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 |
|---|---|---|---|---|---|---|---|---|---|
1964_AHSME_Problems | 35 | 0 | Geometry | Multiple Choice | The sides of a triangle are of lengths $13$, $14$, and $15$. The altitudes of the triangle meet at point $H$. if $AD$ is the altitude to the side of length $14$, the ratio $HD:HA$ is:
$\textbf{(A) }3:11\qquad\textbf{(B) }5:11\qquad\textbf{(C) }1:2\qquad\textbf{(D) }2:3\qquad \textbf{(E) }25:33$
| [
"Using Law of Cosines\nand the fact that the ratio equals cos(a)/[cos(b)cos(c)]\nB 5:11\n\n\n",
"[asy] draw((0,0)--(15,0)--(6.6,11.2)--(0,0)); draw((0,0)--(9.6,7.2)); draw((6.6,0)--(6.6,11.2)); draw((15,0)--(3267/845,5544/845)); label(\"$B$\",(15,0),SE); label(\"$C$\",(6.6,11.2),N); label(\"$E$\",(6.6,0),S); labe... | 3 | ./CreativeMath/AHSME/1964_AHSME_Problems/35.json | AHSME |
1986_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | Suppose $x$ and $y$ are inversely proportional and positive. If $x$ increases by $p\%$, then $y$ decreases by
$\textbf{(A)}\ p\%\qquad \textbf{(B)}\ \frac{p}{1+p}\%\qquad \textbf{(C)}\ \frac{100}{p}\%\qquad \textbf{(D)}\ \frac{p}{100+p}\%\qquad \textbf{(E)}\ \frac{100p}{100+p}\%$
| [
"We see that $x$ is multiplied by $\\frac{100+p}{100}$ when it is increased by $p$%. Therefore, $y$ is multiplied by $\\frac{100}{100+p}$, and so it is decreased by $\\frac{p}{100+p}$ times itself. Therefore, it is decreased by $\\frac{100p}{100+p}$%, and so the answer is $\\boxed{E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/20.json | AHSME |
1986_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | In $\triangle ABC, AB = 8, BC = 7, CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$
is similar to $\triangle PCA$. The length of $PC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B); draw(A--P--C--A--B--C); label("... | [
"Since we are given that $\\triangle{PAB}\\sim\\triangle{PCA}$, we have $\\frac{PC}{PA}=\\frac{6}{8}=\\frac{PA}{PC+7}$.\n\n\n\n\nSolving for $PA$ in $\\frac{PC}{PA}=\\frac{6}{8}=\\frac{3}{4}$ gives us $PA=\\frac{4PC}{3}$.\n\n\n\n\nWe also have $\\frac{PA}{PC+7}=\\frac{3}{4}$. Substituting $PA$ in for our expression... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/16.json | AHSME |
1986_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1. Length $r$ is found to be $32$ inches. After rearranging the blocks as in Figure 2, length $s$ is found to be $28$ inches. How high is the table?
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13pt)); path table =... | [
"Let $h$, $l$, and $w$ represent the height of the table and the length and width of the wood blocks, respectively, in inches. From Figure 1, we have $l+h-w=32$, and from Figure 2, $w+h-l=28$. Adding the equations gives $2h=60 \\implies h=30$, which is $\\boxed{C}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/6.json | AHSME |
1986_AHSME_Problems | 7 | 0 | Number Theory | Multiple Choice | The sum of the greatest integer less than or equal to $x$ and the least integer greater than or equal to $x$ is $5$. The solution set for $x$ is
$\textbf{(A)}\ \Big\{\frac{5}{2}\Big\}\qquad \textbf{(B)}\ \big\{x\ |\ 2 \le x \le 3\big\}\qquad \textbf{(C)}\ \big\{x\ |\ 2\le x < 3\big\}\qquad\\ \textbf{(D)}\ \Big\{x\ |... | [
"If $x \\leq 2$, then $\\lfloor x \\rfloor + \\lceil x \\rceil \\leq 2+2 < 5$, so there are no solutions with $x \\leq 2$. If $x \\geq 3$, then $\\lfloor x \\rfloor + \\lceil x \\rceil \\geq 3+3$, so there are also no solutions here. Finally, if $2<x<3$, then $\\lfloor x \\rfloor + \\lceil x \\rceil = 2 + 3 = 5$, s... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/7.json | AHSME |
1986_AHSME_Problems | 17 | 0 | Counting | Multiple Choice | A drawer in a darkened room contains $100$ red socks, $80$ green socks, $60$ blue socks and $40$ black socks.
A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn.
What is the smallest number of socks that must be selected to guarantee that the selection contains a... | [
"Solution by e_power_pi_times_i\n\n\n\n\nSuppose that you wish to draw one pair of socks from the drawer. Then you would pick $5$ socks (one of each kind, plus one). Notice that in the worst possible situation, you will continue to draw the same sock, until you get $10$ pairs. This is because drawing the same sock ... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/17.json | AHSME |
1986_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | In the configuration below, $\theta$ is measured in radians, $C$ is the center of the circle,
$BCD$ and $ACE$ are line segments and $AB$ is tangent to the circle at $A$.
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair A=(0,-1), E=(0,1), C=(0,0), D=dir(10), F=dir(190), B=(-1/sin(10*pi/180))*dir(10); fill(Arc((... | [
"Well, the shaded sector's area is basically $\\text{(ratio of } \\theta \\text{ to total angle of circle)} \\times \\text{(total area)} = \\frac{\\theta}{2\\pi} \\cdot (\\pi r^2) = \\frac{\\theta}{2} \\cdot (AC)^2$.\n\n\nIn addition, if you let $\\angle{ACB} = \\theta$, then \\[\\tan \\theta = \\frac{AB}{AC}\\]\\[... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/21.json | AHSME |
1986_AHSME_Problems | 10 | 0 | Counting | Multiple Choice | The $120$ permutations of $AHSME$ are arranged in dictionary order as if each were an ordinary five-letter word.
The last letter of the $86$th word in this list is:
$\textbf{(A)}\ \text{A} \qquad \textbf{(B)}\ \text{H} \qquad \textbf{(C)}\ \text{S} \qquad \textbf{(D)}\ \text{M}\qquad \textbf{(E)}\ \text{E}$
| [
"We could list out all of the possible combinations in dictionary order.\n\n\n$73rd:$ MAEHS \n$74th:$ MAESH\n$75th:$ MAHES\n$76th:$ MAHSE\n$77th:$ MASEH\n$78th:$ MASHE\n$79th:$ MEAHS\n$80th:$ MEASH\n$81th:$ MEHAS\n$82th:$ MEHSA\n$83th:$ MESAH\n$84th:$ MESHA\n$85th:$ MHAES\n$86th:$ MHASE\n\n\nWe fi... | 2 | ./CreativeMath/AHSME/1986_AHSME_Problems/10.json | AHSME |
1986_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | It is desired to construct a right triangle in the coordinate plane so that its legs are parallel
to the $x$ and $y$ axes and so that the medians to the midpoints of the legs lie on the lines $y = 3x + 1$
and $y = mx + 2$. The number of different constants $m$ for which such a triangle exists is
$\textbf{(A)}\ 0\qqu... | [
"In \\textit{any} right triangle with legs parallel to the axes, one median to the midpoint of a leg has slope $4$ times that of the other. This can easily be shown with coordinates: any triangle of this sort may be labelled with right angle at $P(a,b)$, other vertices $Q(a,b+2c)$ and $R(a-2d,b)$, and thus midpoint... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/26.json | AHSME |
1986_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | The number of real solutions $(x,y,z,w)$ of the simultaneous equations
$2y = x + \frac{17}{x}, 2z = y + \frac{17}{y}, 2w = z + \frac{17}{z}, 2x = w + \frac{17}{w}$
is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$
| [
"Consider the cases $x>0$ and $x<0$, and also note that by AM-GM, for any positive number $a$, we have $a+\\frac{17}{a} \\geq 2\\sqrt{17}$, with equality only if $a = \\sqrt{17}$. Thus, if $x>0$, considering each equation in turn, we get that $y \\geq \\sqrt{17}, z \\geq \\sqrt{17}, w \\geq \\sqrt{17}$, and finally... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/30.json | AHSME |
1986_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | In the adjoining figure, $AB$ is a diameter of the circle, $CD$ is a chord parallel to $AB$,
and $AC$ intersects $BD$ at $E$, with $\angle AED = \alpha$. The ratio of the area of $\triangle CDE$ to that of $\triangle ABE$ is
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair A=(-1,0), B=(1,0), E=(0,-.4), C=(.6... | [
"$ABE$ and $DCE$ are similar isosceles triangles. It remains to find the square of the ratio of their sides. Draw in $AD$. Because $AB$ is a diameter, $\\angle ADB=\\angle ADE=90^{\\circ}$. Thus, \\[\\frac{DE}{AE}=\\cos\\alpha\\] So \\[\\frac{DE^2}{AE^2}=\\cos^2\\alpha\\] The answer is thus $\\fbox{(C)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/27.json | AHSME |
1986_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | $[x-(y-z)] - [(x-y) - z] =$
$\textbf{(A)}\ 2y \qquad \textbf{(B)}\ 2z \qquad \textbf{(C)}\ -2y \qquad \textbf{(D)}\ -2z \qquad \textbf{(E)}\ 0$
| [
"The expression becomes $(x-y+z)-(x-y-z) = x-y+z-x+y+z = 2z$, which is $\\boxed{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/1.json | AHSME |
1986_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | In $\triangle ABC, AB = 13, BC = 14$ and $CA = 15$. Also, $M$ is the midpoint of side $AB$ and $H$ is the foot of the altitude from $A$ to $BC$.
