}\ 1-\frac{1}{n^2}\qquad\textbf{(E)}\ 1-\frac{1}{n}-\frac{1}{n^2}$
| [
"Just take $\\frac{1(n-1)+(1-\\frac{1}{n})}{n}$. You get $\\frac{n-1+1-\\frac{1}{n}}{n}$, which is just $\\frac{n-\\frac{1}{n}}{n}$, which is just $\\boxed{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/8.json | AHSME |
1962_AHSME_Problems | 22 | 0 | Number Theory | Multiple Choice | The number $121_b$, written in the integral base $b$, is the square of an integer, for
$\textbf{(A)}\ b = 10,\text{ only}\qquad\textbf{(B)}\ b = 10\text{ and }b = 5,\text{ only}\qquad$
$\textbf{(C)}\ 2\leq b\leq 10\qquad\textbf{(D)}\ b > 2\qquad\textbf{(E)}\ \text{no value of }b$
| [
"$121_b$ can be represented in base 10 as $b^2+2b+1$, which factors as $(b+1)^2$.\nNote that $b>2$ because 2 is a digit in the base-b representation, but for any \n$b>2$, $121_b$ is the square of $b+1$. $\\boxed{\\textbf{(D)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/22.json | AHSME |
1962_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | A regular dodecagon ($12$ sides) is inscribed in a circle with radius $r$ inches. The area of the dodecagon, in square inches, is:
$\textbf{(A)}\ 3r^2\qquad\textbf{(B)}\ 2r^2\qquad\textbf{(C)}\ \frac{3r^2\sqrt{3}}{4}\qquad\textbf{(D)}\ r^2\sqrt{3}\qquad\textbf{(E)}\ 3r^2\sqrt{3}$
| [
"The formula for the area of a regular dodecagon is $3r^2$. The answer is $\\boxed{\\textbf{(A)}}$.\n(If you don't know this formula, it's pretty easy to figure out that the area of a square inscribed in a circle is $2r^2$, and all the choices except $3r^2$ are less than $2r^2$. Remember, the more sides a regular p... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/18.json | AHSME |
1962_AHSME_Problems | 38 | 0 | Number Theory | Multiple Choice | The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$, the population was one more than a perfect square. Now, with an additional increase of $100$, the population is again a perfect square.
The original population is a multiple of:
$\textbf{(A)}\ 3\qquad\textbf{... | [
"Let $a^2$ $=$ original population count, $b^2+1$ $=$ the second population count, and $c^2$ $=$ the third population count \nWe first see that $a^2 + 100 = b^2 + 1$ or $99$ $=$ $b^2-a^2$.\nWe then factor the right side getting $99$ $=$ $(b-a)(b+a)$. \nSince we can only have an nonnegative integral population, clea... | 2 | ./CreativeMath/AHSME/1962_AHSME_Problems/38.json | AHSME |
1962_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | If $8^x = 32$, then $x$ equals:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ \frac{5}{3}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{1}{4}$
| [
"Recognizing that $8=2^3$, we know that $2^{3x}=32$. Since $2^5=32$, we have $2^{3x}=2^5$. Therefore, $x=\\dfrac{5}{3}$.\n\n\nSo our answer is $\\boxed{\\textbf{(B)}\\ \\frac{5}{3}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/4.json | AHSME |
1962_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | Let $s$ be the limiting sum of the geometric series $4- \frac83 + \frac{16}{9} - \dots$, as the number of terms increases without bound. Then $s$ equals:
$\textbf{(A)}\ \text{a number between 0 and 1}\qquad\textbf{(B)}\ 2.4\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.6\qquad\textbf{(E)}\ 12$
| [
"The infinite sum of a geometric series with first term $a$ and common ratio $r$ ($-1<r<1$) is $\\frac{a}{1-r}$.\nNow, in this geometric series, $a=4$, and $r=-\\frac23$. Plugging these into the formula, we get \n$\\frac4{1-(-\\frac23)}$, which simplifies to $\\frac{12}5$, or $\\boxed{2.4\\textbf{ (B)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/14.json | AHSME |
1962_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | Given triangle $ABC$ with base $AB$ fixed in length and position. As the vertex $C$ moves on a straight line, the intersection point of the three medians moves on:
$\textbf{(A)}\ \text{a circle}\qquad\textbf{(B)}\ \text{a parabola}\qquad\textbf{(C)}\ \text{an ellipse}\qquad\textbf{(D)}\ \text{a straight line}\qquad\... | [
"Let $CM$ be the median through vertex $C$, and let $G$ be the point of intersection of the triangle's medians. \n\n\nLet $CH$ be the altitude of the triangle through vertex $C$ and $GP$ be the distance from $G$ to $AB$, with the point $P$ laying on $AB$. \n\n\nUsing Thales' intercept theorem, we derive the proport... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/15.json | AHSME |
1962_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | If the radius of a circle is increased by $1$ unit, the ratio of the new circumference to the new diameter is:
$\textbf{(A)}\ \pi+2\qquad\textbf{(B)}\ \frac{2\pi+1}{2}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ \frac{2\pi-1}{2}\qquad\textbf{(E)}\ \pi-2$
| [
"The ratio of a circumference to a diameter always is the same so the answer is obviously C.\n\n\n",
"Let us say that the radius of a circle is $r$. When the radius is increased by $1$, the new radius is $r+1$ so the diameter is $2r+2$. We know that the circumference of a circle is $2\\pi r$ so $2 \\cdot \\pi \\c... | 2 | ./CreativeMath/AHSME/1962_AHSME_Problems/5.json | AHSME |
1962_AHSME_Problems | 39 | 0 | Geometry | Multiple Choice | Two medians of a triangle with unequal sides are $3$ inches and $6$ inches. Its area is $3 \sqrt{15}$ square inches. The length of the third median in inches, is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{6}\qquad\textbf{(D)}\ 6\sqrt{3}\qquad\textbf{(E)}\ 6\sqrt{6}$
