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); label(\"$15$\",(7.5,-0.75),S); label(\"$14$\",(11,5.75),ENE); label(\"$13$\",(3,6),WNW); label(\"$A$\",(0,0),SW); label(\"$D$\",(9.6,7.2),NE); label(\"$H$\",(6.6,3.5),E); [/asy]\nThe reason why we have $HD$ shorter than $HA$ is that all the ratios' left hand side ($HD$) is less than the ratios' right hand side ($HA$).\n\n\nWe label point $A$ as the origin and point $B$, logically, as $(15,0)$. By Heron's Formula, the area of this triangle is $84.$ Thus the height perpendicular to $AB$ has a length of $11.2,$ and by the Pythagorean Theorem, $AE$ and $EB$ have lengths $6.6$ and $8.4,$ respectively. These lengths tell us that $C$ is at $(6.6,11.2)$.\n\n\nThe slope of $BC$ is $\\dfrac{0-11.2}{15-6.6}=-\\dfrac{4}{3},$ and the slope of $AD$ is $\\dfrac{3}{4}$ by taking the negative reciprocal of $-\\dfrac{4}{3}.$ Therefore, the equation of line $AD$ can best be represented by $y=\\dfrac{3}{4}x.$\n\n\nWe next find the intersection of $CE$ and $AD$. We automatically know the $x$-value; it is just $6.6$ because $CE$ is a straight line hitting $(6.6,0).$ Therefore, the $y$-value is at $\\dfrac{3}{4}\\times 6.6=4.95.$ Therefore, the intersection between $CE$ and $AD$ is at $(6.6,4.95)$.\n\n\nWe also need to find the intersection between $BC$ and $AD$. To do that, we know that the line of $AD$ is represented as $y=\\dfrac{3}{4}x,$ and the slope of line $BC$ is $-\\dfrac{4}{3}.$ We just need to find line $BC$'s y-intercept. So far, we have $y=-\\dfrac{4}{3}x+b,$ where $b$ is a real y-intercept. We know that $B$ is located at $(15,0),$ so we plug that into the equation and yield $b=20.$ Therefore, the intersection between the two lines is\n\\[\\dfrac{3}{4}x=-\\dfrac{4}{3}x+20, 9x=-16x+240, 25x=240, x=9.6, y=\\dfrac{3}{4}\\times 9.6, y=7.2.\\]\nAfter that, we use the distance formula: $HA$ has a length of \\[\\sqrt{(6.6-0)^2+(4.95-0)^2}=\\sqrt{\\dfrac{1089}{25}+\\dfrac{9801}{400}}=\\sqrt{\\dfrac{1089*16+9801}{400}}=\\sqrt{\\dfrac{27225}{400}}=\\sqrt{\\dfrac{1089}{16}}=\\dfrac{33}{4}=8.25,\\] and $HD$ has a length of \\[\\sqrt{(9.6-6.6)^2+(7.2-4.95)^2}=\\sqrt{3^2+(\\dfrac{36}{5}-\\dfrac{99}{20})^2}=\\sqrt{9+\\dfrac{81}{16}}=\\sqrt{\\dfrac{225}{16}}=3.75.\\]\nThus, we have that $\\dfrac{3.75}{8.25}=\\dfrac{\\frac{15}{4}}{\\frac{33}{4}}=\\dfrac{15}{33}=\\dfrac{5}{11}=\\boxed{\\bold{B}}.$-OreoChocolate\n\n\n",
"[asy] size(150); real a = 14, b = 15, c = 13; pair C = (0, 0), B = (a, 0); // calculate cos(α) and sin(α) real cos_alpha = (a^2 + c^2 - b^2) / (2 * a * c); real sin_alpha = sqrt(1 - cos_alpha^2); // calculate coordinates of A pair A = (c * cos_alpha, c * sin_alpha); // calculate altitudes pair D = foot(A, B, C); pair E = foot(B, A, C); pair F = foot(C, A, B); // calculate orthocenter pair H = extension(A, D, B, E); // draw triangle and altitudes draw(A--B--C--cycle); draw(A--D, dashed); draw(B--E, dashed); draw(C--F, dashed); // draw right angle markers draw(rightanglemark(A, D, B, 20)); draw(rightanglemark(B, E, A, 20)); draw(rightanglemark(C, F, A, 20)); // label points dot(\"$A$\", A, N); dot(\"$B$\", B, SW); dot(\"$C$\", C, SE); dot(\"$D$\", D, S); dot(\"$E$\", E, NW); dot(\"$F$\", F, NE); dot(\"$H$\", H, SE); // label side lengths label(\"$13$\", (A+C)/2, NW); label(\"$15$\", (A+B)/2, NE); label(\"$14$\", (B+C)/2, S); [/asy]\n\n\nA consequence of Ceva's Theorem that is sometimes attributed to van Aubel states that:\n\n\n\\[\\dfrac{AH}{HD} = \\dfrac{AE}{EC} + \\dfrac{AF}{FB}\\]\n\n\nWe must then find $AE$ and $AF$. To find $AF$ note that $\\triangle AFC$ and $\\triangle BFC$ are both right triangles sharing a common height, $FC$. Thus\n\n\n$AF^2+FC^2=AC^2, \\text{and } BF^2+FC^2=BC^2 \\implies$\n$AF^2+FC^2=13^2, \\text{and } (15-AF)^2+FC^2=14^2$\n\n\nSubtracting the two equations to eliminate the common height term ($FC^2$):\n\n\n$(15-AF)^2-AF^2=27 \\implies AF=\\dfrac{33}{5}$\n\n\nA similar computation using $\\triangle AEB$ and $\\triangle CEB$ gives us:\n\n\n$AE^2+EB^2=AB^2, \\text{and } CE^2+EB^2=BC^2 \\implies$\n$AE^2+EB^2=15^2, \\text{and } (13-AE)^2+EB^2=14^2$\n\n\n$AE^2-(13-AE)^2=29 \\implies AE=\\dfrac{99}{13}$\n\n\nReturning to our original van Aubel equation yields:\n\n\n$\\dfrac{AH}{HD} = \\dfrac{AE}{EC} + \\dfrac{AF}{FB} = \\dfrac{\\dfrac{99}{13}}{13-\\dfrac{99}{13}} + \\dfrac{\\dfrac{33}{5}}{15-\\dfrac{33}{5}}=\\dfrac{99}{70}+\\dfrac{55}{70}=\\dfrac{11}{5}$\n\n\nTherefore $HD:HA = \\boxed{\\textbf{(B) }5:11}$\n\n\n~ proloto\n\n\n"
] | 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("A", A, W); label("B", B, E); label("C", C, NE); label("P", P, NW); label("6", 3*dir(A--C), SE); label("7", B+3*dir(B--C), NE); label("8", (4,0), S); [/asy]
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$
|
[
"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 yields $\\frac{\\frac{4PC}{3}}{PC+7}=\\frac{3}{4}$\n\n\n\n\nWhich we can further simplify to $\\frac{16PC}{3}=3PC+21$\n\n\n\n\n$\\frac{7PC}{3}=21$\n\n\n\n\n$PC=9\\implies\\boxed{\\textbf{(C)}}$\n\n\n"
] | 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 = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle; path block = origin--(3,0)--(3,1.5)--(0,1.5)--cycle; path rotblock = origin--(1.5,0)--(1.5,3)--(0,3)--cycle; draw(table^^shift((14,0))*table); filldraw(shift((7,0))*block^^shift((5.5,7))*rotblock^^shift((21,0))*rotblock^^shift((18,7))*block,gray); draw((7.25,1.75)--(8.5,3.5)--(8.5,8)--(7.25,9.75),Arrows(size=5)); draw((21.25,3.25)--(22,3.5)--(22,8)--(21.25,8.25),Arrows(size=5)); unfill((8,5)--(8,6.5)--(9,6.5)--(9,5)--cycle); unfill((21.5,5)--(21.5,6.5)--(23,6.5)--(23,5)--cycle); label("$r$",(8.5,5.75)); label("$s$",(22,5.75)); [/asy]
$\textbf{(A) }28\text{ inches}\qquad\textbf{(B) }29\text{ inches}\qquad\textbf{(C) }30\text{ inches}\qquad\textbf{(D) }31\text{ inches}\qquad\textbf{(E) }32\text{ inches}$
|
[
"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
$\textbf{(A)}\ \Big\{\frac{5}{2}\Big\}\qquad \textbf{(B)}\ \big\{x\ |\ 2 \le x \le 3\big\}\qquad \textbf{(C)}\ \big\{x\ |\ 2\le x < 3\big\}\qquad\\ \textbf{(D)}\ \Big\{x\ |\ 2 < x\le 3\Big\}\qquad \textbf{(E)}\ \Big\{x\ |\ 2 < x < 3\Big\}$
|
[
"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 $\\lfloor x \\rfloor + \\lceil x \\rceil = 2 + 3 = 5$, so the solution set is $\\boxed{E}$.\n\n\n"
] | 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 at least $10$ pairs?
(A pair of socks is two socks of the same color. No sock may be counted in more than one pair.)
