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 |
|---|---|---|---|---|---|---|---|---|---|
1994_AHSME_Problems | 24 | 0 | Other | Multiple Choice | A sample consisting of five observations has an arithmetic mean of $10$ and a median of $12$. The smallest value that the range (largest observation minus smallest) can assume for such a sample is
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 10$
| [
"The minimum range occurs in the set $\\{7,7,12,12,12\\}$, so the answer is $\\boxed{\\textbf{(C)}\\ 5}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/24.json | AHSME |
1994_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | If $x$ and $y$ are non-zero real numbers such that
\[|x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0,\]
then the integer nearest to $x-y$ is
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5$
| [
"We have two cases to consider: $x$ is positive or $x$ is negative. If $x$ is positive, we have $x+y=3$ and $xy+x^3=0$\n\n\nSolving for $y$ in the top equation gives us $3-x$. Plugging this in gives us: $x^3-x^2+3x=0$. Since we're told $x$ is not zero, we can divide by $x$, giving us: $x^2-x+3=0$\n\n\nThe discrimi... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/25.json | AHSME |
1994_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | Points $A, B$ and $C$ on a circle of radius $r$ are situated so that $AB=AC, AB>r$, and the length of minor arc $BC$ is $r$. If angles are measured in radians, then $AB/BC=$
[asy] draw(Circle((0,0), 13)); draw((-13,0)--(12,5)--(12,-5)--cycle); dot((-13,0)); dot((12,5)); dot((12,-5)); label("A", (-13,0), W); label("B", (12,5), NE); label("C", (12,-5), SE); [/asy]
$\textbf{(A)}\ \frac{1}{2}\csc{\frac{1}{4}} \qquad\textbf{(B)}\ 2\cos{\frac{1}{2}} \qquad\textbf{(C)}\ 4\sin{\frac{1}{2}} \qquad\textbf{(D)}\ \csc{\frac{1}{2}} \qquad\textbf{(E)}\ 2\sec{\frac{1}{2}}$
| [
"First note that arc length equals $r\\theta$, where $\\theta$ is the central angle in radians. Call the center of the circle $O$. Then $\\angle{BOC} = 1$ radian because the minor arc $BC$ has length $r$. Since $ABC$ is isosceles, $\\angle{AOB} = \\pi - \\tfrac{1}{2}$. We use the Law of Cosines to find that \\[\\fr... | 2 | ./CreativeMath/AHSME/1994_AHSME_Problems/29.json | AHSME |
1994_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | How many of the following are equal to $x^x+x^x$ for all $x>0$?
$\textbf{I:}\ 2x^x \qquad\textbf{II:}\ x^{2x} \qquad\textbf{III:}\ (2x)^x \qquad\textbf{IV:}\ (2x)^{2x}$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$
| [
"We look at each statement individually.\n\n\n$\\textbf{I:}\\ 2x^x$. We note that $x^x+x^x=x^x(1+1)=2x^x$. So statement $\\textbf{I}$ is true.\n\n\n$\\textbf{II:}\\ x^{2x}$. We find a counter example which is $x=1$. $2\\neq 1$. So statement $\\textbf{II}$ is false.\n\n\n$\\textbf{III:}\\ (2x)^x$. We see that this s... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/3.json | AHSME |
1994_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | In the polygon shown, each side is perpendicular to its adjacent sides, and all 28 of the sides are congruent. The perimeter of the polygon is $56$. The area of the region bounded by the polygon is
[asy] draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); [/asy]
$\textbf{(A)}\ 84 \qquad\textbf{(B)}\ 96 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 112 \qquad\textbf{(E)}\ 196$
| [
"Since the perimeter is $56$ and all of the sides are congruent, the length of each side is $2$. We break the figure into squares as shown below.\n\n\n[asy] unitsize(0.8cm); draw(shift(3,4)*((0,-1)--(1,-1))); draw(shift(3,4)*((-1,-2)--(2,-2))); draw(shift(3,4)*((-2,-3)--(3,-3))); draw(shift(3,4)*((-2,-4)--(3,-4)));... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/8.json | AHSME |
1994_AHSME_Problems | 22 | 0 | Counting | Multiple Choice | Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the six students and decide to choose their chairs so that each professor will be between two students. In how many ways can Professors Alpha, Beta and Gamma choose their chairs?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 36 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 84 \qquad\textbf{(E)}\ 630$
| [
"Since each professor must sit between two students, they cannot be seated in seats $1$ or $9$ (the seats at either end of the row). Hence, each professor has $7$ seats they can choose from and must be at least $1$ seat apart.\n\n\nThis question is equivalent to choosing $3$ seats from a row of $5$ seats with no re... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/22.json | AHSME |
1994_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | Triangle $ABC$ is inscribed in a circle, and $\angle B = \angle C = 4\angle A$. If $B$ and $C$ are adjacent vertices of a regular polygon of $n$ sides inscribed in this circle, then $n=$
[asy] draw(Circle((0,0), 5)); draw((0,5)--(3,-4)--(-3,-4)--cycle); label("A", (0,5), N); label("B", (-3,-4), SW); label("C", (3,-4), SE); dot((0,5)); dot((3,-4)); dot((-3,-4)); [/asy]
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18$
| [
"We solve for $\\angle A$ as follows: \\[4\\angle A+4\\angle A+\\angle A=180\\implies 9\\angle A=180\\implies \\angle A=20.\\] That means that minor arc $\\widehat{BC}$ has measure $40^\\circ$. We can fit a maximum of $\\frac{360}{40}=\\boxed{\\textbf{(C) }9}$ of these arcs in the circle.\n\n\n--Solution by TheMask... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/18.json | AHSME |
1994_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | In the $xy$-plane, the segment with endpoints $(-5,0)$ and $(25,0)$ is the diameter of a circle. If the point $(x,15)$ is on the circle, then $x=$
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20$
| [
"We see that the center of this circle is at $\\left(\\frac{-5+25}{2},0\\right)=(10,0)$. The radius is $\\frac{30}{2}=15$. So the equation of this circle is \\[(x-10)^2+y^2=225.\\] Substituting $y=15$ yields $(x-10)^2=0$ so $x=\\boxed{\\textbf{(A) }10}$.\n\n\n--Solution by TheMaskedMagician\n\n\n",
"The diameter ... | 2 | ./CreativeMath/AHSME/1994_AHSME_Problems/4.json | AHSME |
1994_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | Find the sum of the arithmetic series
\[20+20\frac{1}{5}+20\frac{2}{5}+\cdots+40\]
$\textbf{(A)}\ 3000 \qquad\textbf{(B)}\ 3030 \qquad\textbf{(C)}\ 3150 \qquad\textbf{(D)}\ 4100 \qquad\textbf{(E)}\ 6000$
| [
"\\subsubsection{Brief Introduction}\nFor those that do not know the formula, the sum of an arithmetic series with first term $a_1$, last term $a_n$ as $n$ terms, is \\[S = \\frac{n(a_1+a_n)}{2}.\\] We can prove this as follows:\n\n\nLet $d$ be the common difference between terms of our series and let $n$ be the nu... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/14.json | AHSME |
1994_AHSME_Problems | 15 | 0 | Number Theory | Multiple Choice | For how many $n$ in $\{1, 2, 3, ..., 100 \}$ is the tens digit of $n^2$ odd?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50$
| [
"Let $n=10a+b$. So $n^2=(10a+b)^2=100a^2+20ab+b^2$. The term $100a^2$ only contributes digits starting at the hundreds place, so this does not affect whether the tens digit is odd. The term $20ab$ only contributes digits starting at the tens place, and the tens digit contributed will be the ones digit of $2ab$ whi... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/15.json | AHSME |
1994_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | Pat intended to multiply a number by $6$ but instead divided by $6$. Pat then meant to add $14$ but instead subtracted $14$. After these mistakes, the result was $16$. If the correct operations had been used, the value produced would have been
$\textbf{(A)}\ \text{less than 400} \qquad\textbf{(B)}\ \text{between 400 and 600} \qquad\textbf{(C)}\ \text{between 600 and 800} \\ \textbf{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}$
| [
"We reverse the operations that he did and then use the correct operations. His end result is $16$. Before that, he subtracted $14$ which means that his number after the first operation was $30$. He divided by $6$ so his number was $180$. \n\n\nNow, we multiply $180$ by $6$ to get $1080$. Finally, $1080+14=1094$. S... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/5.json | AHSME |
1994_AHSME_Problems | 19 | 0 | Other | Multiple Choice | Label one disk "$1$", two disks "$2$", three disks "$3$"$, ...,$ fifty disks "$50$". Put these $1+2+3+ \cdots+50=1275$ labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 51 \qquad\textbf{(C)}\ 415 \qquad\textbf{(D)}\ 451 \qquad\textbf{(E)}\ 501$
| [
"We can solve this problem by thinking of the worst case scenario, essentially an adaptation of the Pigeon-hole principle. \nWe can start by picking up all the disks numbered 1 to 9 since even if we have all those disks we won't have 10 of any one disk. This gives us 45 disks. \n\n\nFrom disks numbered from 10 to 5... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/19.json | AHSME |
1994_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | In the $xy$-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at $(0,0), (0,3), (3,3), (3,1), (5,1)$ and $(5,0)$. The slope of the line through the origin that divides the area of this region exactly in half is
[asy] Label l; l.p=fontsize(6); xaxis("$x$",0,6,Ticks(l,1.0,0.5),EndArrow); yaxis("$y$",0,4,Ticks(l,1.0,0.5),EndArrow); draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));[/asy]
$\textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9}$
| [
"Let the vertices be $A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)$. It is easy to see that the line must pass through $CD$. Let the line intersect $CD$ at the point $G=(3,3-x)$ (i.e. the point $x$ units below $C$). Since the quadrilateral $ABCG$ and pentagon $GDEFA$ must have the same area, we have the equation... | 2 | ./CreativeMath/AHSME/1994_AHSME_Problems/23.json | AHSME |
1994_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | If $\angle A$ is four times $\angle B$, and the complement of $\angle B$ is four times the complement of $\angle A$, then $\angle B=$
$\textbf{(A)}\ 10^{\circ} \qquad\textbf{(B)}\ 12^{\circ} \qquad\textbf{(C)}\ 15^{\circ} \qquad\textbf{(D)}\ 18^{\circ} \qquad\textbf{(E)}\ 22.5^{\circ}$
| [
"Let $\\angle A=x$ and $\\angle B=y$. From the first condition, we have $x=4y$. From the second condition, we have \\[90-y=4(90-x).\\] Substituting $x=4y$ into the previous equation and solving yields \\begin{align*}90-y=4(90-4y)&\\implies 90-y=360-16y\\\\&\\implies 15y=270\\\\&\\implies y=\\boxed{\\textbf{(D) }18^... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/9.json | AHSME |
1981_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | A ray of light originates from point $A$ and travels in a plane, being reflected $n$ times between lines $AD$ and $CD$ before striking a point $B$ (which may be on $AD$ or $CD$) perpendicularly and retracing its path back to $A$ (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for $n=3$). If $\measuredangle CDA=8^\circ$, what is the largest value $n$ can have?