| [
"By the area formula:\n\\[A = \\frac43\\sqrt{s(s-m_1)(s-m_2)(s-m_3)}\\]\nWhere $s = \\frac{m_1+m_2+m_3}{2}$.\nPlugging in the numbers:\n\\[3\\sqrt{15} = \\frac43\\sqrt{\\frac{9+m}2\\cdot\\frac{9-m}2\\cdot\\frac{m+3}2\\cdot\\frac{m-3}2}\\]\nSimplifying and squaring both sides:\n\\[1215 = (9-m)(9+m)(m+3)(m-3)\\]\nNow... | 2 | ./CreativeMath/AHSME/1962_AHSME_Problems/39.json | AHSME |
1962_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | If the parabola $y = ax^2 + bx + c$ passes through the points $( - 1, 12)$, $(0, 5)$, and $(2, - 3)$, the value of $a + b + c$ is:
$\textbf{(A)}\ -4\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$
| [
"Substituting in the $(x, y)$ pairs gives the following system of equations:\n\\[a-b+c=12\\]\n\\[c=5\\]\n\\[4a+2b+c=-3\\]\nWe know $c=5$, so plugging this in reduces the system to two variables:\n\\[a-b=7\\]\n\\[4a+2b=-8\\]\nDividing the second equation by 2 gives $2a+b=-4$, which can be added to the first equation... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/19.json | AHSME |
1962_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | In triangle $ABC$, $CD$ is the altitude to $AB$ and $AE$ is the altitude to $BC$. If the lengths of $AB$, $CD$, and $AE$ are known, the length of $DB$ is:
$\textbf{(A)}\ \text{not determined by the information given} \qquad$
$\textbf{(B)}\ \text{determined only if A is an acute angle} \qquad$
$\textbf{(C)}\ \t... | [
"We can actually determine the length of $DB$ no matter what type of angles $A$ and $B$ are. This can be easily proved through considering all possible cases, though for the purposes of this solution, we'll show that we can determine $DB$ if $A$ is an obtuse angle. \n\n\nLet's see what happens when $A$ is an obtuse... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/23.json | AHSME |
1962_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:
$\textbf{(A)}\ \text{more than 5}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2$
| [
"Obviously, we can factor out an $x$ first to get $x(x^8-1)$.\nNext, we repeatedly factor differences of squares:\n\\[x(x^4+1)(x^4-1)\\]\n\\[x(x^4+1)(x^2+1)(x^2-1)\\]\n\\[x(x^4+1)(x^2+1)(x+1)(x-1)\\]\nNone of these 5 factors can be factored further, so the answer is \n$\\boxed{\\textbf{(B) } 5}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/9.json | AHSME |
1962_AHSME_Problems | 35 | 0 | Algebra | Multiple Choice | A man on his way to dinner short after $6: 00$ p.m. observes that the hands of his watch form an angle of $110^{\circ}$. Returning before $7: 00$ p.m. he notices that again the hands of his watch form an angle of $110^{\circ}$. The number of minutes that he has been away is:
$\textbf{(A)}\ 36\frac{2}3\qquad\textbf{(... | [
"Let $n$ be the number of minutes after 6:00. Let $h(n)=180+\\frac{n}2$ be the angle, in degrees, of the hour hand (with $0^{\\circ}$ at the top and increasing in the clockwise direction); similarly, let $m(n)=6n$ be the angle of the minute hand. We want $|h(n)-m(n)|=110$. This is equivalent to \n\\[180-\\frac{11n}... | 1 | ./CreativeMath/AHSME/1962_AHSME_Problems/35.json | AHSME |
1988_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints
of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown
in the second figure. The ratio of height to base, $XY / YZ$, in this rectang... | [
"Within $WXYZ$, the parallelogram piece has vertical side $BF = \\sqrt{1^2 + 2^2} = \\sqrt{5}$, and diagonal side $FD = 1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $\\triangle AFG$, so we ded... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/20.json | AHSME |
1988_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | [asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, B=(1,-(1/sqrt(3))), C=(-1,-(1/sqrt(3))), A=(0,(2/sqrt(3))), E=(2,-(2/sqrt(3))), F=(-2,-(2/sqrt(3))), D=(0,(4/sqrt(3))); draw(A--B--C--A^^D--E--F--D); label("A'", A, N); label("B'", B, SE); label("C'", C, SW); label("A", D, E); label("B", E, E); label("C", F,... | [
"Let $\\triangle ABC$ have side length $s$ and $\\triangle A'B'C'$ have side length $t$. Thus the altitude of $\\triangle ABC$ is $\\frac{s\\sqrt{3}}{2}$. Now observe that this altitude is made up of three parts: the distance from $BC$ to $B'C'$, plus the altitude of $\\triangle A'B'C'$, plus a top part which is eq... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/16.json | AHSME |
1988_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | A figure is an equiangular parallelogram if and only if it is a
$\textbf{(A)}\ \text{rectangle}\qquad \textbf{(B)}\ \text{regular polygon}\qquad \textbf{(C)}\ \text{rhombus}\qquad \textbf{(D)}\ \text{square}\qquad \textbf{(E)}\ \text{trapezoid}$
| [
"The definition of an equiangular parallelogram is that all the angles are equal, and that pairs of sides are parallel. It may be a rectangle, because all the angles are equal and it is a parallelogram. It is not necessarily a regular polygon, because if the polygon is a pentagon, it is not a parallelogram. It is n... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/6.json | AHSME |
1988_AHSME_Problems | 7 | 0 | Arithmetic | Multiple Choice | Estimate the time it takes to send $60$ blocks of data over a communications channel if each block consists of $512$
"chunks" and the channel can transmit $120$ chunks per second.