$\textbf{(A)}\ 21\qquad \textbf{(B)}\ 23\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 30\qquad \textbf{(E)}\ 50$
|
[
"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 results in a pair every $2$ of that sock, whereas drawing another sock creates another pair. Thus the answer is $5+2\\cdot(10-1) = \\boxed{\\textbf{(B) } 23}$.\n\n\n"
] | 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((0,0),1,10,90)--C--D--cycle,mediumgray); fill(Arc((0,0),1,190,270)--B--F--cycle,mediumgray); draw(unitcircle); draw(A--B--D^^A--E); label("$A$",A,S); label("$B$",B,W); label("$C$",C,SE); label("$\theta$",C,SW); label("$D$",D,NE); label("$E$",E,N); [/asy]
A necessary and sufficient condition for the equality of the two shaded areas, given $0 < \theta < \frac{\pi}{2}$, is
$\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad \textbf{(C)}\ \tan\theta = 4\theta\qquad \textbf{(D)}\ \tan 2\theta =\theta\qquad\\ \textbf{(E)}\ \tan\frac{\theta}{2}=\theta$
|
[
"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}\\]\\[AB = AC\\tan \\theta = r\\tan \\theta\\]\\[[ABC] = \\frac{AB \\cdot AC}{2} = \\frac{r^2\\tan \\theta}{2}\\]Then the area of that shaded thing on the left becomes \\[\\frac{r^2\\tan \\theta}{2} - \\frac{\\theta \\cdot r^2}{2}\\]We want this to be equal to the sector area so \\[\\frac{r^2\\tan \\theta}{2} - \\frac{\\theta \\cdot r^2}{2} = \\frac{\\theta \\cdot r^2}{2}\\]\\[\\frac{r^2\\tan \\theta}{2} = \\theta \\cdot r^2\\]\\[\\tan \\theta = 2\\theta\\]\n\n\n"
] | 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 find that the $86th$ combination ends with the letter E. \nSo the answer is $\\textbf{(E)}\\ E$.\n\n\n",
"We can do this problem without having to list out every single combination.\nThere are $5$ distinct letters, so therefore there are $5!=120$ ways to rearrange the letters.\nWe can divide the $120$ different combinations into 5 groups. Words that start with $A$, words that start with $E$ and so on...\nCombinations $1$-$24$ start with $A$,\ncombinations $25$-$48$ start with $E$,\ncombinations $49$-$72$ start with $H$,\ncombinations $73$-$96$ start with $M$,\nand combinations $97$-$120$ start with $S$.\nWe are only concerned with combination $86$, so we focus on combinations $73$-$96$.\nWe can divide the remaining 24 combinations into 4 groups of 6, based upon the second letter.\nCombinations $73$-$78$ begin with $MA$,\ncombinations $79$-$84$ begin with $ME$,\ncombinations $85$-$90$ begin with $MH$,\nand combinations $91$-$96$ begin with $MS$.\nCombination $86$ begins with $MH$. Now we can fill in the rest of the letters in alphabetical order and get $MHASE$ (as $85$ is $MHAES$). The last letter of the word is $E$, so the answer is $\\textbf{(E)}\\ E$ .\n\n\n"
] | 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
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ \text{more than 3}$
|
[
"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 midpoints $(a,b+c)$ and $(a-d,b)$, so that the slopes are $\\frac{c}{2d}$ and $\\frac{2c}{d} = 4(\\frac{c}{2d})$, thus showing that one is $4$ times the other as required.\n\n\nThus in our problem, $m$ is either $3 \\times 4 = 12$ or $3 \\div 4 = \\frac{3}{4}$. In fact, both are possible, and each for infinitely many triangles. We shall show this for $m=12$, and the argument is analogous for $m=\\frac{3}{4}$. Take any right triangle with legs parallel to the axes and a hypotenuse with slope $12 \\div 2 = 6$, e.g. the triangle with vertices $(0,0)$, $(1,0)$, and $(1,6)$. Then quick calculations show that the medians to the legs have slopes $12$ and $3$. Now translate the triangle (without rotating it) so that its medians intersect at the point where the lines $y=12x+2$ and $y=3x+1$ intersect. This forces the medians to lie on these lines (since their slopes are determined, and now we force them to go through a particular point; a slope and a point uniquely determine a line). Finally, for any central dilation of this triangle (a larger or smaller triangle with the same centroid and sides parallel to this one's sides), the medians will still lie on these lines, showing the \"infinitely many\" part of the result.\n\n\nHence, to sum up, $m$ can in fact be both $12$ or $\\frac{3}{4}$, which is exactly $2$ values, i.e. $\\boxed{C}$.\n\n\n"
] | 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 $x \\geq \\sqrt{17}$. \n\n\nNow suppose $x > \\sqrt{17}$. Then $y - \\sqrt{17} = \\frac{x^{2}+17}{2x} - \\sqrt{17} = (\\frac{x-\\sqrt{17}}{2x})(x-\\sqrt{17}) < \\frac{1}{2}(x-\\sqrt{17})$, so that $x > y$. Similarly, we can get $y > z$, $z > w$, and $w > x$, and combining these gives $x > x$, an obvious contradiction.\n\n\nThus we must have $x \\geq \\sqrt{17}$, but $x \\ngtr \\sqrt{17}$, so if $x > 0$, the only possibility is $x = \\sqrt{17}$, and analogously from the other equations we get $x = y = z = w = \\sqrt{17}$; indeed, by substituting, we verify that this works.\n\n\nAs for the other case, $x < 0$, notice that $(x,y,z,w)$ is a solution if and only if $(-x,-y,-z,-w)$ is a solution, since this just negates both sides of each equation and so they are equivalent. Thus the only other solution is $x = y = z = w = -\\sqrt{17}$, so that we have $2$ solutions in total, and therefore the answer is $\\boxed{B}$.\n\n\n"
] | 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,-.8), D=(-.6,-.8), E=(0,-.8/(1.6)); draw(unitcircle); draw(A--B--D--C--A); draw(Arc(E,.2,155,205)); label("$A$",A,W); label("$B$",B,C); label("$C$",C,C); label("$D$",D,W); label("$\alpha$",E-(.2,0),W); label("$E$",E,N); [/asy]
$\textbf{(A)}\ \cos\ \alpha\qquad \textbf{(B)}\ \sin\ \alpha\qquad \textbf{(C)}\ \cos^2\alpha\qquad \textbf{(D)}\ \sin^2\alpha\qquad \textbf{(E)}\ 1-\sin\ \alpha$
|
[
"$ABE$ 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(A,H,C)); label("A", A, NE); label("B", B, W); label("C", C, E); label("H", H, S); label("M", M, dir(M)); [/asy]
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 6.5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 7.5\qquad \textbf{(E)}\ 8$
|
[
"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, $s=\\frac{13+14+15}{2}=21, [ABC]=\\sqrt{21*(21-13)(21-14)(21-15)}=84$. Using the area formula $A=0.5bh$, $AH=12$. Now using the Pythagorean Theorem, $BH=5, HC=9$.\n\n\nNow $AM=MB=6.5$. Using Stewart's Theorem on $\\triangle{ABH}$, letting $HM=x$:\n\n\n$13*6.5*6.5+13x^2=12*6.5*12+5*6.5*5$ (remember that Stewart's Theorem is $man+dad=bmb+cnc$).\n\n\nThus $x=6.5$ or $x=-6.5$ (reject this solution since $x$ is positive). Thus $HM=6.5$. Select $\\boxed{B}$.\n\n\n~hastapasta\n\n\nP.S.: Although this is torturous, this is a good practice of Heron's formula and Stewart's theorem though.\n\n\n"
] | 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:
$\textbf{(A)}\ y = \frac{1}{3} x + 8 \qquad \textbf{(B)}\ y = \frac{4}{3} x + 2 \qquad \textbf{(C)}\ y =\frac{1}{3}x+4\qquad\\ \textbf{(D)}\ y =\frac{4}{3}x+4\qquad \textbf{(E)}\ y =\frac{1}{3}x+2$
|
[
"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=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label("$O$",O,dir(B)); label("$1$",(O+P)/2,W); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$E$",E,dir(E)); label("$P$",P,dir(P)); label("$Q$",Q,dir(Q)); label("$R$",R,dir(R)); [/asy]
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$
|
[
"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(90 - 4*360/5); O = (0,0); P = (C + D)/2; Q = (A + reflect(B,C)*(A))/2; R = (A + reflect(D,E)*(A))/2; draw((2*R - E)--D--C--(2*Q - B)); draw(A--P); draw(A--Q); draw(A--R); draw(B--A--E); draw((O--B),dashed); draw((O--C),dashed); draw((O--D),dashed); draw((O--E),dashed); label(\"$A$\", A, N); label(\"$B$\", B, dir(0)); label(\"$C$\", C, SE); label(\"$D$\", D, SW); label(\"$E$\", E, W); dot(\"$O$\", O, NE); label(\"$P$\", P, S); label(\"$Q$\", Q, dir(0)); label(\"$R$\", R, W); label(\"$1$\", (O + P)/2, dir(0)); [/asy]\n\n\nIf $s$ is the side length of the regular pentagon, then each of the triangles $AOB$, $BOC$, $COD$, $DOE$, and $EOA$ has base $s$ and height 1, so the area of regular pentagon $ABCDE$ is $5s/2$.\n\n\nNext, we divide regular pentagon $ABCDE$ into triangles $ABC$, $ACD$, and $ADE$.\n\n\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(90 - 4*360/5); O = (0,0); P = (C + D)/2; Q = (A + reflect(B,C)*(A))/2; R = (A + reflect(D,E)*(A))/2; draw((2*R - E)--D--C--(2*Q - B)); draw(A--P); draw(A--Q); draw(A--R); draw(B--A--E); draw(A--C,dashed); draw(A--D,dashed); label(\"$A$\", A, N); label(\"$B$\", B, dir(0)); label(\"$C$\", C, SE); label(\"$D$\", D, SW); label(\"$E$\", E, W); dot(\"$O$\", O, dir(0)); label(\"$P$\", P, S); label(\"$Q$\", Q, dir(0)); label(\"$R$\", R, W); label(\"$1$\", (O + P)/2, dir(0)); [/asy]\nTriangle $ACD$ has base $s$ and height $AP = AO + 1$. Triangle $ABC$ has base $s$ and height $AQ$. Triangle $ADE$ has base $s$ and height $AR$. Therefore, the area of regular pentagon $ABCDE$ is also\n\\[\\frac{s}{2} (AO + AQ + AR + 1).\\]\nHence,\n\\[\\frac{s}{2} (AO + AQ + AR + 1) = \\frac{5s}{2},\\]\nwhich means $AO + AQ + AR + 1 = 5$, or $AO + AQ + AR = \\boxed{4}$. The answer is $\\boxed{(C)}$.\n\n\n",
"Now, we know that angle $D$ has measure $\\frac{180 \\cdot 3}{5} = 108$. Since\n\\[\\sin 54 = \\frac{OP}{DO} = \\frac{1}{DO}, DO = \\frac{1}{\\sin 54}\\]\\[\\tan 54 = \\frac{OP}{DP} = \\frac{1}{DP}, DP = \\frac{1}{\\tan 54}\\]Therefore, $AB = 2DP = \\frac{2}{\\tan 54}$.\n\\[\\sin 72 = \\frac{AQ}{AB} = AQ \\tan 54 \\cdot \\frac{1}{2}, AQ = \\frac{2 \\sin 72}{\\tan 54}\\]Therefore, $AO + AQ + AR = AO + 2AQ = \\frac{1}{\\sin 54}+\\frac{4 \\sin 72}{\\tan 54} = \\frac{1}{\\sin 54} + 8 \\sin 36 \\cos 54 = \\frac{1}{\\cos 36} + 8-8\\cos^2(36)$. Recalling that $\\cos 36 = \\frac{1 + \\sqrt{5}}{4}$ gives a final answer of $\\boxed{4}$.\n\n\n"
] | 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 each problem left unanswered. In the previous scoring system, one started with $30$ points, received $4$ more for each correct answer, lost $1$ point for each wrong answer, and neither gained nor lost points for unanswered questions.)
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 9\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ \text{Not uniquely determined}$
|
[
"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 $\\boxed{B}$.\n\n\n"
] | 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 $2^k$ values of $N$. If $k=10$, $N$ can only have one value ($N=1024$). Thus, the desired sum looks like \\[\\sum_{N=1}^{1024} \\lfloor \\log_{2}N\\rfloor =1(0)+2(1)+4(2)+\\dots+2^k(k)+\\dots+2^{9}(9)+10\\]\n\n\nLet $S$ be the desired sum without the $10$. \\[S=2(1)+4(2)+\\dots+2^{9}(9)\\] Multiplying by $2$ gives \\[2S=4(1)+8(2)+\\dots+2^{10}(9)\\] Subtracting the two equations gives \\[S=2^{10}(9)-(2+4+8+\\dots+2^9)\\] Summing the geometric sequence in parentheses and simplifying, we get \\[S=2^{10}(9)-2^{10}+2=2^{10}(8)+2=8194\\] Finally, adding back the $10$ gives the desired answer $\\fbox{(B) 8204}$\n\n\n"
] | 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 note that (by area equivalence) $\\frac{BC \\cdot h}{2} = \\frac{3x \\cdot 4}{2} = 6x$. Thus, $h = \\frac{12x}{BC}$. We note that to maximize $h$ we must minimize $BC$. Using the triangle inequality, $BC + AB > AC$, thus $BC + x > 3x$ or $BC > 2x$. The minimum value of $BC$ is $2x$, which would output $h = 6$. However, because $BC$ must be larger than $2x$, the minimum integer height must be $5$.\n\n\n$\\fbox{B}$\n\n\n\n\n\n\n",
"The reciprocals of the altitudes of a triangle themselves form a triangle - this can be easily proven. Let our desired altitude be $a$. \n\n\nWe have $\\frac{1}{a}<\\frac{1}{4}+\\frac{1}{12}=\\frac{1}{3}$, which implies $a>3$. We also have $\\frac{1}{a}>\\frac{1}{4}-\\frac{1}{12}=\\frac{1}{6}$, which implies $a<6$. Therefore the maximum integral value of $a$ is 5.\n\n\n$\\fbox{B}$.\n\n\n"
] | 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?