[asy] unitsize(1.5cm); pair D=origin, A=(-6,0), C=6*dir(160), E=3.2*dir(160), F=(-2.1,0), G=1.5*dir(160), B=(-1.4095,0); draw((-6.5,0)--D--C,black); draw(A--E--F--G--B,black); dotfactor=4; dot("$A$",A,S); dot("$C$",C,N); dot("$R_1$",E,N); dot("$R_2$",F,S); dot("$R_3$",G,N); dot("$B$",B,S); markscalefactor=0.015; draw(rightanglemark(G,B,D)); draw(anglemark(C,E,A,12)); draw(anglemark(F,E,G,12)); draw(anglemark(E,F,A)); draw(anglemark(E,F,A,12)); draw(anglemark(B,F,G)); draw(anglemark(B,F,G,12)); draw(anglemark(E,G,F)); draw(anglemark(E,G,F,12)); draw(anglemark(E,G,F,16)); draw(anglemark(B,G,D)); draw(anglemark(B,G,D,12)); draw(anglemark(B,G,D,16)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.}$
| [
"Notice that when we start, we want the smallest angle possible of reflection. The ideal reflection would be $0$, but that would be impossible. Therefore we start by working backwards. Since angle $CDA$ is $8$, the reflection would give us a triangle with angles $16, 90$, and $74$. Then, when we reflect again, we w... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/20.json | AHSME |
1981_AHSME_Problems | 16 | 0 | Number Theory | Multiple Choice | The base three representation of $x$ is
\[12112211122211112222\]
The first digit (on the left) of the base nine representation of $x$ is
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
| [
"Convert $x$ to base 10 then convert the result to base 9.\n\\[12112211122211112222_{3} = 2150029898\\]\n\n\n\\[2150029898 = 5484584488_{9}\\]\n\n\nTherefore, the answer is $\\textbf{(E)}\\ 5.$\n\n\n-edited by coolmath34\n\n\n",
"Every 2 numbers in base 3 represents 1 number in base 9. \nThe first 2 numbers on th... | 2 | ./CreativeMath/AHSME/1981_AHSME_Problems/16.json | AHSME |
1981_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $\frac{x}{x-1} = \frac{y^2 + 2y - 1}{y^2 + 2y - 2},$ then $x$ equals
$\text{(A)} \quad y^2 + 2y - 1$
$\text{(B)} \quad y^2 + 2y - 2$
$\text{(C)} \quad y^2 + 2y + 2$
$\text{(D)} \quad y^2 + 2y + 1$
$\text{(E)} \quad -y^2 - 2y + 1$
| [
"We can cross multiply both sides of the equation and simplify.\n\\[x(y^2 + 2y - 2) = (x-1)(y^2 + 2y - 1)\\]\n\\[xy^2 + 2xy - 2x = xy^2 + 2xy - x - y^2 - 2y + 1\\]\n\\[-x = -y^2 - 2y + 1\\]\n\\[x = y^2 + 2y - 1\\]\n\n\nThe answer is $\\text{A.}$\n\n\n-edited by coolmath34\n\n\n"
] | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/6.json | AHSME |
1981_AHSME_Problems | 7 | 0 | Number Theory | Multiple Choice | How many of the first one hundred positive integers are divisible by all of the numbers $2$, $3$, $4$, and $5$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$
| [
"The least common multiple of 2, 3, 4, and 5 is $2^2 \\cdot 3 \\cdot 5 = 60.$ There is only one multiple of 60 between 0 and 100, so the answer is $\\textbf{(B)}\\ 1.$\n\n\n-edited by coolmath34\n\n\n"
] | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/7.json | AHSME |
1981_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | The function $f$ is not defined for $x=0$, but, for all non-zero real numbers $x$, $f(x)+2f\left(\dfrac{1}x\right)=3x$. The equation $f(x)=f(-x)$ is satisfied by
$\textbf{(A)}\ \text{exactly one real number} \qquad \textbf{(B)}\ \text{exactly two real numbers} \qquad\textbf{(C)}\ \text{no real numbers}\qquad \\ \textbf{(D)}\ \text{infinitely many, but not all, non-zero real numbers} \qquad\textbf{(E)}\ \text{all non-zero real numbers}$
| [
"Substitute $x$ with $\\frac{1}{x}$:\n\n\n$f(\\frac{1}{x})+2f(x)=\\frac{3}{x}$.\n\n\nAdding this to $f(x)+2f\\left(\\dfrac{1}x\\right)=3x$, we get\n\n\n$3f(x)+3f\\left(\\dfrac{1}x\\right)=3x+\\frac{3}{x}$, or\n\n\n$f(x)+f\\left(\\dfrac{1}x\\right)=x+\\frac{1}{x}$.\n\n\nSubtracting this from $f(\\frac{1}{x})+2f(x)=\... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/17.json | AHSME |
1981_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | In a triangle with sides of lengths $a$, $b$, and $c$, $(a+b+c)(a+b-c) = 3ab$. The measure of the angle opposite the side length $c$ is
$\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ$
| [
"We will try to solve for a possible value of the variables. First notice that exchanging $a$ for $b$ in the original equation must also work. Therefore, $a=b$ works. Replacing $b$ for $a$ and expanding/simplifying in the original equation yields $4a^2-c^2=3a^2$, or $a^2=c^2$. Since $a$ and $c$ are positive, $a=c$.... | 2 | ./CreativeMath/AHSME/1981_AHSME_Problems/21.json | AHSME |
1981_AHSME_Problems | 26 | 0 | Probability | Multiple Choice | Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is $\frac{1}{6}$, independent of the outcome of any other toss.)
$\textbf{(A) } \frac{1}{3} \textbf{(B) } \frac{2}{9} \textbf{(C) } \frac{5}{18} \textbf{(D) } \frac{25}{91} \textbf{(E) } \frac{36}{91}$
| [
"The probability that Carol wins during the first cycle through is $\\frac{5}{6}*\\frac{5}{6}*\\frac{1}{6}$, and the probability that Carol wins on the second cycle through is $\\frac{5}{6}*\\frac{5}{6}*\\frac{5}{6}*\\frac{5}{6}*\\frac{5}{6}*\\frac{1}{6}$. It is clear that this is an infinite geometric sequence, an... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/26.json | AHSME |
1981_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | If $a$, $b$, $c$, and $d$ are the solutions of the equation $x^4 - bx - 3 = 0$, then an equation whose solutions are
\[\dfrac {a + b + c}{d^2}, \dfrac {a + b + d}{c^2}, \dfrac {a + c + d}{b^2}, \dfrac {b + c + d}{a^2}\]is
$\textbf{(A)}\ 3x^4 + bx + 1 = 0\qquad \textbf{(B)}\ 3x^4 - bx + 1 = 0\qquad \textbf{(C)}\ 3x^4 + bx^3 - 1 = 0\qquad \\\textbf{(D)}\ 3x^4 - bx^3 - 1 = 0\qquad \textbf{(E)}\ \text{none of these}$
| [
"Using Vieta's formula, we know the sum of the roots is equal to the negative coefficient of the $x^3$ term. Since the coefficient is 0, $a+b+c+d=0$. Thus, $\\frac{a+b+c}{d^2}$ can be rewritten as $\\frac{-d}{d^2}=\\frac{1}{-d}$. Similarly, the other three new roots can be written as $\\frac{1}{-c}$, $\\frac{1}{-b}... | 2 | ./CreativeMath/AHSME/1981_AHSME_Problems/30.json | AHSME |
1981_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If $\sqrt{x+2}=2$, then $(x+2)^2$ equals:
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16$
| [
"If we square both sides of the $\\sqrt{x+2} = 2$, we will get $x+2 = 4$, if we square that again, we get $(x+2)^2 = \\boxed{\\textbf{(E) }16}$\n\n\n",
"We can immediately get that $x = 2$, after we square $(2+2)$, we get $\\boxed{\\textbf{(E) }16}$\n\n\n"
] | 2 | ./CreativeMath/AHSME/1981_AHSME_Problems/1.json | AHSME |
1981_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | The three sides of a right triangle have integral lengths which form an arithmetic progression. One of the sides could have length
$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 81\qquad\textbf{(D)}\ 91\qquad\textbf{(E)}\ 361$
| [
"Let the three sides be $a-d,$ $a,$ and $a+d.$ Because of the Pythagorean Theorem,\n\\[(a-d)^2 + a^2 = (a+d)^2\\]\nThis can be simplified to\n\\[a(a-4d) = 0.\\]\n\n\nSo, $a$ is a multiple of $4d$ and the triangle has sides $3d, 4d, 5d.$ We check the answer choices for anything divisible by 3, 4, or 5. The only one... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/11.json | AHSME |
1981_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | Point $E$ is on side $AB$ of square $ABCD$. If $EB$ has length one and $EC$ has length two, then the area of the square is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5$
| [
"Note that $\\triangle BCE$ is a right triangle. Thus, we do Pythagorean theorem to find that side $BC=\\sqrt{3}$. Since this is the side length of the square, the area of $ABCD$ is $\\boxed{\\textbf{(C)}\\ 3}$. \n\n\n~superagh\n\n\n"
] | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/2.json | AHSME |
1981_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | Consider the set of all equations $x^3 + a_2x^2 + a_1x + a_0 = 0$, where $a_2$, $a_1$, $a_0$ are real constants and $|a_i| < 2$ for $i = 0,1,2$. Let $r$ be the largest positive real number which satisfies at least one of these equations. Then
$\textbf{(A)}\ 1 < r < \dfrac{3}{2}\qquad \textbf{(B)}\ \dfrac{3}{2} < r < 2\qquad \textbf{(C)}\ 2 < r < \dfrac{5}{2}\qquad \textbf{(D)}\ \dfrac{5}{2} < r < 3\qquad \\ \textbf{(E)}\ 3 < r < \dfrac{7}{2}$
| [
"Since $x^3 = -(a_2x^2 + a_1x + a_0)$ and $x$ will be as big as possible, we need $x^3$ to be as big as possible, which means $a_2x^2 + a_1x + a_0$ is as small as possible. Since $x$ is positive (according to the options), it makes sense for all of the coefficients to be $-2$.\n\n\nEvaluating $f(\\frac{5}{2})$ give... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/28.json | AHSME |
1981_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | If $p$, $q$, and $M$ are positive numbers and $q<100$, then the number obtained by increasing $M$ by $p\%$ and decreasing the result by $q\%$ exceeds $M$ if and only if
$\textbf{(A)}\ p>q \qquad\textbf{(B)}\ p>\dfrac{q}{100-q}\qquad\textbf{(C)}\ p>\dfrac{q}{1-q}\qquad \textbf{(D)}\ p>\dfrac{100q}{100+q}\qquad\textbf{(E)}\ p>\dfrac{100q}{100-q}$
| [
"Answer Choice $A$: It is obviously incorrect because if $M$ is $50$ and we increase by $50$% and then decrease $49$%, $M$ will be around $37$. \n\n\nAnswer Choice $B$: If $p$ is $100$ and $q$ is $50$, it should be equal but instead we get $100$ is more than $1$. This is therefore also incorrect.\n\n\nAnswer Choice... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/12.json | AHSME |
1981_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | If $\theta$ is a constant such that $0 < \theta < \pi$ and $x + \dfrac{1}{x} = 2\cos{\theta}$, then for each positive integer $n$, $x^n + \dfrac{1}{x^n}$ equals
$\textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta$
| [
"Multiply both sides by $x$ and rearrange to $x^2-2x\\cos(\\theta)+1=0$. Using the quadratic equation, we can solve for $x$. After some simplifying:\n\n\n\\[x=\\cos(\\theta) + \\sqrt{\\cos^2(\\theta)-1}\\]\n\\[x=\\cos(\\theta) + \\sqrt{(-1)(\\sin^2(\\theta))}\\]\n\\[x=\\cos(\\theta) + i\\sin(\\theta)\\]\n\n\nSubsti... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/24.json | AHSME |
1981_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | In $\triangle ABC$ in the adjoining figure, $AD$ and $AE$ trisect $\angle BAC$. The lengths of $BD$, $DE$ and $EC$ are $2$, $3$, and $6$, respectively. The length of the shortest side of $\triangle ABC$ is
[asy] defaultpen(linewidth(.8pt)); pair A = (0,11); pair B = (2,0); pair D = (4,0); pair E = (7,0); pair C = (13,0); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,S); label("$2$",midpoint(B--D),N); label("$3$",midpoint(D--E),NW); label("$6$",midpoint(E--C),NW); draw(A--B--C--cycle); draw(A--D); draw(A--E); [/asy]
$\textbf{(A)}\ 2\sqrt{10}\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 6\sqrt{6}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ \text{not uniquely determined by the given information}$
| [
"Let $AC=b$, $AB=c$, $AD=d$, and $AE=e$. Then, by the Angle Bisector Theorem, $\\frac{c}{e}=\\frac{2}{3}$ and $\\frac{d}{b}=\\frac12$, thus $e=\\frac{3c}2$ and $d=\\frac b2$.\n\n\nAlso, by Stewart’s Theorem, $198+11d^2=2b^2+9c^2$ and $330+11e^2=5b^2+6c^2$. Therefore, we have the following system of equations using ... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/25.json | AHSME |
1981_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | If $a > 1$, then the sum of the real solutions of
$\sqrt{a - \sqrt{a + x}} = x$
is equal to
$\textbf{(A)}\ \sqrt{a} - 1\qquad \textbf{(B)}\ \dfrac{\sqrt{a}- 1}{2}\qquad \textbf{(C)}\ \sqrt{a - 1}\qquad \textbf{(D)}\ \dfrac{\sqrt{a - 1}}{2}\qquad \textbf{(E)}\ \dfrac{\sqrt{4a- 3} - 1}{2}$
| [
"A solution is available here. Pull up find, and put in \"Since x is the principal\", and you will arrive at the solution.\n\n\nIt's not super clear, and there's some black stuff over it, but its legible. \n\n\nThe solution in the above file/pdf is the following. I tried my best to match it verbatim, but I had to ... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/29.json | AHSME |
1981_AHSME_Problems | 3 | 0 | Number Theory | Multiple Choice | What is the least common multiple of ${\frac{1}{x}}$, $\frac{1}{2x}$, and $\frac{1}{3x}$ is $\frac{1}{6x}$?
