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 |
|---|---|---|---|---|---|---|---|---|---|
1974_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | If $(a, b)$ and $(c, d)$ are two points on the line whose equation is $y=mx+k$, then the distance between $(a, b)$ and $(c, d)$, in terms of $a, c,$ and $m$ is
$\mathrm{(A)\ } |a-c|\sqrt{1+m^2} \qquad \mathrm{(B) \ }|a+c|\sqrt{1+m^2} \qquad \mathrm{(C) \ } \frac{|a-c|}{\sqrt{1+m^2}} \qquad$
$\mathrm{(D) \ } |a-c... | [
"Notice that since $(a, b)$ is on $y=mx+k$, we have $b=am+k$. Similarly, $d=cm+k$. Using the distance formula, the distance between the points $(a, b)$ and $(c, d)$ is\n\n\n$\\sqrt{(a-c)^2+(b-d)^2}=\\sqrt{(a-c)^2+(am+k-cm-k)^2}$\n\n\n$=\\sqrt{(a-c)^2+m^2(a-c)^2}$\n\n\n$=|a-c|\\sqrt{1+m^2}.$\n\n\nAnd so the answer i... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/11.json | AHSME |
1974_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | Let $x_1$ and $x_2$ be such that $x_1\not=x_2$ and $3x_i^2-hx_i=b$, $i=1, 2$. Then $x_1+x_2$ equals
$\mathrm{(A)\ } -\frac{h}{3} \qquad \mathrm{(B) \ }\frac{h}{3} \qquad \mathrm{(C) \ } \frac{b}{3} \qquad \mathrm{(D) \ } 2b \qquad \mathrm{(E) \ }-\frac{b}{3}$
| [
"Notice that $x_1$ and $x_2$ are the distinct solutions to the quadratic $3x^2-hx-b=0$. By Vieta, the sum of the roots of this quadratic is the negation of the coefficient of the linear term divided by the coefficient of the quadratic term, so in this case $-\\frac{-h}{3}=\\frac{h}{3}, \\boxed{\\text{B}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/2.json | AHSME |
1974_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | Which of the following is satisfied by all numbers $x$ of the form
\[x=\frac{a_1}{3}+\frac{a_2}{3^2}+\cdots+\frac{a_{25}}{3^{25}}\]
where $a_1$ is $0$ or $2$, $a_2$ is $0$ or $2$,...,$a_{25}$ is $0$ or $2$?
$\mathrm{(A)\ } 0\le x<1/3 \qquad \mathrm{(B) \ } 1/3\le x<2/3 \qquad \mathrm{(C) \ } 2/3\le x<1 \qquad$
... | [
"We can separate this into two cases.\n\n\n\\textbf{Case 1: $a_1=0$} Here, clearly the minimum sum is when all of the $a_i$ are $0$, so $0\\le x$. The maximum sum is when the rest of the $a_i$ are all $2$. In this case, \\[\\sum_{n=2}^{25}\\frac{2}{3^n}<2\\sum_{n=2}^{\\infty}\\frac{1}{3^n}=2\\left(\\frac{\\frac{1}{... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/28.json | AHSME |
1974_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | If $g(x)=1-x^2$ and $f(g(x))=\frac{1-x^2}{x^2}$ when $x\not=0$, then $f(1/2)$ equals
$\mathrm{(A)\ } 3/4 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } \sqrt{2}/2 \qquad \mathrm{(E) \ }\sqrt{2}$
| [
"Notice that if we can find a $y$ such that $g(y)=\\frac{1}{2}$, then we can plug that into the second equation to get $f\\left(\\frac{1}{2}\\right)=f(g(y))=\\frac{1-y^2}{y^2}$. Thus, we let $1-y^2=\\frac{1}{2}\\implies y^2=\\frac{1}{2}$. Notice that, since our final expression only involves $y^2$, we don't need to... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/12.json | AHSME |
1974_AHSME_Problems | 24 | 0 | Probability | Multiple Choice | A fair die is rolled six times. The probability of rolling at least a five at least five times is
$\mathrm{(A)\ } \frac{13}{729} \qquad \mathrm{(B) \ }\frac{12}{729} \qquad \mathrm{(C) \ } \frac{2}{729} \qquad \mathrm{(D) \ } \frac{3}{729} \qquad \mathrm{(E) \ }\text{none of these}$
| [
"The probability of rolling at least a five on any one roll is $\\frac{2}{6}=\\frac{1}{3}$. If there are exactly $5$ fives or sixes rolled, there are $\\binom{6}{5}=6$ ways to pick which of the rolls are the fives and sixes, and so the probability in this case is $6\\left(\\frac{1}{3}\\right)^5\\left(\\frac{2}{3}\\... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/24.json | AHSME |
1974_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | In parallelogram $ABCD$ of the accompanying diagram, line $DP$ is drawn bisecting $BC$ at $N$ and meeting $AB$ (extended) at $P$. From vertex $C$, line $CQ$ is drawn bisecting side $AD$ at $M$ and meeting $AB$ (extended) at $Q$. Lines $DP$ and $CQ$ meet at $O$. If the area of parallelogram $ABCD$ is $k$, then the area ... | [
"Note that \\[[QPO]=[QAM]+[PBN]+[AMONB]=[AMONB]+[MDC]+[NCD]\\]\n\n\n\\[=[AMONB]+[MDC]+[NOC]+[DOC]=[ABCD]+[DOC]=k+[DOC].\\]\n\n\nAlso, note that $DCNM$ is a parallelogram, and so $[DOC]=\\frac{1}{4}[DCNM]=\\frac{1}{8}[ABCD]=\\frac{k}{8}$.\n\n\nTherefore, $[QPO]=k+\\frac{k}{8}=\\frac{9k}{8}, \\boxed{\\text{C}}$.\n\n\... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/25.json | AHSME |
1974_AHSME_Problems | 13 | 0 | Other | Multiple Choice | Which of the following is equivalent to "If P is true, then Q is false."?
$\mathrm{(A)\ } \text{``P is true or Q is false."} \qquad$
$\mathrm{(B) \ }\text{``If Q is false then P is true."} \qquad$
$\mathrm{(C) \ } \text{``If P is false then Q is true."} \qquad$
$\mathrm{(D) \ } \text{``If Q is true then P is ... | [
"Remember that a statement is logically equivalent to its contrapositive, which is formed by first negating the hypothesis and conclusion and then switching them. In this case, the contrapositive of \"If P is true, then Q is false.\" is \"If Q is true, then P is false.\" $\\boxed{\\text{D}}$\n\n\nThe fact that a st... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/13.json | AHSME |
1974_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | For $p=1, 2, \cdots, 10$ let $S_p$ be the sum of the first $40$ terms of the arithmetic progression whose first term is $p$ and whose common difference is $2p-1$; then $S_1+S_2+\cdots+S_{10}$ is
$\mathrm{(A)\ } 80000 \qquad \mathrm{(B) \ }80200 \qquad \mathrm{(C) \ } 80400 \qquad \mathrm{(D) \ } 80600 \qquad \mathr... | [
"The $40\\text{th}$ term of an arithmetic progression with a first term $p$ and a common difference $2p-1$ is $p+39(2p-1)=79p-39$. Therefore, the sum of the first $40$ terms of such a progression is $\\frac{40}{2}(79p-39+p)=1600p-780$.\n\n\nWe now want to evaluate $\\sum_{p=1}^{10}(1600p-780)$. \\[\\sum_{p=1}^{10}(... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/29.json | AHSME |
1974_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | The coefficient of $x^7$ in the polynomial expansion of
\[(1+2x-x^2)^4\]
is
$\mathrm{(A)\ } -8 \qquad \mathrm{(B) \ }12 \qquad \mathrm{(C) \ } 6 \qquad \mathrm{(D) \ } -12 \qquad \mathrm{(E) \ }\text{none of these}$
| [
"Let's write out the multiplication, so that it becomes easier to see.\n\n\n\\[(1+2x-x^2)(1+2x-x^2)(1+2x-x^2)(1+2x-x^2)\\]\n\n\nWe can now see that the only way to get an $x^7$ is by taking three $-x^2$ and one $2x$. There are $\\binom{4}{1}=4$ ways to pick which term the $2x$ comes from, and the coefficient of eac... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/3.json | AHSME |
1974_AHSME_Problems | 8 | 0 | Number Theory | Multiple Choice | What is the smallest prime number dividing the sum $3^{11}+5^{13}$?
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 3^{11}+5^{13} \qquad \mathrm{(E) \ }\text{none of these}$
| [
"Since we want to find the smallest prime dividing the sum, we start with the smallest prime and move up, so first we try $2$. Notice that $3^{11}$ and $5^{13}$ are both odd, so their sum must be even. This means that $2$ must divide $3^{11}+5^{13}$, and so since $2$ is the smallest prime, our answer must be $2, \\... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/8.json | AHSME |
1974_AHSME_Problems | 22 | 0 | Other | Multiple Choice | The minimum of $\sin\frac{A}{2}-\sqrt{3}\cos\frac{A}{2}$ is attained when $A$ is
$\mathrm{(A)\ } -180^\circ \qquad \mathrm{(B) \ }60^\circ \qquad \mathrm{(C) \ } 120^\circ \qquad \mathrm{(D) \ } 0^\circ \qquad \mathrm{(E) \ }\text{none of these}$
| [
"Define a new number $B$ such that $\\cos(B)=\\frac{1}{2}$ and $\\sin(B)=-\\frac{\\sqrt{3}}{2}$. Notice that $B=\\frac{5\\pi}{3}$. Therefore, we have $\\sin\\frac{A}{2}-\\sqrt{3}\\cos\\frac{A}{2}= 2\\cos(B)\\sin\\left(\\frac{A}{2}\\right)+2\\sin(B)\\cos\\left(\\frac{A}{2}\\right)$. \n\n\nRecognizing the angle sum f... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/22.json | AHSME |
1974_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | If $\log_8{3}=p$ and $\log_3{5}=q$, then, in terms of $p$ and $q$, $\log_{10}{5}$ equals
$\mathrm{(A)\ } pq \qquad \mathrm{(B) \ }\frac{3p+q}{5} \qquad \mathrm{(C) \ } \frac{1+3pq}{p+q} \qquad \mathrm{(D) \ } \frac{3pq}{1+3pq} \qquad \mathrm{(E) \ }p^2+q^2$
| [
"Notice that $\\frac{1}{p}=\\log_{3}{8}$, so it would probably be easier to work in base $3$. From change of base, $\\log_{10}{5}=\\frac{\\log_{3}{5}}{\\log_{3}{10}}$. We're given that $\\log_{3}{5}=q$, so now we just need to find $\\log_{3}{10}$. \n\n\nWe have $\\frac{1}{p}=\\log_{3}{8}=3\\log_{3}{2}$, so $\\log_{... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/18.json | AHSME |
1974_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | What is the remainder when $x^{51}+51$ is divided by $x+1$?