The length of $HM$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A); draw(B--C--A--B^^M--H--A^^rightanglemark(... | [
"In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is $13$, the median must be $6.5$.\n\n\n",
"Warning: this solution is very intensive in calculation. Please do NOT try this on the test!\n\n\nLet's start by finding $AH$. By Heron's Formula, $s... | 2 | ./CreativeMath/AHSME/1986_AHSME_Problems/11.json | AHSME |
1986_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | If the line $L$ in the $xy$-plane has half the slope and twice the $y$-intercept of the line $y = \frac{2}{3} x + 4$, then an equation for $L$ is:
$\textbf{(A)}\ y = \frac{1}{3} x + 8 \qquad \textbf{(B)}\ y = \frac{4}{3} x + 2 \qquad \textbf{(C)}\ y =\frac{1}{3}x+4\qquad\\ \textbf{(D)}\ y =\frac{4}{3}x+4\qquad \text... | [
"The original slope and $y$-intercept are $\\frac{2}{3}$ and $4$, so the new ones are $\\frac{1}{3}$ and $8$ respectively. Thus, using the slope-intercept form ($y = mx+c$), the new equation is $y=\\frac{1}{3}x + 8$, which is $\\boxed{A}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/2.json | AHSME |
1986_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended,
respectively. Let $O$ be the center of the pentagon. If $OP = 1$, then $AO + AQ + AR$ equals
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=... | [
"To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles.\n[asy] unitsize(2 cm); pair A, B, C, D, E, O, P, Q, R; A = dir(90); B = dir(90 - 360/5); C = dir(90 - 2*360/5); D = dir(90 - 3*360/5); E = dir(... | 2 | ./CreativeMath/AHSME/1986_AHSME_Problems/28.json | AHSME |
1986_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | John scores $93$ on this year's AHSME. Had the old scoring system still been in effect, he would score only $84$ for the same answers.
How many questions does he leave unanswered? (In the new scoring system that year, one receives $5$ points for each correct answer,
$0$ points for each wrong answer, and $2$ points for... | [
"Let $c$, $w$, and $u$ be the number of correct, wrong, and unanswered questions respectively. From the old scoring system, we have $30+4c-w=84$, from the new scoring system we have $5c+2u=93$, and since there are $30$ problems in the AHSME, $c+w+u=30$. Solving the simultaneous equations yields $u=9$, which is $\\b... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/12.json | AHSME |
1986_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | Let $p(x) = x^{2} + bx + c$, where $b$ and $c$ are integers.
If $p(x)$ is a factor of both $x^{4} + 6x^{2} + 25$ and $3x^{4} + 4x^{2} + 28x + 5$, what is $p(1)$?
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ 8$
| [
"$p(x)$ must be a factor of $3(x^4+6x^2+25)-(3x^4+4x^2+28x+5)=14x^2-28x+70=14(x^2-2x+5)$.\n\n\nTherefore $p(x)=x^2 -2x+5$ and $p(1)=4$.\n\n\nThe answer is $\\fbox{(D) 4}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/24.json | AHSME |
1986_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | If $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$, then
$\sum_{N=1}^{1024} \lfloor \log_{2}N\rfloor =$
$\textbf{(A)}\ 8192\qquad \textbf{(B)}\ 8204\qquad \textbf{(C)}\ 9218\qquad \textbf{(D)}\ \lfloor\log_{2}(1024!)\rfloor\qquad \textbf{(E)}\ \text{none of these}$
| [
"Because $1 \\le N \\le 1024$, we have $0 \\le \\lfloor \\log_{2}N\\rfloor \\le 10$. We count how many times $\\lfloor \\log_{2}N\\rfloor$ attains a certain value. \n\n\nFor all $k$ except for $k=10$, we have that $\\lfloor \\log_{2}N\\rfloor = k$ is satisfied by all $2^k \\le N<2^{k+1}$, for a total of $2^k$ value... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/25.json | AHSME |
1986_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | A parabola $y = ax^{2} + bx + c$ has vertex $(4,2)$. If $(2,0)$ is on the parabola, then $abc$ equals
$\textbf{(A)}\ -12\qquad \textbf{(B)}\ -6\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 12$
| [
"Consider the quadratic in completed square form: it must be $y=a(x-4)^{2}+2$. Now substitute $x=2$ and $y=0$ to give $a=-\\frac{1}{2}$. Now expanding gives $y=-\\frac{1}{2}x^{2}+4x-6$, so the product is $-\\frac{1}{2} \\cdot 4 \\cdot -6 = 3 \\cdot 4 = 12$, which is $\\boxed{E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/13.json | AHSME |
1986_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | Two of the altitudes of the scalene triangle $ABC$ have length $4$ and $12$.
If the length of the third altitude is also an integer, what is the biggest it can be?
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{none of these}$
| [
"Assume we have a scalene triangle $ABC$. Arbitrarily, let $12$ be the height to base $AB$ and $4$ be the height to base $AC$. Due to area equivalences, the base $AC$ must be three times the length of $AB$. \n\n\nLet the base $AB$ be $x$, thus making $AC = 3x$. Thus, setting the final height to base $BC$ to $h$, we... | 2 | ./CreativeMath/AHSME/1986_AHSME_Problems/29.json | AHSME |
1986_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | $\triangle ABC$ has a right angle at $C$ and $\angle A = 20^\circ$. If $BD$ ($D$ in $\overline{AC}$) is the bisector of $\angle ABC$, then $\angle BDC =$
$\textbf{(A)}\ 40^\circ \qquad \textbf{(B)}\ 45^\circ \qquad \textbf{(C)}\ 50^\circ \qquad \textbf{(D)}\ 55^\circ\qquad \textbf{(E)}\ 60^\circ$
| [
"Since $\\angle C = 90^{\\circ}$ and $\\angle A = 20^{\\circ}$, we have $\\angle ABC = 70^{\\circ}$. Thus $\\angle DBC = 35^{\\circ}$. It follows that $\\angle BDC = 90^{\\circ} - 35^{\\circ} = 55^{\\circ}$, which is $\\boxed{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/3.json | AHSME |
1986_AHSME_Problems | 8 | 0 | Arithmetic | Multiple Choice | The population of the United States in $1980$ was $226,504,825$. The area of the country is $3,615,122$ square miles. There are $(5280)^{2}$
square feet in one square mile. Which number below best approximates the average number of square feet per person?
$\textbf{(A)}\ 5,000\qquad \textbf{(B)}\ 10,000\qquad \textbf... | [
"With about $230$ million people and under $4$ million square miles, there are about $60$ people per square mile. Since a square mile is about $(5000 \\ \\text{ft})^{2} = 25$ million square feet, that gives approximately $\\frac{25}{60}$ of a million square feet per person. $\\frac{25}{60}$ is approximately half, s... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/8.json | AHSME |
1986_AHSME_Problems | 22 | 0 | Probability | Multiple Choice | Six distinct integers are picked at random from $\{1,2,3,\ldots,10\}$. What is the probability that, among those selected, the second smallest is $3$?
$\textbf{(A)}\ \frac{1}{60}\qquad \textbf{(B)}\ \frac{1}{6}\qquad \textbf{(C)}\ \frac{1}{3}\qquad \textbf{(D)}\ \frac{1}{2}\qquad \textbf{(E)}\ \text{none of these}$
... | [
"The total number of ways to choose 6 numbers is ${10\\choose 6} = 210$.\n\n\nAssume $3$ is the second-lowest number. There are $5$ numbers left to choose, $4$ of which must be greater than $3$, and $1$ of which must be less than $3$. This is equivalent to choosing $4$ numbers from the $7$ numbers larger than $3$,... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/22.json | AHSME |
1986_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | A plane intersects a right circular cylinder of radius $1$ forming an ellipse.
If the major axis of the ellipse is $50\%$ longer than the minor axis, the length of the major axis is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ \frac{3}{2}\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ \frac{9}{4}\qquad \textbf{(E)}\ 3$
| [
"We note that we can draw the minor axis to see that because the minor axis is the minimum distance between two opposite points on the ellipse, we can draw a line through two opposite points of the cylinder, and so the minor axis is $2(1) = 2$. Therefore, our answer is $2(1.5) = 3$, and so our answer is $\\boxed{E}... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/18.json | AHSME |
1986_AHSME_Problems | 4 | 0 | Number Theory | Multiple Choice | Let S be the statement
"If the sum of the digits of the whole number $n$ is divisible by $6$, then $n$ is divisible by $6$."
A value of $n$ which shows $S$ to be false is
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ \text{ none of these}$
| [
"For a counterexample, we need a number whose digit sum is divisible by $6$, but which is not itself divisible by $6$. $33$ satisfies these conditions, as $3+3=6$ but $6$ does not divide $33$, so the answer is $\\boxed{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/4.json | AHSME |
1986_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | Suppose hops, skips and jumps are specific units of length. If $b$ hops equals $c$ skips, $d$ jumps equals $e$ hops,
and $f$ jumps equals $g$ meters, then one meter equals how many skips?