$\textbf{(A)}\ 5,000\qquad \textbf{(B)}\ 10,000\qquad \textbf{(C)}\ 50,000\qquad \textbf{(D)}\ 100,000\qquad \textbf{(E)}\ 500,000$
|
[
"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, so the answer is approximately half a million, which is $\\boxed{E}$.\n\n\n"
] | 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 $3$, and $1$ number from the $2$ numbers less than $3$. \\[{7\\choose 4} {2\\choose 1}= 35\\times2.\\]\n\n\nThus, $\\frac{35\\times2}{210} = \\frac{1}{3}$, which is $\\fbox{C}$.\n\n\n"
] | 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 $\\boxed{E}$.\n\n\n"
] | 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?
$\textbf{(A)}\ \frac{bdg}{cef}\qquad \textbf{(B)}\ \frac{cdf}{beg}\qquad \textbf{(C)}\ \frac{cdg}{bef}\qquad \textbf{(D)}\ \frac{cef}{bdg}\qquad \textbf{(E)}\ \frac{ceg}{bdf}$
|
[
"$1$ 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
$\textbf{(A)}\ \text{correct}\quad \textbf{(B)}\ \text{always less than A}\quad \textbf{(C)}\ \text{always greater than A}\quad\\ \textbf{(D)}\ \text{sometimes less than A and sometimes equal to A}\quad\\ \textbf{(E)}\ \text{sometimes greater than A and sometimes equal to A} \quad$
|
[
"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?
$\textbf{(A)}\ \sqrt{13}\qquad \textbf{(B)}\ \sqrt{14}\qquad \textbf{(C)}\ \sqrt{15}\qquad \textbf{(D)}\ \sqrt{16}\qquad \textbf{(E)}\ \sqrt{17}$
|
[
"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 $2 \\cdot \\frac{1}{2} = 1$ km because of the side relationships of an equilateral triangle, then $2$ km because the line is parallel to the $x$-axis, and the remaining distance is $\\frac{1}{2}$ km because she went halfway along and because of the logic for the first part of her route. For her $y$-coordinate, we can use similar logic to find that the coordinate is $\\sqrt{3} + 0 - \\frac{\\sqrt{3}}{2} = \\frac{\\sqrt{3}}{2}$. Therefore, her distance is \\[\\sqrt{\\left(\\frac{7}{2}\\right)^2 + \\left(\\frac{\\sqrt{3}}{2}\\right)^2} = \\sqrt{\\frac{49}{4} + \\frac{3}{4}} = \\sqrt{\\frac{52}{4}} = \\sqrt{13},\\] giving an answer of $\\boxed{A}$.\n\n\n"
] | 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.\n\n\n$6\\cdot6\\cdot6=\\fbox{(E) 216}$.\n\n\n"
] | 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 $[(1-\\frac{1}{2})(1-\\frac{1}{3})(1-\\frac{1}{4})...(1-\\frac{1}{10})] \\cdot [(1+\\frac{1}{2})(1+\\frac{1}{3})(1+\\frac{1}{4})...(1+\\frac{1}{10})] = [\\frac{1}{2} \\frac{2}{3} \\frac{3}{4} ... \\frac{9}{10}][\\frac{3}{2} \\frac{4}{3} \\frac{5}{4} ... \\frac{11}{10}] = \\frac{1}{10} \\cdot \\frac{11}{2} = \\frac{11}{20}$, which is $\\boxed{C}$.\n\n\n"
] | 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?
$\text{(A) For all positive real numbers k, both roots are pure imaginary} \quad\\ \text{(B) For all negative real numbers k, both roots are pure imaginary} \quad\\ \text{(C) For all pure imaginary numbers k, both roots are real and rational} \quad\\ \text{(D) For all pure imaginary numbers k, both roots are real and irrational} \quad\\ \text{(E) For all complex numbers k, neither root is real}$
|
[
"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 $b=c$. In the first case, the roots are pure imaginary and in the second case we have $k = a^2+b^2$, a positive number.\n\n\nWe can therefore conclude that if $k$ is real and negative, it must be the first case and the roots are pure imaginary.\n\n\nIt's possible to rule out the other cases by reasoning through the cases, but this is enough to show that $\\fbox{B}$ is true.\n\n\n"
] | 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 $63 \\div 5$ has a remainder of $3$.\n\n\n$\\fbox{D}$\n\n\n"
] | 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$\\sqrt{256} = 16 = \\boxed{B}$\n\n\n"
] | 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:
$\text{(A) } 10\quad \text{(B) } 11\quad \text{(C) } 12\quad \text{(D) } 13\quad \text{(E) } 15$
|
[
"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 has a 1, but the 10's, 100's and 1,000's place have 0's. The 10,000's place has a 1, but the $10^5$, $10^6$ and $10^7$ places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives $5\\times 3=15$ zeros altogether.\n\n\nThe answer is $\\fbox{E}$\n\n\n"
] | 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), black+linewidth(.75)); draw((1/sqrt(3),-1)--(-1/sqrt(3),1), black+linewidth(.75)); [/asy]
Amy painted a dartboard over a square clock face using the "hour positions" as boundaries.[See figure.] If $t$ is the area of one of the eight triangular regions such as that between 12 o'clock and 1 o'clock, and $q$ is the area of one of the four corner quadrilaterals such as that between 1 o'clock and 2 o'clock, then $\frac{q}{t}=$
$\text{(A) } 2\sqrt{3}-2\quad \text{(B) } \frac{3}{2}\quad \text{(C) } \frac{\sqrt{5}+1}{2}\quad \text{(D) } \sqrt{3}\quad \text{(E) } 2$
|
[
"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 $\\tfrac{1}{\\sqrt3}$.\n\n\n\n\nThis makes the area of each $t = \\tfrac{1}{2}\\cdot b \\cdot h = \\tfrac{1}{2}\\cdot 1 \\cdot \\tfrac{1}{\\sqrt3} = \\tfrac{1}{2\\sqrt3}$\n\n\n\n\nThe total area comprises $4q+8t$, so \\[4q+(8\\cdot \\tfrac{1}{2\\sqrt3}) = 2^2=4\\]\n\n\n\\[4q + \\tfrac{4}{\\sqrt3} = 4\\]\n\n\n\\[4q = 4 - \\tfrac{4}{\\sqrt3}\\]\n\n\n\\[q = 1 - \\tfrac{1}{\\sqrt3} = \\tfrac{\\sqrt3 - 1 }{\\sqrt3}\\]\n\n\n\n\n\\[\\frac{q}{t} = \\frac{\\tfrac{\\sqrt3 - 1 }{\\sqrt3}}{\\tfrac{1}{2\\sqrt3}} = 2\\cdot (\\sqrt3 - 1) = \\boxed{2\\sqrt3-2}\\]\n\n\n\n\n\n\n$\\fbox{A}$\n\n\n"
] | 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 $a_k-a_9=13-7=6=9\\cdot\\frac{2}{3}$.\n\n\nThis means $a_k$ is $9$ terms after $a_9$, so $k=18\\implies\\boxed{B}$\n\n\n"
] | 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 $\\sqrt{8 - x}$ factor is decreasing on the interval. The behavior of the $\\sqrt{x} - \\sqrt{x - 6}$ factor is not immediately clear. But rationalizing the numerator, we find that $\\sqrt{x} - \\sqrt{x - 6} = \\frac{6}{\\sqrt{x} + \\sqrt{x - 6}}$, which is monotonically decreasing. Since both factors are always positive, $f(x)$ is also positive. Therefore, $f(x)$ is decreasing on $[6, 8]$, and the maximum value occurs at $x = 6$. Plugging in, we find that the maximum value is $\\boxed{\\text{(C) } 2\\sqrt{3}}$.\n\n\n",
"Note the form of the function is $f(x) = \\sqrt{ p(x)} - \\sqrt{q(x)}$ where $p(x)$ and $q(x)$ each describe a parabola. Factoring we have $p(x) = x(8-x)$ and $q(x) = (x-6)(8-x)$. \n\n\nThe first term $\\sqrt{p(x)}$ is defined only when $p(x)\\geq 0$ which is the interval $[0,8]$ and the second term $\\sqrt{q(x)}$ is defined only when $q(x)\\geq 0$ which is on the interval $[6,8]$, so the domain of $f(x)$ is $[0,8] \\cap [6,8] = [6,8]$. \n\n\nNow $p(x)$ peaks at the midpoint of its roots at $x=4$ and it decreases to 0 at $x=8$. Thus, $p(x)$ is decreasing over the entire domain of $f(x)$ and it obtains its maximum value at the left boundary $x=6$ and $\\sqrt{p(x)}$ does as well. On the other hand $q(x)$ obtains its minimum value of $q(x)=0$ at the left boundary $x=6$ and $\\sqrt{q(x)}$ does as well. Therefore $\\sqrt{p(x)}-\\sqrt{q(x)}$ is maximized at $x=6$. (If this seems a little unmotivated, a quick sketch of the two parabolic-like curves makes it clear where the distance between them is greatest). \n\n\nThe value at $x=6$ is $\\sqrt{ 6\\cdot 2 } = 2\\sqrt{3}$ and the answer is $\\fbox{C}$.\n\n\n"
] | 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 $x_1 = 2x_0 - 1 = (.a_2 a_3 a_4 \\cdots)_2$.\n\n\nUsing the same logic, we notice that this sequence cycles and that since $x_0 = x_5$ we notice that $a_n = a_{n+5}$. \n\n\nWe have $2$ possibilities for each of $a_1$ to $a_5$ but we can't have $a_1 = a_2 = a_3 = a_4 = a_5 = 1$ so we have $2^5 - 1 = \\boxed{(D)31}$\n\n\n-mathman523\n\n\n"
] | 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 $10$. A circle with center $P$ and radius $1$ rolls around the inside of $\triangle ABC$, always remaining tangent to at least one side of the triangle. When $P$ first returns to its original position, through what distance has $P$ traveled?