$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 49 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ }51$
| [
"From the Remainder Theorem, the remainder when $x^{51}+51$ is divided by $x+1$ is $(-1)^{51}+51=-1+51=50, \\boxed{\\text{D}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/4.json | AHSME |
1974_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | Which statement is correct?
$\mathrm{(A)\ } \text{If } x<0, \text{then } x^2>x. \qquad \mathrm{(B) \ } \text{If } x^2>0, \text{then } x>0.$
$\qquad \mathrm{(C) \ } \text{If } x^2>x, \text{then } x>0. \qquad \mathrm{(D) \ } \text{If } x^2>x, \text{then } x<0.$
$\qquad \mathrm{(E) \ }\text{If } x<1, \text{then }... | [
"We begin by going statement by statement.\n\n\nFor (A), $x^2>0$ for all real $x$, and so $x^2>0>x$. Therefore, $\\boxed{\\text{A}}$ is true.\n\n\nWe can give counterexamples for all other statements:\n\n\nFor (B), $x=-1$\n\n\nFor (C), $x=-1$\n\n\nFor (D), $x=2$\n\n\nFor (E), $x=-1$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/14.json | AHSME |
1974_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | If $x<-2$, then $|1-|1+x||$ equals
$\mathrm{(A)\ } 2+x \qquad \mathrm{(B) \ }-2-x \qquad \mathrm{(C) \ } x \qquad \mathrm{(D) \ } -x \qquad \mathrm{(E) \ }-2$
| [
"Notice that, for $a<0$, $|a|=-a$, and for $a>0$, $|a|=a$.\n\n\nSince $x<-2$, $1+x<1-2<0$, so $|1+x|=-x-1$. Therefore, $1-|1+x|=1+x+1=x+2<-2+2=0$, and so $|1-|1+x||=|x+2|=-2-x, \\boxed{\\text{B}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/15.json | AHSME |
1974_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | Given a quadrilateral $ABCD$ inscribed in a circle with side $AB$ extended beyond $B$ to point $E$, if $\measuredangle BAD=92^\circ$ and $\measuredangle ADC=68^\circ$, find $\measuredangle EBC$.
$\mathrm{(A)\ } 66^\circ \qquad \mathrm{(B) \ }68^\circ \qquad \mathrm{(C) \ } 70^\circ \qquad \mathrm{(D) \ } 88^\circ \... | [
"Since $ABCD$ is cyclic, opposite angles must sum to $180^\\circ$. Therefore, $\\angle ADC+\\angle ABC=180^\\circ$, and $\\angle ABC=180^\\circ-\\angle ADC=180^\\circ-68^\\circ=112^\\circ$. Notice also that $\\angle ABC$ and $\\angle CBE$ form a linear pair, and so they sum to $180^\\circ$. Therefore, $\\angle EBC=... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/5.json | AHSME |
1974_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | In the adjoining figure $ABCD$ is a square and $CMN$ is an equilateral triangle. If the area of $ABCD$ is one square inch, then the area of $CMN$ in square inches is
[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((.82,0)--(1,1)--(0,.76)--cycle); label("A", (0,0), S); label("B", (1,0), S); label("C", (1,1), N); ... | [
"Let $BN=x$ so that $AN=1-x$. From the Pythagorean Theorem on $\\triangle NBC$, we get $CN=\\sqrt{x^2+1}$, and from the Pythagorean Theorem on $\\triangle AMN$, we get $MN=(1-x)\\sqrt{2}$. Since $\\triangle CMN$ is equilateral, we must have $\\sqrt{x^2+1}=(1-x)\\sqrt{2}\\implies x^2+1=2x^2-4x+2\\implies x^2-4x+1=0$... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/19.json | AHSME |
1974_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | In the adjoining figure $TP$ and $T'Q$ are parallel tangents to a circle of radius $r$, with $T$ and $T'$ the points of tangency. $PT''Q$ is a third tangent with $T'''$ as a point of tangency. If $TP=4$ and $T'Q=9$ then $r$ is
[asy] unitsize(45); pair O = (0,0); pair T = dir(90); pair T1 = dir(270); pair T2 = dir(25)... | [
"[asy] unitsize(45); pair O = (0,0); pair T = dir(90); pair T1 = dir(270); pair T2 = dir(25); pair P = (.61,1); pair Q = (1.61, -1); draw(unitcircle); dot(O); label(\"O\",O,W); label(\"T\",T,N); label(\"T'\",T1,S); label(\"T''\",T2,NE); label(\"P\",P,NE); label(\"Q\",Q,S); draw(T--P); label(\"4\",midpoint(T--P),N);... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/23.json | AHSME |
1974_AHSME_Problems | 9 | 0 | Number Theory | Multiple Choice | The integers greater than one are arranged in five columns as follows:
\[\begin{tabular}{c c c c c}\ & 2 & 3 & 4 & 5\\ 9 & 8 & 7 & 6 &\ \\ \ & 10 & 11 & 12 & 13\\ 17 & 16 & 15 & 14 &\ \\ \ & . & . & . & .\\ \end{tabular}\]
(Four consecutive integers appear in each row; in the first, third and other odd numbered row... | [
"We try pairing numbers with the column number they're in.\n\n\n\\[\\begin{tabular}{c c}\\ \\text{Number} & \\text{Column Number} \\\\ 2 & 2\\\\ 3 & 3\\\\ 4 & 4\\\\ 5 & 5\\\\ 6 & 4\\\\ 7 & 3\\\\ 8 & 2\\\\ 9 & 1\\\\ 10 & 2\\\\ 11 & 3\\\\ 12 & 4\\\\ \\end{tabular}\\]\n\n\nNow we can see a pattern. The column number s... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/9.json | AHSME |
1969_AHSME_Problems | 20 | 0 | Arithmetic | Multiple Choice | Let $P$ equal the product of 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in $P$ is:
$\text{(A) } 36\quad \text{(B) } 35\quad \text{(C) } 34\quad \text{(D) } 33\quad \text{(E) } 32$
| [
"Through inspection, we see that the two digit number $33^{2}=1089=4$ digits.\nNotice that any number that has the form $33abcdefg.......$ multiplied by another $33qwertyu.........$ will have its number of digits equal to the sum of the original numbers' digits.\n\n\nIn this case, we see that the first number has $... | 2 | ./CreativeMath/AHSME/1969_AHSME_Problems/20.json | AHSME |
1969_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | When $(a-b)^n,n\ge2,ab\ne0$, is expanded by the binomial theorem, it is found that when $a=kb$, where $k$ is a positive integer, the sum of the second and third terms is zero. Then $n$ equals:
$\text{(A) } \tfrac{1}{2}k(k-1)\quad \text{(B) } \tfrac{1}{2}k(k+1)\quad \text{(C) } 2k-1\quad \text{(D) } 2k\quad \text{(E) ... | [
"Since $a=kb$, we can write $(a-b)^n$ as $(kb-b)^n$.\nExpanding, the second term is $-k^{n-1}b^{n}{{n}\\choose{1}}$, and the third term is $k^{n-2}b^{n}{{n}\\choose{2}}$, so we can write the equation\n\\[-k^{n-1}b^{n}{{n}\\choose{1}}+k^{n-2}b^{n}{{n}\\choose{2}}=0\\]\nSimplifying and multiplying by two to remove th... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/16.json | AHSME |
1969_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | The area of the ring between two concentric circles is $12\tfrac{1}{2}\pi$ square inches. The length of a chord of the larger circle tangent to the smaller circle, in inches, is:
$\text{(A) } \frac{5}{\sqrt{2}}\quad \text{(B) } 5\quad \text{(C) } 5\sqrt{2}\quad \text{(D) } 10\quad \text{(E) } 10\sqrt{2}$
| [
"[asy] draw(circle((0,0),50)); draw(circle((0,0),40)); draw((-30,40)--(30,40),dotted); draw((-30,40)--(0,0)--(0,40),dotted); draw((-5,40)--(-5,35)--(0,35)); dot((-30,40)); dot((0,40)); dot((30,40)); dot((0,0)); [/asy]\nLet $a$ be radius of larger circle, and $b$ be radius of smaller circle. The area of the ring ca... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/6.json | AHSME |
1969_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | If the points $(1,y_1)$ and $(-1,y_2)$ lie on the graph of $y=ax^2+bx+c$, and $y_1-y_2=-6$, then $b$ equals:
$\text{(A) } -3\quad \text{(B) } 0\quad \text{(C) } 3\quad \text{(D) } \sqrt{ac}\quad \text{(E) } \frac{a+c}{2}$
| [
"Because the two points are on the quadratic, $y_1 = a + b + c$ and $y_2 = a - b + c$. Because $y_1 - y_2 = -6$,\n\\[(a+b+c)-(a-b+c)=-6\\]\n\\[2b=-6\\]\n\\[b=-3\\]\nThe answer is $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/7.json | AHSME |
1969_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | The equation $2^{2x}-8\cdot 2^x+12=0$ is satisfied by:
$\text{(A) } log(3)\quad \text{(B) } \tfrac{1}{2}log(6)\quad \text{(C) } 1+log(\tfrac{3}{2})\quad \text{(D) } 1+\frac{log(3)}{log(2)}\quad \text{(E) none of these}$
| [
"Let $2^x=a$. Because $2^{2x}=(2^x)^2$, the given expression can be rewritten as $a^2-8a+12=0$. This can be factored as $(a-6)(a-2)=0$, which has solutions $a=2^x=6$ and $a=2^x=2$. Looking at the answer choices, we see that $x=1$ is absent. Rewriting $2^x=6$ as $x=log_26$ and then applying the logarithm addition id... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/17.json | AHSME |
1969_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | If the graph of $x^2+y^2=m$ is tangent to that of $x+y=\sqrt{2m}$, then:
$\text{(A) m must equal } \tfrac{1}{2}\quad \text{(B) m must equal } \frac{1}{\sqrt{2}}\quad\\ \text{(C) m must equal } \sqrt{2}\quad \text{(D) m must equal } 2\quad\\ \text{(E) m may be any non-negative real number}$