$\textbf{(A)}\ \frac{bdg}{cef}\qquad \textbf{(B)}\ \frac{cdf}{beg}\qquad \textbf{(C)}\ \frac{cdg}{bef}\qquad \textbf{(D)}\ \frac{... | [
"$1$ metre equals $\\frac{f}{g}$ jumps, which is $\\frac{f}{g} \\frac{e}{d}$ hops, and then $\\frac{f}{g} \\frac{e}{d} \\frac{c}{b}$ skips, which becomes $\\frac{cef}{bdg}$, i.e. answer $\\boxed{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/14.json | AHSME |
1986_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | A student attempted to compute the average $A$ of $x, y$ and $z$ by computing the average of $x$ and $y$,
and then computing the average of the result and $z$. Whenever $x < y < z$, the student's final result is
$\textbf{(A)}\ \text{correct}\quad \textbf{(B)}\ \text{always less than A}\quad \textbf{(C)}\ \text{alway... | [
"The true average is $A=\\frac{x+y+z}{3}$, and the student computed $B=\\frac{\\frac{x+y}{2}+z}{2}=\\frac{x+y+2z}{4}$, so $B-A = \\frac{2z-x-y}{12} = \\frac{(z-x)+(z-y)}{12}$, which is always positive as $z>x$ and $z>y$. Thus $B$ is always greater than $A$, i.e. $\\boxed{C}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/15.json | AHSME |
1986_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | Simplify $\left(\sqrt[6]{27} - \sqrt{6 \frac{3}{4} }\right)^2$
$\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{\sqrt 3}{2} \qquad \textbf{(C)}\ \frac{3\sqrt 3}{4}\qquad \textbf{(D)}\ \frac{3}{2}\qquad \textbf{(E)}\ \frac{3\sqrt 3}{2}$
| [
"We have $\\sqrt[6]{27} = (3^{3})^{\\frac{1}{6}} = 3^{\\frac{1}{2}} = \\sqrt{3}$ and $\\sqrt{6 \\frac{3}{4}} = \\sqrt{\\frac{27}{4}} = \\frac{3 \\sqrt{3}}{2}$, so the answer is $(\\sqrt{3} - \\frac{3 \\sqrt{3}}{2})^{2} = (-\\frac{\\sqrt{3}}{2})^{2} = \\frac{3}{4}$, which is $\\boxed{A}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/5.json | AHSME |
1986_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | A park is in the shape of a regular hexagon $2$ km on a side. Starting at a corner,
Alice walks along the perimeter of the park for a distance of $5$ km.
How many kilometers is she from her starting point?
$\textbf{(A)}\ \sqrt{13}\qquad \textbf{(B)}\ \sqrt{14}\qquad \textbf{(C)}\ \sqrt{15}\qquad \textbf{(D)}\ \sqrt... | [
"We imagine this problem on a coordinate plane and let Alice's starting position be the origin. We see that she will travel along two edges and then go halfway along a third. Therefore, her new $x$-coordinate will be $1 + 2 + \\frac{1}{2} = \\frac{7}{2}$ because she travels along a distance of $2 \\cdot \\frac{1}{2... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/19.json | AHSME |
1986_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | Let N = $69^{5} + 5\cdot69^{4} + 10\cdot69^{3} + 10\cdot69^{2} + 5\cdot69 + 1$. How many positive integers are factors of $N$?
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 69\qquad \textbf{(D)}\ 125\qquad \textbf{(E)}\ 216$
| [
"Let $69=a$. Therefore, the equation becomes $a^5+5a^4+10a^3+10a^2+5a+1$. From Pascal's Triangle, we know this equation is equal to $(a+1)^5$. Simplifying, we have the desired sum is equal to $70^5$ which can be prime factorized as $2^5\\cdot5^5\\cdot7^5$. Finally, we can count the number of factors of this number.... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/23.json | AHSME |
1986_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | The product $\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\ldots\left(1-\frac{1}{9^{2}}\right)\left(1-\frac{1}{10^{2}}\right)$ equals
$\textbf{(A)}\ \frac{5}{12}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ \frac{11}{20}\qquad \textbf{(D)}\ \frac{2}{3}\qquad \textbf{(E)}\ \frac{7}{10}$
| [
"Factor each term in the product as a difference of two squares, and group together all the terms that contain a $-$ sign, and all those that contain a $+$ sign. This gives $[(1-\\frac{1}{2})(1-\\frac{1}{3})(1-\\frac{1}{4})...(1-\\frac{1}{10})] \\cdot [(1+\\frac{1}{2})(1+\\frac{1}{3})(1+\\frac{1}{4})...(1+\\frac{1}... | 1 | ./CreativeMath/AHSME/1986_AHSME_Problems/9.json | AHSME |
1993_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | Consider the equation $10z^2-3iz-k=0$, where $z$ is a complex variable and $i^2=-1$. Which of the following statements is true?
$\text{(A) For all positive real numbers k, both roots are pure imaginary} \quad\\ \text{(B) For all negative real numbers k, both roots are pure imaginary} \quad\\ \text{(C) For all pure im... | [
"Let $r_1$ and $r_2$ denote the roots of the polynomial. Then $r_1 + r_2 = 3i$ is pure imaginary, so $r_1$ and $r_2$ have offsetting real parts. Write $r_1 = a + bi$ and $r_2 = -a + ci$. \n\n\nNow $-k = r_1 r_2 = -a^2 -bc + a(c-b)i$. In the case that $k$ is real, then $a(c-b)=0$ so either $a=0$ or that $b=c$. ... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/20.json | AHSME |
1993_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | Consider the non-decreasing sequence of positive integers
\[1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots\]
in which the $n^{th}$ positive integer appears $n$ times. The remainder when the $1993^{rd}$ term is divided by $5$ is
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4$
| [
"The sequence of 1's ends at position 1, and the sequence of 2's ends at position 1+2, and the sequence of $n$'s ends at position $1+2+\\dots+n$.\n\n\nTherefore we want to find the smallest integer $n$ that satisfies $\\frac{n(n+1)}{2}\\geq 1993$.\n\n\nBy trial and error, the value of $n$ is $63$, and $63 \\div 5$ ... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/16.json | AHSME |
1993_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | $\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=$
$\text{(A) } \sqrt{2}\quad \text{(B) } 16\quad \text{(C) } 32\quad \text{(D) } (12)^{\tfrac{2}{3}}\quad \text{(E) } 512.5$
| [
"$\\sqrt{\\frac{ 8^{10}+4^{10} }{8^4 + 4^{11}}} = \\sqrt{\\frac{2^{30}+2^{20}}{2^{12}+2^{22}}}= \\sqrt{\\frac{2^{20}(2^{10}+1)}{2^{12}(1+2^{10})}} = \\sqrt{2^8} = 2^4$\n\n\n$\\fbox{B}$\n\n\n",
"$8^{10}+4^{10} = 1074790400$\n\n\n$8^4 + 4^{11} = 4198400$\n\n\n$\\frac{1074790400}{4198400} = 256$\n\n\n$\\sqrt{256} = ... | 2 | ./CreativeMath/AHSME/1993_AHSME_Problems/6.json | AHSME |
1993_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | The symbol $R_k$ stands for an integer whose base-ten representation is a sequence of $k$ ones. For example, $R_3=111,R_5=11111$, etc. When $R_{24}$ is divided by $R_4$, the quotient $Q=R_{24}/R_4$ is an integer whose base-ten representation is a sequence containing only ones and zeroes. The number of zeros in $Q$ is:
... | [
"Note $R_n = \\sum_{k=0}^{n-1} 10^k = \\frac{10^n - 1}{10-1}$. \n\n\nTherefore $\\frac{R_{24}}{R_4} = \\frac{ 10^{24}-1 }{10^4-1}$. \n\n\nWe can recognize this is also the formula for the sum of a geometric series $1+10^4 + (10^4)^2 + \\dots + (10^4)^5 = 1+ 10^4 + 10^8 + \\dots + 10^{20}$\n\n\nNow the 1's place h... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/7.json | AHSME |
1993_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | [asy] draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle, black+linewidth(.75)); draw((0,-1)--(0,1), black+linewidth(.75)); draw((-1,0)--(1,0), black+linewidth(.75)); draw((-1,-1/sqrt(3))--(1,1/sqrt(3)), black+linewidth(.75)); draw((-1,1/sqrt(3))--(1,-1/sqrt(3)), black+linewidth(.75)); draw((-1/sqrt(3),-1)--(1/sqrt(3),1), blac... | [
"Assume the length of the side of the square is 2, WLOG. This means the side of one t section is 1. As the lines are at clock face positions, each section has a $\\tfrac{360}{12} = 30$ degree angle from the center. So each section t is a $30-60-90$ triangle with a long leg of 1. Therefore, the short leg is $\\t... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/17.json | AHSME |
1993_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | Let $a_1,a_2,\cdots,a_k$ be a finite arithmetic sequence with $a_4 +a_7+a_{10} = 17$ and $a_4+a_5+\cdots+a_{13} +a_{14} = 77$.
If $a_k = 13$, then $k =$
$\text{(A) } 16\quad \text{(B) } 18\quad \text{(C) } 20\quad \text{(D) } 22\quad \text{(E) } 24$
| [
"Note that $a_7-3d=a_4$ and $a_7+3d=a_{10}$ where $d$ is the common difference, so $a_4+a_7+a_{10}=3a_7=17$, or $a_7=\\frac{17}{3}$.\n\n\nLikewise, we can write every term in the second equation in terms of $a_9$, giving us $11a_9=77\\implies a_9=7$.\n\n\nThen the common difference is $\\frac{2}{3}$. Then $a_k-a_9=... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/21.json | AHSME |
1993_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | Let $r$ be the number that results when both the base and the exponent of $a^b$ are tripled, where $a,b>0$. If $r$ equals the product of $a^b$ and $x^b$ where $x>0$, then $x=$
$\text{(A) } 3\quad \text{(B) } 3a^2\quad \text{(C) } 27a^2\quad \text{(D) } 2a^{3b}\quad \text{(E) } 3a^{2b}$
| [
"We have $r=(3a)^{3b}$\n\n\nFrom this we have the equation $(3a)^{3b}=a^bx^b$\n\n\nRaising both sides to the $\\frac{1}{b}$ power we get that $27a^3=ax$ or $x=27a^2$\n\n\n$\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/10.json | AHSME |
1993_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | Find the largest positive value attained by the function
$f(x)=\sqrt{8x-x^2}-\sqrt{14x-x^2-48}$, $x$ a real number.