| [
"The least common multiple of ${\\frac{1}{x}}$, $\\frac{1}{2x}$, and $\\frac{1}{3x}$ is $\\frac{1}{6x}$. \n\n\n$\\frac{1}{x}$ = $\\frac{6}{6x}$, $\\frac{1}{2x}$ = $\\frac{3}{6x}$, $\\frac{1}{3x}$ = $\\frac{2}{6x}$.\n\n\n$\\frac{6}{6x}$ + $\\frac{3}{6x}$ + $\\frac{2}{6x}$ = $\\frac{11}{6x}$\n\n\nThe answer is $\\box... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/3.json | AHSME |
1981_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | For all positive numbers $x$, $y$, $z$, the product
\[(x+y+z)^{-1}(x^{-1}+y^{-1}+z^{-1})(xy+yz+xz)^{-1}[(xy)^{-1}+(yz)^{-1}+(xz)^{-1}]\] equals
$\textbf{(A)}\ x^{-2}y^{-2}z^{-2}\qquad\textbf{(B)}\ x^{-2}+y^{-2}+z^{-2}\qquad\textbf{(C)}\ (x+y+z)^{-1}\qquad \textbf{(D)}\ \dfrac{1}{xyz}\qquad \\ \textbf{(E)}\ \dfrac{1}{xy+yz+xz}$
| [
"Start simplifying:\n\\[(\\frac{1}{x+y+z})(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z})(\\frac{1}{xy+yz+xz})(\\frac{1}{xy}+\\frac{1}{xz}+\\frac{1}{yz})\\]\n\n\n\\[(\\frac{1}{x+y+z})(\\frac{xy+yz+xz}{xyz})(\\frac{1}{xy+yz+xz})(\\frac{x+y+z}{xyz})\\]\n\n\nThe $xy+yz+xz$ and $x+y+z$ cancel out:\n\n\n\\[\\frac{1}{(xyz)^2}\\... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/8.json | AHSME |
1981_AHSME_Problems | 22 | 0 | Counting | Multiple Choice | How many lines in a three dimensional rectangular coordinate system pass through four distinct points of the form $(i, j, k)$, where $i$, $j$, and $k$ are positive integers not exceeding four?
$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 64\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 100$
| [
"No solutions yet!\ninputting...\n\n\n"
] | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/22.json | AHSME |
1981_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | The number of real solutions to the equation \[\dfrac{x}{100}=\sin x\] is
$\textbf{(A)}\ 61\qquad\textbf{(B)}\ 62\qquad\textbf{(C)}\ 63\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 65$
| [
"The answer to this problem is the number of intersections between the graph of $f(x) = \\sin x$ and $f(x) = \\frac{1}{100}x.$ We can do the right side of the coordinate plane first. Each cycle of the sine wave, consisting of 2π, will have 2 intersections (From the positive part of the sine wave) The line $f(x) = \... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/18.json | AHSME |
1981_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | If three times the larger of two numbers is four times the smaller and the difference between the numbers is 8, the the larger of two numbers is:
$\text{(A)}\quad 16 \qquad \text{(B)}\quad 24 \qquad \text{(C)}\quad 32 \qquad \text{(D)}\quad 44 \qquad \text{(E)} \quad 52$
| [
"Let the smaller number be $x$ and the larger number be $y.$ We can use the information given in the problem to write two equations:\n\n\n1) Three times the larger of two numbers is four times the smaller.\n\\[3y = 4x\\]\n\n\n2) The difference between the number is 8.\n\n\n\\[y - x = 8\\]\n\n\nSolving this system o... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/4.json | AHSME |
1981_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | In a geometric sequence of real numbers, the sum of the first $2$ terms is $7$, and the sum of the first $6$ terms is $91$. The sum of the first $4$ terms is
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 84$
| [
"Denote the sum of the first $2$ terms as $x$. Since we know that the sum of the first $6$ terms is $91$ which is $7 \\cdot 13$, we have $x$ + $xy$ + $xy^2$ = $13x$ because it is a geometric series. We can quickly see that $y$ = $3$, and therefore, the sum of the first $4$ terms is $4x = 4 \\cdot 7 = \\boxed {(A) 2... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/14.json | AHSME |
1981_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | If $b>1$, $x>0$, and $(2x)^{\log_b 2}-(3x)^{\log_b 3}=0$, then $x$ is
$\textbf{(A)}\ \dfrac{1}{216}\qquad\textbf{(B)}\ \dfrac{1}{6}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \text{not uniquely determined}$
| [
"$\\boxed {B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/15.json | AHSME |
1981_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | In trapezoid $ABCD$, sides $AB$ and $CD$ are parallel, and diagonal $BD$ and side $AD$ have equal length. If $m\angle DCB=110^\circ$ and $m\angle CBD=30^\circ$, then $m\angle ADB=$
$\textbf{(A)}\ 80^\circ\qquad\textbf{(B)}\ 90^\circ\qquad\textbf{(C)}\ 100^\circ\qquad\textbf{(D)}\ 110^\circ\qquad\textbf{(E)}\ 120^\circ$
| [
"Draw the diagram using the information above. In triangle $DCB,$ note that $m\\angle DCB=110^\\circ$ and $m\\angle CBD=30^\\circ$, so $m\\angle CDB=40^\\circ.$\n\n\nBecause $AB \\parallel CD,$ we have $m\\angle CDB= m\\angle DBA = 40^\\circ.$ Triangle $DAB$ is isosceles, so $m\\angle ADB = 180 - 2(40) = 100^\\circ... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/5.json | AHSME |
1981_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, and $BN\perp AN$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then find $MN$.