| [
"Note that the first equation represents a circle and the second equation represents a line. If a line is tangent to a circle, then it only hits at one point, so there will only be one solution to the system of equations.\n\n\nIn the second equation, $y = -x + \\sqrt{2m}$. Substitution results in \n\\[x^2 + x^2 -... | 2 | ./CreativeMath/AHSME/1969_AHSME_Problems/21.json | AHSME |
1969_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | The number of points equidistant from a circle and two parallel tangents to the circle is:
$\text{(A) } 0\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } \infty$
| [
"[asy] draw(circle((0,0),100)); draw((-300,100)--(300,100),Arrows); draw((-300,-100)--(300,-100),Arrows); draw((-300,0)--(300,0),dotted,Arrows); dot((-200,0)); dot((0,0)); dot((200,0)); draw((-200,100)--(-200,-100),dotted); draw((200,100)--(200,-100),dotted); [/asy]\nThe distance between the two parallel tangents i... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/10.json | AHSME |
1969_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | [asy] draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); draw((-1.7,0)--(0,0)--(1.7,0),dot); draw((0,0)--(0,.98),dot); MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); [/asy]
A parabolic arch has a height of $16$ inches and a span of $40$ inches. The height, in inches, of the arch at the point $5$... | [
"Because the arch has a height of $16$ inches, an equation that models the arch is $y = ax^2 + 16$, where $x$ is the horizontal distance from the center and $y$ is the height. The arch has a span of $40$ inches, so the arch meets the ground $20$ inches from the center. That means $0 = 400a + 16$, so $a = -\\frac{... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/26.json | AHSME |
1969_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | Let $P$ be a point of hypotenuse $AB$ (or its extension) of isosceles right triangle $ABC$. Let $s=AP^2+PB^2$. Then:
$\text{(A) } s<2CP^2 \text{ for a finite number of positions of P}\quad\\ \text{(B) } s<2CP^2 \text{ for an infinite number of positions of P}\quad\\ \text{(C) } s=2CP^2 \text{ only if P is the midpoin... | [
"[asy] pair A=(0,50),B=(50,0),C=(0,0); draw(A--B--C--A); dot(A); label(\"$A$\",A,NE); dot(B); label(\"$B$\",B,NE); dot(C); label(\"$C$\",C,SW); pair P=(14,36); dot(P); label(\"$P$\",P,NE); draw((0,36)--P,dotted); draw((14,0)--P,dotted); label(\"$a$\",(7,36),S); label(\"$a$\",(0,43),W); label(\"$x-a$\",(14,18),E);... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/30.json | AHSME |
1969_AHSME_Problems | 31 | 0 | Geometry | Multiple Choice | Let $OABC$ be a unit square in the $xy$-plane with $O(0,0),A(1,0),B(1,1)$ and $C(0,1)$. Let $u=x^2-y^2$, and $v=xy$ be a transformation of the $xy$-plane into the $uv$-plane. The transform (or image) of the square is:
[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,2)--(1,0)--(0,-2)-... | [
"Each point on the square can be in the form $(0,y)$, $(1,y)$, $(x,0)$, and $(x,1)$, where $0 \\le x,y \\le 1$. Making the appropriate substitutions results in points being $(-y^2,0)$, $(1-y^2,2y)$, $(x^2,0)$, and $(x^2 - 1,2x)$ on the $uv$-plane. \n\n\nNotice that since $v \\ge 0$, none of the points are below t... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/31.json | AHSME |
1969_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile the speed is constant. If the second mile is traversed in $2$ hours, then the time, in hours, needed to traverse the $n$th mile is:
$\text{(A) } \frac{2}{n... | [
"Let $d$ be the distance already traveled and $s_n$ be the speed for the $n^\\text{th}$ mile. Because the speed for the second and subsequent miles varies inversely as the integral number of miles already traveled,\n\\[ds_{d+1} = k\\]\nIf the second mile is traversed in $2$ hours, then the speed in the second mile... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/27.json | AHSME |
1969_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | When $x$ is added to both the numerator and denominator of the fraction
$\frac{a}{b},a \ne b,b \ne 0$, the value of the fraction is changed to $\frac{c}{d}$.
Then $x$ equals:
$\text{(A) } \frac{1}{c-d}\quad \text{(B) } \frac{ad-bc}{c-d}\quad \text{(C) } \frac{ad-bc}{c+d}\quad \text{(D) }\frac{bc-ad}{c-d} \quad \tex... | [
"$\\frac{a+x}{b+x}=\\frac{c}{d}$,\n\n\n$bc+cx=ad+dx$,\n\n\n$(c-d)x=ad-bc$,\n\n\n$x=\\frac{ad-bc}{c-d}$. The answer is $\\fbox{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/1.json | AHSME |
1969_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | Given points $P(-1,-2)$ and $Q(4,2)$ in the $xy$-plane; point $R(1,m)$ is taken so that $PR+RQ$ is a minimum. Then $m$ equals:
$\text{(A) } -\tfrac{3}{5}\quad \text{(B) } -\tfrac{2}{5}\quad \text{(C) } -\tfrac{1}{5}\quad \text{(D) } \tfrac{1}{5}\quad \text{(E) either }-\tfrac{1}{5}\text{ or} \tfrac{1}{5}.$
| [
"[asy] import graph; size(7.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-2.2,xmax=6.2,ymin=-4.2,ymax=4.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=1,gy=1;... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/11.json | AHSME |
1969_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | If an item is sold for $x$ dollars, there is a loss of $15\%$ based on the cost. If, however, the same item is sold for $y$ dollars, there is a profit of $15\%$ based on the cost. The ratio of $y:x$ is:
$\text{(A) } 23:17\quad \text{(B) } 17y:23\quad \text{(C) } 23x:17\quad \\ \text{(D) dependent upon the cost} \quad... | [
"Let $c$ be the cost. Selling the item for $x$ dollars equates to losing $15\\%$ of $c$, so $x=.85c$. Selling the item for $y$ dollars equates to profiting by $15\\%$ of $c$, so $y=1.15c$. Therefore $\\frac{y}{x}=\\frac{1.15c}{.85c}=\\frac{23}{17}$. The answer is $\\fbox{A}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/2.json | AHSME |
1969_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | Let $n$ be the number of points $P$ interior to the region bounded by a circle with radius $1$, such that the sum of squares of the distances from $P$ to the endpoints of a given diameter is $3$. Then $n$ is:
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 4\quad \text{(E) } \infty$
| [
"[asy] draw(circle((0,0),50)); draw((-50,0)--(-10,20)--(50,0)--(-50,0)); draw((-10,20)--(30,40)--(50,0),dotted); dot((-50,0)); label(\"$A$\",(-50,0),W); dot((50,0)); label(\"$B$\",(50,0),E); dot((-10,20)); label(\"$P$\",(-10,20),S); dot((30,40)); label(\"$C$\",(30,40),NE); [/asy]\n\n\nLet $A$ and $B$ be points o... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/28.json | AHSME |
1969_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | Let $F=\frac{6x^2+16x+3m}{6}$ be the square of an expression which is linear in $x$. Then $m$ has a particular value between:
$\text{(A) } 3 \text{ and } 4\quad \text{(B) } 4 \text{ and } 5\quad \text{(C) } 5 \text{ and } 6\quad \text{(D) } -4 \text{ and } -3\quad \text{(E) } -6 \text{ and } -5$
| [
"The quadratic can also be written as\n\\[x^2 + \\frac{8}{3}x + \\frac{m}{2}\\]\nIn order for the quadratic to be a square of a linear expression, the constant term must be the square of the x-term divided by 4. Thus,\n\\[\\frac{m}{2} = (\\frac{8}{3})^2 \\cdot \\frac{1}{4}\\]\n\\[\\frac{m}{2} = \\frac{16}{9}\\]\n\... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/12.json | AHSME |
1969_AHSME_Problems | 32 | 0 | Algebra | Multiple Choice | Let a sequence $\{u_n\}$ be defined by $u_1=5$ and the relationship $u_{n+1}-u_n=3+4(n-1), n=1,2,3\cdots.$If $u_n$ is expressed as a polynomial in $n$, the algebraic sum of its coefficients is:
$\text{(A) 3} \quad \text{(B) 4} \quad \text{(C) 5} \quad \text{(D) 6} \quad \text{(E) 11}$
| [
"Note that the first differences create a linear function, so the sequence ${u_n}$ is quadratic.\n\n\nThe first three terms of the sequence are $5$, $8$, and $15$. From there, a system of equations can be written.\n\\[a+b+c=5\\]\n\\[4a+2b+c=8\\]\n\\[9a+3b+c=15\\]\nSolve the system to get $a=2$, $b=-3$, and $c=6$. ... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/32.json | AHSME |
