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 |
|---|---|---|---|---|---|---|---|---|---|
1975_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | If the first term of an infinite geometric series is a positive integer, the common ratio is the reciprocal of a positive integer, and the sum of the series is $3$, then the sum of the first two terms of the series is
$\textbf{(A)}\ \frac{1}{3} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{8}{3} \qquad... | [
"nothing yet :(\n\n\n"
] | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/16.json | AHSME |
1975_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80$
| [
"Solution by e_power_pi_times_i\n\n\n\n\nWhen the $n$th odd positive integer is subtracted from the $n$th even positive integer, the result is $1$. Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is $80\\cdot1 = \\boxed{\\textbf{(E) } 80... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/6.json | AHSME |
1975_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | For which non-zero real numbers $x$ is $\frac{|x-|x|\-|}{x}$ a positive integer?
$\textbf{(A)}\ \text{for negative } x \text{ only} \qquad \\ \textbf{(B)}\ \text{for positive } x \text{ only} \qquad \\ \textbf{(C)}\ \text{only for } x \text{ an even integer} \qquad \\ \textbf{(D)}\ \text{for all non-zero real number... | [
"Solution by e_power_pi_times_i\n\n\n\n\nNotice that if $x$ is negative, then the whole thing would amount to a negative number. Also notice that if $x$ is positive, then $|x-|x|\\-|$ would be $0$, hence the whole thing would amount to $0$. Therefore, $\\frac{|x-|x|\\-|}{x}$ is positive $\\boxed{\\textbf{(E)}\\ \\t... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/7.json | AHSME |
1975_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | A man can commute either by train or by bus. If he goes to work on the train in the morning, he comes home on the bus in the afternoon; and if he comes home in the afternoon on the train, he took the bus in the morning. During a total of $x$ working days, the man took the bus to work in the morning $8$ times, came home... | [
"The man has three possible combinations of transportation:\n\\[\\text{Morning train, Afternoon bus (m.t., a.b.)}\\]\n\\[\\text{Morning bus, Afternoon train (m.b., a.t.)}\\]\n\\[\\text{Morning bus, Afternoon bus (m.b, a.b.)}\\].\n\n\n\n\nLet $y$ be the number of times the man takes the $\\text{a.t.}$. Then, $9-y$ i... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/17.json | AHSME |
1975_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | Suppose $f(x)$ is defined for all real numbers $x; f(x) > 0$ for all $x;$ and $f(a)f(b) = f(a + b)$ for all $a$ and $b$. Which of the following statements are true?
$I.\ f(0) = 1 \qquad \qquad \ \ \qquad \qquad \qquad II.\ f(-a) = \frac{1}{f(a)}\ \text{for all}\ a \\ III.\ f(a) = \sqrt[3]{f(3a)}\ \text{for all}\ a \... | [
"$I: f(0) = 1$\n\n\nLet $b = 0$. Our equation becomes $f(a)f(0) = f(a)$, so $f(0) = 1$. Therefore $I$ is always true.\n\n\n\n\n$II: f(-a) = \\frac{1}{f(a)} \\text{ for all } a$\n\n\nLet $b = -a$. Our equation becomes $f(a)f(-a) = f(0) = 1 \\longrightarrow f(-a) = \\frac{1}{f(a)}$. Therefore $II$ is always true.\n\n... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/21.json | AHSME |
1975_AHSME_Problems | 10 | 0 | Number Theory | Multiple Choice | The sum of the digits in base ten of $(10^{4n^2+8}+1)^2$, where $n$ is a positive integer, is
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4n \qquad \textbf{(C)}\ 2+2n \qquad \textbf{(D)}\ 4n^2 \qquad \textbf{(E)}\ n^2+n+2$
| [
"We see that the result of this expression will always be in the form $(100\\text{ some number of zeros }001)^2.$ Multiplying these together yields: \\[110\\text{ some number of zeros }011.\\] This works because of the way they are multiplied. Therefore, the answer is $\\boxed{(A) 4}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/10.json | AHSME |
1975_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | In acute $\triangle ABC$ the bisector of $\measuredangle A$ meets side $BC$ at $D$. The circle with center $B$ and radius $BD$
intersects side $AB$ at $M$; and the circle with center $C$ and radius $CD$ intersects side $AC$ at $N$. Then it is always true that
$\textbf{(A)}\ \measuredangle CND+\measuredangle BMD-\mea... | [
"Error\n\n\n"
] | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/26.json | AHSME |
1975_AHSME_Problems | 30 | 0 | Other | Multiple Choice | Let $x=\cos 36^{\circ} - \cos 72^{\circ}$. Then $x$ equals
$\textbf{(A)}\ \frac{1}{3}\qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 3-\sqrt{6} \qquad \textbf{(D)}\ 2\sqrt{3}-3\qquad \textbf{(E)}\ \text{none of these}$
| [
"Using the difference to product identity, we find that\n$x=\\cos 36^{\\circ} - \\cos 72^{\\circ}$ is equivalent to \\[x=\\text{-}2\\sin{\\frac{(36^{\\circ}+72^{\\circ})}{2}}\\sin{\\frac{(36^{\\circ}-72^{\\circ})}{2}} \\implies\\]\n\\[x=\\text{-}2\\sin54^{\\circ}\\sin(\\text{-}18^{\\circ}).\\] \nSince sine is an od... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/30.json | AHSME |
1975_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$
| [
"If $p$ is a root of $x^3 - x^2 + x - 2 = 0$, then $p^3 - p^2 + p - 2 = 0$, or\n\\[p^3 = p^2 - p + 2.\\]\nSimilarly, $q^3 = q^2 - q + 2$, and $r^3 = r^2 - r + 2$, so\n\\[p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.\\]\n\n\nBy Vieta's formulas, $p + q + r = 1$, $pq + pr + qr = 1$, and $pqr = 2$. Squaring t... | 3 | ./CreativeMath/AHSME/1975_AHSME_Problems/27.json | AHSME |
1975_AHSME_Problems | 1 | 0 | Arithmetic | Multiple Choice | The value of $\frac {1}{2 - \frac {1}{2 - \frac {1}{2 - \frac12}}}$ is
$\textbf{(A)}\ 3/4 \qquad \textbf{(B)}\ 4/5 \qquad \textbf{(C)}\ 5/6 \qquad \textbf{(D)}\ 6/7 \qquad \textbf{(E)}\ 6/5$
| [
"Solution by e_power_pi_times_i\n\n\n\n\nCalculating, we find that $\\frac {1}{2 - \\frac {1}{2 - \\frac {1}{2 - \\frac12}}} = \\frac {1}{2 - \\frac {1}{2 - \\frac {2}{3}}} = \\frac {1}{2 - \\frac {3}{4}} = \\frac {1}{\\frac {5}{4}} = \\boxed{\\textbf{(B) } \\dfrac{4}{5}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/1.json | AHSME |
1975_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | Let $P$ be an interior point of circle $K$ other than the center of $K$. Form all chords of $K$ which pass through $P$, and determine their midpoints. The locus of these midpoints is
$\textbf{(A)} \text{ a circle with one point deleted} \qquad \\ \textbf{(B)} \text{ a circle if the distance from } P \text{ to the ce... | [
"nothing yet :(\n\n\n"
] | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/11.json | AHSME |
1975_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | For which real values of m are the simultaneous equations
\begin{align*}y &= mx + 3 \\ y& = (2m - 1)x + 4\end{align*}
satisfied by at least one pair of real numbers $(x,y)$?
$\textbf{(A)}\ \text{all }m\qquad \textbf{(B)}\ \text{all }m\neq 0\qquad \textbf{(C)}\ \text{all }m\neq 1/2\qquad \textbf{(D)}\ \text{al... | [
"Solution by e_power_pi_times_i\n\n\n\n\nSolving the systems of equations, we find that $mx+3 = (2m-1)x+4$, which simplifies to $(m-1)x+1 = 0$. Therefore $x = \\dfrac{1}{1-m}$. \n$x$ is only a real number if $\\boxed{\\textbf{(D) }m\\neq 1}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/2.json | AHSME |
1975_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | In $\triangle ABC$ shown in the adjoining figure, $M$ is the midpoint of side $BC, AB=12$ and $AC=16$. Points $E$ and $F$ are taken on $AC$
and $AB$, respectively, and lines $EF$ and $AM$ intersect at $G$. If $AE=2AF$ then $\frac{EG}{GF}$ equals
[asy] draw((0,0)--(12,0)--(14,7.75)--(0,0)); draw((0,0)--(13,3.875)); d... | [
"Here, we use Mass Points.\nLet $AF = x$. We then have $AE = 2x$, $EC = 16-2x$, and $FB = 12 - x$\nLet $B$ have a mass of $2$. Since $M$ is the midpoint, $C$ also has a mass of $2$. \nLooking at segment $AB$, we have\n\\[2 \\cdot (12-x) = \\text{m}A_{AB} \\cdot x\\]\nSo\n\\[\\text{m}A_{AB} = \\frac{24-2x}{x}\\]\nLo... | 3 | ./CreativeMath/AHSME/1975_AHSME_Problems/28.json | AHSME |
1975_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | If $a \neq b, a^3 - b^3 = 19x^3$, and $a-b = x$, which of the following conclusions is correct?