[asy] size(150); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=14*dir(42), C=intersectionpoint(B--(30,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b); draw(N--B--A--N--M--C--A^^B--M); markscalefactor=0.1; draw(rightanglemark(B,N,A)); pair point=N; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(30)); label(rotate(angle(dir(A--C)))*"$19$", A--C, dir(A--C)*dir(90)); label(rotate(angle(dir(A--B)))*"$14$", A--B, dir(A--B)*dir(90)); [/asy]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ \dfrac{5}{2}\qquad\textbf{(C)}\ \dfrac{5}{2}-\sin\theta\qquad\textbf{(D)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\theta\qquad\textbf{(E)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\left(\dfrac{1}{2}\theta\right)$
| [
"Extend $BN$ to meet $AC$ at $Q$. Then $\\triangle BNM \\sim \\triangle BQC$, so $BN=NQ$ and $QC=19-AQ=2MN$. \n\n\nSince $\\angle ANB=90^\\circ = \\angle ANQ$, $\\angle BAN=\\angle NAQ$ (since $AN$ is an angle bisector) and $\\triangle ANB$ and $\\triangle ANQ$ share side $AN$, $\\triangle ANB \\cong \\triangle ANQ... | 1 | ./CreativeMath/AHSME/1981_AHSME_Problems/19.json | AHSME |
1971_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | The sum of the squares of the roots of the equation $x^2+2hx=3$ is $10$. The absolute value of $h$ is equal to
$\textbf{(A) }-1\qquad \textbf{(B) }\textstyle\frac{1}{2}\qquad \textbf{(C) }\textstyle\frac{3}{2}\qquad \textbf{(D) }2\qquad \textbf{(E) }\text{None of these}$
| [
"We can rewrite the equation as $x^2 + 2hx - 3 = 0.$ By Vieta's Formulas, the sum of the roots is $-2h$ and the product of the roots is $-3.$\n\n\nLet the two roots be $r$ and $s.$ Note that\n\\[r^2 + s^2 = (r+s)^2 - 2rs = (-2h)^2 -2(-3)\\]\n\n\nTherefore, $4h^2 + 6 = 10$ and $h = \\pm 1.$ This doesn't match any of... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/20.json | AHSME |
1971_AHSME_Problems | 16 | 0 | Arithmetic | Multiple Choice | After finding the average of $35$ scores, a student carelessly included the average with the $35$ scores and found the
average of these $36$ numbers. The ratio of the second average to the true average was
$\textbf{(A) }1:1\qquad \textbf{(B) }35:36\qquad \textbf{(C) }36:35\qquad \textbf{(D) }2:1\qquad \textbf{(E) }\text{None of these}$
| [
"Assume the $35$ scores are the first $35$ natural numbers: $1, 2, 3, \\dots 35.$\n\n\nThe average of the scores is $\\frac{\\frac{(35)(36)}{2}}{35} = 18.$\n\n\nIf we add $18$ to the first $35$ numbers, the new average is $\\frac{648}{36} = 18$ still.\n\n\nThe two averages are the same, therefore the answer is $\\t... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/16.json | AHSME |
1971_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | Let $\ast$ be the symbol denoting the binary operation on the set $S$ of all non-zero real numbers as follows:
For any two numbers $a$ and $b$ of $S$, $a\ast b=2ab$. Then the one of the following statements which is not true, is
$\textbf{(A) }\ast\text{ is commutative over }S \qquad \textbf{(B) }\ast\text{ is associative over }S\qquad \\ \textbf{(C) }\frac{1}{2}\text{ is an identity element for }\ast\text{ in }S\qquad \textbf{(D) }\text{Every element of }S\text{ has an inverse for }\ast\qquad \textbf{(E) }\dfrac{1}{2a}\text{ is an inverse for }\ast\text{ of the element }a\text{ of }S$
| [
"$\\textbf{(A) }\\ast\\text{ is commutative over }S$\n\\[a \\ast b = b \\ast a = 2ab\\]\nStatement A is true.\n\n\n\n\n\n\n$\\textbf{(B) }\\ast\\text{ is associative over }S$\n\\[a \\ast (b \\ast c) = a \\ast 2bc = 2abc\\]\n\\[(a \\ast b) \\ast c = 2ab \\ast c = 2abc\\]\nStatement B is true.\n\n\n\n\n$\\textbf{(C) ... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/6.json | AHSME |
1971_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | $2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}$ is equal to
$\textbf{(A) }2^{-2k}\qquad \textbf{(B) }2^{-(2k-1)}\qquad \textbf{(C) }-2^{-(2k+1)}\qquad \textbf{(D) }0\qquad \textbf{(E) }2$
| [
"$Let\\ x\\ equal\\ 2^{-2k}\\ \\\\* From\\ this\\ we\\ get \\frac{x}{2}-(\\frac{-x}{\\frac{-1}{2}})+x\\ by\\ using\\ power\\ rule.\\ \\\\*Now\\ we\\ can\\ see\\ this\\ simplies\\ to\\ \\frac{-x}{2}\\ \\\\*Looking\\ at\\ \\frac{x}{2} we\\ can\\ clearly\\ see\\ that\\ -(\\frac{x}{2}) is\\ equal\\ to\\ -2^{-(2k+1)}\\... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/7.json | AHSME |
1971_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | A circular disk is divided by $2n$ equally spaced radii($n>0$) and one secant line. The maximum number of non-overlapping
areas into which the disk can be divided is
$\textbf{(A) }2n+1\qquad \textbf{(B) }2n+2\qquad \textbf{(C) }3n-1\qquad \textbf{(D) }3n\qquad \textbf{(E) }3n+1$
| [
"We can draw the cases for small values of $n.$\n\\[n = 0 \\rightarrow \\text{areas} = 1\\]\n\\[n = 1 \\rightarrow \\text{areas} = 4\\]\n\\[n = 2 \\rightarrow \\text{areas} = 7\\]\n\\[n = 3 \\rightarrow \\text{areas} = 10\\]\nIt seems that for $2n$ radii, there are $3n+1$ distinct areas. The secant line must pass t... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/17.json | AHSME |
1971_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | If $\log_2(\log_3(\log_4 x))=\log_3(\log_4(\log_2 y))=\log_4(\log_2(\log_3 z))=0$, then the sum $x+y+z$ is equal to
$\textbf{(A) }50\qquad \textbf{(B) }58\qquad \textbf{(C) }89\qquad \textbf{(D) }111\qquad \textbf{(E) }1296$
| [
"If $\\log_{x}{n}=0,$ then $n = 1.$ So, we can rewrite this equation:\n\\[(\\log_3(\\log_4 x))=(\\log_4(\\log_2 y))=(\\log_2(\\log_3 z))=1\\]\n\n\nSolve individually for each variable.\n\\[x = 4^3 = 64\\]\n\\[y = 2^4 = 16\\]\n\\[z = 3^2 = 9\\]\nTherefore, $x + y + z = 89.$\n\n\nThe answer is $\\textbf{(C)}.$\n\n\n-... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/21.json | AHSME |
1971_AHSME_Problems | 10 | 0 | Other | Multiple Choice | Each of a group of $50$ girls is blonde or brunette and is blue eyed of brown eyed. If $14$ are blue-eyed blondes,
$31$ are brunettes, and $18$ are brown-eyed, then the number of brown-eyed brunettes is
$\textbf{(A) }5\qquad \textbf{(B) }7\qquad \textbf{(C) }9\qquad \textbf{(D) }11\qquad \textbf{(E) }13$
| [
"There are $31$ brunettes, so there are $50-31=19$ blondes. We know there are $14$ blue-eyed blondes, so there are $19-14=5$ brown-eyed blondes.\n\n\nNext, we know that $18$ people are brown-eyed, so there are $18-5=13$ brown-eyed brunettes.\n\n\nThe answer is $\\textbf{(E)} \\quad 13.$\n\n\n-edited by coolmath34\n... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/10.json | AHSME |
1971_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | [asy] size(2.5inch); pair A, B, C, E, F, G; A = (0,3); B = (-1,0); C = (3,0); E = (0,0); F = (1,2); G = intersectionpoint(B--F,A--E); draw(A--B--C--cycle); draw(A--E); draw(B--F); label("$A$",A,N); label("$B$",B,W); label("$C$",C,dir(0)); label("$E$",E,S); label("$F$",F,NE); label("$G$",G,SE); //Credit to chezbgone2 for the diagram[/asy]
In $\triangle ABC$, point $F$ divides side $AC$ in the ratio $1:2$. Let $E$ be the point of intersection of side $BC$ and $AG$ where $G$ is the
midpoints of $BF$. The point $E$ divides side $BC$ in the ratio
$\textbf{(A) }1:4\qquad \textbf{(B) }1:3\qquad \textbf{(C) }2:5\qquad \textbf{(D) }4:11\qquad \textbf{(E) }3:8$
| [
"We will use mass points to solve this problem. $AC$ is in the ratio $1:2,$ so we will assign a mass of $2$ to point $A,$ a mass of $1$ to point $C,$ and a mass of $3$ to point $F.$\n\n\nWe also know that $G$ is the midpoint of $BF,$ so $BG:GF=1:1.$ $F$ has a mass of $3,$ so $B$ also has a mass of $3.$\n\n\nIn line... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/26.json | AHSME |
1971_AHSME_Problems | 31 | 0 | Geometry | Multiple Choice | [asy] size(2.5inch); pair A = (-2,0), B = 2dir(150), D = (2,0), C; draw(A..(0,2)..D--cycle); C = intersectionpoint(A..(0,2)..D,Circle(B,arclength(A--B))); draw(A--B--C--D--cycle); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,N); label("$D$",D,E); label("$4$",A--D,S); label("$1$",A--B,E); label("$1$",B--C,SE); //Credit to chezbgone2 for the diagram[/asy]
Quadrilateral $ABCD$ is inscribed in a circle with side $AD$, a diameter of length $4$. If sides $AB$ and $BC$ each have length $1$,
then side $CD$ has length
$\textbf{(A) }\frac{7}{2}\qquad \textbf{(B) }\frac{5\sqrt{2}}{2}\qquad \textbf{(C) }\sqrt{11}\qquad \textbf{(D) }\sqrt{13}\qquad \textbf{(E) }2\sqrt{3}$
Solution
| [
"Note that the length 4 forms a semicircle. We can then use the Law of Cosines. Take the center and form a line segment with the other two points.\n\n\nLet's find the cosine of angle $AOB$. The cosine of that angle is 7/8. We use the double cosine angle to find angle $AOC$: \n\n\n$cos(2x) = 2 cos^2 (x)- 1$\n\n\nThe... | 2 | ./CreativeMath/AHSME/1971_AHSME_Problems/31.json | AHSME |
1971_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips, and at most one third the number of red chips. The number which are white or blue is at least $55$. The minimum number of red chips is
$\textbf{(A) }24\qquad \textbf{(B) }33\qquad \textbf{(C) }45\qquad \textbf{(D) }54\qquad \textbf{(E) }57$
| [
"Let the number of white be $2x$. The number of blue is then $x-y$ for some constant $y$. So we want $2x+x-y=55\\rightarrow 3x-y=55$. We take mod 3 to find y. $55=1\\pmod{3}$, so blue is 1 more than half white. The number of whites is then 36, and the number of blues is 19. So $19*3=\\boxed{57}$\n\n\n\n\n~yofro\n\n... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/27.json | AHSME |
1971_AHSME_Problems | 1 | 0 | Arithmetic | Multiple Choice | The number of digits in the number $N=2^{12}\times 5^8$ is
$\textbf{(A) }9\qquad \textbf{(B) }10\qquad \textbf{(C) }11\qquad \textbf{(D) }12\qquad \textbf{(E) }20$
| [
"$N=2^{12}\\cdot5^8=2^4\\cdot2^8\\cdot5^8=2^4\\cdot(2\\cdot5)^8=2^4\\cdot10^8=16\\cdot100000000$, which has $2+8=10$ digits, so the answer is $(\\textbf{B})$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/1.json | AHSME |
1971_AHSME_Problems | 11 | 0 | Number Theory | Multiple Choice | The numeral $47$ in base a represents the same number as $74$ in base $b$. Assuming that both bases are positive
integers, the least possible value of $a+b$ written as a Roman numeral, is
$\textbf{(A) }\mathrm{XIII}\qquad \textbf{(B) }\mathrm{XV}\qquad \textbf{(C) }\mathrm{XXI}\qquad \textbf{(D) }\mathrm{XXIV}\qquad \textbf{(E) }\mathrm{XVI}$
| [
"$4a+7=7b+4\\implies 7b=4a+3\\implies b=\\frac{4a+3}{7}$. $4a+3\\equiv 0\\implies 4a\\equiv 4\\implies a\\equiv1 \\pmod{7}$. The smallest possible value of $a$ is $8$. Then, $b=\\frac{4\\cdot8+3}{7}=\\frac{35}{7}=5$. However, the digit $7$ is not valid in base $5$, so we have to try a larger value. $a=8+7=15$, give... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/11.json | AHSME |
1971_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | If $b$ men take $c$ days to lay $f$ bricks, then the number of days it will take $c$ men working at the same rate to lay $b$ bricks, is
$\textbf{(A) }fb^2\qquad \textbf{(B) }b/f^2\qquad \textbf{(C) }f^2/b\qquad \textbf{(D) }b^2/f\qquad \textbf{(E) }f/b^2$
| [
"We can use a modified version of the equation $\\text{Distance} = \\text{Rate} \\times {\\text{Time}}$, which is $\\text{Work Done} = \\text{Rate of Work} \\times{ \\text{Time Worked}}$. In this case, the work done is the number of bricks laid, the rate of work is the number of men working, and the time worked is ... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/2.json | AHSME |
1971_AHSME_Problems | 33 | 0 | Algebra | Multiple Choice | If $P$ is the product of $n$ quantities in Geometric Progression, $S$ their sum, and $S'$ the sum of their reciprocals,
then $P$ in terms of $S, S'$, and $n$ is
$\textbf{(A) }(SS')^{\frac{1}{2}n}\qquad \textbf{(B) }(S/S')^{\frac{1}{2}n}\qquad \textbf{(C) }(SS')^{n-2}\qquad \textbf{(D) }(S/S')^n\qquad \textbf{(E) }(S/S')^{\frac{1}{2}(n-1)}$
| [
"We can just look at a very specific case: $1, 2, 4, 8.$ Here, $n=4, P=64, S=15,$ and $S'=\\frac{30}{16}=\\frac{15}{8}.$\n\n\nThen, plug in values of $S, S',$ and $n$ into each of the answer choices and see if it matches the product. \n\n\nAnswer choice $\\textbf{(B)}$ works: $(\\frac{15}{\\frac{15}{8}})^2 = 64.$\n... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/33.json | AHSME |
1971_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | Given the progression $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$.
