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 |
|---|---|---|---|---|---|---|---|---|---|
1977_AHSME_Problems
| 21 | 0 |
Algebra
|
Multiple Choice
|
For how many values of the coefficient a do the equations \begin{align*}x^2+ax+1=0 \\ x^2-x-a=0\end{align*} have a common real solution?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \infty$
|
[
"Subtracting the equations, we get $ax+x+1+a=0$, or $(x+1)(a+1)=0$, so $x=-1$ or $a=-1$. If $x=-1$, then $a=2$, which satisfies the condition. If $a=-1$, then $x$ is nonreal. This means that $a=-1$ is the only number that works, so our answer is $(B)$.\n\n\n~alexanderruan\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/21.json
|
AHSME
|
1977_AHSME_Problems
| 10 | 0 |
Algebra
|
Multiple Choice
|
If $(3x-1)^7 = a_7x^7 + a_6x^6 + \cdots + a_0$, then $a_7 + a_6 + \cdots + a_0$ equals
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 64 \qquad \text{(D)}\ -64 \qquad \text{(E)}\ 128$
|
[
"Solution by e_power_pi_times_i\n\n\nNotice that if $x=1$, then $a_7x^7 + a_6x^6 + \\cdots + a_0 = a_7 + a_6 + \\cdots + a_0$. Therefore the answer is $(3(1)-1)^7) = \\boxed{\\text{(D)}\\ 128}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/10.json
|
AHSME
|
1977_AHSME_Problems
| 30 | 0 |
Geometry
|
Multiple Choice
|
[asy] for (int i=0; i<9; ++i) { draw(dir(10+40*i)--dir(50+40*i)); } draw(dir(50) -- dir(90)); label("$a$", dir(50) -- dir(90), N); draw(dir(10) -- dir(90)); label("$b$", dir(10) -- dir(90), SW); draw(dir(-70) -- dir(90)); label("$d$", dir(-70) -- dir(90), E); //Credit to MSTang for the diagram [/asy]
If $a,b$, and $d$ are the lengths of a side, a shortest diagonal and a longest diagonal, respectively,
of a regular nonagon (see adjoining figure), then
$\textbf{(A) }d=a+b\qquad \textbf{(B) }d^2=a^2+b^2\qquad \textbf{(C) }d^2=a^2+ab+b^2\qquad\\ \textbf{(D) }b=\frac{a+d}{2}\qquad \textbf{(E) }b^2=ad$
|
[
"By the law of cosines we can get the following expressions for $d^{2}$ and $b^{2}$:\n\n\n\\[d^{2}=2b^{2}(1-\\cos (100^\\circ))\\] \\[b^{2}=2a^{2}(1-\\cos (140^\\circ))\\]\n\n\nWe can substitute what we got for $b^2$ into the expression for $d^{2}$:\n\n\n\\[d^{2}=4a^{2}(1-\\cos (140^\\circ))(1-\\cos (100^\\circ))=4a^{2}(1-(\\cos (100^\\circ)+\\cos (140^\\circ))+\\cos (100^\\circ)\\cos (140^\\circ))\\]\n\n\nNow apply sum-to-product and product-to-sum identities:\n\\[d^{2}=4a^{2}(1-(2\\cos (120^\\circ)\\cos (120^\\circ))+\\frac{1}{2}(\\cos (240^\\circ)+\\cos (40^\\circ)))\\]\nSimplifying further gives us: \\[d^{2}=4a^{2}(1+\\cos (20^\\circ)-\\frac{1}{4}+\\frac{1}{2}\\cos (40^\\circ))\\]\nAfter using the fact that $\\cos (40^\\circ)=2\\cos^2 (20^\\circ)-1$, it's not hard to see that the expression in the parentheses is equal to $(\\cos (20^\\circ)+\\frac{1}{2})^{2}$. So we can square-root both sides to find the expression for $d$:\n\n\n\\[d=2a(\\cos (20^\\circ)+\\frac{1}{2})\\]\nNow let's look at the expression for $b^2$. We can apply the reverse of the double angle identity to show that $1-\\cos (140^\\circ)$ equals $2\\sin^2 (70^\\circ))$. So if we square root the entire expression we get that \\[b=2a\\sin (70^\\circ)=2a\\cos (20^\\circ)\\]\nWe now have everything in terms of $a$. Luckily when we consider choice $\\textbf{A}$ we can verify without much work that this must be the answer.\n\n\nSolution by harita19\n\n\n\\begin{itemize}\n\\item NOTE: a much easier solution exists by drawing some lines and recognizing that the nonagon is cyclic, but for those of use who use algebra in every geometry problem, this is the best solution.\n\\end{itemize}\n",
"Using Van Schooten's Theorem, we draw an equilateral triangle in the nonagon, and immediately see that $a+b=d$.\n\n\n"
] | 2 |
./CreativeMath/AHSME/1977_AHSME_Problems/30.json
|
AHSME
|
1977_AHSME_Problems
| 1 | 0 |
Algebra
|
Multiple Choice
|
If $y = 2x$ and $z = 2y$, then $x + y + z$ equals
$\text{(A)}\ x \qquad \text{(B)}\ 3x \qquad \text{(C)}\ 5x \qquad \text{(D)}\ 7x \qquad \text{(E)}\ 9x$
|
[
"Solution by e_power_pi_times_i\n\n\n$x+y+z = x+(2x)+(4x) = \\boxed{\\text{(D)}\\ 7x}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/1.json
|
AHSME
|
1977_AHSME_Problems
| 11 | 0 |
Number Theory
|
Multiple Choice
|
For each real number $x$, let $\textbf{[}x\textbf{]}$ be the largest integer not exceeding $x$
(i.e., the integer $n$ such that $n\le x<n+1$). Which of the following statements is (are) true?