$\textbf{(A)}\ a=3x \qquad \textbf{(B)}\ a=3x \text{ or } a = -2x \qquad \textbf{(C)}\ a=-3x \text{ or } a = 2x \qquad \\ \textbf{(D)}\ a=3x \text{ or } a=2x \qquad \textbf{(E)}\ a=2x$
| [
"We can factor $a^3-b^3=19x^3$ into:\n\\[(a-b)(a^2+ab+b^2)=19x^3.\\]\nSubstituting yields:\n\\[x(a^2+ab+b^2)=19x^3\\]\n\\[a^2+ab+b^2=19x^2.\\]\nThis is equal to:\n\\[(a-b)^2+3ab=19x^2\\]\n\\[x^2+3ab=19x^2\\]\n\\[ab=6x^2.\\]\nChecking with the possible answers, along with $a-b=x$ yields the only answer to be $\\boxe... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/12.json | AHSME |
1975_AHSME_Problems | 24 | 0 | Geometry | Multiple Choice | In triangle $ABC$, $\angle C = \theta$ and $\angle B = 2\theta$, where $0^{\circ} < \theta < 60^{\circ}$. The circle with center $A$ and radius $AB$ intersects $AC$ at $D$ and intersects $BC$, extended if necessary, at $B$ and at $E$ ($E$ may coincide with $B$). Then $EC = AD$
$\textbf{(A)}\ \text{for no values of}\ ... | [
"Since $AD = AE$, we know $EC = AD$ if and only if triangle $ACE$ is isosceles and $\\angle ACE = \\angle CAE$. Letting $\\angle ACE = \\theta$, we want to find when $\\angle CAE = \\theta$. We know $\\angle ABC = \\angle AEB = 2\\theta$, so $\\angle EAB = 180-4\\theta$. We also know $\\angle CAB = 180-3\\theta$, a... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/24.json | AHSME |
1975_AHSME_Problems | 25 | 0 | Other | Multiple Choice | A woman, her brother, her son and her daughter are chess players (all relations by birth). The worst player's twin (who is one of the four players) and the best player are of opposite sex. The worst player and the best player are the same age. Who is the worst player?
$\textbf{(A)}\ \text{the woman} \qquad \textbf{(B... | [
"We know that the worst player's twin and the best player are of opposite sex, and the worst and best players are the same age.\n\n\n\\textbf{Case 1:} Suppose the daughter was the worst player. The son would have to be her twin because nobody else can be her twin. Then, that means the son and the best player are of... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/25.json | AHSME |
1975_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | The equation $x^6 - 3x^5 - 6x^3 - x + 8 = 0$ has
$\textbf{(A)} \text{ no real roots} \\ \textbf{(B)} \text{ exactly two distinct negative roots} \\ \textbf{(C)} \text{ exactly one negative root} \\ \textbf{(D)} \text{ no negative roots, but at least one positive root} \\ \textbf{(E)} \text{ none of these}$
| [
"Let $P(x) = x^6 - 3x^5 - 6x^3 - x + 8$. When $x < 0$, $P(x) > 0$. Therefore, there are no negative roots. \n\n\nNotice that $P(1) = -1$ and $P(0) = 8$. There must be at least one positive root between 0 and 1, therefore the answer is $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/13.json | AHSME |
1975_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$?
$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$
| [
"$(\\sqrt{3}+\\sqrt{2})^6=(5+2\\sqrt{6})^3=(5+2\\sqrt{6})(49+20\\sqrt{6})=(485+198\\sqrt{6})$ Then, find that $\\sqrt{6}$ is about $2.449$. Finally, multiply and add to find that the smallest integer higher is $\\boxed {\\textbf{(C) } 970}$\n\n\n",
"Let's evaluate $(\\sqrt{3}+\\sqrt{2})^6 + (\\sqrt{3}-\\sqrt{2})^... | 2 | ./CreativeMath/AHSME/1975_AHSME_Problems/29.json | AHSME |
1975_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | Which of the following inequalities are satisfied for all real numbers $a, b, c, x, y, z$ which satisfy the conditions $x < a, y < b$, and $z < c$?
$\text{I}. \ xy + yz + zx < ab + bc + ca \\ \text{II}. \ x^2 + y^2 + z^2 < a^2 + b^2 + c^2 \\ \text{III}. \ xyz < abc$
$\textbf{(A)}\ \text{None are satisfied.} \qquad... | [
"Solution by e_power_pi_times_i\n\n\n\n\nNotice if $a$, $b$, and $c$ are $0$, then we can find $x$, $y$, and $z$ to disprove $\\text{I}$ and $\\text{II}$. For example, if $(a, b, c, x, y, z) = (0, 0, 0, -1, -1, -1)$, then $\\text{I}$ and $\\text{II}$ are disproved. If $(a, b, c, x, y, z) = (0, 1, 2, -1, -1, 1)$, th... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/3.json | AHSME |
1975_AHSME_Problems | 8 | 0 | Other | Multiple Choice | If the statement "All shirts in this store are on sale." is false, then which of the following statements must be true?
I. All shirts in this store are at non-sale prices.
II. There is some shirt in this store not on sale.
III. No shirt in this store is on sale.
IV. Not all shirts in this store are on sale.... | [
"Solution by e_power_pi_times_i\n\n\n\n\nIf not all the shirts in the store are on sale, then the answer is $\\boxed{\\textbf{(D) } \\text{II and IV only}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/8.json | AHSME |
1975_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | If $p$ and $q$ are primes and $x^2-px+q=0$ has distinct positive integral roots, then which of the following statements are true?
$I.\ \text{The difference of the roots is odd.} \\ II.\ \text{At least one root is prime.} \\ III.\ p^2-q\ \text{is prime}. \\ IV.\ p+q\ \text{is prime}$
$\\ \textbf{(A)}\ I\ \text{only... | [
"Since the roots are both positive integers, we can say that $x^2-px+q=(x-1)(x-q)$ since $q$ only has $2$ divisors. Thus, the roots are $1$ and $q$ and $p=q+1$. The only two primes which differ by $1$ are $2,3$ so $p=3$ and $q=2$. \n$I$ is true because $3-2=1$.\n$II$ is true because one of the roots is $2$ which is... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/22.json | AHSME |
1975_AHSME_Problems | 18 | 0 | Probability | Multiple Choice | A positive integer $N$ with three digits in its base ten representation is chosen at random,
with each three digit number having an equal chance of being chosen. The probability that $\log_2 N$ is an integer is
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 3/899 \qquad \textbf{(C)}\ 1/225 \qquad \textbf{(D)}\ 1/300 \qqu... | [
"In order for $\\log_2 N$ to be an integer, $N$ has to be an integer power of $2$. Since $N$ is a three-digit number in decimal form, $N$ can only be $2^{7} = 128,\\ 2^{8} = 256,\\ \\text{or}\\ 2^{9} = 512$. From $100$ to $999$, there are $900$ three-digit numbers in total.\n\n\nThe probability of choosing $128$, $... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/18.json | AHSME |
1975_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \sqrt2 \qquad \textbf{(C)}\ 1/2 \qquad \textbf{(D)}\ 2\sqrt2 \qquad \textbf{(E)}\ 4$
| [
"Solution by e_power_pi_times_i\n\n\n\n\nDenote the side of one square as $s$. Then the diagonal of the second square is $s$, so the side of the second square is $\\dfrac{s\\sqrt{2}}{2}$. The area of the second square is $\\dfrac{1}{2}s^2$, so the ratio of the areas is $\\dfrac{s^2}{\\dfrac{1}{2}s^2} = \\boxed{\\te... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/4.json | AHSME |
1975_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | If the $whatsis$ is $so$ when the $whosis$ is $is$ and the $so$ and $so$ is $is \cdot so$,
what is the $whosis \cdot whatsis$ when the $whosis$ is $so$, the $so$ and $so$ is $so \cdot so$ and the $is$ is two
($whatsis, whosis, is$ and $so$ are variables taking positive values)?
$\textbf{(A)}\ whosis \cdot is \cdot... | [
"From the problem, we are given:\n\n\n\\begin{center}\n $whatsis = so$ \\end{center}\n\\begin{center}\n $whosis = is$ \\end{center}\n\\begin{center}\n $so + so = is \\cdot so$ \\end{center}\nWe want to find what $whatsis \\cdot whosis$ is when $whosis = so$, $so + so = so \\cdot so$, and $is = 2$.\n\n\nSince $is = ... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/14.json | AHSME |
1975_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | In the sequence of numbers $1, 3, 2, \ldots$ each term after the first two is equal to the term preceding it minus the term preceding that. The sum of the first one hundred terms of the sequence is
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ -1$
| [
"First, write a few terms of the sequence:\n$1, 3, 2, -1, -3, -2, 1, 3, 2, \\ldots$ Notice how the pattern repeats every six terms and every six terms have a sum of 0. Then, find that the $16*6=96$th term is $-2$ and the sum of the all those previous terms is $0$. Then, write the 97th to the 100th terms down: $1, ... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/15.json | AHSME |
1975_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | The polynomial $(x+y)^9$ is expanded in decreasing powers of $x$. The second and third terms have equal values
when evaluated at $x=p$ and $y=q$, where $p$ and $q$ are positive numbers whose sum is one. What is the value of $p$?
$\textbf{(A)}\ 1/5 \qquad \textbf{(B)}\ 4/5 \qquad \textbf{(C)}\ 1/4 \qquad \textbf{... | [
"Solution by e_power_pi_times_i\n\n\n\n\nThe second and third term of $(x+y)^9$ is $9x^8y$ and $36x^7y^2$, respectively. For them to be equal when $x = p$, $\\dfrac{p}{4} = y$. For them to be equal when $y = q$, $x = 4q$. Then $p+\\dfrac{p}{4} = q+4q$, so $\\dfrac{5p}{4} = 5q$, which simplifies to $p = 4q$. Since $... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/5.json | AHSME |
1975_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Which positive numbers $x$ satisfy the equation $(\log_3x)(\log_x5)=\log_35$?