$\textbf{(A)}\ 0.04 \text{ seconds}\qquad \textbf{(B)}\ 0.4 \text{ seconds}\qquad \textbf{(C)}\ 4 \text{ seconds}\qquad \textbf{(D)}\ 4\te... | [
"We want to figure out the number of chunks in $60$ blocks, so we have $60\\cdot 512 \\approx 30000$. We divide this by $120$ to determine the number of seconds necessary to transmit. $30000/120 \\approx 250$, which means that it takes approximately $4$ minutes to transmit. Thus, the answer is $\\boxed{\\text{D}}$.... | 2 | ./CreativeMath/AHSME/1988_AHSME_Problems/7.json | AHSME |
1988_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | If $|x| + x + y = 10$ and $x + |y| - y = 12$, find $x + y$
$\textbf{(A)}\ -2\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ \frac{18}{5}\qquad \textbf{(D)}\ \frac{22}{3}\qquad \textbf{(E)}\ 22$
| [
"We proceed by casework:\n\n\nCase 1: x, y >= 0\n\n\nThus we have 2x + y = 10, x = 12. This makes y < 0, a contradiction.\n\n\nCase 2: x >= 0 > y\n\n\n2x + y = 10, x - 2y = 12. By multiplying the first equation by two and adding, we have 4x + x = 2*10 + 12 thus x = 32/5. Since the question only specifies 1 solution... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/17.json | AHSME |
1988_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | The complex number $z$ satisfies $z + |z| = 2 + 8i$. What is $|z|^{2}$? Note: if $z = a + bi$, then $|z| = \sqrt{a^{2} + b^{2}}$.
$\textbf{(A)}\ 68\qquad \textbf{(B)}\ 100\qquad \textbf{(C)}\ 169\qquad \textbf{(D)}\ 208\qquad \textbf{(E)}\ 289$
| [
"Let the complex number $z$ equal $a+bi$. Then the preceding equation can be expressed as \\[a+bi+\\sqrt{a^2+b^2} = 2+8i\\] Because $a$ and $b$ must both be real numbers, we immediately have that $bi = 8i$, giving $b = 8$. Plugging this in back to our equation gives us $a+\\sqrt{a^2+64} = 2$. \nRearranging this in... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/21.json | AHSME |
1988_AHSME_Problems | 10 | 0 | Arithmetic | Multiple Choice | In an experiment, a scientific constant $C$ is determined to be $2.43865$ with an error of at most $\pm 0.00312$.
The experimenter wishes to announce a value for $C$ in which every digit is significant.
That is, whatever $C$ is, the announced value must be the correct result when $C$ is rounded to that number of digi... | [
"If added together, we have:\n\\[2.43865+0.00312=2.44177\\]\nThis rounds to $2.44$. If they subtracted, we have:\n\\[2.43865-0.00312=2.43553.\\]\nThis rounds to $2.44$. Therefore, we have the answer to be $\\fbox{\\textbf{(D)} 2.44}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/10.json | AHSME |
1988_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | Suppose that $p$ and $q$ are positive numbers for which \[\operatorname{log}_{9}(p) = \operatorname{log}_{12}(q) = \operatorname{log}_{16}(p+q).\] What is the value of $\frac{q}{p}$?