The least positive integer $n$ such that the product of the first $n$ terms of the progression exceeds $100,000$ is
$\textbf{(A) }7\qquad \textbf{(B) }8\qquad \textbf{(C) }9\qquad \textbf{(D) }10\qquad \textbf{(E) }11$
| [
"The product of the sequence $10^{\\dfrac{1}{11}}, 10^{\\dfrac{2}{11}}, 10^{\\dfrac{3}{11}}, 10^{\\dfrac{4}{11}},\\dots , 10^{\\dfrac{n}{11}}$ is equal to $10^{\\dfrac{1}{11}+\\frac{2}{11}\\dots\\frac{n}{11}}$ since we are looking for the smallest value $n$ that will create $100,000$, or $10^5$. From there, we can ... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/29.json | AHSME |
1971_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | If the point $(x,-4)$ lies on the straight line joining the points $(0,8)$ and $(-4,0)$ in the $xy$-plane, then $x$ is equal to
$\textbf{(A) }-2\qquad \textbf{(B) }2\qquad \textbf{(C) }-8\qquad \textbf{(D) }6\qquad \textbf{(E) }-6$
| [
"Since $(x,-4)$ is on the line $(0,8)$ and $(-4,0)$, the slope between the latter two is the same as the slope between the former two, so $\\frac{-4-8}{x-0}=\\frac{8-0}{0-(-4)}$. Solving for $x$, we get $x=-6$, so the answer is $\\textbf{(E)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/3.json | AHSME |
1971_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | The solution set of $6x^2+5x<4$ is the set of all values of $x$ such that
$\textbf{(A) }\textstyle -2<x<1\qquad \textbf{(B) }-\frac{4}{3}<x<\frac{1}{2}\qquad \textbf{(C) }-\frac{1}{2}<x<\frac{4}{3}\qquad \\ \textbf{(D) }x<\textstyle\frac{1}{2}\text{ or }x>-\frac{4}{3}\qquad \textbf{(E) }x<-\frac{4}{3}\text{ or }x>\frac{1}{2}$
| [
"We are solving the inequality $6x^2 + 5x - 4 < 0.$ This can be factored as\n\\[(2x-1)(3x+4) < 0\\]\n\n\nThe graph of this inequality is a parabola facing upwards, so the area between the roots satisfies the equation. The solution is $x \\in [-\\frac{4}{3}, \\frac{1}{2}]$ and the answer is $\\textbf{(C)}.$\n\n\n-... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/8.json | AHSME |
1971_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | After simple interest for two months at $5$% per annum was credited, a Boy Scout Troop had a total
of $\textdollar 255.31$ in the Council Treasury. The interest credited was a number of dollars plus the following number of cents
$\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }21\qquad \textbf{(E) }31$
| [
"We can use the formula $A = P(1+rt)$ for simple interest, where $P$ is the principal (initial amount), $A$ is the total, $r$ is the interest rate per year, and $t$ is the time in years.\n\n\nWe are solving for $P,$ so we can rearrange the equation:\n\\[P = \\frac{A}{1+rt}\\]\nPlug in all the necessary numbers:\n\\... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/4.json | AHSME |
1971_AHSME_Problems | 14 | 0 | Number Theory | Multiple Choice | The number $(2^{48}-1)$ is exactly divisible by two numbers between $60$ and $70$. These numbers are
$\textbf{(A) }61,63\qquad \textbf{(B) }61,65\qquad \textbf{(C) }63,65\qquad \textbf{(D) }63,67\qquad \textbf{(E) }67,69$
| [
"Factor.\n\\[2^{48}-1 = (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)\\]\n\n\nWe only care about two terms: $2^{6}+1$ and $(2^{3}+1)(2^{3}-1)$. These simplify to $65$ and $63.$\n\n\nThe answer is $\\textbf{(C)}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/14.json | AHSME |
1971_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | Points $A,B,Q,D$, and $C$ lie on the circle shown and the measures of arcs $\widehat{BQ}$ and $\widehat{QD}$
are $42^\circ$ and $38^\circ$ respectively. The sum of the measures of angles $P$ and $Q$ is
$\textbf{(A) }80^\circ\qquad \textbf{(B) }62^\circ\qquad \textbf{(C) }40^\circ\qquad \textbf{(D) }46^\circ\qquad \textbf{(E) }\text{None of these}$
[asy] size(3inch); draw(Circle((1,0),1)); pair A, B, C, D, P, Q; P = (-2,0); B=(sqrt(2)/2+1,sqrt(2)/2); D=(sqrt(2)/2+1,-sqrt(2)/2); Q = (2,0); A = intersectionpoints(Circle((1,0),1),B--P)[1]; C = intersectionpoints(Circle((1,0),1),D--P)[0]; draw(B--P--D); draw(A--Q--C); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SW); label("$D$",D,SE); label("$P$",P,W); label("$Q$",Q,E); //Credit to chezbgone2 for the diagram[/asy]
| [
"We see that the measure of $P$ equals $(\\widehat{BD}-\\widehat{AC})/2$, and that the measure of $Q$ equals $\\widehat{AC}/2$.\nSince $\\widehat{BD} = \\widehat{BQ} + \\widehat{QD} = 42^{\\circ} + 38^{\\circ} = 80^{\\circ}$, the sum of the measures of $P$ and $Q$ is $\\widehat{BD}/2 = 80^{\\circ}/2 = 40^{\\circ} \... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/5.json | AHSME |
1971_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | If the line $y=mx+1$ intersects the ellipse $x^2+4y^2=1$ exactly once, then the value of $m^2$ is
$\textbf{(A) }\textstyle\frac{1}{2}\qquad \textbf{(B) }\frac{2}{3}\qquad \textbf{(C) }\frac{3}{4}\qquad \textbf{(D) }\frac{4}{5}\qquad \textbf{(E) }\frac{5}{6}$
| [
"Plug in $y=mx+1$ into the ellipse's equation to find the intersection points:\n\\[x^2 + 4(mx+1)^2 = 1\\]\nAfter simplifying, we have a quadratic in $x$:\n\\[(4m^2 + 1)x^2 + 8mx +3 = 0\\]\n\n\nBecause there is only one intersection point, then the quadratic has only one solution. This can only happen when the disc... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/19.json | AHSME |
1971_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | An uncrossed belt is fitted without slack around two circular pulleys with radii of $14$ inches and $4$ inches.
If the distance between the points of contact of the belt with the pulleys is $24$ inches, then the distance
between the centers of the pulleys in inches is
$\textbf{(A) }24\qquad \textbf{(B) }2\sqrt{119}\qquad \textbf{(C) }25\qquad \textbf{(D) }26\qquad \textbf{(E) }4\sqrt{35}$
| [
"(I don't know how to use Asymptote, so if you are a visual learner, I apologize)\n\n\nLet the center of the smaller circle be $A$ and the center of the larger circle be $B.$ Draw perpendicular radii $AC$ and $BD.$ The length $CD$ should equal 24.\n\n\nFrom $A,$ draw a perpendicular line $AE,$ where $E$ is the foot... | 1 | ./CreativeMath/AHSME/1971_AHSME_Problems/9.json | AHSME |
1964_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | The sum of the numerical coefficients of all the terms in the expansion of $(x-2y)^{18}$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ -1\qquad \textbf{(E)}\ -19$
| [
"For any polynomial, even a polynomial with more than one variable, the sum of all the coefficients (including the constant, which is the coefficient of $x^0y^0$) is found by setting all variables equal to $1$. Note that we don't have to worry about whether a constant is a coefficient of an \"invisible\" $x^0y^0$ ... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/20.json | AHSME |
1964_AHSME_Problems | 36 | 0 | Geometry | Multiple Choice | In this figure the radius of the circle is equal to the altitude of the equilateral triangle $ABC$. The circle is made to roll along the side $AB$, remaining tangent to it at a variable point $T$ and intersecting lines $AC$ and $BC$ in variable points $M$ and $N$, respectively. Let $n$ be the number of degrees in arc $MTN$. Then $n$, for all permissible positions of the circle:
$\textbf{(A) }\text{varies from }30^{\circ}\text{ to }90^{\circ}$
$\textbf{(B) }\text{varies from }30^{\circ}\text{ to }60^{\circ}$
$\textbf{(C) }\text{varies from }60^{\circ}\text{ to }90^{\circ}$
$\textbf{(D) }\text{remains constant at }30^{\circ}$
$\textbf{(E) }\text{remains constant at }60^{\circ}$
[asy] pair A = (0,0), B = (1,0), C = dir(60), T = (2/3,0); pair M = intersectionpoint(A--C,Circle((2/3,sqrt(3)/2),sqrt(3)/2)), N = intersectionpoint(B--C,Circle((2/3,sqrt(3)/2),sqrt(3)/2)); draw((0,0)--(1,0)--dir(60)--cycle); draw(Circle((2/3,sqrt(3)/2),sqrt(3)/2)); label("$A$",A,dir(210)); label("$B$",B,dir(-30)); label("$C$",C,dir(90)); label("$M$",M,dir(190)); label("$N$",N,dir(75)); label("$T$",T,dir(-90)); [/asy]
| [
"E\n\n\n"
] | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/36.json | AHSME |
1964_AHSME_Problems | 16 | 0 | Number Theory | Multiple Choice | Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0, 1, 2, \dots , 25 \}$.