$\textbf{(A)}\ 3 \text{ and } 5 \text{ only} \qquad \textbf{(B)}\ 3, 5, \text{ and } 15 \text{ only} \qquad \\ \textbf{(C)}\ \text{only numbers of the form } 5^n \cdot 3^m, \text{ where } n \text{ and } m \text{ are positive integers} \qq... | [
"By the change-of-base formula, we can simplify the left side of the equation: $(\\log_3x)(\\log_x5) = (\\frac{\\log_x}{\\log_3})(\\frac{\\log_5}{\\log_x}) = \\frac{\\log_5}{\\log_3}$.\n\n\nWe see that this in fact simplifies to $\\log_35$, which will always equal the right side of the equation, since they are the ... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/19.json | AHSME |
1975_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | In the adjoining figure $AB$ and $BC$ are adjacent sides of square $ABCD$; $M$ is the midpoint of $AB$;
$N$ is the midpoint of $BC$; and $AN$ and $CM$ intersect at $O$. The ratio of the area of $AOCD$ to the area of $ABCD$ is
[asy] draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0)); label("A", (0,0), S); l... | [
"First, let's draw a few auxiliary lines. Drop altitudes from $O$ to $AB$ and from $O$ to $BC$. We can label the points as $J$ and $K$, respectively. This forms square $OKBJ$. Connect $OB$.\n\n\nWithout loss of generality, set the side length of the square $ABCD$ equal to $1$. Let $NK=x$, and since $N$ is the midpo... | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/23.json | AHSME |
1975_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | Let $a_1, a_2, \ldots$ and $b_1, b_2, \ldots$ be arithmetic progressions such that $a_1 = 25, b_1 = 75$, and $a_{100} + b_{100} = 100$.
Find the sum of the first hundred terms of the progression $a_1 + b_1, a_2 + b_2, \ldots$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 100 \qquad \textbf{(C)}\ 10,000 \qquad \textbf{(D)... | [
"Notice that $a_{100}$ and $b_{100}$ are $25+99k_1$ and $75+99k_2$, respectively. Therefore $k_2 = -k_1$. Now notice that $a_n + b_n = 25+k_1(n-1)+75+k_2(n-1) = 100+k_1(n-1)-k_1(n-1) = 100$. The sum of the first $100$ terms is $100\\cdot100 = \\boxed{\\textbf{(C) } 10,000}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/9.json | AHSME |
1960_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | The coefficient of $x^7$ in the expansion of $\left(\frac{x^2}{2}-\frac{2}{x}\right)^8$ is:
$\textbf{(A)}\ 56\qquad \textbf{(B)}\ -56\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ -14\qquad \textbf{(E)}\ 0$
| [
"By the Binomial Theorem, each term of the expansion is $\\binom{8}{n}\\left(\\frac{x^2}{2}\\right)^{8-n}\\left(\\frac{-2}{x}\\right)^n$.\n\n\nWe want the exponent of $x$ to be $7$, so \n\\[2(8-n)-n=7\\]\n\\[16-3n=7\\]\n\\[n=3\\]\n\n\nIf $n=3$, then the corresponding term is\n\\[\\binom{8}{3}\\left(\\frac{x^2}{2}\\... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/20.json | AHSME |
1960_AHSME_Problems | 36 | 0 | Algebra | Multiple Choice | Let $s_1, s_2, s_3$ be the respective sums of $n, 2n, 3n$ terms of the same arithmetic progression with $a$ as the first term and $d$
as the common difference. Let $R=s_3-s_2-s_1$. Then $R$ is dependent on:
$\textbf{(A)}\ a\text{ }\text{and}\text{ }d\qquad \textbf{(B)}\ d\text{ }\text{and}\text{ }n\qquad \textbf{(C)... | [
"The nth term of the sequence with first term $a$ is $a + d(n-1)$. That means the sum of the first $n$ terms is with first term $a$ is $\\frac{n(2a + dn -d)}{2}$.\n\n\nThe 2nth term of the sequence with first term $a$ is $a + d(2n-1)$. That means the sum of the first $2n$ terms is with first term $a$ is $\\frac{2... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/36.json | AHSME |
1960_AHSME_Problems | 16 | 0 | Number Theory | Multiple Choice | In the numeration system with base $5$, counting is as follows: $1, 2, 3, 4, 10, 11, 12, 13, 14, 20,\ldots$.
The number whose description in the decimal system is $69$, when described in the base $5$ system, is a number with:
$\textbf{(A)}\ \text{two consecutive digits} \qquad\textbf{(B)}\ \text{two non-consecutive d... | [
"Since $25<69<125$, divide $69$ by $25$. The quotient is $2$ and the remainder is $19$, so rewrite the number as\n\\[69 = 2 \\cdot 25 + 19\\]\nSimilarly, dividing $19$ by $5$ results in quotient of $3$ and remainder of $4$, so rewrite the number as\n\\[69 = 2 \\cdot 25 + 3 \\cdot 5 + 4 \\cdot 1\\]\nThus, the numbe... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/16.json | AHSME |
1960_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | The circumference of a circle is $100$ inches. The side of a square inscribed in this circle, expressed in inches, is:
$\textbf{(A) }\frac{25\sqrt{2}}{\pi}\qquad \textbf{(B) }\frac{50\sqrt{2}}{\pi}\qquad \textbf{(C) }\frac{100}{\pi}\qquad \textbf{(D) }\frac{100\sqrt{2}}{\pi}\qquad \textbf{(E) }50\sqrt{2}$
| [
"First, find the diameter of the circle. Plug in the circumference for the formula $C = \\pi d$ to solve for $d$.\n\\[100 = \\pi d\\]\n\\[d = \\frac{100}{\\pi}\\]\nSince all of an inscribed square's vertices touch the circle, and one of the angles of the circle is $90^{\\circ}$, the square's diagonal is the diamet... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/6.json | AHSME |
1960_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | Circle $I$ passes through the center of, and is tangent to, circle $II$. The area of circle $I$ is $4$ square inches.
Then the area of circle $II$, in square inches, is:
$\textbf{(A) }8\qquad \textbf{(B) }8\sqrt{2}\qquad \textbf{(C) }8\sqrt{\pi}\qquad \textbf{(D) }16\qquad \textbf{(E) }16\sqrt{2}$
| [
"[asy] draw(circle((0,0),50)); draw(circle((-25,0),25)); [/asy]\n\n",
"Since Circle $I$ is tangent to circle $II$ and touches the center of circle $II$, the diameter of circle $I$ is the radius of circle $II$.\n\n\nThat means circle $II$ is twice as big as circle $I$, so the area of circle $II$ is four times as b... | 3 | ./CreativeMath/AHSME/1960_AHSME_Problems/7.json | AHSME |
1960_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | The formula $N=8 \times 10^{8} \times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars.
The lowest income, in dollars, of the wealthiest $800$ individuals is at least:
$\textbf{(A)}\ 10^4\qquad \textbf{(B)}\ 10^6\qquad \textbf{(C)}\ 10^8\qquad \textbf{(D)}\ 10^{12} \qq... | [
"Plug $800$ for $N$ because $800$ is the number of people who has at least $x$ dollars.\n\\[800 = 8 \\times 10^{8} \\times x^{-3/2}\\]\n\\[10^{-6} = x^{-3/2}\\]\nIn order to undo raising to the $-\\frac{3}{2}$ power, raise both sides to the $-\\frac{2}{3}$ power.\n\\[(10^{-6})^{-2/3} = (x^{-3/2})^{-2/3}\\]\n\\[x = ... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/17.json | AHSME |
1960_AHSME_Problems | 40 | 0 | Geometry | Multiple Choice | Given right $\triangle ABC$ with legs $BC=3, AC=4$. Find the length of the shorter angle trisector from $C$ to the hypotenuse:
$\textbf{(A)}\ \frac{32\sqrt{3}-24}{13}\qquad\textbf{(B)}\ \frac{12\sqrt{3}-9}{13}\qquad\textbf{(C)}\ 6\sqrt{3}-8\qquad\textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad\textbf{(E)}\ \frac{25}{12}\qqu... | [
"Angle $C$ is split into three $30^{\\circ}$ angles. The shorter angle trisector will be the one closer $BC$. Let it intersect $AB$ at point $P$. Let the perpendicular from point $P$ intersect $BC$ at point $R$ and have length $x$. Thus $\\triangle PRC$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle and $RC$ h... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/40.json | AHSME |
1960_AHSME_Problems | 37 | 0 | Geometry | Multiple Choice | The base of a triangle is of length $b$, and the altitude is of length $h$.