$\textbf{(A)}\ \frac{4}{3}\qquad \textbf{(B)}\ \frac{1+\sqrt{3}}{2}\qquad \textbf{(C)}\ \frac{8}{5}\qquad \textbf{(D)}\ \frac{1+\sqrt{5... | [
"We can rewrite the equation as $\\frac{\\log{p}}{\\log{9}} = \\frac{\\log{q}}{\\log{12}} = \\frac{\\log{(p + q)}}{\\log{16}}$. Then, the system can be split into 3 pairs: $\\frac{\\log{p}}{\\log{9}} = \\frac{\\log{q}}{\\log{12}}$, $\\frac{\\log{q}}{\\log{12}} = \\frac{\\log{(p + q)}}{\\log{16}}$, and $\\frac{\\log... | 2 | ./CreativeMath/AHSME/1988_AHSME_Problems/26.json | AHSME |
1988_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | Let $f(x) = 4x - x^{2}$. Give $x_{0}$, consider the sequence defined by $x_{n} = f(x_{n-1})$ for all $n \ge 1$.
For how many real numbers $x_{0}$ will the sequence $x_{0}, x_{1}, x_{2}, \ldots$ take on only a finite number of different values?
$\textbf{(A)}\ \text{0}\qquad \textbf{(B)}\ \text{1 or 2}\qquad \textbf{(... | [
"Note that $x_{0} = 0$ gives the constant sequence $0, 0, ...$, since $f(0) = 4 \\cdot 0 - 0^2 = 0$. Because $f(4)=0, x_{0} = 4$ gives the sequence $4, 0, 0, ...$ with two different values. Similarly, $f(2) = 4$, so $x_{0} = 2$ gives the sequence $2, 4, 0, 0, ...$ with three values. In general, if $x_{0} = a_{n}$ g... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/30.json | AHSME |
1988_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | In the figure, $AB \perp BC, BC \perp CD$, and $BC$ is tangent to the circle with center $O$ and diameter $AD$.
In which one of the following cases is the area of $ABCD$ an integer?
[asy] pair O=origin, A=(-1/sqrt(2),1/sqrt(2)), B=(-1/sqrt(2),-1), C=(1/sqrt(2),-1), D=(1/sqrt(2),-1/sqrt(2)); draw(unitcircle); dot(O);... | [
"Let $E$ and $F$ be the intersections of lines $AB$ and $BC$ with the circle. One can prove that $BCDE$ is a rectangle, so $BE=CD$.\n\n\nIn order for the area of trapezoid $ABCD$ to be an integer, the expression $\\frac{(AB+CD)BC}2=(AB+CD)BF$ must be an integer, so $BF$ must be rational.\n\n\nBy Power of a Point, $... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/27.json | AHSME |
1988_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | $\sqrt{8}+\sqrt{18}=$
\[\text{(A)}\ \sqrt{20}\qquad\text{(B)}\ 2(\sqrt{2}+\sqrt{3})\qquad\text{(C)}\ 7\qquad\text{(D)}\ 5\sqrt{2}\qquad\text{(E)}\ 2\sqrt{13}\]
| [
"$\\sqrt{8} = 2 \\sqrt{2}$\n\n\n$\\sqrt{18} = 3 \\sqrt{2}$\n\n\nSo adding the two terms we get $2 \\sqrt{2} + 3 \\sqrt{2} = 5 \\sqrt{2}$ , which corresponds to answer choice $\\boxed{\\textbf{(D)}}$ .\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/1.json | AHSME |
1988_AHSME_Problems | 11 | 0 | Arithmetic | Multiple Choice | [asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(5,0), B=(7,0), C=(10,0), D=(13,0), E=(16,0); pair F=(4,3), G=(5,3), H=(7,3), I=(10,3), J=(12,3); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(I); dot(J); draw((0,0)--(18,0)^^(0,3)--(18,3)); draw((0,0)--(0,.5)^^(5,0)--(5,.5)^^(10,0)--(10,.5)^^... | [
"We simply have to divide the present population by the original, and whichever one is the largest have the greatest percent change. City $\\text{A}$ is \n$\\frac{50}{40}=1.25$. City $\\text{B}$ is $\\frac{70}{50}=1.4$. City $\\text{C}$ is $\\frac{100}{70} \\approx 1.42$. City $\\text{D}$ is $\\frac{130}{100}=1.3$... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/11.json | AHSME |
1988_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | Triangles $ABC$ and $XYZ$ are similar, with $A$ corresponding to $X$ and $B$ to $Y$. If $AB=3, BC=4$, and $XY=5$, then $YZ$ is:
$\text{(A)}\ 3\frac{3}{4} \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 6\frac{1}{4} \qquad \text{(D)}\ 6\frac{2}{3} \qquad \text{(E)}\ 8$
| [
"Since the triangles are similar, we know that the corresponding sides of the triangle are in ratio to each other. \nWe have $\\frac{\\overline{AB}}{\\overline{XY}}=\\frac{\\overline{BC}}{\\overline{YZ}}$. Plugging in values we have:\n$\\frac{3}{4}=\\frac{5}{\\overline{YZ}}$. Solving for $\\overline{YZ}$, we have $... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/2.json | AHSME |
1988_AHSME_Problems | 28 | 0 | Probability | Multiple Choice | An unfair coin has probability $p$ of coming up heads on a single toss.