$\text{(A) } 10\quad \text{(B) } 12\quad \text{(C) } 14\quad \text{(D) } 15\quad \text{(E) } 17$
|
[
"[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+linewidth(.75)); draw((7,4)--(6.4,4.8),black+linewidth(.75)); MP(\"A\",(0,0),SW);MP(\"B\",(8,0),SE);MP(\"C\",(8,6),NE);MP(\"P\",(4,1),NE);MP(\"E\",(7,1),NE);MP(\"D\",(3,1),SW);MP(\"G\",(3,0),SW);MP(\"H\",(2.4,1.8),NW);MP(\"F\",(7,4),NE);MP(\"I\",(6.4,4.8),NW); MP(\"8\",(4,0),S);MP(\"6\",(8,3),E);MP(\"10\",(4,3),NW); dot((4,1));dot((7,1));dot((3,1));dot((7,4)); [/asy]\n\n\nStart by considering the triangle traced by $P$ as the circle moves around the triangle. It turns out this triangle is similar to the $6-8-10$ triangle (Proof: Realize that the slope of the line made while the circle is on $AC$ is the same as line $AC$ and that it makes a right angle when the circle switches from being on $AB$ to $BC$). Then, drop the perpendiculars as shown.\n\n\nSince the smaller triangle is also a $6-8-10 = 3-4-5$ triangle, we can label the sides $EF,$ $CE,$ and $DF$ as $3x, 4x,$ and $5x$ respectively. Now, it is clear that $GB = DE + 1 = 4x + 1$, so $AH = AG = 8 - GB = 7 - 4x$ since $AH$ and $AG$ are both tangent to the circle P at some point. We can apply the same logic to the other side as well to get $CI = 5 - 3x$. Finally, since we have $HI = DF = 5x$, we have $AC = 10 = (7 - 4x) + (5x) + (5 - 3x) = 12 - 2x$, so $x = 1$ and $3x + 4x + 5x = \\fbox{(B) 12}$\n\n\n-Solution by Someonenumber011\n\n\n"
] | 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 $\triangle ABC$, $\angle A=55^\circ$, $\angle C=75^\circ, D$ is on side $\overline{AB}$ and $E$ is on side $\overline{BC}$. If $DB=BE$, then $\angle{BED} =$
$\text{(A) } 50^\circ\quad \text{(B) } 55^\circ\quad \text{(C) } 60^\circ\quad \text{(D) } 65^\circ\quad \text{(E) } 70^\circ$
|
[
"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 $\\binom{4}{3}=4$ such degenerate triangles. There are $8$ total columns and rows, contributing $32$ invalid triangles. There are also $4$ for both of the diagonals and $1$ for each of the $4$ shorter diagonals. There are a total of $32+8+4=44$ invalid triangles counted in the $560$, so the answer is $560-44=516$ or answer choice $\\fbox{D}$.\n\n\n"
] | 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 $a+b=$
$\text{(A) } 11\quad \text{(B) } 20\quad \text{(C) } 35\quad \text{(D) } 58\quad \text{(E) } 66$
|
[
"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 - - where the dashes can be replaced by 0s. This makes the number of unique combinations 7 choose 3 (if you’re counting w.r.t. shiny coins) OR 7 choose 4 (w.r.t dull coins). Both are equal to 35.\n\n\nNext, observe that, for the event that the third shiny coin is not within your first 4 picks, it has to be within the last three numbers. You can think of this as placing the seven coins in a vertical stack in the box and shuffling that stack randomly. Then, to pick, you extract the first coin on the top and keep repeating. It has the same effect.\n\n\nThe sequence can have 1 shiny coin in the last 3 digits (Case 1), 2 shiny coins in the last 3 digits (Case 2) or 3 shiny coin in the last three digits (Case 3).\n\n\nCase 1:\nLet’s start with the first case of one shiny coin in its last 3 digits.\n\n\nExample: 0110100\n\n\nThe first four numbers has 4 spaces and 2 shiny coins therefore the number of combinations is 4 choose 2 = 6. The last 3 digits has 3 combinations for the same reason.\nSo, probability for Case 1 to occur is:\n$\\dfrac{6*3}{35}=\\dfrac{18}{35}$\n\n\nCase 2:\nUsing the fact that the combinations are uniquely determined by an order of 0s or 1s and you can just fill the rest, in you can ascertain:\n$\\underbracket{0100}_{\\text{4 combinations}}\\, \\underbracket{110}_{\\text{3 combinations}}$\n\n\nSo, P(Case 2)=$12/35$\n\n\nCase 3:\nTrivially, it is 1.\nP(Case 3)=$1/35$\n\n\nAdding all these probabilites together gives you the probability that the third shiny coin will not appear in your first 4 draws:\n$\\dfrac{18}{35}+\\dfrac{12}{35}+\\dfrac{1}{35}=\\dfrac{31}{35}$\n\n\n$\\dfrac{a}{b}=\\dfrac{31}{35}$\n\n\nSince the fraction is irreducible:\n\n\n$a=31$\n,$b=35$\n\n\n\\[a+b=66\\]\n\n\nThe answer is E.\n\n\n\n\n\n\n",
"Using complementary probability, we can reduce the problem into two cases- the third shiny penny is drawn on the third draw, or on the fourth draw. For the first case, there is only one way to have the shiny pennies as the first three coins drawn, out of a possible 35 drawings (7 choose 3). \nFor the second case, the third shiny penny has to be the fourth penny drawn, which leaves three possible orderings for the first three coins drawn (NS S S, S NS S, S S NS), out of 35 (7 choose 4). Adding these two probabilities together gives $\\dfrac{4}{35}$, and subtracting this from one yields $\\dfrac{31}{35}$, which makes a 31 and b 35, which sum to $\\fbox{E}$.\n\n\n"
] | 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 inside the angle
on the angle bisector. Consider all distinct equilateral triangles $PQR$ with $Q$ and $R$ in $S$.
(Points $Q$ and $R$ may be on the same ray, and switching the names of $Q$ and $R$ does not create a distinct triangle.)
There are
$\text{(A) exactly 2 such triangles} \quad\\ \text{(B) exactly 3 such triangles} \quad\\ \text{(C) exactly 7 such triangles} \quad\\ \text{(D) exactly 15 such triangles} \quad\\ \text{(E) more than 15 such triangles}$
|
[
"$\\fbox{E}$\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 $\\triangle OO'P$. We have $m\\angle A = 60^\\circ$ and $m \\angle O = 60^\\circ$, therefore $\\angle A \\cong \\angle O$. By our construction of moving the points the same distance, we have $AA' = OO'$. Also, $AP = OP$ by the original equilateral triangle. Therefore, by SAS congruence, $\\triangle AA'P \\cong \\triangle OO'P$. \n\n\nNow, look at $\\triangle A'PO'$. We have $PA' = PO'$ from the above congruence. We also have the included angle $\\angle A'PO'$ is $60^\\circ$. To prove that, start with the $60^\\circ$ angle $APO$, subtract the angle $APA'$, and add the congruent angle $OPO'$.\n\n\nSince $\\triangle A'PO'$ is an isosceles triangle with vertex of $60^\\circ$, it is equilateral.\n\n\n"
] | 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 forms a right triangle with legs \nof $4$ and $7$. $\\sqrt{4^2+7^2}=\\sqrt{65} \\rightarrow \\boxed{D}$\n\n\n"
] | 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.)
$\text{(A) }\{4,5,6\} \quad \text{(B) } \{4,5,7\} \quad \text{(C) } \{4,6,7\} \quad \text{(D) } \{5,6,7\} \quad \text{(E) } \{5,7,8\}$
|
[
"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 $c$ are positive, then the sum of any two of the squared diagonal lengths must be larger than the square of the third diagonal length. For example, $(a^2 + b^2) + (b^2 + c^2) = (a^2 + c^2) + 2b^2 > a^2 + c^2$ because $2b^2 > 0.$\n\n\nThus, we test each answer choice to see if the sum of the squares of the two smaller numbers is larger than the square of the largest number. Looking at choice (B), we see that $4^2 + 5^2 = 41 < 7^2 = 49,$ so the answer is $\\boxed{\\text{(B) \\{4, 5, 7\\}}}.$\n\n\n--goldentail141\n\n\n",
"Let $a,$ $b,$ and $c$ be the side lengths of the rectangular prism. We then have:\n$a^2+b^2=i^2,$\n$b^2+c^2=j^2,$\n$c^2+a^2=k^2$\nwhere $\\{ i,j,k \\}$ are the given diagonal lengths. WLOG, let $i \\leq j \\leq k.$ It suffices to check if $i^2+j^2 \\geq k.$ We see that for $\\boxed{B}$, $4^2+5^2 = 16 + 25 = 41 < 7^2 = 49$, so this case is impossible.\n\n\n"
] | 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 the tops and encircling downward, and the other touching the bottoms and encircling upward. There are two radius 3 circles passing through the point where $C_1$ and $C_2$ are tangent. For one $C_1$ is internally tangent and $C_2$ is externally tangent, and for the other $C_1$ is externally tangent and $C_2$ is internally tangent.\n\n\n$\\fbox{D}$\n\n\n"