A rectangle of height $x$ is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:
$\textbf{(A)}\ \frac{bx}{h}(h-x)\qquad \textbf{(B)}\ \frac{hx}{b}(b-x)\qquad \textbf{(C)}\ \frac... | [
"Let $AB=b$, $DE=h$, and $WX = YZ = x$.\n[asy] pair A=(0,0),B=(56,0),C=(20,48),D=(20,0),W=(10,0),X=(10,24),Y=(38,24),Z=(38,0); draw(A--B--C--A); draw((10,0)--(10,24)--(38,24)--(38,0)); draw(C--D); dot(A); dot(B); dot(C); dot(D); dot(W); dot(X); dot(Y); dot(Z); dot((20,24)); label(\"$A$\",A,S); label(\"$B$\",B,S); l... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/37.json | AHSME |
1960_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | The diagonal of square $I$ is $a+b$. The area of square $II$ with twice the area of $I$ is:
$\textbf{(A)}\ (a+b)^2\qquad \textbf{(B)}\ \sqrt{2}(a+b)^2\qquad \textbf{(C)}\ 2(a+b)\qquad \textbf{(D)}\ \sqrt{8}(a+b) \qquad \textbf{(E)}\ 4(a+b)$
| [
"Since the diagonal of square $I$ is $a+b$ units long, the side length of square $I$ is $\\frac{a+b}{\\sqrt{2}}$, so the area of square $I$ is $\\frac{(a+b)^2}{2}$.\n\n\nThe area of square $II$ is twice as much as the area of square $I$, so the area of square $II$ is $(a+b)^2$. That means the side length of square... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/21.json | AHSME |
1960_AHSME_Problems | 10 | 0 | Other | Multiple Choice | Given the following six statements:
\[\text{(1) All women are good drivers}\]
\[\text{(2) Some women are good drivers}\]
\[\text{(3) No men are good drivers}\]
\[\text{(4) All men are bad drivers}\]
\[\text{(5) At least one man is a bad driver}\]
\[\text{(6) All men are good drivers.}\]
The statement that negates s... | [
"$\\fbox{(C)3}$\nBecause No men are good negates All men are good.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/10.json | AHSME |
1960_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | Find the set of $x$-values satisfying the inequality $|\frac{5-x}{3}|<2$. [The symbol $|a|$ means $+a$ if $a$ is positive,
$-a$ if $a$ is negative,$0$ if $a$ is zero. The notation $1<a<2$ means that a can have any value between $1$ and $2$, excluding $1$ and $2$. ]
$\textbf{(A)}\ 1 < x < 11\qquad \textbf{(B)}\ -1 < ... | [
"Break up the absolute value into two cases.\n\n\nFor the first case, let $x < 5$, so $\\frac{5-x}{3}$ is positive. That means (for $x<5$)\n\\[\\frac{5-x}{3} < 2\\]\n\\[5-x<6\\]\n\\[-x<1\\]\n\\[x>-1\\]\nFor the second case, let $x \\ge 5$, so $\\frac{5-x}{3}$ is negative. That means (for $x \\ge 5$)\n\\[\\frac{x-... | 2 | ./CreativeMath/AHSME/1960_AHSME_Problems/26.json | AHSME |
1960_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | Given the line $3x+5y=15$ and a point on this line equidistant from the coordinate axes. Such a point exists in:
$\textbf{(A)}\ \text{none of the quadrants}\qquad\textbf{(B)}\ \text{quadrant I only}\qquad\textbf{(C)}\ \text{quadrants I, II only}\qquad$
$\textbf{(D)}\ \text{quadrants I, II, III only}\qquad\textbf{(E)}... | [
"If a point is equidistant from the coordinate axes, then the absolute values of the x-coordinate and y-coordinate are equal. Since the point is on the line $3x+5y=15$, find the intersection point of $y=x$ and $3x+5y=15$ and the intersection point of $y=-x$ and $3x+5y=15$.\n\n\nSubstituting $x$ for $y$ in $3x+5y=1... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/30.json | AHSME |
1960_AHSME_Problems | 31 | 0 | Algebra | Multiple Choice | For $x^2+2x+5$ to be a factor of $x^4+px^2+q$, the values of $p$ and $q$ must be, respectively:
$\textbf{(A)}\ -2, 5\qquad \textbf{(B)}\ 5, 25\qquad \textbf{(C)}\ 10, 20\qquad \textbf{(D)}\ 6, 25\qquad \textbf{(E)}\ 14, 25$
| [
"Let the other quadratic be $x^2 + bx + c$, where $(x^2 + 2x + 5)(x^2 + bx + c) = x^4 + px^2 + q$. Multiply the two quadratics to get\n\\[x^4 + (b+2)x^3 + (c + 2b + 5)x^2 + (2c + 5b)x + 5c\\]\nSince $x^4 + px^2 + q$ have no $x^3$ term and no $x$ term, the coefficients of these terms must be zero.\n\\[b+2=0\\]\n\\[... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/31.json | AHSME |
1960_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | Let $S$ be the sum of the interior angles of a polygon $P$ for which each interior angle is $7\frac{1}{2}$ times the
exterior angle at the same vertex. Then
$\textbf{(A)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may be regular}\qquad \\ \textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P... | [
"Let $a_n$ be the interior angle of the nth vertex, and let $b_n$ be the exterior angle of the nth vertex. From the conditions in the problem,\n\\[a_n = 7.5b_n\\]\nThat means\n\\[a_1 + a_2 \\cdots a_n = 7.5(b_1 + b_2 \\cdots b_n)\\]\nSince the sum of the exterior angles of a polygon is $360^{\\circ}$, the equation... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/27.json | AHSME |
1960_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If $2$ is a solution (root) of $x^3+hx+10=0$, then $h$ equals:
$\textbf{(A) }10\qquad \textbf{(B) }9 \qquad \textbf{(C) }2\qquad \textbf{(D) }-2\qquad \textbf{(E) }-9$
| [
"Substitute $2$ for $x$. We are given that this equation is true. Solving for $h$ gives $h=-9$. The answer is $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/1.json | AHSME |
1960_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | For a given value of $k$ the product of the roots of $x^2-3kx+2k^2-1=0$
is $7$. The roots may be characterized as:
$\textbf{(A) }\text{integral and positive} \qquad\textbf{(B) }\text{integral and negative} \qquad \\ \textbf{(C) }\text{rational, but not integral} \qquad\textbf{(D) }\text{irrational} \qquad\textbf{(E) ... | [
"If the product of the roots are $7$, then by Vieta's formulas, \n\\[2k^2-1=7\\]\nSolve for $k$ in the resulting equation to get\n\\[2k^2=8\\]\n\\[k^2=4\\]\n\\[k=\\pm 2\\]\nThat means the two quadratics are $x^2-6x+7=0$ and $x^2+6x+7=0$. Since $b^2$, $a$, and $c$ are the same, the discriminant of both is\n$36-(4 \... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/11.json | AHSME |
1960_AHSME_Problems | 2 | 0 | Arithmetic | Multiple Choice | It takes $5$ seconds for a clock to strike $6$ o'clock beginning at $6:00$ o'clock precisely. If the strikings are uniformly spaced, how long, in seconds, does it take to strike $12$ o'clock?
$\textbf{(A)}9\frac{1}{5}\qquad \textbf{(B )}10\qquad \textbf{(C )}11\qquad \textbf{(D )}14\frac{2}{5}\qquad \textbf{(E )}\tex... | [
"Between six strikes, there are five intervals of space. Since it takes five seconds total, each interval is one second long. Therefore, $12$ strikes will take $11$ seconds. $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/2.json | AHSME |
1960_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | The equation $x-\frac{7}{x-3}=3-\frac{7}{x-3}$ has:
$\textbf{(A)}\ \text{infinitely many integral roots}\qquad\textbf{(B)}\ \text{no root}\qquad\textbf{(C)}\ \text{one integral root}\qquad$
$\textbf{(D)}\ \text{two equal integral roots}\qquad\textbf{(E)}\ \text{two equal non-integral roots}$
| [
"Both terms have a $-\\frac{7}{x-3}$ term, so add $\\frac{7}{x-3}$ to both sides. This results in $x = 3$.\n\n\nHowever, note that if $3$ is plugged back into the original equation, it results in\n\\[3-\\frac{7}{0}=3-\\frac{7}{0}\\]\n\n\nSince dividing by zero is undefined, $3$ is an extraneous solution. That mea... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/28.json | AHSME |
1960_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | The locus of the centers of all circles of given radius $a$, in the same plane, passing through a fixed point, is:
$\textbf{(A) }\text{a point}\qquad \textbf{(B) }\text{a straight line}\qquad \textbf{(C) }\text{two straight lines}\qquad \textbf{(D) }\text{a circle}\qquad \textbf{(E) }\text{two circles}$
| [
"[asy] draw(circle((0,0),50)); dot((0,0)); dot((-30,-40)); draw(circle((-30,-40),50),dotted); dot((50,0)); draw(circle((50,0),50),dotted); draw((-30,-40)--(0,0)--(50,0)); [/asy]\n\n\nIf a circle passes through a point, then the point is $a$ units away from the center. That means that all of the centers are $a$ uni... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/12.json | AHSME |
1960_AHSME_Problems | 32 | 0 | Geometry | Multiple Choice | In this figure the center of the circle is $O$. $AB \perp BC$, $ADOE$ is a straight line, $AP = AD$, and $AB$ has a length twice the radius. Then:
[asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); real e=350,c=55; pair O=origin,E=dir(e),C=dir(c),B=dir(180+c),D=dir(180+e), rot=rotate(90,B)*O,A=extension(E,D,B,... | [
"We claim that $\\fbox{A}$ is the right answer.\n\n\nLet the radius of circle $O$ be $r$, and let the length of $AD=x$. Since $AB \\perp BC$, $AB$ is a tangent to circle $O$. Thus, by the tangent-secant theorem, we have $AB^2=AD\\times AE$, or, $(2r)^2=x(x+2r)$. Through some algebraic manipulation, we find \\begin{... | 2 | ./CreativeMath/AHSME/1960_AHSME_Problems/32.json | AHSME |
1960_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | If $\log_{2x}216 = x$, where $x$ is real, then $x$ is:
$\textbf{(A)}\ \text{A non-square, non-cube integer}\qquad$
$\textbf{(B)}\ \text{A non-square, non-cube, non-integral rational number}\qquad$
$\textbf{(C)}\ \text{An irrational number}\qquad$
$\textbf{(D)}\ \text{A perfect square}\qquad$
$\textbf{(E)}\ \text{A pe... | [
"Rewrite the equation to $(2x)^x = 216$. Since $2x$ is the base of the logarithm, we only need to check x-values that are positive.\n\n\nWith trial and error, $3$ is a solution because $(2 \\cdot 3)^3 = 6^3 = 216$. Since $(2x)^x$ gets larger as x gets larger, $3$ is the only solution, so the answer is $\\boxed{\\... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/24.json | AHSME |
1960_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$.