$\text{(A) } \sqrt{7}-1\quad \text{(B) } 3\quad \text{(C) } 2\sqrt{3}\quad \text{(D) } 4\quad \text{(E) } \sqrt{55}-\sqrt{5}$
| [
"We can rewrite the function as $f(x) = \\sqrt{x (8 - x)} - \\sqrt{(x - 6) (8 - x)}$ and then factor it to get $f(x) = \\sqrt{8 - x} \\left(\\sqrt{x} - \\sqrt{x - 6}\\right)$. From the expressions under the square roots, it is clear that $f(x)$ is only defined on the interval $[6, 8]$.\n\n\nThe $\\sqrt{8 - x}$ fact... | 2 | ./CreativeMath/AHSME/1993_AHSME_Problems/26.json | AHSME |
1993_AHSME_Problems | 30 | 0 | Number Theory | Multiple Choice | Given $0\le x_0<1$, let
\[x_n=\left\{ \begin{array}{ll} 2x_{n-1} &\text{ if }2x_{n-1}<1 \\ 2x_{n-1}-1 &\text{ if }2x_{n-1}\ge 1 \end{array}\right.\]
for all integers $n>0$. For how many $x_0$ is it true that $x_0=x_5$?
$\text{(A) 0} \quad \text{(B) 1} \quad \text{(C) 5} \quad \text{(D) 31} \quad \text{(E) }\infty$
... | [
"We are going to look at this problem in binary. \n\n\n$x_0 = (0.a_1 a_2 \\cdots )_2$\n\n\n$2x_0 = (a_1.a_2 a_3 \\cdots)_2$\n\n\nIf $2x_0 < 1$, then $x_0 < \\frac{1}{2}$ which means that $a_1 = 0$ and so $x_1 = (.a_2 a_3 a_4 \\cdots)_2$\n\n\nIf $2x_0 \\geq 1$ then $x \\geq \\frac{1}{2}$ which means that $x_1 = 2x_0... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/30.json | AHSME |
1993_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | [asy] draw(circle((4,1),1),black+linewidth(.75)); draw((0,0)--(8,0)--(8,6)--cycle,black+linewidth(.75)); MP("A",(0,0),SW);MP("B",(8,0),SE);MP("C",(8,6),NE);MP("P",(4,1),NW); MP("8",(4,0),S);MP("6",(8,3),E);MP("10",(4,3),NW); MP("->",(5,1),E); dot((4,1)); [/asy]
The sides of $\triangle ABC$ have lengths $6,8,$ and $10$.... | [
"[asy] draw(circle((4,1),1),black+linewidth(.75)); draw((0,0)--(8,0)--(8,6)--cycle,black+linewidth(.75)); draw((3,1)--(7,1)--(7,4)--cycle,black+linewidth(.75)); draw((3,1)--(3,0),black+linewidth(.75)); draw((3,1)--(2.4,1.8),black+linewidth(.75)); draw((7,1)--(8,1),black+linewidth(.75)); draw((7,1)--(7,0),black+line... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/27.json | AHSME |
1993_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | For integers $a,b,$ and $c$ define $\fbox{a,b,c}$ to mean $a^b-b^c+c^a$. Then $\fbox{1,-1,2}$ equals:
$\text{(A) } -4\quad \text{(B) } -2\quad \text{(C) } 0\quad \text{(D) } 2\quad \text{(E) } 4$
| [
"Plug in the values for $a,b,c$ and you get $1^{-1} - (-1)^2 + 2^1 \\Rightarrow 1-1+2 \\Rightarrow \\fbox{2}$\n\n\n$\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/1.json | AHSME |
1993_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | If $\log_2(\log_2(\log_2(x)))=2$, then how many digits are in the base-ten representation for x?
$\text{(A) } 5\quad \text{(B) } 7\quad \text{(C) } 9\quad \text{(D) } 11\quad \text{(E) } 13$
| [
"Taking successive exponentials $\\log_2(\\log_2(x)) = 2^2 = 4$ and $\\log_2(x) = 2^4=16$ and $x = 2^{16}$. Now $2^{10} = 1024 \\approx 10^3$ and $2^6 = 64$ so we can approximate $2^{16} \\approx 64000$ which has 5 digits. In general, $2^n$ has approximately $n/3$ digits.\n\n\n$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/11.json | AHSME |
1993_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | [asy] draw((-5,0)--(5,0)--(2,14)--cycle,black+linewidth(.75)); draw((-2.25,5.5)--(4,14/3),black+linewidth(.75)); MP("A",(-5,0),S);MP("C",(5,0),S);MP("B",(2,14),N);MP("E",(4,14/3),E);MP("D",(-2.25,5.5),W); MP("55^\circ",(-4.5,0),NE);MP("75^\circ",(5,0),NW); [/asy]
In $\triangle ABC$, $\angle A=55^\circ$, $\angle C=75... | [
"We first consider $\\angle CBA$. Because $\\angle A = 55$ and $\\angle C = 75$, $\\angle B = 180 - 55 - 75 = 50$. Then, because $\\triangle BED$ is isosceles, we have the equation $2 \\angle BED + 50 = 180$. Solving this equation gives us $\\angle BED = 65 \\rightarrow \\fbox{\\textbf{(D)}65}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/2.json | AHSME |
1993_AHSME_Problems | 28 | 0 | Counting | Multiple Choice | How many triangles with positive area are there whose vertices are points in the $xy$-plane whose coordinates are integers $(x,y)$ satisfying $1\le x\le 4$ and $1\le y\le 4$?
$\text{(A) } 496\quad \text{(B) } 500\quad \text{(C) } 512\quad \text{(D) } 516\quad \text{(E) } 560$
| [
"The vertices of the triangles are limited to a $4\\times4$ grid, with $16$ points total. Every triangle is determined by $3$ points chosen from these $16$ for a total of $\\binom{16}{3}=560$. However, triangles formed by collinear points do not have positive area. For each column or row, there are $\\binom{4}{3}=4... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/28.json | AHSME |
1993_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | If $f(2x)=\frac{2}{2+x}$ for all $x>0$, then $2f(x)=$
$\text{(A) } \frac{2}{1+x}\quad \text{(B) } \frac{2}{2+x}\quad \text{(C) } \frac{4}{1+x}\quad \text{(D) } \frac{4}{2+x}\quad \text{(E) } \frac{8}{4+x}$
| [
"As $f(2x)=\\frac{2}{2+x}$, we have that $f(x)=\\frac{2}{2+\\frac{x}{2}}$. This also means that $2f(x)=\\frac{4}{2+\\frac{x}{2}}$ which implies that the answer is $\\fbox{E}$. ~ samrocksnature\n\n\nNote: Wait what\n\n\n"
] | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/12.json | AHSME |
1993_AHSME_Problems | 24 | 0 | Probability | Multiple Choice | A box contains $3$ shiny pennies and $4$ dull pennies. One by one, pennies are drawn at random from the box and not replaced. If the probability is $a/b$ that it will take more than four draws until the third shiny penny appears and $a/b$ is in lowest terms, then $a+b=$
$\text{(A) } 11\quad \text{(B) } 20\quad \text{... | [
"First let’s try to find the number of possible unique combinations. I’ll denote shiny coins as 1 and dull coins as 0. \n\n\nNow, each configuration can be represented by a string of 1s and 0s e.g. 0110100. Notice that a combination can be uniquely determined solely by the placement of their 0s OR 1s e.g. 1 - - 1 1... | 2 | ./CreativeMath/AHSME/1993_AHSME_Problems/24.json | AHSME |
1993_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | [asy] draw((0,0)--(1,sqrt(3)),black+linewidth(.75),EndArrow); draw((0,0)--(1,-sqrt(3)),black+linewidth(.75),EndArrow); draw((0,0)--(1,0),dashed+black+linewidth(.75)); dot((1,0)); MP("P",(1,0),E); [/asy]
Let $S$ be the set of points on the rays forming the sides of a $120^{\circ}$ angle, and let $P$ be a fixed point i... | [
"$\\fbox{E}$\n\n\nTake the \"obvious\" equilateral triangle $OAP$, where $O$ is the vertex, $A$ is on the upper ray, and $P$ is our central point. Slide $A$ down on the top ray to point $A'$, and slide $O$ down an equal distance on the bottom ray to point $O'$.\n\n\nNow observe $\\triangle AA'P$ and $\\triangle OO... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/25.json | AHSME |
1993_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | A square of perimeter 20 is inscribed in a square of perimeter 28. What is the greatest distance between a vertex of the inner square and a vertex of the outer square?
$\text{(A) } \sqrt{58}\quad \text{(B) } \frac{7\sqrt{5}}{2}\quad \text{(C) } 8\quad \text{(D) } \sqrt{65}\quad \text{(E) } 5\sqrt{3}$
| [
"Assume one of the segments bisected by the inscribed square has length $x$. Thus, the alternate segment has length $7-x$. Applying Pythagorean's Theorem, $x^2+(x-7)^2=5^2$. Simplifying, $(x-3)(x-4)=0$, so $x=3$ or $x=4$ (it does not matter, as rotations produce the same figure). The longest line that can be made f... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/13.json | AHSME |
1993_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | Which of the following could NOT be the lengths of the external diagonals of a right regular prism [a "box"]? (An $\textit{external diagonal}$ is a diagonal of one of the rectangular faces of the box.)