$\textbf{I. [}x+1\textbf{]}=\textbf{[}x\textbf{]}+1\text{ for all }x \\ \textbf{II. [}x+y\textbf{]}=\textbf{[}x\textbf{]}+\textbf{[}y\textbf{]}\text{ for all }x\text{ and }y \\ \textbf{III. [}xy\textbf{]}=\textbf{[}x\textbf{]}\textbf{[}y\textbf{]}\text{ for all }x\text{ and }y$
$\textbf{(A) }\text{none}\qquad \textbf{(B) }\textbf{I }\text{only}\qquad \textbf{(C) }\textbf{I}\text{ and }\textbf{II}\text{ only}\qquad \textbf{(D) }\textbf{III }\text{only}\qquad \textbf{(E) }\text{all}$
|
[
"Solution by e_power_pi_times_i\n\n\nNotice that $\\textbf{[}x\\textbf{]}$ is just $\\left \\lfloor x \\right \\rfloor$. We see that $\\textbf{I}$ is true, as adding by one does not change the fraction part of the number. Similarly, $\\textbf{II}$ is false, because $\\textbf{[}x+y\\textbf{]}$ does not always equal $\\textbf{[}x\\textbf{]}+\\textbf{[}y\\textbf{]}$ (If both numbers were $1.5$, $\\textbf{[}1.5+1.5\\textbf{]} = 3$, $\\textbf{[}1.5\\textbf{]}+\\textbf{[}1.5\\textbf{]} = 2$). Because $\\textbf{I}$ is true and $\\textbf{II}$ is false, the answer is $\\boxed{\\textbf{(B) }\\textbf{I }\\text{only}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/11.json
|
AHSME
|
1977_AHSME_Problems
| 2 | 0 |
Geometry
|
Multiple Choice
|
Which one of the following statements is false? All equilateral triangles are
$\textbf{(A)}\ \text{ equiangular}\qquad \textbf{(B)}\ \text{isosceles}\qquad \textbf{(C)}\ \text{regular polygons }\qquad\\ \textbf{(D)}\ \text{congruent to each other}\qquad \textbf{(E)}\ \text{similar to each other}$
|
[
"Solution by e_power_pi_times_i\n\n\nEquilateral triangles can be in different sizes, therefore they are not $\\boxed{\\textbf{(D)}\\ \\text{congruent to each other}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/2.json
|
AHSME
|
1977_AHSME_Problems
| 28 | 0 |
Algebra
|
Multiple Choice
|
Let $g(x)=x^5+x^4+x^3+x^2+x+1$. What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$?
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$
\subsubsection{Solution 1}
Let $r(x)$ be the remainder when $g(x^{12})$ is divided by $g(x)$. Then $r(x)$ is the unique polynomial such that
\[g(x^{12}) - r(x) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 - r(x)\]
is divisible by $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1$, and $\deg r(x) < 5$.
Note that $(x - 1)(x^5 + x^4 + x^3 + x^2 + 1) = x^6 - 1$ is a multiple of $g(x)$. Also,
\[g(x^{12}) - 6 = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} - 5 \\ = (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).\]
Each term is a multiple of $x^6 - 1$. For example,
\[x^{60} - 1 = (x^6 - 1)(x^{54} + x^{48} + \dots + x^6 + 1).\]
Hence, $g(x^{12}) - 6$ is a multiple of $x^6 - 1$, which means that $g(x^{12}) - 6$ is a multiple of $g(x)$. Therefore, the remainder is $\boxed{6}$. The answer is (A).
\subsubsection{Solution 2}
We express the quotient and remainder as follows.
\[g(x^{12}) = Q(x) g(x) + R(x)\]
Note that the solutions to $g(x)$ correspond to the 6th roots of unity, excluding $1$. Hence, we have $x^6 = 1$, allowing us to set:
\[g(x^{12}) = 6\]
\[g(x) = 0\]
We have $5$ values of $x$ that return $R(x) = 6$. However, $g(x)$ is quintic, implying the remainder is of degree at most $4$. Since there are $5$ solutions, the only possibility is that the remainder is a constant $\boxed{6}$.