The largest integer which divides all possible numbers of the form $m^2-n^2$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$
| [
"First, factor the difference of squares.\n\\[(m+n)(m-n)\\]\nSince $m$ and $n$ are odd numbers, let $m=2a+1$ and $n=2b+1$, where $a$ and $b$ can be any integer.\n\\[(2a+2b+2)(2a-2b)\\]\nFactor the resulting expression.\n\\[4(a+b+1)(a-b)\\]\nIf $a$ and $b$ are both even, then $a-b$ is even. If $a$ and $b$ are both ... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/25.json | AHSME |
1960_AHSME_Problems | 33 | 0 | Number Theory | Multiple Choice | You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2 \times 3 \times 5 \times\ldots \times 61$
of all prime numbers less than or equal to $61$, and $n$ takes, successively, the values $2, 3, 4,\ldots, 59$.
Let $N$ be the number of primes appearing in this sequence. T... | [
"First, note that $n$ does not have a prime number larger than $61$ as one of its factors. Also, note that $n$ does not equal $1$.\n\n\nTherefore, since the prime factorization of $n$ only has primes from $2$ to $59$, $n$ and $P$ share at least one common factor other than $1$. Therefore $P+n$ is not prime for an... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/33.json | AHSME |
1960_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | The polygon(s) formed by $y=3x+2, y=-3x+2$, and $y=-2$, is (are):
$\textbf{(A) }\text{An equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad \\ \textbf{(D) }\text{a triangle and a trapezoid}\qquad\textbf{(E) }\text{a quadrilateral}$
| [
"[asy]import graph; size(10.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.2,xmax=4.2,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/1960_AHSME_Problems/13.json | AHSME |
1960_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | Five times $A$'s money added to $B$'s money is more than $$51.00$. Three times $A$'s money minus $B$'s money is $$21.00$.
If $a$ represents $A$'s money in dollars and $b$ represents $B$'s money in dollars, then:
$\textbf{(A)}\ a>9, b>6 \qquad \textbf{(B)}\ a>9, b<6 \qquad \textbf{(C)}\ a>9, b=6\qquad \textbf{(D)}\ a>... | [
"Use math symbols to write an equation and an inequality based on the conditions in the problem.\n\\[5a+b>51\\]\n\\[3a-b=21\\]\nFrom the second equation, $b = 3a-21$ and $a = 7 + b/3$.\n\n\nSubstituting $b$ in the inequality results in\n\\[5a + 3a - 21 > 51\\]\n\\[8a > 72\\]\n\\[a > 9\\]\n\n\nSubstituting $a$ in th... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/29.json | AHSME |
1960_AHSME_Problems | 3 | 0 | Arithmetic | Multiple Choice | Applied to a bill for $\textdollar{10,000}$ the difference between a discount of $40$% and two successive discounts of $36$% and $4$%,
expressed in dollars, is:
$\textbf{(A)}0\qquad \textbf{(B)}144\qquad \textbf{(C)}256\qquad \textbf{(D)}400\qquad \textbf{(E)}416$
| [
"Taking the discount of $40$% means you're only paying $60$% of the bill. That results in $10,000\\cdot0.6=\\textdollar{6,000}$.\n\n\nLikewise, taking two discounts of $36$% and $4$% means taking $64$% of the original amount and then $96$% of the result. That results in $10,000\\cdot0.64\\cdot0.96=\\textdollar{6,... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/3.json | AHSME |
1960_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | Two swimmers, at opposite ends of a $90$-foot pool, start to swim the length of the pool,
one at the rate of $3$ feet per second, the other at $2$ feet per second.
They swim back and forth for $12$ minutes. Allowing no loss of times at the turns, find the number of times they pass each other.
$\textbf{(A)}\ 24\qqu... | [
"First, note that it will take $30$ seconds for the first swimmer to reach the other side and $45$ seconds for the second swimmer to reach the other side. Also, note that after $180$ seconds (or $3$ minutes), both swimmers will complete an even number of laps, essentially returning to their starting point.\n\n\n[a... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/34.json | AHSME |
1960_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | The number $2.5252525\ldots$ can be written as a fraction.
When reduced to lowest terms the sum of the numerator and denominator of this fraction is:
$\textbf{(A) }7\qquad \textbf{(B) }29\qquad \textbf{(C) }141\qquad \textbf{(D) }349\qquad \textbf{(E) }\text{none of these}$
| [
"Let $x = 2.5252525\\ldots$ , so\n\\[100x = 252.52525\\ldots\\]\n\\[x = 2.5252525\\ldots\\]\n\\[99x = 250\\]\n\\[x = \\frac{250}{99}\\]\n\n\nThe sum of the numerator and the denominator is $349$, so the answer is $\\boxed{\\textbf{(D)}}$ .\n\n",
"The number $2.5252525\\ldots$ can also be written as\n\\[2+\\frac{5... | 2 | ./CreativeMath/AHSME/1960_AHSME_Problems/8.json | AHSME |
1960_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | The equality $(x+m)^2-(x+n)^2=(m-n)^2$, where $m$ and $n$ are unequal non-zero constants, is satisfied by $x=am+bn$, where:
$\textbf{(A)}\ a = 0, b \text{ } \text{has a unique non-zero value}\qquad \\ \textbf{(B)}\ a = 0, b \text{ } \text{has two non-zero values}\qquad \\ \textbf{(C)}\ b = 0, a \text{ } \text{has a u... | [
"Expand binomials, combine like terms, and subtract terms from both sides.\n\\[x^2 + 2xm + m^2 - x^2 - 2xn - n^2 = m^2 - 2mn + n^2\\]\n\\[2xm + m^2 - 2xn - n^2 = m^2 - 2mn + n^2\\]\n\\[2xm - 2xn - n^2 = -2mn + n^2\\]\nGet all the x-terms on one side and factor to solve for x.\n\\[2xm - 2xn = -2mn + 2n^2\\]\n\\[2x(m... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/22.json | AHSME |
1960_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | The pair of equations $3^{x+y}=81$ and $81^{x-y}=3$ has:
$\textbf{(A)}\ \text{no common solution} \qquad \\ \textbf{(B)}\ \text{the solution} \text{ } x=2, y=2\qquad \\ \textbf{(C)}\ \text{the solution} \text{ } x=2\frac{1}{2}, y=1\frac{1}{2} \qquad \\ \textbf{(D)}\text{ a common solution in positive and negative int... | [
"Rewrite the equations so both sides have a common base.\n\\[3^{x+y}=3^4\\]\n\\[3^{4(x-y)}=3^1\\]\nTaking the logarithm base 3, we get a linear system of equations.\n\\[x+y=4\\]\n\\[4x-4y=1\\]\nSolve the system to get $x=\\frac{17}{8}$ and $y=\\frac{15}{8}$. The answer is $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/18.json | AHSME |
1960_AHSME_Problems | 38 | 0 | Geometry | Multiple Choice | In this diagram $AB$ and $AC$ are the equal sides of an isosceles $\triangle ABC$, in which is inscribed equilateral $\triangle DEF$.
Designate $\angle BFD$ by $a$, $\angle ADE$ by $b$, and $\angle FEC$ by $c$. Then:
[asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); pair A=(5,12),B=origin,C=(10,0),D=(5/3,4),... | [
"Since $\\triangle DEF$ is an equilateral triangle, all of the angles are $60^{\\circ}$.\nThe angles in a line add up to $180^{\\circ}$, so\n\\[\\angle FDB = 120 - b\\]\n\\[\\angle EFC = 120 - a\\]\nThe angles in a triangle add up to $180^{\\circ}$, so\n\\[\\angle ABC = 60 + b - a\\]\n\\[\\angle ACB = 60 - c + a\\]... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/38.json | AHSME |
1960_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | Each of two angles of a triangle is $60^{\circ}$ and the included side is $4$ inches. The area of the triangle, in square inches, is:
$\textbf{(A)} 8\sqrt{3}\qquad \textbf{(B)} 8\qquad \textbf{(C)} 4\sqrt{3}\qquad \textbf{(D)} 4\qquad \textbf{(E)} 2\sqrt{3}$
| [
"If two of the angles are $60^{\\circ}$, then the other angle is $60^{\\circ}$ because angles in triangle add up to $180^{\\circ}$. That makes the triangle an equilateral triangle, so all sides are $4$ inches long.\n\n\n[asy] draw((0,0)--(50,0)--(25,43.301)--cycle); label(\"$4$\",(10,25)); label(\"$2$\",(12.5,-5))... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/4.json | AHSME |
1960_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | If $a$ and $b$ are real numbers, the equation $3x-5+a=bx+1$ has a unique solution $x$ [The symbol $a \neq 0$ means that $a$ is different from zero]:
$\text{(A) for all a and b} \qquad \text{(B) if a }\neq\text{2b}\qquad \text{(C) if a }\neq 6\qquad \\ \text{(D) if b }\neq 0\qquad \text{(E) if b }\neq 3$
| [
"If the coefficients of the x-terms are equal on both sides, then when the x-terms are subtracted from both sides, the equation results in a number equals 1.\n\n\nThis means the equation has either infinite or no solutions, so the x-terms can not be equal on both sides. Thus, $b \\neq 3$, so the answer is $\\boxed... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/14.json | AHSME |
1960_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | Triangle $I$ is equilateral with side $A$, perimeter $P$, area $K$, and circumradius $R$ (radius of the circumscribed circle).