1969_AHSME_Problems | 24 | 0 | Number Theory | Multiple Choice | When the natural numbers $P$ and $P'$, with $P>P'$, are divided by the natural number $D$, the remainders are $R$ and $R'$, respectively. When $PP'$ and $RR'$ are divided by $D$, the remainders are $r$ and $r'$, respectively. Then:
$\text{(A) } r>r' \text{ always}\quad \text{(B) } r<r' \text{ always}\quad\\ \text{(... | [
"The divisors are the same, so take each variable modulo $D$.\n\\[P \\equiv R \\pmod{D}\\]\n\\[P' \\equiv R’ \\pmod{D}\\]\nThat means\n\\[PP’ \\equiv RR’ \\pmod{D}\\]\nThus, $PP’$ and $RR’$ have the same remainder when divided by $D$, so the answer is $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/24.json | AHSME |
1969_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | If it is known that $\log_2(a)+\log_2(b) \ge 6$, then the least value that can be taken on by $a+b$ is:
$\text{(A) } 2\sqrt{6}\quad \text{(B) } 6\quad \text{(C) } 8\sqrt{2}\quad \text{(D) } 16\quad \text{(E) none of these}$
| [
"We use the logarithm property of addition:\n\\begin{align*} \\log_2(a)+\\log_2(b) \\ge 6 &= \\log_2(ab) \\ge 6\\\\ &\\Rightarrow 2^{log_2(ab)} \\ge 2^6\\\\ &= ab \\ge 64 \\end{align*}\nDue to the Quadratic Optimization or the AM-GM Inequality, the least value is obtained when $a = b$.\nTherefore, $a = b = 8 \\Righ... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/25.json | AHSME |
1969_AHSME_Problems | 33 | 0 | Algebra | Multiple Choice | Let $S_n$ and $T_n$ be the respective sums of the first $n$ terms of two arithmetic series. If $S_n:T_n=(7n+1):(4n+27)$ for all $n$, the ratio of the eleventh term of the first series to the eleventh term of the second series is:
$\text{(A) } 4:3\quad \text{(B) } 3:2\quad \text{(C) } 7:4\quad \text{(D) } 78:71\quad \... | [
"Let $S$ be the first arithmetic sequence and $T$ be the second arithmetic sequence. If $n = 1$, then $S_1:T_1 = 8:31$. Since $S_1$ and $T_1$ are just the first term, the first term of $S$ is $8a$ and the first term of $T$ is $31a$ for some $a$. If $n = 2$, then $S_2:T_2 = 15:35 = 3:7$, so the sum of the first t... | 3 | ./CreativeMath/AHSME/1969_AHSME_Problems/33.json | AHSME |
1969_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | A circle with radius $r$ is contained within the region bounded by a circle with radius $R$. The area bounded by the larger circle is $\frac{a}{b}$ times the area of the region outside the smaller circle and inside the larger circle. Then $R:r$ equals:
$\text{(A) }\sqrt{a}:\sqrt{b} \quad \text{(B) } \sqrt{a}:\sqrt{a-... | [
"The area of the larger circle is $\\pi R^2$, and the area of the region outside the smaller circle and inside the larger circle $\\pi R^2 - \\pi r^2$. Thus,\n\\[\\pi R^2 = \\frac{a}{b} \\cdot (\\pi R^2 - \\pi r^2)\\]\n\\[R^2 = \\frac{a}{b} \\cdot R^2 - \\frac{a}{b} \\cdot r^2\\]\n\\[\\frac{a}{b} \\cdot r^2 = (\\f... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/13.json | AHSME |
1969_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | If $x=t^{1/(t-1)}$ and $y=t^{t/(t-1)},t>0,t \ne 1$, a relation between $x$ and $y$ is:
$\text{(A) } y^x=x^{1/y}\quad \text{(B) } y^{1/x}=x^{y}\quad \text{(C) } y^x=x^y\quad \text{(D) } x^x=y^y\quad \text{(E) none of these}$
| [
"Plug in $t = 2$ and test each expression (if the relation works, then both sides are equal). After testing, options A, B, and D are out, but for option C, both sides are equal. To check that C is a valid option, substitute the values of $x$ and $y$ and use exponent properties.\n\\[x^y = (t^{\\frac{1}{t-1}})^{t^{... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/29.json | AHSME |
1969_AHSME_Problems | 3 | 0 | Number Theory | Multiple Choice | If $N$, written in base $2$, is $11000$, the integer immediately preceding $N$, written in base $2$, is:
$\text{(A) } 10001\quad \text{(B) } 10010\quad \text{(C) } 10011\quad \text{(D) } 10110\quad \text{(E) } 10111$
| [
"$11000_2-1_2=10111_2$. $\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/3.json | AHSME |
1969_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | The remainder $R$ obtained by dividing $x^{100}$ by $x^2-3x+2$ is a polynomial of degree less than $2$. Then $R$ may be written as:
$\text{(A) }2^{100}-1 \quad \text{(B) } 2^{100}(x-1)-(x-2)\quad \text{(C) } 2^{200}(x-3)\quad\\ \text{(D) } x(2^{100}-1)+2(2^{99}-1)\quad \text{(E) } 2^{100}(x+1)-(x+2)$
| [
"Let the polynomial $Q(x)$ be the quotient when $x^{100}$ is divided by $x^2-3x+2$, and let the remainder $R=ax+b$, for some real $a$ and $b$. Then we can write: $x^{100}=(x^2-3x+2)Q(x)+ax+b$. Since it is hard to deal with $Q(x)$ (it is of degree 98!), we factor $x^2-3x+2$ as $(x-2)(x-1)$ so we can eliminate $Q(x)$... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/34.json | AHSME |
1969_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | Triangle $ABC$ is inscribed in a circle. The measure of the non-overlapping minor arcs $AB$, $BC$ and $CA$ are, respectively, $x+75^{\circ} , 2x+25^{\circ},3x-22^{\circ}$. Then one interior angle of the triangle is:
$\text{(A) } 57\tfrac{1}{2}^{\circ}\quad \text{(B) } 59^{\circ}\quad \text{(C) } 60^{\circ}\quad \text... | [
"[asy] draw(circle((0,0),65)); draw((25,60)--(39,-52)--(-52,-39)--(25,60)); dot((25,60)); dot((39,-52)); dot((-52,-39)); dot((0,0)); draw((0,0)--(-52,-39)); draw((0,0)--(39,-52)); draw((0,0)--(25,60)); label(\"A\",(-52,-39),SW); label(\"B\",(25,60),NE); label(\"C\",(39,-52),SE); [/asy]\nBecause the triangle is insc... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/8.json | AHSME |
1969_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | Let $K$ be the measure of the area bounded by the $x$-axis, the line $x=8$, and the curve defined by
\[f={(x,y)\quad |\quad y=x \text{ when } 0 \le x \le 5, y=2x-5 \text{ when } 5 \le x \le 8}.\]
Then $K$ is:
$\text{(A) } 21.5\quad \text{(B) } 36.4\quad \text{(C) } 36.5\quad \text{(D) } 44\quad \text{(E) less... | [
"[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=9.2,ymin=-5.2,ymax=13.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=1,gy=1... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/22.json | AHSME |
1969_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | The number of points common to the graphs of
$(x-y+2)(3x+y-4)=0 \text{ and } (x+y-2)(2x-5y+7)=0$
is:
$\text{(A) } 2\quad \text{(B) } 4\quad \text{(C) } 6\quad \text{(D) } 16\quad \text{(E) } \infty$
| [
"By the Zero Product Property, $x-y+2=0$ or $3x+y-4=0$ in the first equation and $x+y-2=0$ or $2x-5y+7=0$ in the second equation. Thus, from the first equation, $y = x+2$ or $y =-3x+4$, and from the second equation, $y=-x+2$ or $y = \\frac{2}{5}x + \\frac{7}{5}$.\n\n\nIf a point is common to the two graphs, then t... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/18.json | AHSME |
1969_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | Let a binary operation $\star$ on ordered pairs of integers be defined by $(a,b)\star (c,d)=(a-c,b+d)$. Then, if $(3,3)\star (0,0)$ and $(x,y)\star (3,2)$ represent identical pairs, $x$ equals:
$\text{(A) } -3\quad \text{(B) } 0\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 6$
| [
"Performing the operation based on the definition, $(3,3)\\star(0,0) = (3,3)$ and $(x,y)\\star(3,2)=(x-3,y+2)$. Because the outputs are identical pairs, they must equal each other, so $3 = x-3$. Solving for x yields $x = 6$, which is answer choice $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/4.json | AHSME |
1969_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | The complete set of $x$-values satisfying the inequality $\frac{x^2-4}{x^2-1}>0$ is the set of all $x$ such that:
$\text{(A) } x>2 \text{ or } x<-2 \text{ or} -1<x<1\quad \text{(B) } x>2 \text{ or } x<-2\quad \\ \text{(C) } x>1 \text{ or } x<-2\qquad\qquad\qquad\quad \text{(D) } x>1 \text{ or } x<-1\quad \\ \text{(E... | [
"Factor the difference of squares.\n\\[\\frac{(x+2)(x-2)}{(x+1)(x-1)}>0\\]\nNote that the graph intersects the x-axis at when $x = \\pm2$ or $x \\pm 1$, so check the sign of the result to see if it is positive. After testing, $x<-2$ or $-1<x<1$ or $x>2$, so the answer is $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/14.json | AHSME |
1969_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | In a circle with center $O$ and radius $r$, chord $AB$ is drawn with length equal to $r$ (units). From $O$, a perpendicular to $AB$ meets $AB$ at $M$. From $M$ a perpendicular to $OA$ meets $OA$ at $D$. In terms of $r$ the area of triangle $MDA$, in appropriate square units, is:
$\text{(A) } \frac{3r^2}{16}\quad \tex... | [
"[asy] pair O = (0,0); pair A = (-8.660,5); pair B = (-8.660,-5); pair M = (-8.660,0); pair D = (-8.660*0.75,5*0.75); draw(circle(O,10)); dot(O); label(\"$O$\",O,E); dot(A); label(\"$A$\",A,NW); dot(B); label(\"$B$\",B,SW); dot(M); label(\"$M$\",M,SE); dot(D); label(\"$D$\",D,NE); draw(A--B--O--A); draw(O--M); draw... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/15.json | AHSME |
1969_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | If a number $N,N \ne 0$, diminished by four times its reciprocal, equals a given real constant $R$, then, for this given $R$, the sum of all such possible values of $N$ is
$\text{(A) } \frac{1}{R}\quad \text{(B) } R\quad \text{(C) } 4\quad \text{(D) } \frac{1}{4}\quad \text{(E) } -R$
| [
"Write an equation from the given information.\n\\[N - \\frac{4}{N} = R\\]\n\\[N^2 - 4 = RN\\]\n\\[N^2 - RN - 4 = 0\\]\nBy Vieta's Formulas, the sum of all possible values of $N$ for a given $R$ is $R$, so the answer is $\\boxed{\\textbf{(B)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/5.json | AHSME |
1969_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | The number of distinct ordered pairs $(x,y)$ where $x$ and $y$ have positive integral values satisfying the equation $x^4y^4-10x^2y^2+9=0$ is:
$\text{(A) } 0\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 12\quad \text{(E) } \infty$
| [
"Let $(xy)^2=a$. The expression given is equal to $a^2-10a+9=0$, which can be factored as $(a-9)(a-1)=0$. Thus, we have $a=(xy)^2=9$ and $a=(xy)^2=1$. Because $x$ and $y$ are positive, we can eliminate the possibilities where $xy$ is negative. From here it is easy to see that the only integral pairs of $x$ and $y$ ... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/19.json | AHSME |
1969_AHSME_Problems | 23 | 0 | Number Theory | Multiple Choice | For any integer $n>1$, the number of prime numbers greater than $n!+1$ and less than $n!+n$ is:
$\text{(A) } 0\quad\qquad \text{(B) } 1\quad\\ \text{(C) } \frac{n}{2} \text{ for n even, } \frac{n+1}{2} \text{ for n odd}\quad\\ \text{(D) } n-1\quad \text{(E) } n$
| [
"Observe that for all $k \\in 1< k< n$, since $k$ divides $n!$, $k$ also divides $n!+k$. Therefore, all numbers $a$ in the range $n!+1<a<n!+n$ are composite. Therefore there are 0 primes in that range. $\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/23.json | AHSME |
1969_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | The arithmetic mean (ordinary average) of the fifty-two successive positive integers beginning at 2 is:
$\text{(A) } 27\quad \text{(B) } 27\tfrac{1}{4}\quad \text{(C) } 27\tfrac{1}{2}\quad \text{(D) } 28\quad \text{(E) } 27\tfrac{1}{2}$
| [
"To solve the problem, find the sum of the first $52$ terms of an arithmetic sequence with first term $2$ and common difference $1$ and divide that by $52$. The $52^\\text{nd}$ term of the sequence is $2+51=53$, so the sum of the first $52$ terms of the sequence is $\\frac{52(2+53)}{2} = 1430$. Thus, the arithmet... | 1 | ./CreativeMath/AHSME/1969_AHSME_Problems/9.json | AHSME |
1969_AHSME_Problems | 35 | 0 | Algebra | Multiple Choice | Let $L(m)$ be the $x$ coordinate of the left end point of the intersection of the graphs of $y=x^2-6$ and $y=m$, where $-6<m<6$. Let $r=[L(-m)-L(m)]/m$. Then, as $m$ is made arbitrarily close to zero, the value of $r$ is:
$\text{(A) arbitrarily close to } 0\quad\\ \text{(B) arbitrarily close to }\frac{1}{\sqrt{6}}\qu... | [
"Since $L(a)$ is the $x$ coordinate of the left end point of the intersection of the graphs of $y=x^2-6$ and $y=a$, we can substitute $a$ for $y$ and find the lowest solution $x$.\n\\[x^2 - 6 = a\\]\n\\[x = \\pm \\sqrt{a+6}\\]\nThat means $L(m) = -\\sqrt{m+6}$ and $L(-m) = -\\sqrt{-m+6}$. That means\n\\[r = \\frac... | 2 | ./CreativeMath/AHSME/1969_AHSME_Problems/35.json | AHSME |
1996_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | In the xy-plane, what is the length of the shortest path from $(0,0)$ to $(12,16)$ that does not go inside the circle $(x-6)^{2}+(y-8)^{2}= 25$?
$\text{(A)}\ 10\sqrt 3\qquad\text{(B)}\ 10\sqrt 5\qquad\text{(C)}\ 10\sqrt 3+\frac{ 5\pi}{3}\qquad\text{(D)}\ 40\frac{\sqrt{3}}{3}\qquad\text{(E)}\ 10+5\pi$
| [
"The pathway from $A(0,0)$ to $D(12,16)$ will consist of three segments:\n\n\n1) $\\overline{AB}$, where $AB$ is tangent to the circle at point $B$.\n\n\n2) $\\overline{CD}$, where $CD$ is tangent to the circle at point $C$.\n\n\n3) $\\widehat {BC}$, where $BC$ is an arc around the circle.\n\n\nThe actual path w... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/20.json | AHSME |
1996_AHSME_Problems | 16 | 0 | Probability | Multiple Choice | A fair standard six-sided dice is tossed three times. Given that the sum of the first two tosses equal the third, what is the probability that at least one "2" is tossed?
$\text{(A)}\ \frac{1}{6}\qquad\text{(B)}\ \frac{91}{216}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{8}{15}\qquad\text{(E)}\ \frac{7}{12}$
... | [
"The third toss cannot be $1$, since the minimal sum on the other two tosses is $2$.\n\n\nIf the third roll is $2$, then the first two rolls must be $(1,1)$.\n\n\nIf the third roll is $3$, then the first two rolls must be $(1,2)$ or $(2,1)$.\n\n\nIf the third roll is $4$, then the first two rolls must be $(1,3)$, $... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/16.json | AHSME |
1996_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $f(x) = x^{(x+1)}(x+2)^{(x+3)}$, then $f(0)+f(-1)+f(-2)+f(-3) =$
$\text{(A)}\ -\frac{8}{9}\qquad\text{(B)}\ 0\qquad\text{(C)}\ \frac{8}{9}\qquad\text{(D)}\ 1\qquad\text{(E)}\ \frac{10}{9}$
| [
"Plugging in $x=0$ into the function will give $0^1\\cdot 2^3$. Since $0^1 = 0$, this gives $0$.\n\n\nPlugging in $x=-1$ into the function will give $(-1)^0 \\cdot 1^2$. Since $(-1)^0 = 1$ and $1^2 = 1$, this gives $1$.\n\n\nPlugging in $x=-2$ will give a $0^1$ factor as the second term, giving an answer of $0$.\... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/6.json | AHSME |
1996_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | A father takes his twins and a younger child out to dinner on the twins' birthday. The restaurant charges $4.95$ for the father and $0.45$ for each year of a child's age, where age is defined as the age at the most recent birthday. If the bill is $9.45$, which of the following could be the age of the youngest child?
... | [
"The bill for the three children is $9.45 - 4.95 = 4.50$. Since the charge is $0.45$ per year for the children, they must have $\\frac{4.50}{0.45} = 10$ years among the three of them.\n\n\nThe twins must have an even number of years in total (presuming that they did not dine in the 17 minutes between the time when... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/7.json | AHSME |
1996_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | In rectangle $ABCD$, angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$, where $E$ is on $\overline{AB}$, $F$ is on $\overline{AD}$, $BE=6$ and $AF=2$. Which of the following is closest to the area of the rectangle $ABCD$?
[asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--c... | [
"Since $\\angle C = 90^\\circ$, each of the three smaller angles is $30^\\circ$, and $\\triangle BEC$ and $\\triangle CDF$ are both $30-60-90$ triangles.\n\n\n[asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label(\"$A... | 2 | ./CreativeMath/AHSME/1996_AHSME_Problems/17.json | AHSME |
1996_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | Triangles $ABC$ and $ABD$ are isosceles with $AB=AC=BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC$, then $\angle C+\angle D$ is
[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); marksca... | [
"Redraw the figure as a concave pentagon $ADECB$:\n\n\n[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--E--C--B--A); markscalefactor=0.005; draw(rightanglemark(D, E, C)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); l... | 2 | ./CreativeMath/AHSME/1996_AHSME_Problems/21.json | AHSME |
1996_AHSME_Problems | 10 | 0 | Counting | Multiple Choice | How many line segments have both their endpoints located at the vertices of a given cube?