Let $w$ be the probability that, in $5$ independent toss of this coin,
heads come up exactly $3$ times. If $w = 144 / 625$, then
$\textbf{(A)}\ p\text{ must be }\tfrac{2}{5}\qquad \textbf{(B)}\ p\text{ must be }\tfrac{3}{5}\qquad\\ \textbf{(C)... | [
"We have $w = {5\\choose3}p^{3}(1-p)^{2} = 10p^{3}(1-p)^{2}$, so we need to solve $10p^{3}(1-p)^{2} = \\frac{144}{625} \\implies p^{3}(1-p)^{2} = \\frac{72}{3125}$. Now observe that when $p=0$, the left-hand side evaluates to $0$, which is less than $\\frac{72}{3125}$; when $p=\\frac{1}{2}$, it evaluates to $\\frac... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/28.json | AHSME |
1988_AHSME_Problems | 12 | 0 | Probability | Multiple Choice | Each integer $1$ through $9$ is written on a separate slip of paper and all nine slips are put into a hat.
Jack picks one of these slips at random and puts it back. Then Jill picks a slip at random.
Which digit is most likely to be the units digit of the sum of Jack's integer and Jill's integer?
$\textbf{(A)}\ 0\qq... | [
"We can draw a sample space diagram, and find that of the $9^2 = 81$ possibilities, $9$ of them give a sum of $0$, and each other sum mod $10$ (from $1$ to $9$) is given by $8$ of the possibilities (and indeed we can check that $9 + 8 \\times 9 = 81$). Thus $0$ is the most likely, so the answer is $\\boxed{\\text{A... | 2 | ./CreativeMath/AHSME/1988_AHSME_Problems/12.json | AHSME |
1988_AHSME_Problems | 24 | 0 | Geometry | Multiple Choice | An isosceles trapezoid is circumscribed around a circle. The longer base of the trapezoid is $16$,
and one of the base angles is $\arcsin(.8)$. Find the area of the trapezoid.
$\textbf{(A)}\ 72\qquad \textbf{(B)}\ 75\qquad \textbf{(C)}\ 80\qquad \textbf{(D)}\ 90\qquad \textbf{(E)}\ \text{not uniquely determined}$
... | [
"Let the trapezium have diagonal legs of length $x$ and a shorter base of length $y$. Drop altitudes from the endpoints of the shorter base to the longer base to form two right-angled triangles, which are congruent since the trapezium is isosceles. Thus using the base angle of $\\arcsin(0.8)$ gives the vertical sid... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/24.json | AHSME |
1988_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | $X, Y$ and $Z$ are pairwise disjoint sets of people. The average ages of people in the sets
$X, Y, Z, X \cup Y, X \cup Z$ and $Y \cup Z$ are $37, 23, 41, 29, 39.5$ and $33$ respectively.
Find the average age of the people in set $X \cup Y \cup Z$.
$\textbf{(A)}\ 33\qquad \textbf{(B)}\ 33.5\qquad \textbf{(C)}\ 33.6\... | [
"Let the variables $X$, $Y$, and $Z$ represent the sums of the ages of the people in sets $X$, $Y$, and $Z$ respectively. Let $x$, $y$, and $z$ represent the numbers of people who are in sets $X$, $Y$, and $Z$ respectively. Since the sets are disjoint, we know, for example, that the number of people in the set $X \... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/25.json | AHSME |
1988_AHSME_Problems | 13 | 0 | Other | Multiple Choice | If $\sin(x) =3 \cos(x)$ then what is $\sin(x) \cdot \cos(x)$?
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{5}\qquad \textbf{(C)}\ \frac{2}{9}\qquad \textbf{(D)}\ \frac{1}{4}\qquad \textbf{(E)}\ \frac{3}{10}$
| [
"In the problem we are given that $\\sin{(x)}=3\\cos{(x)}$, and we want to find $\\sin{(x)}\\cos{(x)}$. We can divide both sides of the original equation by $\\cos{(x)}$ to get \\[\\frac{\\sin{(x)}}{\\cos{(x)}}=\\tan{(x)}=3.\\]\nWe can now use right triangle trigonometry to finish the problem.\n[asy] pair A,B,C; A ... | 2 | ./CreativeMath/AHSME/1988_AHSME_Problems/13.json | AHSME |
1988_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | You plot weight $(y)$ against height $(x)$ for three of your friends and obtain the points
$(x_{1},y_{1}), (x_{2},y_{2}), (x_{3},y_{3})$. If $x_{1} < x_{2} < x_{3}$ and $x_{3} - x_{2} = x_{2} - x_{1}$,
which of the following is necessarily the slope of the line which best fits the data?