Triangle $II$ is equilateral with side $a$, perimeter $p$, area $k$, and circumradius $r$. If $A$ is different from $a$, then:
$\textbf{(A)}\ P:p = R:r \text{ } \text{only sometimes} \qquad... | [
"[asy] pair A=(0,50),O=(0,0),B=(43.301,-25),C=(-43.301,-25); draw(A--B--C--A); draw(circle(O,50)); draw(B--O--C); draw(anglemark(C,O,B,200)); draw((0,0)--(0,-25)); label(\"$60^{\\circ}$\",(-7,-12)); [/asy]\n\n\nFirst, find $P$, $K$, and $R$ in terms of $A$. Since all sides of an equilateral triangle are the same, ... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/15.json | AHSME |
1960_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | The number of distinct points common to the graphs of $x^2+y^2=9$ and $y^2=9$ is:
$\textbf{(A) }\text{infinitely many}\qquad \textbf{(B) }\text{four}\qquad \textbf{(C) }\text{two}\qquad \textbf{(D) }\text{one}\qquad \textbf{(E) }\text{none}$
| [
"Solve the second equation by taking the square root of both sides.\n\\[y^2=9\\]\n\\[y=\\pm3\\]\n\n\nSolve the first equation by substituting $y^2$ into the first equation then solving for $x$.\n\\[x^2+9=9\\]\n\\[x^2=0\\]\n\\[x=0\\]\n\n\nThe two solutions are $(0,3)$ and $(0,-3)$, so the answer is $\\boxed{\\textbf... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/5.json | AHSME |
1960_AHSME_Problems | 39 | 0 | Algebra | Multiple Choice | To satisfy the equation $\frac{a+b}{a}=\frac{b}{a+b}$, $a$ and $b$ must be:
$\textbf{(A)}\ \text{both rational}\qquad\textbf{(B)}\ \text{both real but not rational}\qquad\textbf{(C)}\ \text{both not real}\qquad$
$\textbf{(D)}\ \text{one real, one not real}\qquad\textbf{(E)}\ \text{one real, one not real or both not r... | [
"First, note that $a \\neq 0$ and $a \\neq -b$. Cross multiply both sides to get\n\\[a^2 + 2ab + b^2 = ab\\]\nSubtract both sides by $ab$ to get\n\\[a^2 + ab + b^2 = 0\\]\nFrom the quadratic formula,\n\\[a = \\frac{-b \\pm \\sqrt{b^2 - 4b^2}}{2}\\]\n\\[a = \\frac{-b \\pm \\sqrt{-3b^2}}{2}\\]\nIf $b$ is real, then ... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/39.json | AHSME |
1960_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Consider equation $I: x+y+z=46$ where $x, y$, and $z$ are positive integers, and equation $II: x+y+z+w=46$,
where $x, y, z$, and $w$ are positive integers. Then
$\textbf{(A)}\ \text{I can be solved in consecutive integers} \qquad \\ \textbf{(B)}\ \text{I can be solved in consecutive even integers} \qquad \\ \textbf{... | [
"Consider each option, one at a time.\n\n\nFor option A, let $x=y-1$ and $z=y+1$. That means $3y=46$, so $y=\\frac{46}{3}$. That is not an integer, so option A is eliminated.\n\n\nFor option B, let $x=y-2$ and $z=y+2$. That also means $3y=46$, so $y=\\frac{46}{3}$. That is also not an integer, so option B is el... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/19.json | AHSME |
1960_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | The radius $R$ of a cylindrical box is $8$ inches, the height $H$ is $3$ inches.
The volume $V = \pi R^2H$ is to be increased by the same fixed positive amount when $R$
is increased by $x$ inches as when $H$ is increased by $x$ inches. This condition is satisfied by:
$\textbf{(A)}\ \text{no real value of x} \qquad ... | [
"Since increasing the height by $x$ inches should result in the same volume as increasing the radius by $x$ inches, write an equation with the two cylinders (one with height increased, one with radius increased).\n\\[\\pi (8+x)^2 \\cdot 3 = \\pi \\cdot 8^2 \\cdot (3+x)\\]\n\\[3(x^2+16x+64) = 64(x+3)\\]\n\\[3x^2+48x... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/23.json | AHSME |
1960_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | The fraction $\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}$ is (with suitable restrictions of the values of a, b, and c):
$\text{(A) irreducible}\qquad$
$\text{(B) reducible to negative 1}\qquad$
$\text{(C) reducible to a polynomial of three terms}\qquad$
$\text{(D) reducible to} \frac{a-b+c}{a+b-c}\qquad$
$\text{... | [
"Use the commutative property to get\n\\[\\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\\]\nFactor perfect square trinomials to get\n\\[\\frac{(a+b)^2-c^2}{(a+c)^2-b^2}\\]\nFactor difference of squares to get\n\\[\\frac{(a+b+c)(a+b-c)}{(a+b+c)(a-b+c)}\\]\nCancel out like terms (with suitable restrictions of a, b, c) to g... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/9.json | AHSME |
1960_AHSME_Problems | 35 | 0 | Geometry | Multiple Choice | From point $P$ outside a circle, with a circumference of $10$ units, a tangent is drawn.
Also from $P$ a secant is drawn dividing the circle into unequal arcs with lengths $m$ and $n$.
It is found that $t_1$, the length of the tangent, is the mean proportional between $m$ and $n$.
If $m$ and $t$ are integers, then $... | [
"By definition of mean proportional, $t = \\sqrt{mn}$. Since $m+n=10$, $t = \\sqrt{m(10-m)}$.\n\n\nWith trial and error, note that when $m=9$, $t=3$ and when $m=2$, $t=4$. These values work since another tangent line can be drawn from $P$, and the angle between the tangent and secant can decrease to match the val... | 1 | ./CreativeMath/AHSME/1960_AHSME_Problems/35.json | AHSME |
1990_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | [asy] pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0); draw(A--B--C--D--cycle,dot); draw(A--E--F--C,dot); draw(D--E--F--B,dot); markscalefactor = 0.1; draw(rightanglemark(B, A, D)); draw(rightanglemark(D, E, C)); draw(rightanglemark(B, F, A)); draw(rightanglemark(D, C, B)); MP("A",(0,0),W);... | [
"Label the angles as shown in the diagram. Since $\\angle DEC$ forms a linear pair with $\\angle DEA$, $\\angle DEA$ is a right angle.\n\n\n[asy] pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0); draw(A--B--C--D--cycle,dot); draw(A--E--F--C,dot); draw(D--E--F--B,dot); markscalefactor = ... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/20.json | AHSME |
1990_AHSME_Problems | 16 | 0 | Counting | Multiple Choice | At one of George Washington's parties, each man shook hands with everyone except his spouse, and no handshakes took place between women. If $13$ married couples attended, how many handshakes were there among these $26$ people?
$\text{(A) } 78\quad \text{(B) } 185\quad \text{(C) } 234\quad \text{(D) } 312\quad \text{(... | [
"We split this problem into two cases: A) The number of ways that men can shake hands with other men B) The number of ways that the men can shake hands with the other women (excluding their spouse).\n\n\nA) Since there are $13$ men, the number of handshakes between only men is $\\frac{13 \\cdot 12}{2}=78$.\n\n\nB) ... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/16.json | AHSME |
1990_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | Points $A$ and $B$ are $5$ units apart. How many lines in a given plane containing $A$ and $B$ are $2$ units from $A$ and $3$ units from $B$?
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) more than }3$
| [
"The lines have to be tangent to both of these circles.\n[asy] dot((0,0));dot((5,0)); label(\"$A$\",(0,0),S);label(\"$B$\",(5,0),S); draw(Circle((0,0),2));draw(Circle((5,0),3)); real m = sqrt(6)/12; path p = (-.4-3,4*sqrt(6)/5-3*m)--(4.4+4,6*sqrt(6)/5+4*m); draw(p,dotted);draw(reflect((0,0),(1,0))*p,dotted);draw((2... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/6.json | AHSME |
1990_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | A triangle with integral sides has perimeter $8$. The area of the triangle is
$\text{(A) } 2\sqrt{2}\quad \text{(B) } \frac{16}{9}\sqrt{3}\quad \text{(C) }2\sqrt{3} \quad \text{(D) } 4\quad \text{(E) } 4\sqrt{2}$
| [
"The shortest side must be $1$ or $2$. However none of $(1,1,6),(1,2,5),(1,3,4)$ form triangles, so the shortest side must be $2$. Then $(2,2,4)$ is degenerate, so the sides must be $(2,3,3)$.\n\n\nThis can be cut in half and reassembled into a rectangle with one side $1$ and diagonal $3$. By Pythagoras its area is... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/7.json | AHSME |
1990_AHSME_Problems | 17 | 0 | Counting | Multiple Choice | How many of the numbers, $100,101,\cdots,999$ have three different digits in increasing order or in decreasing order?