The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:
$\textbf{(A)}\ 25\qquad \textbf{(B)}\ 22\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 17$
| [
"Note that for all polynomials $f(x)$, $f(x + 6) \\equiv f(x) \\pmod 6$. \n\n\nProof:\nIf $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$, then $f(x+6) = a_n(x+6)^n + a_{n-1}(x+6)^{n-1} +...+ a_0$. In the second equation, we can use the binomial expansion to expand every term, and then subtract off all terms that ha... | 2 | ./CreativeMath/AHSME/1964_AHSME_Problems/16.json | AHSME |
1964_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $x, 2x+2, 3x+3, \dots$ are in geometric progression, the fourth term is:
$\textbf{(A)}\ -27 \qquad \textbf{(B)}\ -13\frac{1}{2} \qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 13\frac{1}{2}\qquad \textbf{(E)}\ 27$
| [
"Since we know the sequence is a geometric sequence, the ratio of consecutive terms is always the same number. Thus, we can set up an equation:\n\n\n$\\frac{2x+2}{x} = \\frac{3x+3}{2x+2}$.\n\n\nSolving it, we get:\n\n\n$\\frac{2x+2}{x} = \\frac{3x+3}{2x+2}$\n\n\n$(2x+2)(2x+2) = (3x+3)(x)$\n\n\n$4x^2+8x+4 = 3x^2+3x... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/6.json | AHSME |
1964_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | Let n be the number of real values of $p$ for which the roots of
$x^2-px+p=0$
are equal. Then n equals:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \text{a finite number greater than 2}\qquad \textbf{(E)}\ \infty$
| [
"If the roots of the quadratic $Ax^2 + Bx + C = 0$ are equal, then $B^2 - 4AC = 0$. Plugging in $A=1, B=-p, C = p$ into the equation gives $p^2 - 4p = 0$. This leads to $p = 0, 4$, so there are two values of $p$ that work, giving answer $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/7.json | AHSME |
1964_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | Given the distinct points $P(x_1, y_1), Q(x_2, y_2)$ and $R(x_1+x_2, y_1+y_2)$.
Line segments are drawn connecting these points to each other and to the origin $O$.
Of the three possibilities: (1) parallelogram (2) straight line (3) trapezoid, figure $OPRQ$,
depending upon the location of the points $P, Q$, and $R$, can be:
$\textbf{(A)}\ \text{(1) only}\qquad \textbf{(B)}\ \text{(2) only}\qquad \textbf{(C)}\ \text{(3) only}\qquad \textbf{(D)}\ \text{(1) or (2) only}\qquad \textbf{(E)}\ \text{all three}$
| [
"Using vector addition can help solve this problem quickly. Note that algebraically, adding $\\overrightarrow{OP}$ to $\\overrightarrow{OQ}$ will give $\\overrightarrow{OR}$. One method of vector addition is literally known as the \"parallelogram rule\" - if you are given $\\overrightarrow{OP}$ and $\\overrightar... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/17.json | AHSME |
1964_AHSME_Problems | 40 | 0 | Arithmetic | Multiple Choice | A watch loses $2\frac{1}{2}$ minutes per day. It is set right at $1$ P.M. on March 15. Let $n$ be the positive correction, in minutes, to be added to the time shown by the watch at a given time. When the watch shows $9$ A.M. on March 21, $n$ equals:
$\textbf{(A) }14\frac{14}{23}\qquad\textbf{(B) }14\frac{1}{14}\qquad\textbf{(C) }13\frac{101}{115}\qquad\textbf{(D) }13\frac{83}{115}\qquad \textbf{(E) }13\frac{13}{23}$
| [
"From March 15 $1$ P.M. on the watch to March 21 $9$ A.M. on the watch, the watch passed $20 + 5 \\times 24 = 140$ hours. \n\n\nSince $1$ watch hour equals $\\frac{24}{23 + \\frac{57.5}{60}} = \\frac{576}{575}$ real hour, the difference between the watch time and the actual time passed is $140 \\times \\left( \\fra... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/40.json | AHSME |
1964_AHSME_Problems | 37 | 0 | Algebra | Multiple Choice | Given two positive numbers $a$, $b$ such that $a<b$. Let $A.M.$ be their arithmetic mean and let $G.M.$ be their positive geometric mean. Then $A.M.$ minus $G.M.$ is always less than:
$\textbf{(A) }\dfrac{(b+a)^2}{ab}\qquad\textbf{(B) }\dfrac{(b+a)^2}{8b}\qquad\textbf{(C) }\dfrac{(b-a)^2}{ab}$
$\textbf{(D) }\dfrac{(b-a)^2}{8a}\qquad \textbf{(E) }\dfrac{(b-a)^2}{8b}$
| [
"$\\boxed{ \\textbf{(D)} }$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/37.json | AHSME |
1964_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | If $\log_{b^2}x+\log_{x^2}b=1, b>0, b \neq 1, x \neq 1$, then $x$ equals:
$\textbf{(A)}\ 1/b^2 \qquad \textbf{(B)}\ 1/b \qquad \textbf{(C)}\ b^2 \qquad \textbf{(D)}\ b \qquad \textbf{(E)}\ \sqrt{b}$
| [
"Using natural log as a \"neutral base\", and applying the change of base formula to each term, we get:\n\n\n$\\frac{\\ln x}{\\ln b^2} + \\frac{\\ln b}{\\ln x^2} = 1$\n\n\n\n\n$\\frac{\\ln x}{2\\ln b} + \\frac{\\ln b}{2\\ln x} = 1$\n\n\n\n\n$\\frac{\\ln x \\ln x + \\ln b \\ln b}{2\\ln b \\ln x} = 1$\n\n\n\n\n$\\ln ... | 2 | ./CreativeMath/AHSME/1964_AHSME_Problems/21.json | AHSME |
1964_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | Given a square side of length $s$. On a diagonal as base a triangle with three unequal sides is constructed so that its area
equals that of the square. The length of the altitude drawn to the base is:
$\textbf{(A)}\ s\sqrt{2} \qquad \textbf{(B)}\ s/\sqrt{2} \qquad \textbf{(C)}\ 2s \qquad \textbf{(D)}\ 2\sqrt{s} \qquad \textbf{(E)}\ 2/\sqrt{s}$
| [
"The area of the square is $s^2$. The diagonal of a square with side $s$ bisects the square into two $45-45-90$ right triangles, so the diagonal has length $s\\sqrt{2}$.\n\n\nThe area of the triangle is $\\frac{1}{2}bh$. The base $b$ of the triangle is the diagonal of the square, which is $b = s\\sqrt{2}$. If th... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/10.json | AHSME |
1964_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | In a ten-mile race $\textit{First}$ beats $\textit{Second}$ by $2$ miles and
$\textit{First}$ beats $\textit{Third}$ by $4$ miles.
If the runners maintain constant speeds throughout the race,
by how many miles does $\textit{Second}$ beat $\textit{Third}$?
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2\frac{1}{4}\qquad \textbf{(C)}\ 2\frac{1}{2}\qquad \textbf{(D)}\ 2\frac{3}{4}\qquad \textbf{(E)}\ 3$
| [
"Let the speeds of the runners in miles per hour be $a, b, c$, with $a>b>c$. If person $a$ reaches the finish line in $h$ hours, then we have $10 = ah, 8 = bh, 6 = bh$.\n\n\nThus, the speeds of the second and third place people are $\\frac{8}{h}$ and $\\frac{6}{h}$. The ratio of those speeds is $\\frac{\\frac{8}{... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/26.json | AHSME |
1964_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | The larger root minus the smaller root of the equation \[(7+4\sqrt{3})x^2+(2+\sqrt{3})x-2=0\] is
$\textbf{(A) }-2+3\sqrt{3}\qquad\textbf{(B) }2-\sqrt{3}\qquad\textbf{(C) }6+3\sqrt{3}\qquad\textbf{(D) }6-3\sqrt{3}\qquad \textbf{(E) }3\sqrt{3}+2$
| [
"Dividing the quadratic by $7 + 4\\sqrt{3}$ to obtain a monic polynomial will give a linear coefficient of $\\frac{2 + \\sqrt{3}}{7 + 4\\sqrt{3}}$. Rationalizing the denominator gives:\n\n\n$\\frac{(2 + \\sqrt{3})(7 - 4\\sqrt{3})}{7^2 - 4^2 \\cdot 3}$\n\n\n$=\\frac{14 - 12 - \\sqrt{3}}{49-48}$\n\n\n$=2 - \\sqrt{3}... | 2 | ./CreativeMath/AHSME/1964_AHSME_Problems/30.json | AHSME |
1964_AHSME_Problems | 31 | 0 | Algebra | Multiple Choice | Let \[f(n)=\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\left(\dfrac{1-\sqrt{5}}{2}\right)^n.\]
Then $f(n+1)-f(n-1)$, expressed in terms of $f(n)$, equals:
$\textbf{(A) }\frac{1}{2}f(n)\qquad\textbf{(B) }f(n)\qquad\textbf{(C) }2f(n)+1\qquad\textbf{(D) }f^2(n)\qquad \textbf{(E) }$
$\frac{1}{2}(f^2(n)-1)$
| [
"We compute $f(n+1)$ and $f(n-1)$, while pulling one copy of the exponential part outside:\n\n\n$f(n+1) = \\dfrac{5+3\\sqrt{5}}{10}\\cdot \\frac{1 + \\sqrt{5}}{2}\\cdot\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n+\\dfrac{5-3\\sqrt{5}}{10}\\cdot \\frac{1 - \\sqrt{5}}{2}\\cdot\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n$\n\n... | 2 | ./CreativeMath/AHSME/1964_AHSME_Problems/31.json | AHSME |
1964_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | If $x$ is a real number and $|x-4|+|x-3|<a$ where $a>0$, then:
$\textbf{(A)}\ 0<a<.01\qquad \textbf{(B)}\ .01<a<1 \qquad \textbf{(C)}\ 0<a<1\qquad \\ \textbf{(D)}\ 0<a \le 1\qquad \textbf{(E)}\ a>1$
| [
"Let $A$ be point $3$ on a number line, and let $B$ be point $4$. Let $X$ be a mobile point at $x$. Geometrically, $|x-4| + |x-3|$ represents $AX + BX$. If $X$ is between $A$ and $B$, then $AX + BX = AB = 1$. Otherwise, if $X$ is to the left of $A$, then $AX + BX = AX + (AX + AB) = 2AX + 1$, which is greater th... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/27.json | AHSME |
1964_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | What is the value of $[\log_{10}(5\log_{10}100)]^2$?
$\textbf{(A)}\ \log_{10}50 \qquad \textbf{(B)}\ 25\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 1$
| [
"$[\\log_{10}(5\\log_{10}100)]^2$\n\n\nSince $10^2 = 100$, we have $\\log_{10} 100 = 2$, so:\n\n\n$[\\log_{10}(5\\cdot 2)]^2$\n\n\nMultiply:\n\n\n$[\\log_{10}(10)]^2$\n\n\nSince $10^1 = 10$, we have $\\log_{10} 10 = 1$, so:\n\n\n$[1]^2$\n\n\n$1$\n\n\n$\\fbox{E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/1.json | AHSME |
1964_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | Given $2^x=8^{y+1}$ and $9^y=3^{x-9}$, find the value of $x+y$
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$
| [
"Since $8^{y + 1} = 2^{3(y+1)}$ and $9^y = 3^{2y}$, we have:\n\n\n$2^x = 2^{3(y+1)}$ and $3^{2y} = 3^{x - 9}$\n\n\n\n\nNote that if $a^b = a^c$, then $b=c$. Setting the exponents equal gives $x = 3y + 3$ and $2y = x - 9$. Plugging the first equation into the second equation gives:\n\n\n$2y = (3y + 3) - 9$\n\n\n$2... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/11.json | AHSME |
1964_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | The graph of $x^2-4y^2=0$ is:
$\textbf{(A)}\ \text{a parabola} \qquad \textbf{(B)}\ \text{an ellipse} \qquad \textbf{(C)}\ \text{a pair of straight lines}\qquad \\ \textbf{(D)}\ \text{a point}\qquad \textbf{(E)}\ \text{None of these}$
| [
"In the equation $x^2 - 4y^2 = k$, because the coefficients of $x^2$ and $y^2$ are of opposite sign, the graph is typically a hyperbola for most real values of $k$. However, there is one exception. When $k=0$, the equation can be factored as $(x - 2y)(x+2y) = 0$. This gives the graph of two lines passing though ... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/2.json | AHSME |
1964_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | The sum of $n$ terms of an arithmetic progression is $153$, and the common difference is $2$.