] | 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), linewidth(1)); draw((0,5)--(1,7)--(6,7), linewidth(1)); draw((3.5,7)--(4.5,9)--(9.5,9)--(14.5,9)--(19.5,9)--(18.5,7)--(19.5,9)--(19.5,7), linewidth(1)); draw((8.5,7)--(9.5,9), linewidth(1)); draw((13.5,7)--(14.5,9), linewidth(1)); draw((7,9)--(8,11)--(13,11)--(18,11)--(17,9)--(18,11)--(18,9), linewidth(1)); draw((12,9)--(13,11), linewidth(1)); draw((10.5,11)--(11.5,13)--(16.5,13)--(16.5,11)--(16.5,13)--(15.5,11), linewidth(1)); draw((25,0)--(30,0)--(30,5)--(25,5)--(25,0), dashed); draw((30,0)--(35,0)--(40,0)--(45,0)--(45,5)--(40,5)--(35,5)--(30,5)--(31,7)--(36,7)--(41,7)--(46,7)--(46,2)--(45,0), dashed); draw((35,0)--(35,5)--(36,7), dashed); draw((40,0)--(40,5)--(41,7), dashed); draw((45,0)--(45,5)--(46,7), dashed); draw((25,5)--(26,7)--(31,7), dashed); draw((28.5,7)--(29.5,9)--(34.5,9)--(39.5,9)--(44.5,9)--(43.5,7)--(44.5,9)--(44.5,7), dashed); draw((33.5,7)--(34.5,9), dashed); draw((38.5,7)--(39.5,9), dashed); draw((32,9)--(33,11)--(38,11)--(43,11)--(42,9)--(43,11)--(43,9), dashed); draw((37,9)--(38,11), dashed); draw((35.5,11)--(36.5,13)--(41.5,13)--(41.5,11)--(41.5,13)--(40.5,11), dashed); draw((50,0)--(55,0)--(55,5)--(50,5)--(50,0), dashed); draw((55,0)--(60,0)--(65,0)--(70,0)--(70,5)--(65,5)--(60,5)--(55,5)--(56,7)--(61,7)--(66,7)--(71,7)--(71,2)--(70,0), dashed); draw((60,0)--(60,5)--(61,7), dashed); draw((65,0)--(65,5)--(66,7), dashed); draw((70,0)--(70,5)--(71,7), dashed); draw((50,5)--(51,7)--(56,7), dashed); draw((53.5,7)--(54.5,9)--(59.5,9)--(64.5,9)--(69.5,9)--(68.5,7)--(69.5,9)--(69.5,7), dashed); draw((58.5,7)--(59.5,9), dashed); draw((63.5,7)--(64.5,9), dashed); draw((57,9)--(58,11)--(63,11)--(68,11)--(67,9)--(68,11)--(68,9), dashed); draw((62,9)--(63,11), dashed); draw((60.5,11)--(61.5,13)--(66.5,13)--(66.5,11)--(66.5,13)--(65.5,11), dashed); draw((75,0)--(80,0)--(80,5)--(75,5)--(75,0), dashed); draw((80,0)--(85,0)--(90,0)--(95,0)--(95,5)--(90,5)--(85,5)--(80,5)--(81,7)--(86,7)--(91,7)--(96,7)--(96,2)--(95,0), dashed); draw((85,0)--(85,5)--(86,7), dashed); draw((90,0)--(90,5)--(91,7), dashed); draw((95,0)--(95,5)--(96,7), dashed); draw((75,5)--(76,7)--(81,7), dashed); draw((78.5,7)--(79.5,9)--(84.5,9)--(89.5,9)--(94.5,9)--(93.5,7)--(94.5,9)--(94.5,7), dashed); draw((83.5,7)--(84.5,9), dashed); draw((88.5,7)--(89.5,9), dashed); draw((82,9)--(83,11)--(88,11)--(93,11)--(92,9)--(93,11)--(93,9), dashed); draw((87,9)--(88,11), dashed); draw((85.5,11)--(86.5,13)--(91.5,13)--(91.5,11)--(91.5,13)--(90.5,11), dashed); draw((28,6)--(33,6)--(38,6)--(43,6)--(43,11)--(38,11)--(33,11)--(28,11)--(28,6), linewidth(1)); draw((28,11)--(29,13)--(34,13)--(39,13)--(44,13)--(43,11)--(44,13)--(44,8)--(43,6), linewidth(1)); draw((33,6)--(33,11)--(34,13)--(39,13)--(38,11)--(38,6), linewidth(1)); draw((31,13)--(32,15)--(37,15)--(36,13)--(37,15)--(42,15)--(41,13)--(42,15)--(42,13), linewidth(1)); draw((34.5,15)--(35.5,17)--(40.5,17)--(39.5,15)--(40.5,17)--(40.5,15), linewidth(1)); draw((53,6)--(58,6)--(63,6)--(68,6)--(68,11)--(63,11)--(58,11)--(53,11)--(53,6), dashed); draw((53,11)--(54,13)--(59,13)--(64,13)--(69,13)--(68,11)--(69,13)--(69,8)--(68,6), dashed); draw((58,6)--(58,11)--(59,13)--(64,13)--(63,11)--(63,6), dashed); draw((56,13)--(57,15)--(62,15)--(61,13)--(62,15)--(67,15)--(66,13)--(67,15)--(67,13), dashed); draw((59.5,15)--(60.5,17)--(65.5,17)--(64.5,15)--(65.5,17)--(65.5,15), dashed); draw((78,6)--(83,6)--(88,6)--(93,6)--(93,11)--(88,11)--(83,11)--(78,11)--(78,6), dashed); draw((78,11)--(79,13)--(84,13)--(89,13)--(94,13)--(93,11)--(94,13)--(94,8)--(93,6), dashed); draw((83,6)--(83,11)--(84,13)--(89,13)--(88,11)--(88,6), dashed); draw((81,13)--(82,15)--(87,15)--(86,13)--(87,15)--(92,15)--(91,13)--(92,15)--(92,13), dashed); draw((84.5,15)--(85.5,17)--(90.5,17)--(89.5,15)--(90.5,17)--(90.5,15), dashed); draw((56,12)--(61,12)--(66,12)--(66,17)--(61,17)--(56,17)--(56,12), linewidth(1)); draw((61,12)--(61,17)--(62,19)--(57,19)--(56,17)--(57,19)--(67,19)--(66,17)--(67,19)--(67,14)--(66,12), linewidth(1)); draw((59.5,19)--(60.5,21)--(65.5,21)--(64.5,19)--(65.5,21)--(65.5,19), linewidth(1)); draw((81,12)--(86,12)--(91,12)--(91,17)--(86,17)--(81,17)--(81,12), dashed); draw((86,12)--(86,17)--(87,19)--(82,19)--(81,17)--(82,19)--(92,19)--(91,17)--(92,19)--(92,14)--(91,12), dashed); draw((84.5,19)--(85.5,21)--(90.5,21)--(89.5,19)--(90.5,21)--(90.5,19), dashed); draw((84,18)--(89,18)--(89,23)--(84,23)--(84,18)--(84,23)--(85,25)--(90,25)--(89,23)--(90,25)--(90,20)--(89,18), linewidth(1));[/asy]
Twenty cubical blocks are arranged as shown. First, 10 are arranged in a triangular pattern; then a layer of 6, arranged in a triangular pattern, is centered on the 10; then a layer of 3, arranged in a triangular pattern, is centered on the 6; and finally one block is centered on top of the third layer. The blocks in the bottom layer are numbered 1 through 10 in some order. Each block in layers 2,3 and 4 is assigned the number which is the sum of numbers assigned to the three blocks on which it rests. Find the smallest possible number which could be assigned to the top block.
$\text{(A) } 55\quad \text{(B) } 83\quad \text{(C) } 114\quad \text{(D) } 137\quad \text{(E) } 144$
|
[
"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 block. Thus, if we think about how the weights propagate down the block pyramid, its sort of like a three dimensional Pascal's triangle.\n\n\nThe first layer is $1$\n\n\nThe second layer is \n\n\n$| \\, \\, 1 \\\\ |1 \\, 1$\n\n\nThe third layer is \n\n\n$| \\,\\,\\,\\, 1 \\\\| \\,\\, 2 \\, 2 \\\\| 1 \\,2 \\, 1$\n\n\nThe fourth layer is \n\n\n$| \\,\\,\\,\\,\\,\\, 1 \\\\| \\,\\,\\,\\, 3 \\, 3 \\\\| \\,\\, 3 \\,6 \\, 3 \\\\| 1\\, 3\\, 3\\, 1$\n\n\nThe top level sum is minimized if we associate the block with weight 6 with the smallest number 1, the blocks with weight 1 with the largest numbers 8, 9 and 10, and the rest of the blocks with weight 3 with the rest of the numbers. The sum is $6\\cdot 1 + 3 \\cdot (2+3+4+5+6+7 ) + 1\\cdot (8+9+10) = 114$ so the answer is $\\fbox{C}$\n\n\n"
] | 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?
$\text{(A) } 48\quad \text{(B) } 50\quad \text{(C) } 72\quad \text{(D) } 75\quad \text{(E) } 100$
|
[
"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 to figure out how many coinciding rest days they have in the first 20 days, and multiply that number by $\\frac{1000}{20}=50$ to determine the total number of coinciding rest days. These first 20 days are written out below:\n\n\nDay 1: Al works; Barb works\n\n\nDay 2: Al works; Barb works\n\n\nDay 3: Al works; Barb works\n\n\nDay 4: Al rests; Barb works\n\n\nDay 5: Al works; Barb works\n\n\nDay 6: Al works; Barb works\n\n\nDay 7: Al works; Barb works\n\n\nDay 8: Al rests; Barb rests\n\n\nDay 9: Al works; Barb rests\n\n\nDay 10: Al works; Barb rests\n\n\nDay 11: Al works; Barb works\n\n\nDay 12: Al rests; Barb works\n\n\nDay 13: Al works; Barb works\n\n\nDay 14: Al works; Barb works\n\n\nDay 15: Al works; Barb works\n\n\nDay 16: Al rests; Barb works\n\n\nDay 17: Al works; Barb works\n\n\nDay 18: Al works; Barb rests\n\n\nDay 19: Al works; Barb rests\n\n\nDay 20: Al rests; Barb rests\n\n\nIn this 20-day cycle, Al and Barb's rest days coincide twice: on day 8 and day 20. As a result (as discussed above), the answer is $2\\cdot50=\\fbox{100\\text{ (E)}}$. ~cw357\n\n\n"
] | 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((-1-sqrt(2),sqrt(2)));dot((0,sqrt(2)+sqrt(13-2*sqrt(2)))); [/asy]
The convex pentagon $ABCDE$ has $\angle{A}=\angle{B}=120^\circ,EA=AB=BC=2$ and $CD=DE=4$. What is the area of ABCDE?
$\text{(A) } 10\quad \text{(B) } 7\sqrt{3}\quad \text{(C) } 15\quad \text{(D) } 9\sqrt{3}\quad \text{(E) } 12\sqrt{5}$
|
[
"[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\",(1,0),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((-1-sqrt(2),sqrt(2)));dot((0,sqrt(2)+sqrt(13-2*sqrt(2)))); [/asy]\n\n\nFirst, drop perpendiculars from points $A$ and $B$ to segment $EC$.\n\n\nSince $\\angle EAB = 120^{\\circ}$, $\\angle EAF = 30^{\\circ}$.\n\n\nThis implies that $\\triangle EAF$ is a 30-60-90 Triangle, so $EF = 1$ and $AF = \\sqrt3$.\n\n\nSimilarly, $GB = \\sqrt3$ and $GC = 1$.\n\n\nSince $FABG$ is a rectangle, $FG = AB = 2$.\n\n\nNow, notice that since $EC = EF + FG + GC = 1+2+1=4$, triangle $DEC$ is equilateral.\n\n\nThus, $[ABCDE] = [EABC]+[DCE] = \\frac{2+4}{2}(\\sqrt3)+\\frac{4^2\\cdot\\sqrt3}{4} = 3\\sqrt3+4\\sqrt3=\\boxed{7\\sqrt3 (B)}$\n\n\n-AOPS81619\n\n\n"
] | 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 the number of divisors is 4*3*2 = 24. But we're not done yet. We cannot have a 1 or 2 sided polygon so we subtract off two bringing us to our final answer of 22 $\\fbox{D}$.\n\n\n"
] | 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:
$\text{(A) } 6\%\quad \text{(B) } 7\%\quad \text{(C) } 7.5\%\quad \text{(D) } 8\%\quad \text{(E) } 15\%$
|
[
"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, which are 1,2,4, and 8. This results in four pairs, $(5,10), (6,6), (8,4),$ and $(12,3)$, which is answer \n$\\fbox{D}$.\n\n\n"
] | 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",(-10,0),W);MP("D",(10,0),E);MP("B",(9,sqrt(19)),E);MP("C",(9,-sqrt(19)),E); [/asy]
Points $A,B,C$ and $D$ are on a circle of diameter $1$, and $X$ is on diameter $\overline{AD}.$
If $BX=CX$ and $3\angle{BAC}=\angle{BXC}=36^\circ$, then $AX=$
$\text{(A) } \cos(6^\circ)\cos(12^\circ)\sec(18^\circ)\quad\\ \text{(B) } \cos(6^\circ)\sin(12^\circ)\csc(18^\circ)\quad\\ \text{(C) } \cos(6^\circ)\sin(12^\circ)\sec(18^\circ)\quad\\ \text{(D) } \sin(6^\circ)\sin(12^\circ)\csc(18^\circ)\quad\\ \text{(E) } \sin(6^\circ)\sin(12^\circ)\sec(18^\circ)$
|
[
"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 Sines).\n\n\nWe get:\n\n\n$\\frac{AB}{\\sin(\\angle{AXB})} =\\frac{AX}{\\sin(\\angle{ABX})}$\n\n\nThat's equal to\n\n\n$\\frac{\\cos(6)}{\\sin(180-18)} =\\frac{AX}{\\sin(12)}$\n\n\nTherefore, our answer is equal to:\n$\\fbox{B}$\n\n\nNote that $\\sin(162) = \\sin(18)$, and don't accidentally put $\\fbox{C}$ because you thought $\\frac{1}{\\sin}$ was $\\sec$!\n\n\n"
] | 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 of $A$ to the wealth of a citizen of $B$.