$\text{(A) } 120\quad \text{(B) } 168\quad \text{(C) } 204\quad \text{(D) } 216\quad \text{(E) } 240$
| [
"For decreasing order, we just need to choose any three digits from the ten: $\\tbinom{10}3$\n\n\nFor increasing order, the number cannot start with $0$, so choose from nine: $\\tbinom93$\n\n\nThe sum is $204$, giving $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/17.json | AHSME |
1990_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | Consider a pyramid $P-ABCD$ whose base $ABCD$ is square and whose vertex $P$ is equidistant from $A,B,C$ and $D$. If $AB=1$ and $\angle{APB}=2\theta$, then the volume of the pyramid is
$\text{(A) } \frac{\sin(\theta)}{6}\quad \text{(B) } \frac{\cot(\theta)}{6}\quad \text{(C) } \frac{1}{6\sin(\theta)}\quad \text{(D) }... | [
"As the base has area $1$, the volume will be one third of the height. Drop a line from $P$ to $AB$, bisecting it at $Q$.\n[asy] import three;unitsize(1cm);size(200);real h = 0.7; //currentprojection=perspective(1/3,-1,1/2); triple P = (.5,.5,h); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--P--(1,... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/21.json | AHSME |
1990_AHSME_Problems | 10 | 0 | Counting | Multiple Choice | An $11\times 11\times 11$ wooden cube is formed by gluing together $11^3$ unit cubes. What is the greatest number of unit cubes that can be seen from a single point?
$\text{(A) 328} \quad \text{(B) 329} \quad \text{(C) 330} \quad \text{(D) 331} \quad \text{(E) 332}$
| [
"[asy]import three; unitsize(1cm);size(100); draw((0,0,0)--(0,1,0),red); draw((0,0,1)--(0,0,0)--(1,0,0)--(1,0,1),red); draw((0,0,1)--(1,0,1),red); draw((0,0,1)--(0,1,1)--(0,1,0)--(1,1,0)--(1,0,0),red); draw((1,1,0)--(1,1,1),red); draw((0,1,1)--(1,1,1)--(1,0,1),red); [/asy]\nThe best angle for cube viewing is center... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/10.json | AHSME |
1990_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to them in the circle. Then each person computes and announces the average of the numbers of their two neighbors. The figure shows the average announced by each person (\textit{not} the original number the person picked.)
[asy] u... | [
"For $i\\in\\{1,2,3,\\ldots,10\\},$ suppose Person $i$ picks the number $a_i$ and announces the number $i.$ We wish to find $a_6.$\n\n\nTaking the indices modulo $10,$ we are given that $\\frac{a_{i-1}+a_{i+1}}{2}=i,$ from which $a_{i-1}+a_{i+1}=2i.$\n\n\nWe have ten equations: five with odd-numbered indices and fi... | 2 | ./CreativeMath/AHSME/1990_AHSME_Problems/26.json | AHSME |
1990_AHSME_Problems | 30 | 0 | Number Theory | Multiple Choice | If $R_n=\tfrac{1}{2}(a^n+b^n)$ where $a=3+2\sqrt{2}$ and $b=3-2\sqrt{2}$, and $n=0,1,2,\cdots,$ then $R_{12345}$ is an integer. Its units digit is
$\text{(A) } 1\quad \text{(B) } 3\quad \text{(C) } 5\quad \text{(D) } 7\quad \text{(E) } 9$
| [
"$(a+b)R_n=\\tfrac12(a^{n+1}+b^{n+1})+ab\\cdot\\tfrac12(a^{n-1}+b^{n-1})=R_{n+1}+abR_{n-1}$\nbut $a+b=6$ and $ab=1$, so this means that $R_{n+1}=6R_n-R_{n-1}$. Since $R_0=1$ and $R_1=3$, all terms are integers and we can continue the recurrence $\\rm{mod}\\ 10$ to get the repeating sequence $1,3,7,9,7,3$. The numbe... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/30.json | AHSME |
1990_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | Which of these triples could $\underline{not}$ be the lengths of the three altitudes of a triangle?
$\text{(A) } 1,\sqrt{3},2\quad \text{(B) } 3,4,5\quad \text{(C) } 5,12,13\quad \text{(D) } 7,8,\sqrt{113}\quad \text{(E) } 8,15,17$
| [
"Let $a$, $b$, and $c$ be the side lengths of the triangle such that $a<b<c$. We are given $a+b>c$ by the triangle inequality.\n\n\nLet $h_a$, $h_b$, and $h_c$ be the altitudes to sides $a$, $b$, and $c$ respectively. We see that $h_c<h_b<h_a$. By computing the areas using $a$, $b$, and $c$ as bases we get \\[\\fra... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/27.json | AHSME |
1990_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If $\dfrac{\frac{x}{4}}{2}=\dfrac{4}{\frac{x}{2}}$, then $x=$
$\text{(A)}\ \pm\frac{1}{2}\qquad\text{(B)}\ \pm 1\qquad\text{(C)}\ \pm 2\qquad\text{(D)}\ \pm 4\qquad\text{(E)}\ \pm 8$
| [
"Cross-multiplying leaves \n\n\n\\begin{align*}\\dfrac{x^2}{8} &= 8\\\\ x^2 &= 64\\\\ \\sqrt{x^2} &= \\sqrt{64}\\\\ x &= \\pm 8\\end{align*}\n\n\nSo the answer is $\\boxed{\\text{(E)} \\, \\pm 8}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/1.json | AHSME |
1990_AHSME_Problems | 11 | 0 | Number Theory | Multiple Choice | How many positive integers less than $50$ have an odd number of positive integer divisors?
$\text{(A) } 3\quad \text{(B) } 5\quad \text{(C) } 7\quad \text{(D) } 9\quad \text{(E) } 11$
| [
"Divisors come in pairs, unless there is an integer square root, so we just need the perfect squares below $50$. There are $7$, so $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/11.json | AHSME |
1990_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | $\left(\frac{1}{4}\right)^{-\tfrac{1}{4}}=$
$\text{(A) } -16\quad \text{(B) } -\sqrt{2}\quad \text{(C) } -\frac{1}{16}\quad \text{(D) } \frac{1}{256}\quad \text{(E) } \sqrt{2}$
| [
"$\\left(\\frac14\\right)^{-\\tfrac14}=4^{\\tfrac14}=\\sqrt[4]4=\\sqrt{\\sqrt4}=\\sqrt2$ which is $\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/2.json | AHSME |
1990_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | A quadrilateral that has consecutive sides of lengths $70,90,130$ and $110$ is inscribed in a circle and also has a circle inscribed in it. The point of tangency of the inscribed circle to the side of length 130 divides that side into segments of length $x$ and $y$. Find $|x-y|$.
$\text{(A) } 12\quad \text{(B) } 13\q... | [
"Let $A$, $B$, $C$, and $D$ be the vertices of this quadrilateral such that $AB=70$, $BC=110$, $CD=130$, and $DA=90$. Let $O$ be the center of the incircle. Draw in the radii from the center of the incircle to the points of tangency. Let these points of tangency $X$, $Y$, $Z$, and $W$ be on $AB$, $BC$, $CD$, and $D... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/28.json | AHSME |
1990_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | Let $f$ be the function defined by $f(x)=ax^2-\sqrt{2}$ for some positive $a$. If $f(f(\sqrt{2}))=-\sqrt{2}$ then $a=$
$\text{(A) } \frac{2-\sqrt{2}}{2}\quad \text{(B) } \frac{1}{2}\quad \text{(C) } 2-\sqrt{2}\quad \text{(D) } \frac{\sqrt{2}}{2}\quad \text{(E) } \frac{2+\sqrt{2}}{2}$
| [
"If $f(w)=-\\sqrt2$, then $aw^2=0\\implies w=0$. Therefore $f(\\sqrt2)=0\\implies 2a=\\sqrt2$, so $\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/12.json | AHSME |
1990_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | All students at Adams High School and at Baker High School take a certain exam. The average scores for boys, for girls, and for boys and girls combined, at Adams HS and Baker HS are shown in the table, as is the average for boys at the two schools combined. What is the average score for the girls at the two schools com... | [
"Let the numbers of boys and girls at Adams be $(A,a)$ and the numbers of boys and girls at Baker be $(B,b)$.\n\n\nThen, reading down, we have $71A+76a=74(A+a)$ and $81B+90b=84(B+b)$.\nReading across, we have $71A+81B=79(A+B)$.\n\n\nSimplifying each in turn, $3A-2a=0 \\implies A=2a/3$ and $3B-6b=0 \\implies B=2b$ a... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/24.json | AHSME |
1990_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | Nine congruent spheres are packed inside a unit cube in such a way that one of them has its center at the center of the cube and each of the others is tangent to the center sphere and to three faces of the cube. What is the radius of each sphere?