$\text{(A) }\{4,5,6\} \quad \text{(B) } \{4,5,7\} \quad \text{(C) } \{4,6,7\} \quad \text{(D) } \{5,6,7\} \quad \te... | [
"Let $a,$ $b,$ and $c$ be the side lengths of the rectangular prism. By Pythagoras, the lengths of the external diagonals are $\\sqrt{a^2 + b^2},$ $\\sqrt{b^2 + c^2},$ and $\\sqrt{a^2 + c^2}.$ If we square each of these to obtain $a^2 + b^2,$ $b^2 + c^2,$ and $a^2 + c^2,$ we observe that since each of $a,$ $b,$ and... | 2 | ./CreativeMath/AHSME/1993_AHSME_Problems/29.json | AHSME |
1993_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | $\frac{15^{30}}{45^{15}} =$
$\text{(A) } \left(\frac{1}{3}\right)^{15}\quad \text{(B) } \left(\frac{1}{3}\right)^{2}\quad \text{(C) } 1\quad \text{(D) } 3^{15}\quad \text{(E) } 5^{15}$
| [
"First we must convert these to the same bases. We can rewrite $45^{15}$ as $15^{15} \\cdot 3^{15}$ Now\n\n\n$\\frac{15^{30}}{15^{15} \\cdot 3^{15}}$\n\n\nCanceling.... \n\n\n$\\frac{15^{15}}{3^{15}} \\Rightarrow (\\frac{15}{3})^{15} \\Rightarrow 5^{15}$\n\n\n$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/3.json | AHSME |
1993_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | Let $C_1$ and $C_2$ be circles of radius 1 that are in the same plane and tangent to each other. How many circles of radius 3 are in this plane and tangent to both $C_1$ and $C_2$?
$\text{(A) } 2\quad \text{(B) } 4\quad \text{(C) } 5\quad \text{(D) } 6\quad \text{(E) } 8$
| [
"There are two radius 3 circles to which $C_1$ and $C_2$ are both externally tangent. One touches the tops of $C_1$ and $C_2$ and extends upward, and the other the other touches the bottoms and extends downward. There are also two radius 3 circles to which $C_1$ and $C_2$ are both internally tangent, one touching... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/8.json | AHSME |
1993_AHSME_Problems | 22 | 0 | Counting | Multiple Choice | [asy] size((400)); draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0), linewidth(1)); draw((5,0)--(10,0)--(15,0)--(20,0)--(20,5)--(15,5)--(10,5)--(5,5)--(6,7)--(11,7)--(16,7)--(21,7)--(21,2)--(20,0), linewidth(1)); draw((10,0)--(10,5)--(11,7), linewidth(1)); draw((15,0)--(15,5)--(16,7), linewidth(1)); draw((20,0)--(20,5)--(21,7), ... | [
"The value assigned at the top is the weighted sum of the values in the cubes of a given level. The weight for a given cube is the sum of the weights of the blocks which touch it above, since the number in a cube will get incorporated into the overall sum via each of those blocks, according to the weight of each b... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/22.json | AHSME |
1993_AHSME_Problems | 18 | 0 | Number Theory | Multiple Choice | Al and Barb start their new jobs on the same day. Al's schedule is 3 work-days followed by 1 rest-day. Barb's schedule is 7 work-days followed by 3 rest-days. On how many of their first 1000 days do both have rest-days on the same day?
$\text{(A) } 48\quad \text{(B) } 50\quad \text{(C) } 72\quad \text{(D) } 75\quad \... | [
"We note that Al and Barb's work schedules clearly form a cycle, with Al's work cycle being 4 days long and Barb's work schedule being 10 days long. Since $\\text{LCM}\\left(4,10\\right)=20$, we know that Al's and Barb's work cycles coincide, and thus collectively repeat, every 20 days. As a result, we only need ... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/18.json | AHSME |
1993_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | Define the operation "$\circ$" by $x\circ y=4x-3y+xy$, for all real numbers $x$ and $y$. For how many real numbers $y$ does $3\circ y=12$?
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) more than 4}$
| [
"Note that $3 \\circ y = 4 \\cdot 3 - 3y + 3y = 12$, so $3 \\circ y = 12$ is true for all values of $y$. Thus there are more than four solutions, and the answer is $\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/4.json | AHSME |
1993_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | [asy] draw((-1,0)--(1,0)--(1+sqrt(2),sqrt(2))--(0,sqrt(2)+sqrt(13-2*sqrt(2)))--(-1-sqrt(2),sqrt(2))--cycle,black+linewidth(.75)); MP("A",(-1,0),SW);MP("B",(1,0),SE);MP("C",(1+sqrt(2),sqrt(2)),E);MP("D",(0,sqrt(2)+sqrt(13-2*sqrt(2))),N);MP("E",(-1-sqrt(2),sqrt(2)),W); dot((-1,0));dot((1,0));dot((1+sqrt(2),sqrt(2)));dot(... | [
"[asy] draw((-1,0)--(1,0)--(1+sqrt(2),sqrt(2))--(0,sqrt(2)+sqrt(13-2*sqrt(2)))--(-1-sqrt(2),sqrt(2))--cycle,black+linewidth(.75)); draw((1+sqrt(2),sqrt(2))--(-1-sqrt(2),sqrt(2))); draw((-1,0)--(-1,sqrt(2))); draw((1,0)--(1,sqrt(2))); MP(\"F\",(-1,sqrt(2)),N);MP(\"G\",(1,sqrt(2)),N); MP(\"A\",(-1,0),SW);MP(\"B\",(1,... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/14.json | AHSME |
1993_AHSME_Problems | 15 | 0 | Number Theory | Multiple Choice | For how many values of $n$ will an $n$-sided regular polygon have interior angles with integral measures?
$\text{(A) } 16\quad \text{(B) } 18\quad \text{(C) } 20\quad \text{(D) } 22\quad \text{(E) } 24$
| [
"Start with the facts that all polygons have their exterior angles sum to 360 and the exterior and interior angles make a linear pair of angles. So our goal is to find the number of divisors of 360 to make both the interior and exterior angles integers. The prime factorization of 360 is $2^3 * 3^2 * 5$. That means ... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/15.json | AHSME |
1993_AHSME_Problems | 5 | 0 | Arithmetic | Multiple Choice | Last year a bicycle cost $160 and a cycling helmet $40. This year the cost of the bicycle increased by $5\%$, and the cost of the helmet increased by $10\%$. The percent increase in the combined cost of the bicycle and the helmet is:
$\text{(A) } 6\%\quad \text{(B) } 7\%\quad \text{(C) } 7.5\%\quad \text{(D) } 8\%\qu... | [
"Since the bicycle originally cost $160, the new cost will be 160 * 5% = 160 * 1/20 = 8, and 160 + 8 = 168. The new cost of the helmet will be 40 * 10% = 40 * 1/10 = 4, and 40 + 4 = 44. 160 + 40 = 200, and 168 + 44 = 212, so the total percent increase is 12/200 = 6%, or A.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/5.json | AHSME |
1993_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | How many ordered pairs $(m,n)$ of positive integers are solutions to
\[\frac{4}{m}+\frac{2}{n}=1?\]
$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } \text{more than }6$
| [
"Multiply both sides by $mn$ to clear the denominator. Moving all the terms to the right hand side, the equation becomes $0 = mn-4n-2m$. Adding 8 to both sides allows us to factor the equation as follows: $(m-4)(n-2) = 8$. Since the problem only wants integer pairs $(m,n)$, the pairs are given by the factors of 8, ... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/19.json | AHSME |
1993_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | [asy] draw(circle((0,0),10),black+linewidth(.75)); draw((-10,0)--(10,0),black+linewidth(.75)); draw((-10,0)--(9,sqrt(19)),black+linewidth(.75)); draw((-10,0)--(9,-sqrt(19)),black+linewidth(.75)); draw((2,0)--(9,sqrt(19)),black+linewidth(.75)); draw((2,0)--(9,-sqrt(19)),black+linewidth(.75)); MP("X",(2,0),N);MP("A",(-10... | [
"We have all the angles we need, but most obviously, we see that right angle in triangle $ABD$.\n\n\nNote also that angle $BAD$ is 6 degrees, so length $AB = cos(6)$ because the diameter, $AD$, is 1.\n\n\nNow, we can concentrate on triangle $ABX$ (after all, now we can decipher all angles easily and use Law of Sine... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/23.json | AHSME |
1993_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | Country $A$ has $c\%$ of the world's population and $d\%$ of the worlds wealth. Country $B$ has $e\%$ of the world's population and $f\%$ of its wealth. Assume that the citizens of $A$ share the wealth of $A$ equally,and assume that those of $B$ share the wealth of $B$ equally. Find the ratio of the wealth of a citizen... | [
"Let $W$ be the wealth of the world and $P$ be the population of the world. Hence the wealth of each citizen of $A$ is $w_A = \\frac{0.01d W}{0.01cP}=\\frac{dW}{cP}$. Similarly the wealth of each citizen of $B$ is $w_B =\\frac{eW}{fP}$. We divide $\\frac{w_A}{w_B} = \\frac{de}{cf}$ and see the answer is $\\fbox{... | 1 | ./CreativeMath/AHSME/1993_AHSME_Problems/9.json | AHSME |
1979_AHSME_Problems | 20 | 0 | Other | Multiple Choice | If $a=\tfrac{1}{2}$ and $(a+1)(b+1)=2$ then the radian measure of $\arctan a + \arctan b$ equals
$\textbf{(A) }\frac{\pi}{2}\qquad \textbf{(B) }\frac{\pi}{3}\qquad \textbf{(C) }\frac{\pi}{4}\qquad \textbf{(D) }\frac{\pi}{5}\qquad \textbf{(E) }\frac{\pi}{6}$
| [
"Solution by e_power_pi_times_i\n\n\nSince $a=\\frac{1}{2}$, $b=\\frac{1}{3}$. Now we evaluate $\\arctan a$ and $\\arctan b$. Denote $x$ and $\\theta$ such that $\\arctan x = \\theta$. Then $\\tan(\\arctan(x)) = \\tan(\\theta)$, and simplifying gives $x = \\tan(\\theta)$. So $a = \\tan(\\theta_a) = \\frac{1}{2}$ an... | 2 | ./CreativeMath/AHSME/1979_AHSME_Problems/20.json | AHSME |
1979_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | A circle with area $A_1$ is contained in the interior of a larger circle with area $A_1+A_2$. If the radius of the larger circle is $3$,
and if $A_1 , A_2, A_1 + A_2$ is an arithmetic progression, then the radius of the smaller circle is
$\textbf{(A) }\frac{\sqrt{3}}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }\frac{... | [