$\text{(A) } \frac{cd}{ef}\quad \text{(B) } \frac{ce}{ef}\quad \text{(C) } \frac{cf}{de}\quad \text{(D) } \frac{de}{cf}\quad \text{(E) } \frac{df}{ce}$
|
[
"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 $\\fbox{D}$\n\n\n"
] | 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 $a = \\tan(\\theta_a) = \\frac{1}{2}$ and $b = \\tan(\\theta_b) = \\frac{1}{3}$. The question asks for $\\theta_a + \\theta_b$, so we try to find $\\tan(\\theta_a + \\theta_b)$ in terms of $\\tan(\\theta_a)$ and $\\tan(\\theta_b)$. Using the angle addition formula for $\\tan(\\alpha+\\beta)$, we get that $\\tan(\\theta_a + \\theta_b) = \\frac{\\tan(\\theta_a)+\\tan(\\theta_b)}{1-\\tan(\\theta_a)\\tan(\\theta_b)}$. Plugging $\\tan(\\theta_a) = \\frac{1}{2}$ and $\\tan(\\theta_b) = \\frac{1}{3}$ in, we have $\\tan(\\theta_a + \\theta_b) = \\frac{\\frac{1}{2}+\\frac{1}{3}}{1-(\\frac{1}{2})(\\frac{1}{3})}$. Simplifying, $\\tan(\\theta_a + \\theta_b) = 1$, so $\\theta_a + \\theta_b$ in radians is $\\boxed{\\textbf{(C) } \\frac{\\pi}{4}}$.\n\n\n",
"Thinking through the problem, we can see $\\arctan a$ is the angle whose tangent is $0.5$. $\\arctan b$ is the angle whose tangent is $\\frac{1}{3}$. Call the former angle $\\theta_1$, the latter $\\theta_2$. So we are trying to find $\\theta_1+\\theta_2$. So given two tangent measures, it is natural for us to think about the sum of tangent measures (what else can we try? Remember: we are not allowed to use calculators). Plug in as above and continue on.\n\n\n~hastapasta\n\n\n"
] | 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
$\textbf{(A) }\frac{\sqrt{3}}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }\frac{2}{\sqrt{3}}\qquad \textbf{(D) }\frac{3}{2}\qquad \textbf{(E) }\sqrt{3}$
|
[
"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 $\\boxed{\\textbf{(E) } \\sqrt{3}}$.\n\n\n"
] | 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,S); label("$D$",D,S); label("x",X,fontsize(5pt)); //Credit to TheMaskedMagician for the diagram [/asy]
Points $A , B, C$, and $D$ are distinct and lie, in the given order, on a straight line.
Line segments $AB, AC$, and $AD$ have lengths $x, y$, and $z$, respectively.
If line segments $AB$ and $CD$ may be rotated about points $B$ and $C$, respectively,
so that points $A$ and $D$ coincide, to form a triangle with positive area,
then which of the following three inequalities must be satisfied?
$\textbf{I. }x<\frac{z}{2}\qquad\\ \textbf{II. }y<x+\frac{z}{2}\qquad\\ \textbf{III. }y<\frac{z}{2}\qquad$
$\textbf{(A) }\textbf{I. }\text{only}\qquad \textbf{(B) }\textbf{II. }\text{only}\qquad \textbf{(C) }\textbf{I. }\text{and }\textbf{II. }\text{only}\qquad \textbf{(D) }\textbf{II. }\text{and }\textbf{III. }\text{only}\qquad \textbf{(E) }\textbf{I. },\textbf{II. },\text{and }\textbf{III. }$
|
[
"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 $2y>z$, so we have to prove $z<2y<2x+z$. $2x<z$, so we have to prove that $z<2z$, which is true for all positive $z$. Therefore the answer is $\\boxed{\\textbf{(C) } \\textbf{I. }\\text{and }\\textbf{II. }\\text{only}}$.\n\n\n"
] | 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
$\textbf{(A) }\frac{\pi r}{h+2r}\qquad \textbf{(B) }\frac{\pi r}{h+r}\qquad \textbf{(C) }\frac{\pi}{2h+r}\qquad \textbf{(D) }\frac{\pi r^2}{r^2+h^2}\qquad \textbf{(E) }\text{none of these}$
|
[
"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 the incircle meets the triangle as $X,Y,Z$, where $O$ is the incenter, and denote $AX = AY = z, BX = BZ = y, CY = CZ = x$. Since $XOZB$ is a square (tangents are perpendicular to radius), $r = BX = BZ = y$. The perimeter can be expressed as $2(x+y+z)$, so the semiperimeter is $x+y+z$. The hypotenuse is $AY+CY = z+x$. Thus we have $s = x+y+z = (z+x)+y = h+r$. The answer is $\\boxed{\\textbf{(B) } \\frac{\\pi r}{h+r}}$.\n\n\n"
] | 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
$\textbf{(A) }6\qquad \textbf{(B) }2\sqrt{6}\qquad \textbf{(C) }\frac{8\sqrt{3}}{3}\qquad \textbf{(D) }3\sqrt{3}\qquad \textbf{(E) }4\sqrt{3}$
|
[
"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 the form $ax^{2} + bx + c$. To solve for $a, b,$ and $c$, we can first work backwards to solve for the values of f(0) and f(-1):\n\n\n\\[f(1) = f(0) + 0 + 2 \\Rightarrow 1 = f(0) + 2 \\Rightarrow f(0) = -1\\]\n\\[f(0) = f(-1) -1 + 2 \\Rightarrow -1 = f(-1) + 1 \\Rightarrow f(-1) = -2\\]\n\n\nSince $f(0) = -1$:\n\\[f(0) = a(0)^{2} + b(0) + c = -1 \\Rightarrow c = -1\\] \nSince $f(1) = 1$: \n\\[f(1) = a(1)^{2} + b(1)+ c = a + b - 1 = 1 \\Rightarrow a + b = 2\\] \nSimilarly, since $f(-1) = -2$:\n\\[f(-1) = a(-1)^{2} + b(-1)+ c = a - b - 1 = -2 \\Rightarrow a - b = -1\\]\n\n\nThus we have the system of equations:\n\n\n\\[a + b = 2\\]\n\\[a - b = -1\\]\n\n\nWhich can be solved to yield $a = \\frac{1}{2}$, $b = \\frac{3}{2}$. Therefore, $f(x) = \\frac{1}{2} x^{2} + \\frac{3}{2} x - 1$.\n\n\nSince we are searching for values for which $f(x) = x$, we have the equation $\\frac{1}{2} x^{2} + \\frac{3}{2} x - 1 = x$. Subtracting $x$ yields $\\frac{1}{2} x^{2} + \\frac{1}{2} x - 1 = 0$, which we can simplify by dividing both sides by $\\frac{1}{2}$: $x^{2} + x - 2 = 0$. This factors into $(x + 2) (x - 1) = 0$, so therefore there are two solutions to $f(x) = x$: $-2$ and $1$. Since the problem asks only for solutions that do not equal $1$, the answer is $\\fbox{(B) 1}$.\n\n\n"
] | 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",8,EE,C,D,0.3,2,shift(1,3)*C); draw(arc(shift(-0.1,0.05)*C,0.25,100,180),arrow =ArcArrow()); MA("100^\circ",8,A,B,C,0.1,0); MA("60^\circ",8,C,A,B,0.1,0); //Credit to TheMaskedMagician for the diagram [/asy]
In $\triangle ABC$, $E$ is the midpoint of side $BC$ and $D$ is on side $AC$.
If the length of $AC$ is $1$ and $\measuredangle BAC = 60^\circ, \measuredangle ABC = 100^\circ, \measuredangle ACB = 20^\circ$ and
$\measuredangle DEC = 80^\circ$, then the area of $\triangle ABC$ plus twice the area of $\triangle CDE$ equals
$\textbf{(A) }\frac{1}{4}\cos 10^\circ\qquad \textbf{(B) }\frac{\sqrt{3}}{8}\qquad \textbf{(C) }\frac{1}{4}\cos 40^\circ\qquad \textbf{(D) }\frac{1}{4}\cos 50^\circ\qquad \textbf{(E) }\frac{1}{8}$
|
[
"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 $BC=2(EC)$. Also $\\triangle FGC$ is congruent to $\\triangle ABC$. Therefore, $[\\triangle ACF] = ([\\triangle ABC] + [\\triangle GCF]) + [\\triangle BCG]$. Plugging in the values that we know and then dividing by 2 results in an answer of $\\boxed{B) \\frac{\\sqrt{3}}{8}.}$\n\n\nThis solution is from the solution manual but was typed here by alpha_2.\n\n\n"
] | 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?
$\textbf{(A) }\frac{106}{121}\qquad \textbf{(B) }\frac{108}{121}\qquad \textbf{(C) }\frac{110}{121}\qquad \textbf{(D) }\frac{112}{121}\qquad \textbf{(E) }\text{none of these}$
|
[
"$\\boxed{E}$\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 lastly, for both roots to be positive we\nneed $-b>\\sqrt{b^2-4c}$. Combining these inequalities we determine that $\\frac{b^2}{4}>c>0$ and $b<0$. \n\n\nThis leaves just $10$ cases $\\{(-5,1),(-5,2)(-5,3),(-5,4),(-5,5),(-4,1),(-4,2),(-4,3),(-3,1),(-3,2)\\}$ which can easily be checked by hand.\n\n\n\n\nThe answer is $\\frac{111}{121}$\n\n\n"
] | 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 rectangle ABCD has area 72 square meters and E and G are the midpoints of sides AD and CD, respectively, then the area of rectangle DEFG in square meters is
$\textbf{(A) }8\qquad \textbf{(B) }9\qquad \textbf{(C) }12\qquad \textbf{(D) }18\qquad \textbf{(E) }24$
|
[
"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 final answer.\n\n\n",
"Notice that we can do $\\frac{x-y}{xy} = \\frac{xy}{xy}$. We are left with $\\frac{1}{y} - \\frac{1}{x} = 1$. Multiply by $-1$ to achieve $\\frac{1}{x} - \\frac{1}{y} = \\boxed{-1}$.\n\n\n"
] | 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); //Credit to TheMaskedMagician for the diagram[/asy]
Circles with centers $A ,~ B$, and $C$ each have radius $r$, where $1 < r < 2$.
The distance between each pair of centers is $2$. If $B'$ is the point of intersection of circle $A$ and circle $C$
which is outside circle $B$, and if $C'$ is the point of intersection of circle $A$ and circle $B$ which is outside circle $C$,
then length $B'C'$ equals
$\textbf{(A) }3r-2\qquad \textbf{(B) }r^2\qquad \textbf{(C) }r+\sqrt{3(r-1)}\qquad\\ \textbf{(D) }1+\sqrt{3(r^2-1)}\qquad \textbf{(E) }\text{none of these}$
|
[
"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 gives $y=\\frac{\\sqrt{3}\\pm \\sqrt{r^2-1}}{2}$.\n\n\nThe larger root $y=\\frac{\\sqrt{3}+\\sqrt{r^2-1}}{2}$ is the point B' described in the question. This root corresponds to $x=-\\frac{1+\\sqrt{3(r^2-1)}}{2}$. \n\n\nBy symmetry across the y-axis the length of the line segment $B'C'$ is $1+\\sqrt{3(r^2-1)}$ which is $\\boxed{D}$.\n\n\n",
"Suppose $B’B$ and $AC$ intersect at $P$. By the Pythagorean Theorem, $B’P = \\sqrt{r^2 - 1}$ and by a $30-60-90$ triangle, $PB = \\sqrt{3}$. Using Ptolemy’s Theorem on isosceles trapezoid $BCB’C’$, we get that \\[2(B’C’) + r^2 = (\\sqrt{3} + \\sqrt{r^2 - 1})^2.\\] After a little algebra, we get that $B’C’ = \\boxed{1 + \\sqrt{3(r^2 - 1)}}$ as desired.\nSolasky (talk) 12:29, 27 May 2023 (EDT)\n\n\n"
] | 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("$45^\circ$",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$. Point $A$ lies on the extension of $DC$ past $C$;
point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle.