$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$
| [
"There are $8$ choices for the first endpoint of the line segment, and $7$ choices for the second endpoint, giving a total of $8\\cdot 7 = 56$ segments. However, both $\\overline{AB}$ and $\\overline{BA}$ were counted, while they really are the same line segment. Every segment got double counted in a similar mann... | 2 | ./CreativeMath/AHSME/1996_AHSME_Problems/10.json | AHSME |
1996_AHSME_Problems | 26 | 0 | Probability | Multiple Choice | An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:
(a) the selection of four red marbles;
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbl... | [
"Let the bag contain $n$ marbles total, with $r, w, b, g$ representing the number of red, white, blue, and green marbles, respectively. Note that $r + w + b + g = n$.\n\n\nThe number of ways to select four red marbles out of the set of marbles without replacement is:\n\n\n\\[\\binom{r}{4} = \\frac{r!}{24\\cdot (r ... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/26.json | AHSME |
1996_AHSME_Problems | 1 | 0 | Arithmetic | Multiple Choice | The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?
$\begin{tabular}{rr}&\ \texttt{6 4 1}\\ &\texttt{8 5 2}\\ &+\texttt{9 7 3}\\ \hline &\texttt{2 4 5 6}\end{tabular}$
$\text{(A)}\ 4\qquad\text{(B)}\ 5\qquad\text{(C)}\ 6\qquad\text{(D)}\ 7\qquad\text{(E)}\ 8... | [
"Doing the addition as is, we get $641 + 852 + 973 = 2466$. This number is $10$ larger than the desired sum of $2456$. Therefore, we must make one of the three numbers $10$ smaller.\n\n\nWe may either change $641 \\rightarrow 631$, $852 \\rightarrow 842$, or $973 \\rightarrow 963$. Either change results in a val... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/1.json | AHSME |
1996_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | Given a circle of radius $2$, there are many line segments of length $2$ that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.
$\text{(A)}\ \frac{\pi} 4\qquad\text{(B)}\ 4-\pi\qquad\text{(C)}\ \frac{\pi} 2\qquad\text{(D)}\ \pi\qquad\text{(E)}\ 2\pi$
| [
"Let line segment $AB = 2$, and let it be tangent to circle $O$ at point $P$, with radius $OP = 2$. Let $AP = PB = 1$, so that $P$ is the midpoint of $AB$.\n\n\n$\\triangle OAP$ is a right triangle with right angle at $P$, because $AB$ is tangent to circle $O$ at point $P$, and $OP$ is a radius.\n\n\nSince $AP^2 +... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/11.json | AHSME |
1996_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ ... | [
"If Walter had done his chores for $10$ days without doing any of them well, he would have earned $3 \\cdot 10 = 30$ dollars. He got $6$ dollars more than this.\n\n\nHe gets a $5 - 3 = 2$ dollar bonus every day he does his chores well. Thus, he did his chores exceptionally well $\\frac{6}{2} = 3$ days, and the an... | 3 | ./CreativeMath/AHSME/1996_AHSME_Problems/2.json | AHSME |
1996_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | On a $4\times 4\times 3$ rectangular parallelepiped, vertices $A$, $B$, and $C$ are adjacent to vertex $D$. The perpendicular distance from $D$ to the plane containing
$A$, $B$, and $C$ is closest to
[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), ... | [
"By placing the cube in a coordinate system such that $D$ is at the origin, $A(0,0,3)$, $B(4,0,0)$, and $C(0,4,0)$, we find that the equation of plane $ABC$ is:\n\n\n\\[\\frac{x}{4} + \\frac{y}{4} + \\frac{z}{3} = 1,\\] so $3x + 3y + 4z - 12 = 0.$ The equation for the distance of a point $(a,b,c)$ to a plane $Ax + ... | 2 | ./CreativeMath/AHSME/1996_AHSME_Problems/28.json | AHSME |
1996_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | A function $f$ from the integers to the integers is defined as follows:
\[f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ n/2 &\text{if n is even}\end{cases}\]
Suppose $k$ is odd and $f(f(f(k))) = 27$. What is the sum of the digits of $k$?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(... | [
"\\subsubsection{First iteration}\nTo get $f(k) = 27$, you could either have $f(27 - 3)$ and add $3$, or $f(27\\cdot 2)$ and divide by $2$.\n\n\nIf you had the former, you would have $f(24)$, and the function's rule would have you divide. Thus, $k=54$ is the only number for which $f(k) = 27$.\n\n\n\\subsubsection{... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/12.json | AHSME |
1996_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$’s separated by blocks of $2$’s with $n$ $2$’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\te... | [
"The sum of the first $1$ numbers is $1$\n\n\nThe sum of the next $2$ numbers is $2 + 1$\n\n\nThe sum of the next $3$ numbers is $2 + 2 + 1$\n\n\nIn general, we can write \"the sum of the next $n$ numbers is $1 + 2(n-1)$\", where the word \"next\" follows the pattern established above.\n\n\nThus, we first want to f... | 2 | ./CreativeMath/AHSME/1996_AHSME_Problems/24.json | AHSME |
1996_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | Given that $x^2 + y^2 = 14x + 6y + 6$, what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$
| [
"Complete the square to get \n\\[(x-7)^2 + (y-3)^2 = 64.\\]\nApplying Cauchy-Schwarz directly,\n\\[64\\cdot25=(3^2+4^2)((x-7)^2 + (y-3)^2) \\ge (3(x-7)+4(y-3))^2.\\]\n\\[40 \\ge 3x+4y-33\\]\n\\[3x+4y \\le 73.\\]\nThus our answer is $\\boxed{(B)}$.\n\n\n",
"The first equation is a circle, so we find its center and... | 8 | ./CreativeMath/AHSME/1996_AHSME_Problems/25.json | AHSME |
1996_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | Sunny runs at a steady rate, and Moonbeam runs $m$ times as fast, where $m$ is a number greater than 1. If Moonbeam gives Sunny a head start of $h$ meters, how many meters must Moonbeam run to overtake Sunny?
$\text{(A)}\ hm\qquad\text{(B)}\ \frac{h}{h+m}\qquad\text{(C)}\ \frac{h}{m-1}\qquad\text{(D)}\ \frac{hm}{m-... | [
"If Sunny runs at a rate of $s$ for $h$. Then the distance covered is $sh$. Now we know that Moonbeam runs $m$ times as fast than Sunny, so Moonbeam runs at the rate of $ms$. Now Moonbeam gave Sunny a headstart of $h$ meters, so he will catch on Sunny at the rate of $s(m-1)$ . At time $\\frac{h}{m-1}$ Moon beam w... | 3 | ./CreativeMath/AHSME/1996_AHSME_Problems/13.json | AHSME |
1996_AHSME_Problems | 29 | 0 | Number Theory | Multiple Choice | If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$
| [
"Working with the second part of the problem first, we know that $3n$ has $30$ divisors. We try to find the various possible prime factorizations of $3n$ by splitting $30$ into various products of $1, 2$ or $3$ integers.\n\n\n$30 \\rightarrow p^{29}$\n\n\n$2 \\cdot 15 \\rightarrow pq^{14}$\n\n\n$3\\cdot 10 \\right... | 3 | ./CreativeMath/AHSME/1996_AHSME_Problems/29.json | AHSME |
1996_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | $\frac{(3!)!}{3!}=$
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120$
| [
"The numerator is $(3!)! = 6!$.\n\n\nThe denominator is $3! = 6$.\n\n\nUsing the property that $6! = 6 \\cdot 5!$ in the numerator, the sixes cancel, leaving $5! = 120$, which is answer $\\boxed{E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/3.json | AHSME |
1996_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | If $3 = k\cdot 2^r$ and $15 = k\cdot 4^r$, then $r =$
$\text{(A)}\ -\log_{2}5\qquad\text{(B)}\ \log_{5}2\qquad\text{(C)}\ \log_{10}5\qquad\text{(D)}\ \log_{2}5\qquad\text{(E)}\ \frac{5}{2}$
| [
"We want to find $r$, so our strategy is to eliminate $k$.\n\n\nThe first equation gives $k = \\frac{3}{2^r}$.\n\n\nThe second equation gives $k = \\frac{15}{4^r}$\n\n\nSetting those two equal gives $\\frac{3}{2^r} = \\frac{15}{4^r}$\n\n\nCross-multiplying and dividing by $3$ gives $5\\cdot 2^r = 4^r$.\n\n\nWe know... | 2 | ./CreativeMath/AHSME/1996_AHSME_Problems/8.json | AHSME |
1996_AHSME_Problems | 22 | 0 | Counting | Multiple Choice | Four distinct points, $A$, $B$, $C$, and $D$, are to be selected from $1996$ points
evenly spaced around a circle. All quadruples are equally likely to be chosen.
What is the probability that the chord $AB$ intersects the chord $CD$?
$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{1}{3}\qquad\text{(C)}\ \frac{1}{2... | [
"Let $WXYZ$ be a convex cyclic quadrilateral inscribed in a circle. There are $\\frac{\\binom{4}{2}}{2} = 3$ ways to divide the points into two groups of two.\n\n\nIf you pick $WX$ and $YZ$, you have two sides of the quadrilateral, which do not intersect.\n\n\nIf you pick $XY$ and $ZW$, you have the other two side... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/22.json | AHSME |
1996_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | A circle of radius $2$ has center at $(2,0)$. A circle of radius $1$ has center at $(5,0)$. A line is tangent to the two circles at points in the first quadrant. Which of the following is closest to the $y$-intercept of the line?