"Solution by e_power_pi_times_i\n\n\nThe area of the larger circle is $A_1 + A_2 = 9\\pi$. Then $A_1 , 9\\pi-A_1 , 9\\pi$ are in an arithmetic progression. Thus $9\\pi-(9\\pi-A_1) = 9\\pi-A_1-A_1$. This simplifies to $3A_1 = 9\\pi$, or $A_1 = 3\\pi$. The radius of the smaller circle is $\\boxed{\\textbf{(E) } \\sqr... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/16.json | AHSME |
1979_AHSME_Problems | 6 | 0 | Arithmetic | Multiple Choice | $\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7=$
$\textbf{(A) }-\frac{1}{64}\qquad \textbf{(B) }-\frac{1}{16}\qquad \textbf{(C) }0\qquad \textbf{(D) }\frac{1}{16}\qquad \textbf{(E) }\frac{1}{64}$
| [
"Solution by e_power_pi_times_i\n\n\nSimplifying, we have $\\frac{96}{64}+\\frac{80}{64}+\\frac{72}{64}+\\frac{68}{64}+\\frac{66}{64}+\\frac{65}{64}-7$, which is $\\frac{96+80+72+68+66+65}{64}-7 = \\frac{447}{64}-7 = 6\\frac{63}{64}-7 = \\boxed{\\textbf{(A) } -\\frac{1}{64}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/6.json | AHSME |
1979_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | The square of an integer is called a perfect square. If $x$ is a perfect square, the next larger perfect square is
$\textbf{(A) }x+1\qquad \textbf{(B) }x^2+1\qquad \textbf{(C) }x^2+2x+1\qquad \textbf{(D) }x^2+x\qquad \textbf{(E) }x+2\sqrt{x}+1$
| [
"Solution by e_power_pi_times_i\n\n\nSince $x$ is a perfect square, denote $k$ such that $k^2 = x$. Then the next perfect square is $(k+1)^2 = k^2+2k+1$. Substituting back in the equation $k = \\sqrt{x}$, the next square is $\\boxed{\\textbf{(E) }x+2\\sqrt{x}+1}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/7.json | AHSME |
1979_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | [asy] size(200); dotfactor=3; pair A=(0,0),B=(1,0),C=(2,0),D=(3,0),X=(1.2,0.7); draw(A--D); dot(A);dot(B);dot(C);dot(D); draw(arc((0.4,0.4),0.4,180,110),arrow = Arrow(TeXHead)); draw(arc((2.6,0.4),0.4,0,70),arrow = Arrow(TeXHead)); draw(B--X,dotted); draw(C--X,dotted); label("$A$",A,SW); label("$B$",B,S); label("$C$",C... | [
"Solution by e_power_pi_times_i\n\n\nWe know that this triangle has lengths of $x$, $y-x$, and $z-y$. Using the Triangle Inequality, we get $3$ inequalities: $2y>z, z>2x, 2x+z>0$. Therefore, we know that $\\textbf{I}$ is true and $\\textbf{III}$ is false. In $\\textbf{II}$, we have to prove $2y<2x+z$. We know that ... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/17.json | AHSME |
1979_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | The length of the hypotenuse of a right triangle is $h$ , and the radius of the inscribed circle is $r$.
The ratio of the area of the circle to the area of the triangle is
$\textbf{(A) }\frac{\pi r}{h+2r}\qquad \textbf{(B) }\frac{\pi r}{h+r}\qquad \textbf{(C) }\frac{\pi}{2h+r}\qquad \textbf{(D) }\frac{\pi r^2}{r^2+h... | [
"Solution by e_power_pi_times_i\n\n\nSince $rs = A$, where $r$ is the inradius, $s$ is the semiperimeter, and $A$ is the area, we have that the ratio of the area of the circle to the area of the triangle is $\\frac{\\pi r^2}{rs} = \\frac{\\pi r}{s}$. Now we try to express $s$ as $h$ and $r$. Denote the points where... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/21.json | AHSME |
1979_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | If $P_1P_2P_3P_4P_5P_6$ is a regular hexagon whose apothem (distance from the center to midpoint of a side) is $2$,
and $Q_i$ is the midpoint of side $P_iP_{i+1}$ for $i=1,2,3,4$, then the area of quadrilateral $Q_1Q_2Q_3Q_4$ is
$\textbf{(A) }6\qquad \textbf{(B) }2\sqrt{6}\qquad \textbf{(C) }\frac{8\sqrt{3}}{3}\qqua... | [
"Solution by e_power_pi_times_i\n\n\nNotice that quadrilateral $Q_1Q_2Q_3Q_4$ consists of $3$ equilateral triangles with side length $2$. Thus the area of the quadrilateral is $3\\cdot(\\frac{2^2\\cdot\\sqrt{3}}{4}) = 3\\cdot\\sqrt{3} = \\boxed{\\textbf{(D) } 3\\sqrt{3}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/10.json | AHSME |
1979_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | The function $f$ satisfies the functional equation $f(x) +f(y) = f(x + y ) - xy - 1$ for every pair $x,~ y$
of real numbers. If $f( 1) = 1$, then the number of integers $n \neq 1$ for which $f ( n ) = n$ is
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad \textbf{(E) }\infty$
| [
"We are given that $f(x) +f(y) = f(x + y ) - xy - 1$ and $f( 1) = 1$, so we can let $y = 1$. Thus we have:\n\n\n\\[f(x) + 1 = f(x + 1) - x - 1\\]\n\n\nRearranging gives a recursive formula for $f$:\n\n\n\\[f(x + 1) = f(x) + x + 2\\]\n\n\nWe notice that this is the recursive form for a quadratic, so f(x) must be of... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/26.json | AHSME |
1979_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | [asy] size(200); import cse5; pathpen=black; anglefontpen=black; pointpen=black; anglepen=black; dotfactor=3; pair A=(0,0),B=(0.5,0.5*sqrt(3)),C=(3,0),D=(1.7,0),EE; EE=(B+C)/2; D(MP("$A$",A,W)--MP("$B$",B,N)--MP("$C$",C,E)--cycle); D(MP("$E$",EE,N)--MP("$D$",D,S)); D(D);D(EE); MA("80^\circ",8,D,EE,C,0.1); MA("20^\circ"... | [
"Let $F$ be the point on the extension of side $AB$ past $B$ for which $AF=1$. Since $AF=AC$ and $\\measuredangle FAC = 60^\\circ$,$\\triangle ACF$ is equilateral. Let $G$ be the point on line segment $BF$ for which $\\measuredangle BCG=20^\\circ$. Then $\\triangle BCG$ is similar to $\\triangle DCE$ and $BC=2(EC)$... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/30.json | AHSME |
1979_AHSME_Problems | 27 | 0 | Probability | Multiple Choice | An ordered pair $( b , c )$ of integers, each of which has absolute value less than or equal to five, is chosen at random, with each
such ordered pair having an equal likelihood of being chosen. What is the probability that the equation $x^ 2 + bx + c = 0$ will
not have distinct positive real roots?
$\textbf{(A) }\fr... | [
"$\\boxed{E}$\n\n\nThere are $10$ cases where the roots are real, positive and distinct.\nThe roots to the quadratic equation are $\\frac{-b\\pm\\sqrt{b^2-4c}}{2}$. \n\n\nIn order for $-b-\\sqrt{b^2-4c}>0$ we certainly need $b<0$.\nAdditionally, for the roots to be both real and distinct we need $b^2-4c>0$. And las... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/27.json | AHSME |
1979_AHSME_Problems | 1 | 0 | Geometry | Multiple Choice | [asy] draw((-2,1)--(2,1)--(2,-1)--(-2,-1)--cycle); draw((0,0)--(0,-1)--(-2,-1)--(-2,0)--cycle); label("$F$",(0,0),E); label("$A$",(-2,1),W); label("$B$",(2,1),E); label("$C$", (2,-1),E); label("$D$",(-2,-1),WSW); label("$E$",(-2,0),W); label("$G$",(0,-1),S); //Credit to TheMaskedMagician for the diagram [/asy]
If rec... | [
"Solution by e_power_pi_times_i\n\n\nSince the dimensions of $DEFG$ are half of the dimensions of $ABCD$, the area of $DEFG$ is $\\dfrac{1}{2}\\cdot\\dfrac{1}{2}$ of $ABCD$, so the area of $ABCD$ is $\\dfrac{1}{4}\\cdot72 = \\boxed{\\textbf{(D) } 18}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/1.json | AHSME |
1979_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | Find a positive integral solution to the equation
$\frac{1+3+5+\dots+(2n-1)}{2+4+6+\dots+2n}=\frac{115}{116}$
$\textbf{(A) }110\qquad \textbf{(B) }115\qquad \textbf{(C) }116\qquad \textbf{(D) }231\qquad\\ \textbf{(E) }\text{The equation has no positive integral solutions.}$
| [
"Solution by e_power_pi_times_i\n\n\nNotice that the numerator and denominator are the sum of the first $n$ odd and even numbers, respectively. Then the numerator is $n^2$, and the denominator is $n(n+1)$. Then $\\frac{n}{n+1} = \\frac{115}{116}$, so $n = \\boxed{\\textbf{(B) } 115}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/11.json | AHSME |
1979_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals
$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$
| [
"Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into \\[(x+1)(y-1) = -1\\] Plugging in $-1$ and $1$ as the $x$ and $y$ sides respectively, we get $x = -2$ and $y = 2$. Plugging this in to $\\frac{1}{x}-\\frac{1}{y}$ gives us $\\boxed{-1}$ as our ... | 2 | ./CreativeMath/AHSME/1979_AHSME_Problems/2.json | AHSME |
1979_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | [asy] import cse5; pathpen=black; pointpen=black; dotfactor=3; pair A=(1,2),B=(2,0),C=(0,0); D(CR(A,1.5)); D(CR(B,1.5)); D(CR(C,1.5)); D(MP("$A$",A)); D(MP("$B$",B)); D(MP("$C$",C)); pair[] BB,CC; CC=IPs(CR(A,1.5),CR(B,1.5)); BB=IPs(CR(A,1.5),CR(C,1.5)); D(BB[0]--CC[1]); MP("$B'$",BB[0],NW);MP("$C'$",CC[1],NE); //Credi... | [
"The circles can be described in the cartesian plane as being centered at $(-1,0),(1,0)$ and $(0,\\sqrt{3})$ with radius $r$ by the equations \n\n\n$x^2+(y-\\sqrt{3})^2=r^2$\n\n\n$(x+1)^2+y^2=r^2$\n\n\n$(x-1)^2+y^2=r^2$.\n\n\nSolving the first 2 equations gives $x=1-\\sqrt{3}\\cdot y$ which when substituted back in... | 2 | ./CreativeMath/AHSME/1979_AHSME_Problems/28.json | AHSME |
1979_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | [asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\ci... | [
"Solution by e_power_pi_times_i\n\n\nBecause $AB = OD$, triangles $ABO$ and $BOE$ are isosceles. Denote $\\measuredangle BAO = \\measuredangle AOB = \\theta$. Then $\\measuredangle ABO = 180^\\circ-2\\theta$, and $\\measuredangle EBO = \\measuredangle OEB = 2\\theta$, so $\\measuredangle BOE = 180^\\circ-4\\theta$.... | 2 | ./CreativeMath/AHSME/1979_AHSME_Problems/12.json | AHSME |
1979_AHSME_Problems | 24 | 0 | Geometry | Multiple Choice | Sides $AB,~ BC$, and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5$, and $20$, respectively.