"Best fits" means that the su... | [
"Apply one of the standard formulae for the gradient of the line of best fit, e.g. $\\frac{\\frac{\\sum {x_i y_i}}{n} - \\bar{x} \\bar{y}}{\\frac{\\sum {x_{i}^2}}{n} - \\bar{x}^2}$, and substitute in the given condition $x_3 - x_2 = x_2 - x_1$. The answer is $\\boxed{\\text{A}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/29.json | AHSME |
1988_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | [asy] draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)--(0,1)--(-1,1)--(-1,2)); draw((-1,2)--(0,2)--(0,4)--(-1,4)--(-1,5)--(1,5)--(1,6)--(0,6)); draw((0,6)--(0,5)--(3,5)--(3,6)--(4,6)--(4,2)--(5,2)); draw((5,2)--(5,1)--(1,1)--(3,1)--(3,0)--(4,0)--(4,1)); draw((1,4)--(3,4)--(3,2)--(1,2)--(4,2)--(3,2)--(3,6)); draw((3,6)--(4,6)--(... | [
"We first notice that the paper strips cover up part of the others. Since the width of the overlap is $1$ and the length of the overlap is $1$, the area of \neach of the strips with the overlap is $(10\\cdot 1)-1=9$. Since there are 4 strips, $4\\cdot 9=36 \\implies \\boxed{\\text{A}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/3.json | AHSME |
1988_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | If $\frac{b}{a} = 2$ and $\frac{c}{b} = 3$, what is the ratio of $a + b$ to $b + c$?
$\textbf{(A)}\ \frac{1}{3}\qquad \textbf{(B)}\ \frac{3}{8}\qquad \textbf{(C)}\ \frac{3}{5}\qquad \textbf{(D)}\ \frac{2}{3}\qquad \textbf{(E)}\ \frac{3}{4}$
| [
"Since we are finding ratios, it would be helpful to put everything in terms of one variable. Since $b$ is in both equations, that would be a place to start. \nWe manipulate the equations yielding $\\frac{b}{2}=a$ and $c=3b$. Since we are asked to find the ratio of $a+b$ to $b+c$, we need to find $\\frac{a+b}{b+c}$... | 2 | ./CreativeMath/AHSME/1988_AHSME_Problems/8.json | AHSME |
1988_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | For how many integers $x$ does a triangle with side lengths $10, 24$ and $x$ have all its angles acute?
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{more than } 7$
| [
"We first notice that the sides $10$ and $24$, can be part of $2$ different right triangles, one with sides $10,24,26$, and the other with a leg\nsomewhere between $21$ and $22$. We now notice that if $x$ is less than or equal to $21$, one of the angles is obtuse, and that the same is the same for \nany value of $x... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/22.json | AHSME |
1988_AHSME_Problems | 18 | 0 | Counting | Multiple Choice | At the end of a professional bowling tournament, the top 5 bowlers have a playoff.
First #5 bowls #4. The loser receives $5$th prize and the winner bowls #3 in another game.
The loser of this game receives $4$th prize and the winner bowls #2.
The loser of this game receives $3$rd prize and the winner bowls #1.
The ... | [
"We have $2$ choices for who wins the first game, and that uniquely determines $5^{\\text{th}}$ place. Then there are $2$ choices for a next game and that uniquely determines $4^{\\text{th}}$ place, followed by $2$ choices for the next game that uniquely determines $3^{\\text{rd}}$ place. Finally, there are $2$ cho... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/18.json | AHSME |
1988_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | The slope of the line $\frac{x}{3} + \frac{y}{2} = 1$ is
$\textbf{(A)}\ -\frac{3}{2}\qquad \textbf{(B)}\ -\frac{2}{3}\qquad \textbf{(C)}\ \frac{1}{3}\qquad \textbf{(D)}\ \frac{2}{3}\qquad \textbf{(E)}\ \frac{3}{2}$
| [
"To find the slope, all we have to do is put the equation into slope-intercept form. We subtract $\\frac{x}{3}$ from both sides and then multiple all \nterms by $2$. This yields $y=-\\frac{2}{3}x+1$, so the slope is $-\\frac{2}{3} \\implies \\boxed{\\text{B}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/4.json | AHSME |
1988_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | For any real number a and positive integer k, define
${a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}$
What is
${-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}$?