$\text{(A)}\ \sqrt{2}/4\qquad\text{(B)}\ 8/3\qquad\text{(C)}\ 1+\sqrt 3\qquad\text{(D)}\... | [
"The two circles are tangent to each other at the point $(4,0)$, since it is both $2$ units from $(2,0)$ and $1$ unit from $(5,0)$.\n\n\nLabel the x-intercept of the common tangent line $A$, and label the y-intercept of the common tangent $B$. Triangle $\\triangle OAB$ is a right triangle at the origin.\n\n\nLabel... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/18.json | AHSME |
1996_AHSME_Problems | 4 | 0 | Other | Multiple Choice | Six numbers from a list of nine integers are $7,8,3,5,9$ and $5$. The largest possible value of the median of all nine numbers in this list is
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$
| [
"First, put the six numbers we have in order, since we are concerned with the median: $3, 5, 5, 7, 8, 9$.\n\n\nWe have three more numbers to insert into the list, and the median will be the $5^{th}$ highest (and $5^{th}$ lowest) number on the list. If we top-load the list by making all three of the numbers greate... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/4.json | AHSME |
1996_AHSME_Problems | 14 | 0 | Arithmetic | Multiple Choice | Let $E(n)$ denote the sum of the even digits of $n$. For example, $E(5681) = 6+8 = 14$. Find $E(1)+E(2)+E(3)+\cdots+E(100)$
$\text{(A)}\ 200\qquad\text{(B)}\ 360\qquad\text{(C)}\ 400\qquad\text{(D)}\ 900\qquad\text{(E)}\ 2250$
| [
"The problem is asking for the sum of all the even digits in the numbers $1$ to $100$. We can remove $100$ from the list, add $00$ to the list, and tack on some leading zeros to the single digit numbers without changing the sum of the even digits. This gives the list:\n\n\n$00, 01, 02, 03, ..., 10, 11, ..., 98, 9... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/14.json | AHSME |
1996_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | Two opposite sides of a rectangle are each divided into $n$ congruent segments, and the endpoints of one segment are joined to the center to form triangle $A$. The other sides are each divided into $m$ congruent segments, and the endpoints of one of these segments are joined to the center to form triangle $B$. [See fig... | [
"Place the rectangle on a coordinate grid, with diagonal vertices $(0,0)$ and $(x,y)$. Each horizontal segment of the rectangle will have length $\\frac{x}{m}$, while each vertical segment of the rectangle will have length $\\frac{y}{n}$.\n\n\nThe center of this rectangle will be $(\\frac{x}{2}, \\frac{y}{2})$.\n\... | 2 | ./CreativeMath/AHSME/1996_AHSME_Problems/15.json | AHSME |
1996_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | Given that $0 < a < b < c < d$, which of the following is the largest?
$\text{(A)}\ \frac{a+b}{c+d} \qquad\text{(B)}\ \frac{a+d}{b+c} \qquad\text{(C)}\ \frac{b+c}{a+d} \qquad\text{(D)}\ \frac{b+d}{a+c} \qquad\text{(E)}\ \frac{c+d}{a+b}$
| [
"Assuming that one of the above fractions is indeed always the largest, try plugging in $a=1, b=2, c=3, d=4$, since those are valid values for the variables given the constraints of the problem. The options become:\n\n\n$\\text{(A)}\\ \\frac{1+2}{3+4} \\qquad\\text{(B)}\\ \\frac{1+4}{2+3} \\qquad\\text{(C)}\\ \\... | 2 | ./CreativeMath/AHSME/1996_AHSME_Problems/5.json | AHSME |
1996_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon?
[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(24... | [
"[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); pair O=(0,0); dot(A^^B^^C^^D^^E^^F^^O); label(\"$A$\", A, dir(origin--A)); label(\"$B$\", B, dir(ori... | 2 | ./CreativeMath/AHSME/1996_AHSME_Problems/19.json | AHSME |
1996_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | The sum of the lengths of the twelve edges of a rectangular box is $140$, and
the distance from one corner of the box to the farthest corner is $21$. The total
surface area of the box is
$\text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812$
| [
"Let $x, y$, and $z$ be the unique lengths of the edges of the box. Each box has $4$ edges of each length, so:\n\\[4x + 4y + 4z = 140 \\ \\Longrightarrow \\ x + y + z = 35.\\]\nThe spacial diagonal (longest distance) is given by $\\sqrt{x^2 + y^2 + z^2}$. Thus, we have $\\sqrt{x^2 + y^2 + z^2} = 21$, so $x^2 + y^... | 1 | ./CreativeMath/AHSME/1996_AHSME_Problems/23.json | AHSME |
1996_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA = 3, PB = 4$ and $AB = 5$, what is $PD$?
[asy] real r=sqrt(2)/2; draw(origin--(8,0)--(8,-1)--(0,-1)--cycle); draw(origin--(8,0)--(8+r, r)--(r,r)--cycle); filldraw(origin--(-6*r, -6*r)--(8-6*r, -6*r)--(8, 0)--cycle, white, black); filldraw(ori... | [
"Since the two planes are perpendicular, it follows that $\\triangle PAD$ is a right triangle. Thus, $PD = \\sqrt{PA^2 + AD^2} = \\sqrt{PA^2 + AB^2} = \\sqrt{34}$, which is option $\\boxed{\\text{B}}$.\n\n",
"Place the points on a coordinate grid, and let the $xy$ plane (where $z=0$) contain triangle $PAB$. Squa... | 2 | ./CreativeMath/AHSME/1996_AHSME_Problems/9.json | AHSME |
1983_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | If $\tan{\alpha}$ and $\tan{\beta}$ are the roots of $x^2 - px + q = 0$, and $\cot{\alpha}$ and $\cot{\beta}$
are the roots of $x^2 - rx + s = 0$, then $rs$ is necessarily
$\textbf{(A)} \ pq \qquad \textbf{(B)} \ \frac{1}{pq} \qquad \textbf{(C)} \ \frac{p}{q^2} \qquad \textbf{(D)}\ \frac{q}{p^2}\qquad \textbf{(E)... | [
"By Vieta's Formulae, we have $\\tan(\\alpha)\\tan(\\beta)=q$ and $\\cot(\\alpha)\\cot(\\beta)=s$. Recalling that $\\cot\\theta=\\frac{1}{\\tan\\theta}$, we have $\\frac{1}{\\tan(\\alpha)\\tan(\\beta)}=\\frac{1}{q}=s$. \n\n\nAlso by Vieta's Formulae, we have $\\tan(\\alpha)+\\tan(\\beta)=p$ and $\\cot(\\alpha)+\\co... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/20.json | AHSME |
1983_AHSME_Problems | 16 | 0 | Arithmetic | Multiple Choice | Let $x = .123456789101112....998999$, where the digits are obtained by writing the integers $1$ through $999$ in order.
The $1983$\textsuperscript{rd} digit to the right of the decimal point is
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$
| [
"We consider the first $1983$ digits, letting the $1983$\\textsuperscript{rd} digit be $z$. We can break the string of digits into three segments: let $A$ denote $123456789$ (the $1$-digit numbers), let $B$ denote $1011...9899$ (the $2$-digit numbers), and let $C$ denote $100101...z$ (the $3$-digit numbers). Clearl... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/16.json | AHSME |
1983_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | When $x^5, x+\frac{1}{x}$ and $1+\frac{2}{x} + \frac{3}{x^2}$ are multiplied, the product is a polynomial of degree.
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$
| [
"We have $x^5\\left(x+\\frac{1}{x}\\right)\\left(1+\\frac{2}{x}+\\frac{3}{x^2}\\right) = (x^6+x^4)\\left(1+\\frac{2}{x}+\\frac{3}{x^2}\\right) = x^6 + \\text{lower order terms}$, where we know that the $x^6$ will not get cancelled out by e.g. a $-x^6$ term since all the terms inside the brackets are positive. Thus ... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/6.json | AHSME |
1983_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | Alice sells an item at $$10$ less than the list price and receives $10\%$ of her selling price as her commission.
Bob sells the same item at $$20$ less than the list price and receives $20\%$ of his selling price as his commission.
If they both get the same commission, then the list price is
$\textbf{(A) } $20\qqua... | [
"If $x$ is the list price, then $10\\%(x-10)=20\\%(x-20)$. Solving this equation gives $x=30$, so the answer is $\\boxed{\\textbf{(B) }$30}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/7.json | AHSME |
1983_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | Find the smallest positive number from the numbers below.
$\textbf{(A)} \ 10-3\sqrt{11} \qquad \textbf{(B)} \ 3\sqrt{11}-10 \qquad \textbf{(C)}\ 18-5\sqrt{13}\qquad \textbf{(D)}\ 51-10\sqrt{26}\qquad \textbf{(E)}\ 10\sqrt{26}-51$
| [
"Notice that $3\\sqrt{11} - 10 = \\sqrt{99} - \\sqrt{100} < 0$, $18 - 5\\sqrt{13} = \\sqrt{324} - \\sqrt{325} < 0$, and similarly, $10\\sqrt{26} - 51 = \\sqrt{2600} - \\sqrt{2601} < 0$, so these numbers can be excluded immediately as they are negative and we seek the smallest \\textbf{positive} number. The remainin... | 2 | ./CreativeMath/AHSME/1983_AHSME_Problems/21.json | AHSME |
1983_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | Segment $AB$ is both a diameter of a circle of radius $1$ and a side of an equilateral triangle $ABC$.
The circle also intersects $AC$ and $BC$ at points $D$ and $E$, respectively. The length of $AE$ is
$\textbf{(A)} \ \frac{3}{2} \qquad \textbf{(B)} \ \frac{5}{3} \qquad \textbf{(C)} \ \frac{\sqrt 3}{2} \qquad \... | [
"Note that since $AB$ is a diameter, $\\angle AEB = 90^{\\circ}$, which means $AB$ is an altitude of equilateral triangle $ABC$. It follows that $\\triangle ABE$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, and so $AE = AB \\cdot \\frac{\\sqrt{3}}{2} = (2 \\cdot 1) (\\frac{\\sqrt{3}}{2}) = \\boxed{\\textbf{... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/10.json | AHSME |
1983_AHSME_Problems | 26 | 0 | Probability | Multiple Choice | The probability that event $A$ occurs is $\frac{3}{4}$; the probability that event B occurs is $\frac{2}{3}$.