If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5}$, then side $AD$ has length
\begin{itemize}
\item A polygon is called “simple” if it is not self intersecting.
\end{itemize}
$\textbf{(... | [
"We know that $\\sin(C)=-\\cos(B)=\\frac{3}{5}$. Since $B$ and $C$ are obtuse, we have $\\sin(180-C)=\\cos(180-B)=\\frac{3}{5}$. It is known that $\\sin(x)=\\cos(90-x)$, so $180-C=90-(180-C)=180-B$. We simplify this as follows:\n\n\n\\[-90+C=180-B\\]\n\n\n\\[B+C=270^{\\circ}\\]\n\n\nSince $B+C=270^{\\circ}$, we kno... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/24.json | AHSME |
1979_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | If $q_1 ( x )$ and $r_ 1$ are the quotient and remainder, respectively, when the polynomial $x^ 8$ is divided by
$x + \tfrac{1}{2}$ , and if $q_ 2 ( x )$ and $r_2$ are the quotient and remainder, respectively,
when $q_ 1 ( x )$ is divided by $x + \tfrac{1}{2}$, then $r_2$ equals
$\textbf{(A) }\frac{1}{256}\qquad \t... | [
"Solution by e_power_pi_times_i\n\n\nFirst, we divide $x^8$ by $x+\\frac{1}{2}$ using synthetic division or some other method. The quotient is $x^7-\\frac{1}{2}x^6+\\frac{1}{4}x^5-\\frac{1}{8}x^4+\\frac{1}{16}x^3-\\frac{1}{32}x^2+\\frac{1}{64}x-\\frac{1}{128}$, and the remainder is $\\frac{1}{256}$. Then we plug th... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/25.json | AHSME |
1979_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | The inequality $y-x<\sqrt{x^2}$ is satisfied if and only if
$\textbf{(A) }y<0\text{ or }y<2x\text{ (or both inequalities hold)}\qquad \textbf{(B) }y>0\text{ or }y<2x\text{ (or both inequalities hold)}\qquad \textbf{(C) }y^2<2xy\qquad \textbf{(D) }y<0\qquad \textbf{(E) }x>0\text{ and }y<2x$
| [
"Solution by e_power_pi_times_i\n\n\n$\\sqrt{x^2} = \\pm x$, so the inequality is just $y-x<\\pm x$. Therefore we get the two inequalities $y<0$ and $y<2x$. Checking the answer choices, we find that $\\boxed{\\textbf{(A) } y<0\\text{ or }y<2x\\text{ (or both inequalities hold)}}$ is the answer.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/13.json | AHSME |
1979_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | For each positive number $x$, let
$f(x)=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2} {\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}$.
The minimum value of $f(x)$ is
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }6$
| [
"Let $a = \\left( x + \\frac{1}{x} \\right)^3$ and $b = x^3 + \\frac{1}{x^3}$. Then\n\\begin{align*} f(x) &= \\frac{\\left( x + \\frac{1}{x} \\right)^6 - (x^6 + \\frac{1}{x^6}) - 2}{\\left( x + \\frac{1}{x} \\right)^3 + (x^3 + \\frac{1}{x^3})} \\\\ &= \\frac{\\left( x + \\frac{1}{x} \\right)^6 - (x^6 + 2 + \\frac{1... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/29.json | AHSME |
1979_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | [asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $ABCD$ ... | [
"Solution by e_power_pi_times_i\n\n\nNotice that $\\measuredangle DAE = 90^\\circ+60^\\circ = 150^\\circ$ and that $AD = AE$. Then triangle $ADE$ is isosceles, so $\\measuredangle AED = \\dfrac{180^\\circ-150^\\circ}{2} = \\boxed{\\textbf{(C) } 15}$.\n\n\n",
"WLOG, let the side length of the square and the equila... | 2 | ./CreativeMath/AHSME/1979_AHSME_Problems/3.json | AHSME |
1979_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | Find the area of the smallest region bounded by the graphs of $y=|x|$ and $x^2+y^2=4$.
$\textbf{(A) }\frac{\pi}{4}\qquad \textbf{(B) }\frac{3\pi}{4}\qquad \textbf{(C) }\pi\qquad \textbf{(D) }\frac{3\pi}{2}\qquad \textbf{(E) }2\pi$
| [
"Solution by e_power_pi_times_i\n\n\nThe graph of $x^2+y^2 = 4$ is a circle with radius $2$ centered at the origin. The graph of $y=|x|$ is the combined graphs of $y=x$ and $y=-x$ with a nonnegative y. Because the arguments of $y=x$ and $y=-x$ are $135^\\circ$ and $45^\\circ$ respectively, the angle between the gra... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/8.json | AHSME |
1979_AHSME_Problems | 22 | 0 | Number Theory | Multiple Choice | Find the number of pairs $(m, n)$ of integers which satisfy the equation $m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1$.
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }3\qquad \textbf{(D) }9\qquad \textbf{(E) }\infty$
| [
"The equation is equivalent to $m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1$. Taking mod 3, we get\n$m(m+1)(m+2)=1 (\\bmod 3)$. However, $m(m+1)(m+2)$ is always divisible by $3$ for any integer $m$. Thus, the answer is $\\boxed{\\textbf{(A)} 0}$\nSolution by mickyboy789\n\n\n"
] | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/22.json | AHSME |
1979_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | To the nearest thousandth, $\log_{10}2$ is $.301$ and $\log_{10}3$ is $.477$.
Which of the following is the best approximation of $\log_5 10$?
$\textbf{(A) }\frac{8}{7}\qquad \textbf{(B) }\frac{9}{7}\qquad \textbf{(C) }\frac{10}{7}\qquad \textbf{(D) }\frac{11}{7}\qquad \textbf{(E) }\frac{12}{7}$
| [
"Solution by e_power_pi_times_i\n\n\nNotice that $\\log_5 10 = \\log_5 2 + 1 = \\frac{1}{\\log_2 5} + 1$. So we are trying to find $\\log_2 5$. Denote $\\log_{10}2$ as $x$. Then $\\frac{1}{x} = \\log_2 10 = \\log_2 5 + 1$. Therefore $\\log_2 5 = \\frac{1}{x}-1$, and plugging this in gives $\\log_5 10 = \\frac{1}{\\... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/18.json | AHSME |
1979_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | In a certain sequence of numbers, the first number is $1$, and, for all $n\ge 2$, the product of the first $n$ numbers in the sequence is $n^2$.
The sum of the third and the fifth numbers in the sequence is
$\textbf{(A) }\frac{25}{9}\qquad \textbf{(B) }\frac{31}{15}\qquad \textbf{(C) }\frac{61}{16}\qquad \textbf{(D)... | [
"Solution by e_power_pi_times_i\n\n\nSince the product of the first $n$ numbers in the sequence is $n^2$, the product of the first $n+1$ numbers in the sequence is $(n+1)^2$. Therefore the $n+1$th number in the sequence is $\\frac{(n+1)^2}{n^2}$. Therefore the third and the fifth numbers are $\\frac{9}{4}$ and $\\f... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/14.json | AHSME |
1979_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | Two identical jars are filled with alcohol solutions, the ratio of the volume of alcohol to the volume of water being $p : 1$
in one jar and $q : 1$ in the other jar. If the entire contents of the two jars are mixed together,
the ratio of the volume of alcohol to the volume of water in the mixture is
$\textbf{(A) }... | [
"Solution by e_power_pi_times_i\n\n\nThe amount of alcohol in the jars are $\\frac{p}{p+1}$ and $\\frac{q}{q+1}$, and the amount of water in the jars are $\\frac{1}{p+1}$ and $\\frac{1}{q+1}$. Then the total amount of alcohol is $\\frac{p}{p+1} + \\frac{q}{q+1} = \\frac{p+q+2pq}{(p+1)(q+1)}$, and the total amount o... | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/15.json | AHSME |
1979_AHSME_Problems | 5 | 0 | Arithmetic | Multiple Choice | Find the sum of the digits of the largest even three digit number (in base ten representation)
which is not changed when its units and hundreds digits are interchanged.