$\textbf{(A)}\ -199\qquad \textbf{(B)}\ -197\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 197\qquad \textbf{(E)}\ 1... | [
"We expand both the numerator and the denominator.\n\n\n\\begin{align*} \\binom{-\\frac{1}{2}}{100}\\div\\binom{\\frac{1}{2}}{100} &= \\frac{ \\dfrac{ (-\\frac{1}{2}) (-\\frac{1}{2} - 1) (-\\frac{1}{2} - 2) \\cdots (-\\frac{1}{2} - (100 - 1)) }{\\cancel{(100)(99)\\cdots(1)}} }{ \\dfrac{ ... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/14.json | AHSME |
1988_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | If $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1$, then $b$ is
$\textbf{(A)}\ -2\qquad \textbf{(B)}\ -1\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2$
| [
"Using polynomial division, we find that the remainder is $(2a+b)x+(a+b+1)$, so for the condition to hold, we need this remainder to be $0$. This gives $2a+b=0$ and $a+b+1=0$, so $b=-2a$ and $a-2a+1=0 \\implies a=1 \\implies b=-2$, which is $\\boxed{\\text{A}}.$\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/15.json | AHSME |
1988_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | If $b$ and $c$ are constants and $(x + 2)(x + b) = x^2 + cx + 6$, then $c$ is
$\textbf{(A)}\ -5\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 5$
| [
"We first start out by expanding the left side of the equation, $(x+2)(x+b)=x^{2}+bx+2x+2b=x^2+(2+b)x+2b=x^2+cx+6$. We know the constant \nterms have to be equal so we have $2b=6$, so $b=3$. Plugging $b=3$ back in yields $x^2+(2+3)x+6=x^2+cx+6$. Thus, $c=5 \\implies \\boxed{\\text{E}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/5.json | AHSME |
1988_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Simplify
$\frac{bx(a^2x^2 + 2a^2y^2 + b^2y^2) + ay(a^2x^2 + 2b^2x^2 + b^2y^2)}{bx + ay}$
$\textbf{(A)}\ a^2x^2 + b^2y^2\qquad \textbf{(B)}\ (ax + by)^2\qquad \textbf{(C)}\ (ax + by)(bx + ay)\qquad\\ \textbf{(D)}\ 2(a^2x^2+b^2y^2)\qquad \textbf{(E)}\ (bx+ay)^2$
| [
"We can multiply each answer choice by $bx + ay$ and then compare with the numerator. This gives $\\boxed{\\text{B}}$.\n\n\n",
"Expanding everything in the brackets, we get\n$\\frac{ba^2x^3 + 2ba^2xy^2 + b^3xy^2 + a^3x^2y + 2ab^2x^2y + ab^2y^3}{bx + ay}$. We can then group numbers up in pairs so they equal $n(bx ... | 4 | ./CreativeMath/AHSME/1988_AHSME_Problems/19.json | AHSME |
1988_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | The six edges of a tetrahedron $ABCD$ measure $7, 13, 18, 27, 36$ and $41$ units. If the length of edge $AB$ is $41$, then the length of edge $CD$ is
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$
| [
"By the triangle inequality in $\\triangle ABC$, we find that $BC$ and $CA$ must sum to greater than $41$, so they must be (in some order) $7$ and $36$, $13$ and $36$, $18$ and $27$, $18$ and $36$, or $27$ and $36$. We try $7$ and $36$, and now by the triangle inequality in $\\triangle ABD$, we must use the remaini... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/23.json | AHSME |
1988_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | [asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,0), B=(16,0), C=(16,16), D=(0,16), E=(32,0), F=(48,0), G=(48,16), H=(32,16), I=(0,8), J=(10,8), K=(10,16), L=(32,6), M=(40,6), N=(40,16); draw(A--B--C--D--A^^E--F--G--H--E^^I--J--K^^L--M--N); label("S", (18,8)); label("S", (50,8)); label("Figure 1", (A+B)/2, S); ... | [
"Small correction: The writer below has maximized the area of the rectangle (with sides parallel to the walls) that fits around the table, but there is a larger single dimension we can find in the table. The height or width is maximized when the diagonal of the table is horizontal or vertical. By the Pythagorean Th... | 1 | ./CreativeMath/AHSME/1988_AHSME_Problems/9.json | AHSME |
1963_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | Two men at points $R$ and $S$, $76$ miles apart, set out at the same time to walk towards each other.
The man at $R$ walks uniformly at the rate of $4\tfrac{1}{2}$ miles per hour; the man at $S$ walks at the constant
rate of $3\tfrac{1}{4}$ miles per hour for the first hour, at $3\tfrac{3}{4}$ miles per hour for the ... | [
"First, find the number of hours it takes for the two to meet together. After $h$ hours, the person at $R$ walks $4.5h$ miles. In the same amount of time, the person at $S$ has been walking at $3.25+0.5(h-1)$ mph for the past hour, so the person walks $\\frac{h(6.5+0.5(h-1))}{2}$ miles.\n\n\nIn order for both to ... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/20.json | AHSME |
1963_AHSME_Problems | 36 | 0 | Probability | Multiple Choice | A person starting with $$64$ and making $6$ bets, wins three times and loses three times,
the wins and losses occurring in random order. The chance for a win is equal to the chance for a loss.