Let $p$ be the probability that both $A$ and $B$ occur. The smallest interval necessarily containing $p$ is the interval
$\textbf{(A)}\ \Big[\frac{1}{12},\frac{1}{2}\Big]\qquad \textbf{(B)}\ \Big[\frac{5}{12... | [
"Firstly note that $p \\leq \\frac{3}{4}$ and $p \\leq \\frac{2}{3}$, as clearly the probability that both $A$ and $B$ occur cannot be more than the probability that $A$ or $B$ alone occurs. The more restrictive condition is $p \\leq \\frac{2}{3}$, since $\\frac{2}{3} < \\frac{3}{4}$. \n\n\nFurthermore, by the Incl... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/26.json | AHSME |
1983_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | Distinct points $A$ and $B$ are on a semicircle with diameter $MN$ and center $C$.
The point $P$ is on $CN$ and $\angle CAP = \angle CBP = 10^{\circ}$. If $\stackrel{\frown}{MA} = 40^{\circ}$, then $\stackrel{\frown}{BN}$ equals
[asy] import geometry; import graph; unitsize(2 cm); pair A, B, C, M, N, P; M = (-1,0... | [
"Since $\\angle CAP = \\angle CBP = 10^\\circ$, quadrilateral $ABPC$ is cyclic (as shown below) by the converse of the theorem \"angles inscribed in the same arc are equal\".\n\n\n[asy] import geometry; import graph; unitsize(2 cm); pair A, B, C, M, N, P; M = (-1,0); N = (1,0); C = (0,0); A = dir(140); B = dir(2... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/30.json | AHSME |
1983_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | A large sphere is on a horizontal field on a sunny day. At a certain time the shadow of the sphere reaches out a distance
of $10$ m from the point where the sphere touches the ground. At the same instant a meter stick
(held vertically with one end on the ground) casts a shadow of length $2$ m. What is the radius of t... | [
"Consider the angle that the shadow makes with the ground. Since the sun's rays are parallel, it's the same as the angle made with the shadow of the stick. We know that the angle of the stick is $\\arctan \\left(\\frac{1}{2}\\right )$, since the stick is $1$ m high and its shadow is $2$ m long, so by considering a ... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/27.json | AHSME |
1983_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If $x \neq 0, \frac x{2} = y^2$ and $\frac{x}{4} = 4y$, then $x$ equals
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 32\qquad \textbf{(D)}\ 64\qquad \textbf{(E)}\ 128$
| [
"From $\\frac{x}{4} = 4y$, we get $x=16y$. Plugging in the other equation, $\\frac{16y}{2} = y^2$, so $y^2-8y=0$. Factoring, we get $y(y-8)=0$, so the solutions are $0$ and $8$. Since $x \\neq 0$, we also have $y \\neq 0$, so $y=8$. Hence $x=16\\cdot 8 = \\boxed{\\textbf{(E)}\\ 128}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/1.json | AHSME |
1983_AHSME_Problems | 11 | 0 | Other | Multiple Choice | Simplify $\sin (x-y) \cos y + \cos (x-y) \sin y$.
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ \sin x\qquad \textbf{(C)}\ \cos x\qquad \textbf{(D)}\ \sin x \cos 2y\qquad \textbf{(E)}\ \cos x\cos 2y$
| [
"By the addition formula for $\\sin$, this becomes $\\sin{((x-y)+y)} = \\sin{x}$, so the answer is $\\boxed{\\textbf{(B)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/11.json | AHSME |
1983_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | Point $P$ is outside circle $C$ on the plane. At most how many points on $C$ are $3$ cm from $P$?
$\textbf{(A)} \ 1 \qquad \textbf{(B)} \ 2 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 4 \qquad \textbf{(E)} \ 8$
| [
"The points $3$ cm away from $P$ can be represented as a circle centered at $P$ with radius $3$ cm. The maximum number of intersection points of two circles is $\\boxed{\\textbf{(B)} \\ 2}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/2.json | AHSME |
1983_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | Triangle $\triangle ABC$ in the figure has area $10$. Points $D, E$ and $F$, all distinct from $A, B$ and $C$,
are on sides $AB, BC$ and $CA$ respectively, and $AD = 2, DB = 3$. If triangle $\triangle ABE$ and quadrilateral $DBEF$
have equal areas, then that area is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pa... | [
"Let $G$ be the intersection point of $AE$ and $DF$. Since $[DBEF] = [ABE]$, we have $[DBEG] + [EFG] = [DBEG] + [ADG]$, i.e. $[EFG] = [ADG]$. It therefore follows that $[ADG] + [AGF] = [EFG] + [AGF]$, so $[ADF] = [AFE]$. Now, taking $AF$ as the base of both $\\triangle ADF$ and $\\triangle AFE$, and using the fact ... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/28.json | AHSME |
1983_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | If $\log_7 \Big(\log_3 (\log_2 x) \Big) = 0$, then $x^{-1/2}$ equals
$\textbf{(A)} \ \frac{1}{3} \qquad \textbf{(B)} \ \frac{1}{2 \sqrt 3} \qquad \textbf{(C)}\ \frac{1}{3\sqrt 3}\qquad \textbf{(D)}\ \frac{1}{\sqrt{42}}\qquad \textbf{(E)}\ \text{none of these}$
| [
"Because $\\log_7 \\Big(\\log_3 (\\log_2 x) \\Big) = 0$, we deduce $\\log_3 (\\log_2 x) =1$, and thus $\\log_2 x=3$. Therefore, $x=8$, which means $x^{-1/2}=\\frac{1}{2\\sqrt{2}}$. Since this does not match any of the answer choices, the answer is $\\fbox{{\\bf(E)} \\text{none of these}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/12.json | AHSME |
1983_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | How many non-congruent right triangles are there such that the perimeter in $\text{cm}$ and the area in $\text{cm}^2$ are numerically equal?
$\textbf{(A)} \ \text{none} \qquad \textbf{(B)} \ 1 \qquad \textbf{(C)} \ 2 \qquad \textbf{(D)} \ 4 \qquad \textbf{(E)} \ \text{infinitely many}$
| [
"Let the triangle have legs of length $a$ and $b$, so by the Pythagorean Theorem, the hypotenuse has length $\\sqrt{a^2+b^2}$. Therefore we require \\begin{align*} &a + b + \\sqrt{a^2+b^2} = \\frac{1}{2} ab \\\\ \\Rightarrow \\quad &2 \\sqrt{a^2+b^2} = ab - 2a - 2b \\\\ \\Rightarrow \\quad &4a^2 + 4b^2 = a^{2}b^{2}... | 2 | ./CreativeMath/AHSME/1983_AHSME_Problems/24.json | AHSME |
1983_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | If $60^a=3$ and $60^b=5$, then $12^{(1-a-b)/\left(2\left(1-b\right)\right)}$ is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad$
| [
"We have that $12=\\frac{60}{5}$. We can substitute our value for 5, to get\n\\[12=\\frac{60}{60^b}=60\\cdot 60^{-b}=60^{1-b}.\\]\nHence\n\\[12^{(1-a-b)/\\left(2\\left(1-b\\right)\\right)}=60^{(1-b)(1-a-b)/\\left(2\\left(1-b\\right)\\right)}=60^{(1-a-b)/2}.\\]\nSince $4=\\frac{60}{3\\cdot 5}$, we have\n\\[4=\\frac{... | 2 | ./CreativeMath/AHSME/1983_AHSME_Problems/25.json | AHSME |
1983_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | If $xy = a, xz =b,$ and $yz = c$, and none of these quantities is $0$, then $x^2+y^2+z^2$ equals
$\textbf{(A)}\ \frac{ab+ac+bc}{abc}\qquad \textbf{(B)}\ \frac{a^2+b^2+c^2}{abc}\qquad \textbf{(C)}\ \frac{(a+b+c)^2}{abc}\qquad \textbf{(D)}\ \frac{(ab+ac+bc)^2}{abc}\qquad \textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}$
... | [
"From the equations, we deduce $x = \\frac{a}{y}, z = \\frac{b}{x},$ and $y = \\frac{c}{z}$. Substituting these into the expression $x^2+y^2+z^2$ thus gives $\\frac{a^2}{y^2} + \\frac{b^2}{x^2} + \\frac{c^2}{z^2} = \\frac{a^2x^2z^2+b^2y^2z^2+c^2y^2x^2}{x^2y^2z^2} = \\frac{a^2b^2+b^2c^2+c^2a^2}{x^2y^2z^2}$, so the a... | 2 | ./CreativeMath/AHSME/1983_AHSME_Problems/13.json | AHSME |
1983_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | A point $P$ lies in the same plane as a given square of side $1$. Let the vertices of the square,
taken counterclockwise, be $A, B, C$ and $D$. Also, let the distances from $P$ to $A, B$ and $C$, respectively, be $u, v$ and $w$.
What is the greatest distance that $P$ can be from $D$ if $u^2 + v^2 = w^2$?
$\textbf{(... | [
"Place the square in the $xy$-plane with $A$ as the origin, so that $B=(1,0), C=(1,1),$ and $D=(0,1).$ We are given that $PA^2+PB^2=PC^2,$ so \n\n\n\\begin{align*}&(x^2+y^2)+((x-1)^2+y^2)=(x-1)^2+(y-1)^2\\\\ \\Rightarrow \\quad &2x^2+2y^2-2x+1=x^2+y^2-2x-2y+2\\\\ \\Rightarrow \\quad &x^2+y^2=-2y+1\\\\ \\Rightarrow ... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/29.json | AHSME |
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