If the first term is an integer, and $n>1$, then the number of possible values for $n$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 6$
| [
"Let the progression start at $a$, have common difference $2$, and end at $a + 2(n-1)$.\n\n\n\n\nThe average term is $\\frac{a + (a + 2(n-1))}{2}$, or $a + n - 1$. Since the number of terms is $n$, and the sum of the terms is $153$, we have:\n\n\n$n(a+n-1) = 153$\n\n\nSince $n$ is a positive integer, it must be a ... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/28.json | AHSME |
1964_AHSME_Problems | 12 | 0 | Other | Multiple Choice | Which of the following is the negation of the statement: For all $x$ of a certain set, $x^2>0$?
$\textbf{(A)}\ \text{For all x}, x^2 < 0\qquad \textbf{(B)}\ \text{For all x}, x^2 \le 0\qquad \textbf{(C)}\ \text{For no x}, x^2>0\qquad \\ \textbf{(D)}\ \text{For some x}, x^2>0\qquad \textbf{(E)}\ \text{For some x}, x^2 \le 0$
| [
"In general, the negation of a universal (\"for all\") quantifier will use an existential (\"there exists\") quantifier, and negate the statement inside. \n\n\nIn this case, we change the \"For all\" to \"There exists\", and negate the inner statement from $x^2 > 0$ to $x^2 \\le 0$.\n\n\nSo, the negation of the or... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/12.json | AHSME |
1964_AHSME_Problems | 32 | 0 | Algebra | Multiple Choice | If $\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}$, then:
$\textbf{(A) }a \text{ must equal }c\qquad\textbf{(B) }a+b+c+d\text{ must equal zero}\qquad$
$\textbf{(C) }\text{either }a=c\text{ or }a+b+c+d=0\text{, or both}\qquad$
$\textbf{(D) }a+b+c+d\ne 0\text{ if }a=c\qquad$
$\textbf{(E) }a(b+c+d)=c(a+b+d)$
| [
"Cross-multiplying gives:\n\n\n$(a+b)(a+d) = (b+c)(c+d)$\n\n\n$a^2 + ad + ab + bd = bc + bd + c^2 + cd$\n\n\n$a^2 + ad + ab - bc - c^2 - cd = 0$\n\n\n$a(a + b + d) - c(b+c+d)= 0$\n\n\nThis looks close to turning into option C, but we don't have a $c$ term in the first parentheses, and we don't have an $a$ term in... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/32.json | AHSME |
1964_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | Let $y=(x-a)^2+(x-b)^2, a, b$ constants. For what value of $x$ is $y$ a minimum?
$\textbf{(A)}\ \frac{a+b}{2} \qquad \textbf{(B)}\ a+b \qquad \textbf{(C)}\ \sqrt{ab} \qquad \textbf{(D)}\ \sqrt{\frac{a^2+b^2}{2}}\qquad \textbf{(E)}\ \frac{a+b}{2ab}$
| [
"Expanding the quadratic and collecting terms gives $y = 2x^2 - (2a + 2b)x + (a^2+b^2)$. For a quadratic of the form $y = Ax^2 + Bx + C$ with $A>0$, $y$ is minimized when $x = -\\frac{B}{2A}$, which is the average of the roots.\n\n\nThus, the quadratic is minimized when $x = \\frac{2a+2b}{2\\cdot 2} = \\frac{a+b}{... | 2 | ./CreativeMath/AHSME/1964_AHSME_Problems/24.json | AHSME |
1964_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | The set of values of $m$ for which $x^2+3xy+x+my-m$ has two factors, with integer coefficients, which are linear in $x$ and $y$, is precisely:
$\textbf{(A)}\ 0, 12, -12\qquad \textbf{(B)}\ 0, 12\qquad \textbf{(C)}\ 12, -12\qquad \textbf{(D)}\ 12\qquad \textbf{(E)}\ 0$
| [
"Since there are only $3$ candidate values for $m$, we test $0, 12, -12$.\n\n\nIf $m=0$, then the expression is $x^2 + 3xy + x$. The term $x$ appears in each monomial, giving $x(x + 3y + 1)$, which has two factors that are linear in $x$ and $y$ with integer coefficients. This eliminates $C$ and $D$.\n\n\nWe next ... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/25.json | AHSME |
1964_AHSME_Problems | 33 | 0 | Geometry | Multiple Choice | $P$ is a point interior to rectangle $ABCD$ and such that $PA=3$ inches, $PD=4$ inches, and $PC=5$ inches. Then $PB$, in inches, equals:
$\textbf{(A) }2\sqrt{3}\qquad\textbf{(B) }3\sqrt{2}\qquad\textbf{(C) }3\sqrt{3}\qquad\textbf{(D) }4\sqrt{2}\qquad \textbf{(E) }2$
[asy] pair A, B, C, D, P; A = (0, 0); B = (6.5, 0); C = (6.5, 4.5); D = (0, 4.5); P = (2.5, 1.5); draw(A--B--C--D--cycle); draw(A--P); draw(C--P); draw(D--P); draw(B--P, dashed); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$P$", P, S); label("$3$", midpoint(A--P), NW); label("$4$", midpoint(D--P), NE); label("$5$", midpoint(C--P), NW); [/asy]
| [
"From point $P$, create perpendiculars to all four sides, labeling them $a, b, c, d$ starting from going north and continuing clockwise. Label the length $PB$ as $x$:\n\n\n[asy] unitsize(1cm); pair A, B, C, D, P, ABfoot, BCfoot, CDfoot, DAfoot; A = (0, 0); B = (6.5, 0); C = (6.5, 4.5); D = (0, 4.5); P = (2.5, 1.5... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/33.json | AHSME |
1964_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | A circle is inscribed in a triangle with side lengths $8, 13$, and $17$. Let the segments of the side of length $8$,
made by a point of tangency, be $r$ and $s$, with $r<s$. What is the ratio $r:s$?
$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 2:5 \qquad \textbf{(C)}\ 1:2 \qquad \textbf{(D)}\ 2:3 \qquad \textbf{(E)}\ 3:4$
| [
"Label our triangle $ABC$ where that $AB=17$, $AC=13$, and $BC=8$. Let $L$, $J$, and $K$ be the tangency points of $BC$, $AC$, and $AB$ respectively. Let $AK=x$, which implies $AJ=x$. Thus $KB=BL=17-x$ and $JC=LC=13-x$. \n\n\nSince $BL+LC=(17-x)+(13-x)=8$, $x=11$. Thus $r:s=CL:BL=13-x:17-x=2:6=1:3$, hence our answe... | 2 | ./CreativeMath/AHSME/1964_AHSME_Problems/13.json | AHSME |
1964_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | In this figure $\angle RFS=\angle FDR$, $FD=4$ inches, $DR=6$ inches, $FR=5$ inches, $FS=7\dfrac{1}{2}$ inches. The length of $RS$, in inches, is:
$\textbf{(A) }\text{undetermined}\qquad\textbf{(B) }4\qquad\textbf{(C) }5\dfrac{1}{2}\qquad\textbf{(D) }6\qquad\textbf{(E) }6\dfrac{1}{2}\qquad$
[asy] import olympiad; draw((0,0)--7.5*dir(0)); draw((0,0)--5*dir(55)); draw((0,0)--4*dir(140)); draw(4*dir(140)--5*dir(55)--7.5*dir(0)); markscalefactor=0.1; draw(anglemark((0,0),4*dir(140),5*dir(55))); draw(anglemark(7.5*dir(0),(0,0),5*dir(55))); label("$F$",(0,0),dir(240)); label("$D$",4*dir(140),dir(180)); label("$R$",5*dir(55),dir(80)); label("$S$",7.5*dir(0),dir(-25)); label("$4$",2*dir(140),dir(230)); label("$5$",2.5*dir(55),dir(155)); label("$6$",(4*dir(140)+5*dir(55))/2,dir(100)); label("$7\frac{1}{2}$",3.75,dir(-90)); [/asy]
| [
"We examine $\\triangle RDF$ and $\\triangle SFR$. We are given $\\angle RDF \\cong \\angle SFR$. Also note that $\\frac{SF}{RD} = \\frac{7.5}{6} = 1.25$ and $\\frac{FR}{DF} = \\frac{5}{4} = 1.25$, so $\\frac{SF}{RD} = \\frac{FR}{DF}$.\n\n\nIf two triangles have two pairs of sides that are proportional, and the in... | 2 | ./CreativeMath/AHSME/1964_AHSME_Problems/29.json | AHSME |
1964_AHSME_Problems | 3 | 0 | Number Theory | Multiple Choice | When a positive integer $x$ is divided by a positive integer $y$, the quotient is $u$ and the remainder is $v$, where $u$ and $v$ are integers.
What is the remainder when $x+2uy$ is divided by $y$?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 2u \qquad \textbf{(C)}\ 3u \qquad \textbf{(D)}\ v \qquad \textbf{(E)}\ 2v$
| [
"\\begin{itemize}\n\\item We can solve this problem by elemetary modular arthmetic,\n\\end{itemize}\n$x \\equiv v\\ (\\textrm{mod}\\ y)$ $=>$ $x+2uy \\equiv v\\ (\\textrm{mod}\\ y)$\n\n\n~GEOMETRY-WIZARD\n\n\n",
"By the definition of quotient and remainder, problem states that $x = uy + v$.\n\n\nThe problem asks ... | 3 | ./CreativeMath/AHSME/1964_AHSME_Problems/3.json | AHSME |
1964_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | If $n$ is a multiple of $4$, the sum $s=1+2i+3i^2+\cdots+(n+1)i^n$, where $i=\sqrt{-1}$, equals:
$\textbf{(A) }1+i\qquad\textbf{(B) }\frac{1}{2}(n+2)\qquad\textbf{(C) }\frac{1}{2}(n+2-ni)\qquad$
$\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)$
| [
"The real part is $1 + 3i^2 + 5i^4 + ...$, which is $1 - 3 + 5 - 7 + ...$. If $n$ is a multiple of $4$, then we have an odd number of terms in total: we start with $1$ at $n=0$, then add two more terms to get $1 - 3 + 5$ at $n=4$, etc. With each successive addition, we're really adding a total of $2$, since $-3 ... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/34.json | AHSME |
1964_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | The smaller root of the equation
$\left(x-\frac{3}{4}\right)\left(x-\frac{3}{4}\right)+\left(x-\frac{3}{4}\right)\left(x-\frac{1}{2}\right) =0$ is:
$\textbf{(A)}\ -\frac{3}{4}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ \frac{5}{8}\qquad \textbf{(D)}\ \frac{3}{4}\qquad \textbf{(E)}\ 1$
| [
"Note that this equation is of the form $a^2 + ab = 0$, which factors to $a(a + b) = 0$. Plugging in $a = x - \\frac{3}{4}$ and $b = x - \\frac{1}{2}$ gives:\n\n\n$(x - \\frac{3}{4})(x - \\frac{3}{4} + x - \\frac{1}{2}) = 0$\n\n\n$(x - \\frac{3}{4})(2x - \\frac{5}{4}) = 0$\n\n\nThe roots are $x = \\frac{3}{4}$ nad... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/8.json | AHSME |
1964_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | Given parallelogram $ABCD$ with $E$ the midpoint of diagonal $BD$. Point $E$ is connected to a point $F$ in $DA$ so that
$DF=\frac{1}{3}DA$. What is the ratio of the area of $\triangle DFE$ to the area of quadrilateral $ABEF$?