$\text{(A) } 1-\frac{\sqrt{3}}{2}\quad \text{(B) } \frac{2\sqrt{3}-3}{2... | [
"Let $r$ be the radius, let $C$ be the center of the cube, and let $P$ be the center of one of the eight outer spheres, noting that $PC=2r$.\n\n\nBack, in the corner of the unit cube, a smaller cube whose inner corner coincides with $P$. The cube is of dimensions $1\\times 1\\times 1$ and its space diagonal is of l... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/25.json | AHSME |
1990_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | If the following instructions are carried out by a computer, which value of $X$ will be printed because of instruction $5$?
\begin{verbatim}
1. START $X$ AT $3$ AND $S$ AT $0$.
2. INCREASE THE VALUE OF $X$ BY $2$.
3. INCREASE THE VALUE OF $S$ BY THE VALUE OF $X$.
4. IF $S$ IS AT LEAST $10000$,
THEN GO... | [
"Looking at the first few values, it becomes clear that the program stops when \\[5+7+9+11+13+\\ldots+(x-2)+x\\ge 10000\\] which is to say \\[1+3+5+7+9+11+13+\\ldots+(x-2)+x\\ge 10004\\]\nHowever, the left hand side is now simply the square $\\frac{(x+1)^2}4$. Multiplying out, we get \\[x+1\\ge \\sqrt{40016}\\appro... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/13.json | AHSME |
1990_AHSME_Problems | 29 | 0 | Counting | Multiple Choice | A subset of the integers $1,2,\cdots,100$ has the property that none of its members is 3 times another. What is the largest number of members such a subset can have?
$\text{(A) } 50\quad \text{(B) } 66\quad \text{(C) } 67\quad \text{(D) } 76\quad \text{(E) } 78$
| [
"Notice that inclusion of the integers from $34$ to $100$ is allowed as long as no integer between $11$ and $33$ inclusive is within the set. This provides a total of $100 - 34 + 1$ = 67 solutions. \n\n\nFurther analyzation of the remaining integers between $1$ and $10$, we notice that we can include all the number... | 2 | ./CreativeMath/AHSME/1990_AHSME_Problems/29.json | AHSME |
1990_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | The consecutive angles of a trapezoid form an arithmetic sequence. If the smallest angle is $75^\circ$, then the largest angle is
$\text{(A) } 95^\circ\quad \text{(B) } 100^\circ\quad \text{(C) } 105^\circ\quad \text{(D) } 110^\circ\quad \text{(E) } 115^\circ$
| [
"A trapezoid is a quadrilateral; therefore the interior angles sum to $360^\\circ$.\nThus $75+(75+x)+(75+2x)+(75+3x)=360$, so $x=10$ and the largest angle is $105^\\circ$ which is $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/3.json | AHSME |
1990_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | The number of real solutions of the equation
\[|x-2|+|x-3|=1\]
is
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) more than } 3$
| [
"For $2\\le x\\le 3$, the left-hand side is $(x-2)-(x-3)$ which is $1$ for all $x$ in the interval. $\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/8.json | AHSME |
1990_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | If the six solutions of $x^6=-64$ are written in the form $a+bi$, where $a$ and $b$ are real, then the product of those solutions with $a>0$ is
$\text{(A) } -2\quad \text{(B) } 0\quad \text{(C) } 2i\quad \text{(D) } 4\quad \text{(E) } 16$
| [
"This equation is $r^6e^{6\\theta i}=2^6e^{(\\pi\\pm 2k\\pi) i}$. Solving in the usual way, $r=2$ and $\\theta\\in\\{\\pm 30^\\circ,\\pm 90^\\circ,\\pm 150^\\circ\\}$.\n\n\nThus there are only two solutions with positive real part, and they are conjugates, so their product is $r^2=4$ which is $\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/22.json | AHSME |
1990_AHSME_Problems | 18 | 0 | Probability | Multiple Choice | First $a$ is chosen at random from the set $\{1,2,3,\cdots,99,100\}$, and then $b$ is chosen at random from the same set. The probability that the integer $3^a+7^b$ has units digit $8$ is
$\text{(A) } \frac{1}{16}\quad \text{(B) } \frac{1}{8}\quad \text{(C) } \frac{3}{16}\quad \text{(D) } \frac{1}{5}\quad \text{(E) }... | [
"The units digits of the powers of $3$ and $7$ both cycle through $1,3,9,7$ in opposite directions, and as $4\\mid 100$ each power's units digit is equally probable. There are $16$ ordered pairs of units digits, and three of them $(1,7),(7,1),(9,9)$ have a sum with units digit $8$.\n\n\nThus the probability is $\\f... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/18.json | AHSME |
1990_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | [asy] draw((0,0)--(16,0)--(21,5*sqrt(3))--(5,5*sqrt(3))--cycle,dot); draw((5,5*sqrt(3))--(1,5*sqrt(3))--(16,0),dot); MP("A",(0,0),S);MP("B",(16,0),S);MP("C",(21,5sqrt(3)),NE);MP("D",(5,5sqrt(3)),N);MP("E",(1,5sqrt(3)),N); MP("16",(8,0),S);MP("10",(18.5,5sqrt(3)/2),E);MP("4",(3,5sqrt(3)),N); dot((4,4sqrt(3))); MP("F",(4... | [
"$DFE$ and $AFB$ are similar triangles, so $FD$ is one quarter the length of the corresponding side $AF$.\nThus it is one fifth of the length of $AD$, which means $\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/4.json | AHSME |
1990_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | [asy] draw(circle((0,0),1),black); draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot); draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot); draw(arc((0,1),.25,230,310)); MP("A",(0,1),N);MP("B",(cos(pi/14),-sin(pi/14)),E);MP("C",(-cos(pi/14),-sin(pi/14)),W);MP("D",(0... | [
"We can make two equations (assume angle D is y): $y+2x=180$ and $4y+x=180$. We find that $x=\\dfrac{540}{7}$. Now we have to convert this to radians. 360 degrees is $2\\pi$ radians, so since we have $\\dfrac{540}{7}$ degrees, the answer is $\\dfrac{3\\pi}{7}$ which is $\\fbox{A}.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/14.json | AHSME |
1990_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | Four whole numbers, when added three at a time, give the sums $180,197,208$ and $222$. What is the largest of the four numbers?
$\text{(A) } 77\quad \text{(B) } 83\quad \text{(C) } 89\quad \text{(D) } 95\quad \text{(E) cannot be determined from the given information}$
| [
"Let $S$ be the sum of the four numbers. Then $180+197+208+222=807=3S\\therefore S=269$, and subtracting $180$ gives $89$ for $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/15.json | AHSME |
1990_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | Which of these numbers is largest?
$\text{(A) } \sqrt{\sqrt[3]{5\cdot 6}}\quad \text{(B) } \sqrt{6\sqrt[3]{5}}\quad \text{(C) } \sqrt{5\sqrt[3]{6}}\quad \text{(D) } \sqrt[3]{5\sqrt{6}}\quad \text{(E) } \sqrt[3]{6\sqrt{5}}$
| [
"Putting them all in the form $\\sqrt{\\sqrt[3]N}$ (the order of the radicals doesn't matter), the resulting interior numbers are \\[30, 1080, 750, 150, 180\\] so the answer is $\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/5.json | AHSME |
1990_AHSME_Problems | 19 | 0 | Number Theory | Multiple Choice | For how many integers $N$ between $1$ and $1990$ is the improper fraction $\frac{N^2+7}{N+4}$ $\underline{not}$ in lowest terms?
$\text{(A) } 0\quad \text{(B) } 86\quad \text{(C) } 90\quad \text{(D) } 104\quad \text{(E) } 105$
| [
"What we want to know is for how many $n$ is \\[\\gcd(n^2+7, n+4) > 1.\\] We start by setting \\[n+4 \\equiv 0 \\mod m\\] for some arbitrary $m$. This shows that $m$ evenly divides $n+4$. Next we want to see under which conditions $m$ also divides $n^2 + 7$. We know from the previous statement that \\[n \\equiv -4... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/19.json | AHSME |
1990_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | If $x,y>0, \log_y(x)+\log_x(y)=\frac{10}{3} \text{ and } xy=144,\text{ then }\frac{x+y}{2}=$
$\text{(A) } 12\sqrt{2}\quad \text{(B) } 13\sqrt{3}\quad \text{(C) } 24\quad \text{(D) } 30\quad \text{(E) } 36$
| [
"Rewrite the first equation as \\[\\frac{\\log y}{\\log x}+\\frac{\\log x}{\\log y}=3\\tfrac13\\] and this is of the form $u+\\tfrac1u =3\\tfrac13$; by inspection it is easy to see that $u=3$ or $\\tfrac13$. Therefore $\\log y=3\\log x$ (or vice versa—it doesn't matter here) so $y=x^3$. Substituting this into $xy=1... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/23.json | AHSME |
1990_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | Each edge of a cube is colored either red or black. Every face of the cube has at least one black edge. The smallest number possible of black edges is
$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 5\quad \text{(E) } 6$
| [
"Each black edge can only take care of two adjoining faces, so we know at least three will be needed. Once the first black edge is placed, it is easy to see that three will be sufficient, if they are separated and go in different directions:\n[asy]import three; unitsize(1cm);size(100); draw((0,0,0)--(0,1,0),linewid... | 1 | ./CreativeMath/AHSME/1990_AHSME_Problems/9.json | AHSME |
1985_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | A wooden cube with edge length $n$ units (where $n$ is an integer $>2$) is painted black all over. By slices parallel to its faces, the cube is cut into $n^3$ smaller cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free o... | [
"Observe that if we remove the outer layer of unit cubes from the entire cube, what remains is a smaller cube of side length $(n-2)$, which contains all of the unpainted cubes and no others. This shows that there are exactly $(n-2)^3$ unpainted cubes. Similarly, taking one face of the cube and removing the outer ed... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/20.json | AHSME |
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