If each wager is for half the money remaining at the time of the bet, then the final result is:
$\textbf{(A)}\text{ a loss ... | [
"If the person wins the bet, the person has $\\frac{3}{2}$ of the previous amount. If the person loses the bet, the person only has $\\frac{1}{2}$ of the previous amount.\n\n\nBecause of the Commutative Property, the order of multiplying the multipliers does not matter. Thus, the person walks away with $64 \\cdot... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/36.json | AHSME |
1963_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | Three numbers $a,b,c$, none zero, form an arithmetic progression. Increasing $a$ by $1$ or increasing $c$ by $2$ results
in a geometric progression. Then $b$ equals:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$
| [
"Let $d$ be the common difference of the arithmetic sequence, so $a = b-d$ and $c = b+d$.\n\n\nSince increasing $a$ by $1$ or $c$ by $2$ results in a geometric sequence,\n\\[\\frac{b}{b-d+1} = \\frac{b+d}{b}\\]\n\\[\\frac{b}{b-d} = \\frac{b+d+2}{b}\\]\nCross-multiply in both equations to get a system of equations.\... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/16.json | AHSME |
1963_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | $\triangle BAD$ is right-angled at $B$. On $AD$ there is a point $C$ for which $AC=CD$ and $AB=BC$. The magnitude of $\angle DAB$ is:
$\textbf{(A)}\ 67\tfrac{1}{2}^{\circ}\qquad \textbf{(B)}\ 60^{\circ}\qquad \textbf{(C)}\ 45^{\circ}\qquad \textbf{(D)}\ 30^{\circ}\qquad \textbf{(E)}\ 22\tfrac{1}{2}^{\circ}$
| [
"[asy] draw((0,0)--(0,10)--(17.321,0)--(0,0)); draw((0,0)--(8.661,5)); dot((0,0)); label(\"$B$\",(0,0),SW); dot((0,10)); label(\"$A$\",(0,10),NW); dot((17.321,0)); label(\"$D$\",(17.321,0),SE); dot((8.661,5)); label(\"$C$\",(8.661,5),NE); [/asy]\nNote that because $C$ is the midpoint of $AD$, $C$ is the circumcente... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/6.json | AHSME |
1963_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | Given the four equations:
$\textbf{(1)}\ 3y-2x=12 \qquad\textbf{(2)}\ -2x-3y=10 \qquad\textbf{(3)}\ 3y+2x=12 \qquad\textbf{(4)}\ 2y+3x=10$
The pair representing the perpendicular lines is:
$\textbf{(A)}\ \text{(1) and (4)}\qquad \textbf{(B)}\ \text{(1) and (3)}\qquad \textbf{(C)}\ \text{(1) and (2)}\qquad \text... | [
"Write each equation in slope-intercept from since the slopes are easier to compare.\n\n\nEquation $(1)$ in slope-intercept form is $y = \\frac{2}{3}x+4$.\n\n\nEquation $(2)$ in slope-intercept form is $y = -\\frac{2}{3}x - \\frac{10}{3}$.\n\n\nEquation $(3)$ in slope-intercept form is $y = -\\frac{2}{3}x + 4$.\n\n... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/7.json | AHSME |
1963_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | The expression $\dfrac{\dfrac{a}{a+y}+\dfrac{y}{a-y}}{\dfrac{y}{a+y}-\dfrac{a}{a-y}}$, $a$ real, $a\neq 0$, has the value $-1$ for:
$\textbf{(A)}\ \text{all but two real values of }y \qquad \\ \textbf{(B)}\ \text{only two real values of }y \qquad \\ \textbf{(C)}\ \text{all real values of }y\qquad \\ \textbf{(D)}\ \te... | [
"First, note that $y \\ne \\pm a$ because that would make the denominator $0$.\n\n\nCreate common denominators in the complex fraction.\n\\[\\frac{\\frac{a^2-ay}{a^2-y^2} + \\frac{ay+y^2}{a^2-y^2}}{\\frac{ay-y^2}{a^2-y^2} - \\frac{a^2+ay}{a^2-y^2}}\\]\n\\[\\frac{\\frac{a^2+y^2}{a^2-y^2}}{\\frac{-a^2-y^2}{a^2-y^2}}\... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/17.json | AHSME |
1963_AHSME_Problems | 40 | 0 | Algebra | Multiple Choice | If $x$ is a number satisfying the equation $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$, then $x^2$ is between:
$\textbf{(A)}\ 55\text{ and }65\qquad \textbf{(B)}\ 65\text{ and }75\qquad \textbf{(C)}\ 75\text{ and }85\qquad \textbf{(D)}\ 85\text{ and }95\qquad \textbf{(E)}\ 95\text{ and }105$
| [
"Let $a = \\sqrt[3]{x + 9}$ and $b = \\sqrt[3]{x - 9}$. Cubing these equations, we get $a^3 = x + 9$ and $b^3 = x - 9$, so $a^3 - b^3 = 18$. The left-hand side factors as\n\\[(a - b)(a^2 + ab + b^2) = 18.\\]\n\n\nHowever, from the given equation $\\sqrt[3]{x + 9} - \\sqrt[3]{x - 9} = 3$, we get $a - b = 3$. Then $3... | 4 | ./CreativeMath/AHSME/1963_AHSME_Problems/40.json | AHSME |
1963_AHSME_Problems | 37 | 0 | Geometry | Multiple Choice | Given points $P_1, P_2,\cdots,P_7$ on a straight line, in the order stated (not necessarily evenly spaced).
Let $P$ be an arbitrarily selected point on the line and let $s$ be the sum of the undirected lengths
$PP_1, PP_2, \cdots , PP_7$. Then $s$ is smallest if and only if the point $P$ is:
$\textbf{(A)}\ \text{mid... | [
"By the Triangle Inequality, $P_1P + P_7P \\ge P_1P_7$, with equality happening when $P$ is between $P_1$ and $P_7$. Using similar logic, $P$ must be between $P_3$ and $P_5$ in order for the distance to be minimized.\n\n\nThe only point left to deal with is $P_4$ (which is also between $P_3$ and $P_5$). The minimu... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/37.json | AHSME |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.