If length $AB$ equals length $OD$, and the measure of $\measuredangle EOD$ is $45^\circ$, then the
measure of $\measuredangle BAO$ is
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$
|
[
"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 $\\measuredangle BOE = 180^\\circ-4\\theta$. Notice that $\\measuredangle AOB + \\measuredangle BOE + 45^\\circ = 180^\\circ$. Therefore $\\theta+180-4\\theta = 135^\\circ$, and $\\theta = \\boxed{\\textbf{(B) } 15^\\circ}$.\n\n\n",
"Draw $BO$. Let $y = \\angle BAO$. Since $AB = OD = BO$, triangle $ABO$ is isosceles, so $\\angle BOA = \\angle BAO = y$. Angle $\\angle EBO$ is exterior to triangle $ABO$, so $\\angle EBO = \\angle BAO + \\angle BOA = y + y = 2y$.\n\n\nTriangle $BEO$ is isosceles, so $\\angle BEO = \\angle EBO = 2y$. Then $\\angle EOD$ is external to triangle $AEO$, so $\\angle EOD = \\angle EAO + \\angle AEO = y + 2y = 3y$. But $\\angle EOD = 45^\\circ$, so $\\angle BAO = y = 45^\\circ/3 = \\boxed{15^\\circ}$. That means the answer is $\\boxed{\\textbf{(B) } 15^\\circ}$.\n\n\n"
] | 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}
$\textbf{(A) }24\qquad \textbf{(B) }24.5\qquad \textbf{(C) }24.6\qquad \textbf{(D) }24.8\qquad \textbf{(E) }25$
|
[
"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 know that $A+D=360-(B+C)=90^{\\circ}$. Now extend $AB$ and $CD$ as follows:\n\n\n[asy] size(10cm); label(\"A\",(-1,0)); dot((0,0)); label(\"B\",(-1,4)); dot((0,4)); label(\"E\",(-1,7)); dot((0,7)); label(\"C\",(4,8)); dot((4,7)); label(\"D\",(24,8)); dot((24,7)); draw((0,0)--(0,4)); draw((0,4)--(4,7)); draw((4,7)--(24,7)); draw((24,7)--(0,0)); draw((0,4)--(0,7), dashed); draw((0,7)--(4,7), dashed); //diagram by WannabeCharmander [/asy]\n\n\nLet $AB$ and $CD$ intersect at $E$. We know that $\\angle AED=90^{\\circ}$ because $\\angle E = 180 - (A+D)=180-90 = 90^{\\circ}$.\n\n\nSince $\\sin BCD = \\frac{3}{5}$, we get $\\sin ECB=\\sin(180-BCD)=\\sin BCD = \\frac{3}{5}$. Thus, $EB=3$ and $EC=4$ from simple sin application.\n\n\n$AD$ is the hypotenuse of right $\\triangle AED$, with leg lengths $AB+BE=7$ and $EC+CD=24$. Thus, $AD=\\boxed{\\textbf{(E)}25}$\n\n\n-WannabeCharmander\n\n\n"
] | 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
$\textbf{(A) }\frac{1}{256}\qquad \textbf{(B) }-\frac{1}{16}\qquad \textbf{(C) }1\qquad \textbf{(D) }-16\qquad \textbf{(E) }256$
|
[
"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 the solution to $x+\\frac{1}{2} = 0$ into the quotient to find the remainder. Notice that every term in the quotient, when $x=-\\frac{1}{2}$, evaluates to $-\\frac{1}{128}$. Thus $r_2 =-\\frac{8}{128} = \\boxed{\\textbf{(B) } -\\frac{1}{16}}$.\n\n\n"
] | 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}{x^6})}{\\left( x + \\frac{1}{x} \\right)^3 + (x^3 + \\frac{1}{x^3})} \\\\ &= \\frac{\\left( x + \\frac{1}{x} \\right)^6 - (x^3 + \\frac{1}{x^3})^2}{\\left( x + \\frac{1}{x} \\right)^3 + (x^3 + \\frac{1}{x^3})} \\\\ &= \\frac{a^2 - b^2}{a + b}. \\end{align*}\n\n\nBy difference of squares,\n\\begin{align*} f(x) &= \\frac{(a - b)(a + b)}{a + b} \\\\ &= a - b \\\\ &= \\left( x + \\frac{1}{x} \\right)^3 - \\left( x^3 + \\frac{1}{x^3} \\right) \\\\ &= \\left( x^3 + 3x + \\frac{3}{x} + \\frac{1}{x^3} \\right) - \\left( x^3 + \\frac{1}{x^3} \\right) \\\\ &= 3x + \\frac{3}{x} \\\\ &= 3 \\left( x + \\frac{1}{x} \\right). \\end{align*}\n\n\nBy the AM-GM inequality,\n\\[x + \\frac{1}{x} \\ge 2,\\]\nso $f(x) \\ge 6$. Furthermore, when $x = 1$, $f(1) = 6$, so the minimum value of $f(x)$ is $\\boxed{6}$.\n\n\n"
] | 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, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$.
What is the measure of $\measuredangle AED$ in degrees?
$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$
|
[
"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 equilateral triangle be $1$. $\\angle{DAE}=90^\\circ+60^\\circ=150^\\circ$. Apply the law of cosines then the law of sines, we find that $\\angle{AED}=15^\\circ$. Select $\\boxed{C}$.\n\n\nPlease don't try this solution. This is just obnoxiously tedious and impractical.\n\n\n~hastapasta\n\n\n"
] | 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 graphs of $y=x$ and $y=-x$ is $90^\\circ$. Thus, the smallest region bounded by the graphs is $\\frac{1}{4}\\cdot2^2\\cdot\\pi = \\boxed{\\textbf{(C) } \\pi}$.\n\n\n"
] | 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 $\\log_5 10 = \\frac{1}{\\frac{1}{x}-1}+1 = \\frac{x}{1-x}+1 = \\frac{1}{1-x}$. Since $x$ is around $\\frac{3}{10}$, we substitute and get $\\boxed{\\textbf{(C) }\\frac{10}{7}}$.\n\n\n"
] | 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
$\textbf{(A) }\frac{25}{9}\qquad \textbf{(B) }\frac{31}{15}\qquad \textbf{(C) }\frac{61}{16}\qquad \textbf{(D) }\frac{576}{225}\qquad \textbf{(E) }34$
|
[
"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 $\\frac{25}{16}$ respectively. The sum of those numbers is $\\frac{36}{16}+\\frac{25}{16} = \\boxed{\\textbf{(C) } \\frac{61}{16}}$.\n\n\n"
] | 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
$\textbf{(A) }\frac{p+q}{2}\qquad \textbf{(B) }\frac{p^2+q^2}{p+q}\qquad \textbf{(C) }\frac{2pq}{p+q}\qquad \textbf{(D) }\frac{2(p^2+pq+q^2)}{3(p+q)}\qquad \textbf{(E) }\frac{p+q+2pq}{p+q+2}$
|
[
"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 of water is $\\frac{1}{p+1} and \\frac{1}{q+1} = \\frac{p+q+2}{(p+1)(q+1)}$. The ratio of the volume of alcohol to the volume of water in the mixture is $\\frac{\\frac{p+q+2pq}{(p+1)(q+1)}}{\\frac{p+q+2}{(p+1)(q+1)}} = \\boxed{\\textbf{(E) } \\frac{p+q+2pq}{p+q+2}}$.\n\n\n"
] | 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 equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are $\\pm1$, and dilating by $2$ gives $\\pm2$. The sum of the squares is $(2)^2+(-2)^2 = \\boxed{\\textbf{(A) } 8}$\n\n\n",
"Notice that we can change the equation to be $x^{256}=256^{32}$.\n\n\nThe RHS can be simplified to $x^{256}=2^{256}$\n\n\nHence, the only real solutions are $x=-2,2$\n\n\nHence, $(-2)^2+(2)^2=8=\\boxed{A}$.\n\n\n"
] | 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; add("hatch",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; aux=midpoint(AA--BB); draw(BB--DD); P=midpoint(AA--aux); aux=midpoint(CC--DD); Q=midpoint(CC--aux); draw(AA--CC,dashed); dot(P); dot(Q); fill(DD--BB--CC--cycle,pattern("hatch")); label("$A$",AA,W); label("$B$",BB,S); label("$C$",CC,E); label("$D$",DD,N); label("$P$",P,S); label("$Q$",Q,E); //Credit to TheMaskedMagician for the diagram[/asy]
$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }\frac{3}{4}\qquad \textbf{(C) }\frac{\sqrt{2}}{2}\qquad \textbf{(D) }\frac{\sqrt{3}}{2}\qquad \textbf{(E) }\frac{\sqrt{3}}{3}$
|
[
"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 $\\triangle ABC$, $CP=\\frac{\\sqrt{3}}{2}$.\n\n\n[asy] size(150); import patterns; import olympiad; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux; add(\"hatch\",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; draw(BB--DD); P=midpoint(AA--BB); Q=midpoint(CC--DD); draw(P--Q,dashed); draw(P--CC,dashed); draw(AA--CC,dashed); dot(P); dot(Q); label(\"$A$\",AA,W); label(\"$B$\",BB,S); label(\"$C$\",CC,E); label(\"$D$\",DD,N); label(\"$P$\",P,S); label(\"$Q$\",Q,E); //Credit to TheMaskedMagician for the diagram //Changes made by Treetor10145[/asy]\n\n\nNext, we need to find $\\cos(\\angle PCQ)$ in order to find $PQ$ by the Law of Cosines. To do so, drop down $D$ onto $\\triangle ABC$ to get the point $D^\\prime$.\n\n\n$\\angle PCD$ is congruent to $\\angle D^\\prime CD$, since $P$, $D^\\prime$, and $C$ are collinear. Therefore, we can just find $\\cos(\\angle D^\\prime CD)$.\n\n\nNote that $\\triangle CD^\\prime D$ is a right triangle with $\\angle CD^\\prime D$ as a right angle. \n\n\n\n\n[asy] size(150); import patterns; import olympiad; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux,R; add(\"hatch\",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; draw(BB--DD); P=midpoint(AA--BB); Q=midpoint(CC--DD); R=midpoint(AA--CC); pair X=intersectionpoints(P--CC,BB--R)[0]; draw(AA--CC,dashed); draw(DD--X,dashed); draw(X--CC,dashed); draw(rightanglemark(CC,X,DD)); dot(P); dot(Q); dot(X); label(\"$A$\",AA,W); label(\"$B$\",BB,S); label(\"$C$\",CC,E); label(\"$D$\",DD,N); label(\"$P$\",P,S); label(\"$Q$\",Q,E); label(\"$D^\\prime$\",X,W); //Credit to TheMaskedMagician for the diagram //Changes made by Treetor10145[/asy]\n\n\n\n\nAs given by the problem, $CD=1$.\n\n\nNote that $D^\\prime$ is the centroid of equilateral $\\triangle ABC$. Additionally, since $\\triangle ABC$ is equilateral, $D^\\prime$ is also the orthocenter. Due to this, the distance from $C$ to $D^\\prime$ is $\\frac{2}{3}$ of the altitude of $\\triangle ABC$. Therefore, $CD^\\prime=\\frac{\\sqrt{3}}{3}$. \n\n\nSince $\\cos(\\angle D^\\prime CD)=\\cos(\\angle PCQ)=\\frac{CD^\\prime}{CD}$, $\\cos(\\angle PCQ)=\\frac{\\frac{\\sqrt{3}}{3}}{1}=\\frac{\\sqrt{3}}{3}$\n\n\n\\[PQ^2=CP^2+CQ^2-2(CP)(CQ)\\cos(\\angle PCQ)\\]\n\\[PQ^2=\\frac{3}{4}+\\frac{1}{4}-2\\left(\\frac{\\sqrt{3}}{4}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{\\sqrt{3}}{3}\\right)\\]\nSimplifying, $PQ^2=\\frac{1}{2}$.\nTherefore, $PQ=\\frac{\\sqrt{2}}{2}\\Rightarrow$ $\\boxed{\\textbf{C}}$\n\n\nSolution by treetor10145\n\n\n",