$\textbf{(A)}\ 1:2 \qquad \textbf{(B)}\ 1:3 \qquad \textbf{(C)}\ 1:5 \qquad \textbf{(D)}\ 1:6 \qquad \textbf{(E)}\ 1:7$
| [
"If it works for a parallelogram $ABCD$, it should also work for a unit square, with $A(0, 0), B(0, 1), C(1, 1), D(1, 0)$. We are given that $E$ is the midpoint of $BD$, so $E(0.5, 0.5)$. If $F$ is on $DA$, then $F(x, 0)$. We note that $DF = 1-x$ and $DA = 1$, so $DF = \\frac{1}{3}DA$ means $1-x = \\frac{1}{3}$,... | 2 | ./CreativeMath/AHSME/1964_AHSME_Problems/22.json | AHSME |
1964_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | Let $n$ be the number of pairs of values of $b$ and $c$ such that $3x+by+c=0$ and $cx-2y+12=0$ have the same graph. Then $n$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ \text{finite but more than 2}\qquad \textbf{(E)}\ \infty$
| [
"For two lines to be the same, their slopes must be equal and their intercepts must be equal. This is a necessary and sufficient condition.\n\n\nThe slope of the first line is $\\frac{-3}{b}$, while the slope of the second line is $\\frac{c}{2}$. Thus, $\\frac{-3}{b} = \\frac{c}{2}$, or $bc = -6$.\n\n\nThe interc... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/18.json | AHSME |
1964_AHSME_Problems | 38 | 0 | Geometry | Multiple Choice | The sides $PQ$ and $PR$ of triangle $PQR$ are respectively of lengths $4$ inches, and $7$ inches. The median $PM$ is $3\frac{1}{2}$ inches. Then $QR$, in inches, is:
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$
| [
"By the Median Formula, $PM = \\frac12\\sqrt{2PQ^2+2PR^2-QR^2}$\n\n\nPlugging in the numbers given in the problem, we get\n\\[\\frac72=\\frac12\\sqrt{2\\cdot4^2+2\\cdot7^2-QR^2}\\]\n\n\nSolving,\n\\[7=\\sqrt{2(16)+2(49)-QR^2}\\]\n\\[49=32+98-QR^2\\]\n\\[QR^2=81\\]\n\\[QR=9=\\boxed{D}\\]\n\n\n-AOPS81619\n\n\n"
] | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/38.json | AHSME |
1964_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | The expression
\[\frac{P+Q}{P-Q}-\frac{P-Q}{P+Q}\]
where $P=x+y$ and $Q=x-y$, is equivalent to:
$\textbf{(A)}\ \frac{x^2-y^2}{xy}\qquad \textbf{(B)}\ \frac{x^2-y^2}{2xy}\qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{x^2+y^2}{xy}\qquad \textbf{(E)}\ \frac{x^2+y^2}{2xy}$
| [
"Simplifying before substituting by obtaining a common denominator gives: \n\n\n$\\frac{(P+Q)^2 - (P - Q)^2}{(P-Q)(P+Q)}$\n\n\n$\\frac{(P^2 + 2PQ + Q^2) - (P^2 - 2PQ + Q^2)}{P^2 - Q^2}$\n\n\n$\\frac{4PQ}{P^2 - Q^2}$\n\n\nSubstituting gives:\n\n\n$\\frac{4(x+y)(x-y)}{(x+y)^2 - (x - y)^2}$\n\n\n$\\frac{4(x^2 - y^2)}{... | 3 | ./CreativeMath/AHSME/1964_AHSME_Problems/4.json | AHSME |
1964_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | A farmer bought $749$ sheep. He sold $700$ of them for the price paid for the $749$ sheep.
The remaining $49$ sheep were sold at the same price per head as the other $700$.
Based on the cost, the percent gain on the entire transaction is:
$\textbf{(A)}\ 6.5 \qquad \textbf{(B)}\ 6.75 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 7.5 \qquad \textbf{(E)}\ 8$
| [
"Let us say each sheep cost $x$ dollars. The farmer paid $749x$ for the sheep. He sold $700$ of them for $749x$, so each sheep sold for $\\frac{749}{700} = 1.07x$.\n\n\nSince every sheep sold for the same price per head, and since every sheep cost $x$ and sold for $1.07x$, there is an increase of $\\frac{1.07x - ... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/14.json | AHSME |
1964_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | A line through the point $(-a,0)$ cuts from the second quadrant a triangular region with area $T$. The equation of the line is:
$\textbf{(A)}\ 2Tx+a^2y+2aT=0 \qquad \textbf{(B)}\ 2Tx-a^2y+2aT=0 \qquad \textbf{(C)}\ 2Tx+a^2y-2aT=0 \qquad \\ \textbf{(D)}\ 2Tx-a^2y-2aT=0 \qquad \textbf{(E)}\ \text{none of these}$
| [
"The right triangle has area $T$ and base $a$, so the height $h$ satisfies $\\frac{1}{2}ha = T$. This means $h = \\frac{2T}{a}$. Becuase the triangle is in the second quadrant, the coordinates are the origin, $(-a, 0)$, and $(0, h)$, which means the third coordinate is $(0, \\frac{2T}{a})$.\n\n\nSo, we want the e... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/15.json | AHSME |
1964_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | If $y$ varies directly as $x$, and if $y=8$ when $x=4$, the value of $y$ when $x=-8$ is:
$\textbf{(A)}\ -16 \qquad \textbf{(B)}\ -4 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 4k, k= \pm1, \pm2, \dots \qquad \\ \textbf{(E)}\ 16k, k=\pm1,\pm2,\dots$
| [
"If $y$ varies directly as $x$, then $\\frac{x}{y}$ will be a constant for every pair of $(x, y)$. Thus, we have:\n\n\n$\\frac{4}{8} = \\frac{-8}{y}$\n\n\n$y \\cdot \\frac{1}{2} = -8$\n\n\n$y = -16$\n\n\n$\\boxed{\\textbf{(A)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/5.json | AHSME |
1964_AHSME_Problems | 39 | 0 | Geometry | Multiple Choice | The magnitudes of the sides of triangle $ABC$ are $a$, $b$, and $c$, as shown, with $c\le b\le a$. Through interior point $P$ and the vertices $A$, $B$, $C$, lines are drawn meeting the opposite sides in $A'$, $B'$, $C'$, respectively. Let $s=AA'+BB'+CC'$. Then, for all positions of point $P$, $s$ is less than:
$\textbf{(A) }2a+b\qquad\textbf{(B) }2a+c\qquad\textbf{(C) }2b+c\qquad\textbf{(D) }a+2b\qquad \textbf{(E) }$ $a+b+c$
[asy] import math; pair A = (0,0), B = (1,3), C = (5,0), P = (1.5,1); pair X = extension(B,C,A,P), Y = extension(A,C,B,P), Z = extension(A,B,C,P); draw(A--B--C--cycle); draw(A--X); draw(B--Y); draw(C--Z); dot(P); dot(A); dot(B); dot(C); label("$A$",A,dir(210)); label("$B$",B,dir(90)); label("$C$",C,dir(-30)); label("$A'$",X,dir(-100)); label("$B'$",Y,dir(65)); label("$C'$",Z,dir(20)); label("$P$",P,dir(70)); label("$a$",X,dir(80)); label("$b$",Y,dir(-90)); label("$c$",Z,dir(110)); [/asy]
| [
"We know that in a $\\triangle DEF$, if $\\angle D \\le \\angle E$ then $EF \\le DF$, we can use this fact in the different triangles to form inequalities, and then add the inequalities.\n\n\nIn $\\triangle ABC$, since $c \\le b \\le a$, we have $\\angle C \\le \\angle B \\le \\angle A$ by the above argument.\n\n\n... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/39.json | AHSME |
1964_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | If $2x-3y-z=0$ and $x+3y-14z=0, z \neq 0$, the numerical value of $\frac{x^2+3xy}{y^2+z^2}$ is:
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ -20/17\qquad \textbf{(E)}\ -2$
| [
"If the value of $\\frac{x^2+3xy}{y^2+z^2}$ is constant, as the answers imply, we can pick a value of $z$, and then solve the two linear equations for the corresponding $(x, y)$. We can then plug in $(x,y, z)$ into the expression to get the answer.\n\n\nIf $z=1$, then $2x - 3y = 1$ and $x + 3y = 14$. We can solve... | 2 | ./CreativeMath/AHSME/1964_AHSME_Problems/19.json | AHSME |
1964_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | Two numbers are such that their difference, their sum, and their product are to one another as $1:7:24$. The product of the two numbers is:
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96$
| [
"Set the two numbers as $x$ and $y$. Therefore, \n$x+y=7(x-y), xy=24(x-y)$, and\n$24(x+y)=7xy$. Simplifying the first equation gives $y=\\frac{3}{4}x$. Substituting for $y$ in the second equation gives $\\frac{3}{4}x^2=6x.$ Solving yields $x=8$ or $x=0$. Substituting $x=0$ back into the first equation yields $1=-7... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/23.json | AHSME |
1964_AHSME_Problems | 9 | 0 | Arithmetic | Multiple Choice | A jobber buys an article at $$24$ less $12\frac{1}{2}\%$. He then wishes to sell the article at a gain of $33\frac{1}{3}\%$ of his cost
after allowing a $20\%$ discount on his marked price. At what price, in dollars, should the article be marked?
$\textbf{(A)}\ 25.20 \qquad \textbf{(B)}\ 30.00 \qquad \textbf{(C)}\ 33.60 \qquad \textbf{(D)}\ 40.00 \qquad \textbf{(E)}\ \text{none of these}$
| [
"The item is bought for $24$ minus $12\\frac{1}{2}\\%$ of $24$. Since $12\\frac{1}{2}\\% = \\frac{1}{8}$, the item was purchased for $24 - 24 \\cdot \\frac{1}{8}$, or $21$.\n\n\nHe wants to make $33\\frac{1}{3}\\%$ of the purchase price, or $21 \\cdot \\frac{1}{3} = 7$. This means the price must actually sell for... | 1 | ./CreativeMath/AHSME/1964_AHSME_Problems/9.json | AHSME |
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