$\textbf{(A) }22\qquad \textbf{(B) }23\qquad \textbf{(C) }24\qquad \textbf{(D) }25\qquad \textbf{(E) }26$
| [
"Solution by e_power_pi_times_i\n\n\nSince the number doesn't change when the units and hundreds digits are switched, the number must be of the form $aba$. We want to create the largest even $3$-digit number, so $a = 8$ and $b = 9$. The sum of the digits is $8+9+8 = \\boxed{\\textbf{(D) } 25}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/5.json | AHSME |
1979_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Find the sum of the squares of all real numbers satisfying the equation $x^{256}-256^{32}=0$.
$\textbf{(A) }8\qquad \textbf{(B) }128\qquad \textbf{(C) }512\qquad \textbf{(D) }65,536\qquad \textbf{(E) }2(256^{32})$
| [
"Solution by e_power_pi_times_i\n\n\nNotice that the solutions to the equation $x^{256}-1=0$ are the $256$ roots of unity. Then the solutions to the equation $x^{256}-256^{32}=0$ are the $256$ roots of unity dilated by $\\sqrt[256]{256^{32}} = \\sqrt[256]{2^{256}} = 2$. However, the only real solutions to the equat... | 2 | ./CreativeMath/AHSME/1979_AHSME_Problems/19.json | AHSME |
1979_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | The edges of a regular tetrahedron with vertices $A ,~ B,~ C$, and $D$ each have length one.
Find the least possible distance between a pair of points $P$ and $Q$, where $P$ is on edge $AB$ and $Q$ is on edge $CD$.
[asy] size(150); import patterns; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux; ad... | [
"Note that the distance $PQ$ will be minimized when $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$.\n\n\nTo find this distance, consider triangle $\\triangle PCQ$. $Q$ is the midpoint of $CD$, so $CQ=\\frac{1}{2}$. Additionally, since $CP$ is the altitude of equilateral $\\triangle ABC$, $CP=\\frac{\\s... | 2 | ./CreativeMath/AHSME/1979_AHSME_Problems/23.json | AHSME |
1979_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | The product of $\sqrt[3]{4}$ and $\sqrt[4]{8}$ equals
$\textbf{(A) }\sqrt[7]{12}\qquad \textbf{(B) }2\sqrt[7]{12}\qquad \textbf{(C) }\sqrt[7]{32}\qquad \textbf{(D) }\sqrt[12]{32}\qquad \textbf{(E) }2\sqrt[12]{32}$
| [
"Solution by e_power_pi_times_i\n\n\n$\\sqrt[3]{4}$ and $\\sqrt[4]{8}$ can be expressed as $2^{\\frac{2}{3}}$ and $2^{\\frac{3}{4}}$, so their product is $2^{\\frac{17}{12}} = \\boxed{\\textbf{(E) } 2\\sqrt[12]{32}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1979_AHSME_Problems/9.json | AHSME |
1999_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | The sequence $a_{1},a_{2},a_{3},\ldots$ satisfies $a_{1} = 19,a_{9} = 99$, and, for all $n\geq 3$, $a_{n}$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2$.
$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$
| [
"Let $m$ be the arithmetic mean of $a_1$ and $a_2$. We can then write $a_1=m-x$ and $a_2=m+x$ for some $x$.\n\n\nBy definition, $a_3=m$.\n\n\nNext, $a_4$ is the mean of $m-x$, $m+x$ and $m$, which is again $m$. \n\n\nRealizing this, one can easily prove by induction that $\\forall n\\geq 3;~ a_n=m$.\n\n\nIt follows... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/20.json | AHSME |
1999_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | What is the radius of a circle inscribed in a rhombus with diagonals of length $10$ and $24$?
$\mathrm{(A) \ }4 \qquad \mathrm{(B) \ }\frac {58}{13} \qquad \mathrm{(C) \ }\frac{60}{13} \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6$
| [
"Let $d_1=10$ and $d_2=24$ be the lengths of the diagonals, $a$ the side, and $r$ the radius of the inscribed circle.\n\n\nUsing Pythagorean theorem we can compute $a=\\sqrt{ (d_1/2)^2 + (d_2/2)^2 }=13$.\n\n\nWe can now express the area of the rhombus in two different ways: as $d_1 d_2 / 2$, and as $2ar$. Solving $... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/16.json | AHSME |
1999_AHSME_Problems | 6 | 0 | Arithmetic | Multiple Choice | What is the sum of the digits of the decimal form of the product $2^{1999}\cdot 5^{2001}$?
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10$
| [
"$2^{1999}\\cdot5^{2001}=2^{1999}\\cdot5^{1999}\\cdot5^{2}=25\\cdot10^{1999}$, a number with the digits \"25\" followed by 1999 zeros. The sum of the digits in the decimal form would be $2+5=7$, thus making the answer $\\boxed{\\text{D}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/6.json | AHSME |
1999_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | What is the largest number of acute angles that a convex hexagon can have?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$
| [
"The sum of the interior angles of a hexagon is $720$ degrees. In a convex polygon, each angle must be strictly less than $180$ degrees. \n\n\nSix acute angles can only sum to less than $90\\cdot 6 = 540$ degrees, so six acute angles could not form a hexagon.\n\n\nFive acute angles and one obtuse angle can only s... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/7.json | AHSME |
1999_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-19$, the remainder is $99$, and when $P(x)$ is divided by $x - 99$, the remainder is $19$. What is the remainder when $P(x)$ is divided by $(x-19)(x-99)$?
$\mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \ma... | [
"According to the problem statement, there are polynomials $Q(x)$ and $R(x)$ such that $P(x) = Q(x)(x-19) + 99 = R(x)(x-99) + 19$.\n\n\nFrom the last equality we get $Q(x)(x-19) + 80 = R(x)(x-99)$.\n\n\nThe value $x=99$ is a root of the polynomial on the right hand side, therefore it must be a root of the one on th... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/17.json | AHSME |
1999_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | A circle is circumscribed about a triangle with sides $20,21,$ and $29,$ thus dividing the interior of the circle into four regions. Let $A,B,$ and $C$ be the areas of the non-triangular regions, with $C$ be the largest. Then
$\mathrm{(A) \ }A+B=C \qquad \mathrm{(B) \ }A+B+210=C \qquad \mathrm{(C) \ }A^2+B^2=C^2 \qqu... | [
"$20^2 + 21^2 = 841 = 29^2$. Therefore the triangle is a right triangle. But then its hypotenuse is a diameter of the circumcircle, and thus $C$ is exactly one half of the circle. Moreover, the area of the triangle is $\\frac{20\\cdot 21}{2} = 210$. Therefore the area of the other half of the circumcircle can be ex... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/21.json | AHSME |
1999_AHSME_Problems | 10 | 0 | Other | Multiple Choice | A sealed envelope contains a card with a single digit on it. Three of the following statements are true, and the other is false.
I. The digit is 1.
II. The digit is not 2.
III. The digit is 3.
IV. The digit is not 4.
Which one of the following must necessarily be correct?
$\textbf{(A)}\ \text{I is true.} \qquad ... | [
"Three of the statements are correct, and only one digit is on the card. Thus, one of I and III are false. Therefore, II and IV must both be true. The answer is therefore $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/10.json | AHSME |
1999_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1$. The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$. Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest... | [
"We are looking for three regular polygons such that the sum of their internal angle sizes is exactly $360^{\\circ}$. \n\n\nLet the number of sides in our polygons be $3\\leq a,b,c$. From each of the polygons, two sides touch the other two, and the remaining sides are on the perimeter. Therefore the answer to our p... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/26.json | AHSME |
1999_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | The number of ordered pairs of integers $(m,n)$ for which $mn \ge 0$ and
\begin{center}
$m^3 + n^3 + 99mn = 33^3$\end{center}
is equal to
$\mathrm{(A) \ }2 \qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 33\qquad \mathrm{(D) \ }35 \qquad \mathrm{(E) \ } 99$
| [
"We recall the factorization (see elementary symmetric sums)\n\n\n\\[x^3 + y^3 + z^3 - 3xyz = \\frac12\\cdot(x + y + z)\\cdot((x - y)^2 + (y - z)^2 + (z - x)^2)\\]\n\n\nSetting $x = m,y = n,z = - 33$, we have that either $m + n - 33 = 0$ or $m = n = - 33$ (by the Trivial Inequality). Thus, there are $35 \\Longright... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/30.json | AHSME |
1999_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | In triangle $ABC$, $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$. Then $\angle C$ in degrees is
$\mathrm{(A) \ }30 \qquad \mathrm{(B) \ }60 \qquad \mathrm{(C) \ }90 \qquad \mathrm{(D) \ }120 \qquad \mathrm{(E) \ }150$
| [
"Square the given equations and add (simplifying with the Pythagorean identity $\\sin^2 x + \\cos^2 x = 1$):\n\n\n\\begin{align*} 9\\sin^2 A + 16\\cos^2 B + 24 \\sin A \\cos B & = 36 \\\\ + 9\\cos^2 A + 16\\sin^2 B + 24 \\sin B \\cos A & = 1 \\\\ \\Longrightarrow 25 + 24(\\sin A \\cos B + \\sin B \\cos A ) & = 37 \... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/27.json | AHSME |
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