"Notice, like above said, that $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. \n\n\n\n\nTo find the length of $PQ$, first draw in lines $CP$ and $DP$. Notice that $DP$ is an altitude of $\\triangle ADP$. We find that $\\angle{DAP} = 60 ^{\\circ}$ (since $\\triangle ABD$ is equilateral), and $AD=\\frac{1}{2}$. Use the properties of 30-60-90 triangles to get $DP=\\frac{\\sqrt{3}}{2}$. Since $CP$ is an altitude of a congruent equilateral triangle, $CP=DP=\\frac{\\sqrt{3}}{2}$.\n\n\n\n\n\n\nNotice that $\\triangle CDP$ is isosceles with $CP=DP$. Also, since $Q$ is the midpoint of base $CD$, we can conclude that $PQ$ is an altitude. We can use Pythagorean theorem to get the following (taking into consideration $DQ=\\frac{1}{2}$):\n\n\n\n\n\n\n\\[DQ^2+PQ^2=PD^2\\]\n\n\n\n\n\\[\\left(\\frac{1}{2}\\right)^2+PQ^2 = \\left(\\frac{\\sqrt{3}}{2}\\right)^2\\]\n\n\n\n\n\\[PQ^2=\\frac{3}{4}-\\frac{1}{4}=\\frac{1}{2}\\]\n\n\n\n\n\\[PQ=\\frac{\\sqrt{2}}{2}\\Rightarrow \\boxed{\\textbf{C}}\\]\n\n\n\n\n-WannabeCharmander\n\n\n"
] | 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 that $m=a_9=99$. From $19=a_1=m-x$ we get that $x=80$. And thus $a_2 = m+x = \\boxed{(E) 179}$.\n\n\n",
"Let $a_1=a$ and $a_2=b$. Then, $a_3=\\frac{a+b}{2}$, $a_4=\\frac{a+b+\\frac{a+b}{2}}{3}=\\frac{a+b}{2},$ and so on.\n\n\nSince $a_3=a_4$, $a_n=a_3$ for all $n\\geq3.$\n\n\nHence, $a_9=\\frac{a_1+a_2}{2}=\\frac{a+b}{2}=99, a+b=198.$ We also know that $a_1=a=19.$\n\n\nSubtracting $a_1$ from $198,$ we get $b=a_2=\\boxed{(E) 179}.$\n\n\n~Benedict T (countmath1)\n\n\n"
] | 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 $d_1 d_2 / 2 = 2ar$ for $r$ we get $r=\\boxed{\\frac{60}{13}}$.\n\n\n(The first formula computes the area as one half of the circumscribed rectangle whose sides are parallel to the diagonals. The second one comes from the fact that we can divide the rhombus into $4$ equal triangles, and in those the height on the side $a$ is equal to $r$. See pictures below.)\n\n\n\\begin{center}\n[asy] unitsize(1cm); pair A=(-3,0), B=(0,-1.25), C=(3,0), D=(0,1.25); pair E=(-3,-1.25), F=(-3,1.25), G=(3,1.25), H=(3,-1.25); fill ( E -- F -- G -- H -- cycle, lightgray ); draw ( E -- F -- G -- H -- cycle ); draw ( A -- B -- C -- D -- cycle ); label(\"$d_1$\",(E+F)*0.5,W); label(\"$d_2$\",(F+G)*0.5,N); [/asy]\\end{center}\n\\begin{center}\n[asy] unitsize(1cm); pair A=(-3,0), B=(0,-1.25), C=(3,0), D=(0,1.25); pair P = intersectionpoint( A--B--C--D--cycle, circle( (0,0), 15/13 ) ); fill ( A -- B -- (0,0) -- cycle, lightgray ); draw ( A -- B -- (0,0) -- cycle ); draw ( A -- B -- C -- D -- cycle ); draw ( circle( (0,0), 15/13 ) ); draw ( (0,0) -- P ); label(\"$a$\",(A+B)*0.5,SW); label(\"$r$\",P*0.5,0.5*NW); [/asy]\\end{center}\n"
] | 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 sum to less than $90\\cdot 5 + 180 = 630$ degrees, so these angles could not form a hexagon.\n\n\nFour acute angles and two obtuse angles can only sum to less than $90\\cdot 4 + 180\\cdot 2 = 720$ degrees. This is a strict inequality, so these angles could not form a hexagon. (The limiting figure would be four right angles and two straight angles, which would really be a square with two \"extra\" points on two sides to form the straight angles.)\n\n\nThree acute angles and three obtuse angles work. For example, if you pick three acute angles of $80$ degrees, the three obtuse angles would be $160$ degrees and give a sum of $80\\cdot 3 + 160\\cdot 3 = 720$ degrees, which is a genuine hexagon. Thus, the answer is $\\boxed{(B) 3}$\n\n\n"
] | 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 $(x-19)(x-99)$?
$\mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0$
|
[
"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 the left hand side as well. Substituting, we get $Q(99)(99-19) + 80 = 0$, from which $Q(99)=-1$. This means that $99$ is a root of the polynomial $Q(x)+1$. In other words, there is a polynomial $S(x)$ such that $Q(x)+1 = S(x)(x-99)$.\n\n\nSubstituting this into the original formula for $P(x)$ we get \n\\[P(x) = Q(x)(x-19) + 99 = (S(x)(x-99) - 1)(x-19) + 99 =\\]\n\\[= S(x)(x-99)(x-19) - (x-19) + 99\\]\n\n\nTherefore when $P(x)$ is divided by $(x-19)(x-99)$, the remainder is $\\boxed{-x + 118}$.\n\n\n",
"Since the divisor $(x-19)(x-99)$ is a quadratic, the degree of the remainder is at most linear. We can write $P(x)$ in the form\n\\[P(x) = Q(x)(x-19)(x-99) + cx+d\\] \nwhere $cx+d$ is the remainder.\nBy the Remainder Theorem, plugging in $19$ and $99$ gives us a system of equations.\n\\[99c+d = 19\\]\n\\[19c+d = 99\\]\n\n\nSolving gives us $c=-1$ and $d = 118$, thus, our answer is $\\boxed{ \\mathrm{(C) \\ }-x+118}$\n\n\n"
] | 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
$\mathrm{(A) \ }A+B=C \qquad \mathrm{(B) \ }A+B+210=C \qquad \mathrm{(C) \ }A^2+B^2=C^2 \qquad \mathrm{(D) \ }20A+21B=29C \qquad \mathrm{(E) \ } \frac 1{A^2}+\frac 1{B^2}= \frac 1{C^2}$
|
[
"$20^2 + 21^2 = 841 = 29^2$. 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 expressed as $A+B+210$. Thus the answer is $\\boxed{\\mathrm{(B)}}$.\n\n\nTo complete the solution, note that $\\mathrm{(A)}$ is clearly false. As $A+B < C$, we have $A^2 + B^2 < (A+B)^2 < C^2$ and thus $\\mathrm{(C)}$ is false. Similarly $20A + 21B < 21(A+B) < 21C < 29C$, thus $\\mathrm{(D)}$ is false. And finally, since $0<A<C$, $\\frac 1{C^2} < \\frac1{A^2} < \\frac 1{A^2} + \\frac 1{B^2}$, thus $\\mathrm{(E)}$ is false as well.\n\n\n",
"Copy the link into the browser to see a diagram (I don't know how to link it and I tried using the brackets, but it didn't work)https://geogebra.org/classic/psg3ugzm\n\n\nLet $\\circ{O}$ be the circumcircle of $\\triangle{XYZ}$ in the problem, and let the circle have a radius of $r$. Let $XY=20, YZ=21, XZ=29$.\n\n\nUsing the law of cosines: $29^2=20^2+21^2-2*20*21*\\cos{XYZ}$.\n\n\nThus $\\cos{XYZ}=0$, thus the triangle is right. Thus follow as above: \"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 expressed as $A+B+210$. Thus the answer is $\\boxed{\\mathrm{(B)}}$\" (I quoted the solution above to show you where to continue).\n\n\n~hastapasta\n\n\n\\hrule\nFixed the link ~Yiyj1\n\n\n"
] | 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?
$\textbf{(A)}\ \text{I is true.} \qquad \textbf{(B)}\ \text{I is false.}\qquad \textbf{(C)}\ \text{II is true.} \qquad \textbf{(D)}\ \text{III is true.} \qquad \textbf{(E)}\ \text{IV is false.}$
|
[
"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 possible perimeter that this polygon can have?
$\mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24$
|
[
"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 problem is the value $(a-2)+(b-2)+(c-2) = (a+b+c)-6$.\n\n\nThe integral angle of a regular $k$-gon is $180 \\frac{k-2}k$. Therefore we are looking for integer solutions to:\n\n\n\\[360 = 180\\left( \\frac{a-2}a + \\frac{b-2}b + \\frac{c-2}c \\right)\\]\n\n\nWhich can be simplified to:\n\n\n\\[2 = \\left( \\frac{a-2}a + \\frac{b-2}b + \\frac{c-2}c \\right)\\]\n\n\nFurthermore, we know that two of the polygons are congruent, thus WLOG $a=c$. Our equation now becomes\n\n\n\\[2 = \\left( 2\\cdot\\frac{a-2}a + \\frac{b-2}b \\right)\\]\n\n\nMultiply both sides by $ab$ and simplify to get $ab - 4b - 2a = 0$.\n\n\nUsing the standard technique for Diophantine equations, we can add $8$ to both sides and rewrite the equation as $(a-4)(b-2)=8$.\n\n\nRemembering that $a,b\\geq 3$ the only valid options for $(a-4,b-2)$ are: $(1,8)$, $(2,4)$, $(4,2)$, and $(8,1)$.\n\n\nThese correspond to the following pairs $(a,b)$: $(5,10)$, $(6,6)$, $(8,4)$, and $(12,3)$. \n\n\nThe perimeters of the resulting polygon for these four cases are $14$, $12$, $14$, and $\\boxed{21}$.\n\n\n",
"We want to maximize the number of sides of the two congruent polygons, so we need to make the third polygon have the fewest number of sides possible, i.e. a triangle. The interior angle measure of the two congruent polygons is therefore $\\frac{360-60}{2}=150$ degrees, so they are dodecagons. Of all the $12 + 12 + 3 = 27$ sides, six of them are not part of the perimeter of the resulting polygon, so the resulting polygon has $\\boxed{21}$ sides.\n\n\n"
] | 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 $35 \\Longrightarrow \\mathrm{(D)}$ solutions satisfying $mn \\ge 0$.\n\n\n"
] | 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 \\end{align*}\n\n\nThus $\\frac 12 = \\sin A \\cos B + \\sin B \\cos A$. This is the sine addition identity, so $\\frac 12 = \\sin (A + B) = \\sin (180 - C) = \\sin C$. Thus either $C = 30^{\\circ}, 150^{\\circ}$. \n\n\nIf $C = 150$, then $A + B = 30 \\Longrightarrow A,B < 30$, and $\\sin A < \\frac 12, \\cos A < 1$. The first equation implies $6 = 3 \\sin A + 4\\cos B < 3\\left(\\frac 12\\right) + 4(1) = 5.5 < 6$, which is a contradiction; thus $C = 30 \\Longrightarrow \\mathrm{(A)}$.\n\n\n"
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./CreativeMath/AHSME/1999_AHSME_Problems/27.json
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AHSME
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