\subsubsection{Solution 3}
We can use the Chinese remainder theorem over $\mathbb{Q}[x].$ Since $g(x)=(x+1)(x^2-x+1)(x^2+x+1)$, $\mathbb{Q}[x]/g\cong \mathbb{Q}[x]/(x+1)\times \mathbb{Q}[x]/(x^2-x+1)\times \mathbb{Q}[x]/(x^2+x+1).$ This means that if we can find the remainder of $g(x^{12})$ modulo $x+1,x^2-x+1,x^2+x+1$, we can reconstruct the remainder modulo $g.$ We can further use that each factor is irreducible and that if $p(x)$ is an irreducible polynomial over $\mathbb{Q}$ with root $\alpha$, $\mathbb{Q}[x]/p\cong \mathbb{Q}(\alpha)$ so to evaluate the remainders of $g(x^{12})$, we just need to evaluate it on one of the roots of the irreducible factors. The first factor has root $-1$, the second has roots the primitive sixth roots of unity, and the third as roots the primitive cube roots of unity (this is easily seen as $g(x)(x-1)=x^6-1$). Evaluating $g(x^{12})$ on each of these values yields $g(1)=6$ so the remainder is $6$ on each factor on the right of the isomorphism. Hence, by the Chinese remainder theorem, the remainder modulo $g$ must be $\boxed{6}$ as well.
|
[
"Let $r(x)$ be the remainder when $g(x^{12})$ is divided by $g(x)$. Then $r(x)$ is the unique polynomial such that\n\\[g(x^{12}) - r(x) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 - r(x)\\]\nis divisible by $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1$, and $\\deg r(x) < 5$.\n\n\nNote that $(x - 1)(x^5 + x^4 + x^3 + x^2 + 1) = x^6 - 1$ is a multiple of $g(x)$. Also,\n\\[g(x^{12}) - 6 = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} - 5 \\\\ = (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).\\]\nEach term is a multiple of $x^6 - 1$. For example,\n\\[x^{60} - 1 = (x^6 - 1)(x^{54} + x^{48} + \\dots + x^6 + 1).\\]\nHence, $g(x^{12}) - 6$ is a multiple of $x^6 - 1$, which means that $g(x^{12}) - 6$ is a multiple of $g(x)$. Therefore, the remainder is $\\boxed{6}$. The answer is (A).\n\n",
"We express the quotient and remainder as follows.\n\\[g(x^{12}) = Q(x) g(x) + R(x)\\]\nNote that the solutions to $g(x)$ correspond to the 6th roots of unity, excluding $1$. Hence, we have $x^6 = 1$, allowing us to set:\n\\[g(x^{12}) = 6\\]\n\\[g(x) = 0\\]\nWe have $5$ values of $x$ that return $R(x) = 6$. However, $g(x)$ is quintic, implying the remainder is of degree at most $4$. Since there are $5$ solutions, the only possibility is that the remainder is a constant $\\boxed{6}$.\n\n",
"We can use the Chinese remainder theorem over $\\mathbb{Q}[x].$ Since $g(x)=(x+1)(x^2-x+1)(x^2+x+1)$, $\\mathbb{Q}[x]/g\\cong \\mathbb{Q}[x]/(x+1)\\times \\mathbb{Q}[x]/(x^2-x+1)\\times \\mathbb{Q}[x]/(x^2+x+1).$ This means that if we can find the remainder of $g(x^{12})$ modulo $x+1,x^2-x+1,x^2+x+1$, we can reconstruct the remainder modulo $g.$ We can further use that each factor is irreducible and that if $p(x)$ is an irreducible polynomial over $\\mathbb{Q}$ with root $\\alpha$, $\\mathbb{Q}[x]/p\\cong \\mathbb{Q}(\\alpha)$ so to evaluate the remainders of $g(x^{12})$, we just need to evaluate it on one of the roots of the irreducible factors. The first factor has root $-1$, the second has roots the primitive sixth roots of unity, and the third as roots the primitive cube roots of unity (this is easily seen as $g(x)(x-1)=x^6-1$). Evaluating $g(x^{12})$ on each of these values yields $g(1)=6$ so the remainder is $6$ on each factor on the right of the isomorphism. Hence, by the Chinese remainder theorem, the remainder modulo $g$ must be $\\boxed{6}$ as well.\n\n\n"
] | 3 |
./CreativeMath/AHSME/1977_AHSME_Problems/28.json
|
AHSME
|
1977_AHSME_Problems
| 12 | 0 |
Algebra
|
Multiple Choice
|
Al's age is $16$ more than the sum of Bob's age and Carl's age, and the square of Al's age is $1632$ more than the square of the sum of
Bob's age and Carl's age. What is the sum of the ages of Al, Bob, and Carl?
$\text{(A)}\ 64 \qquad \text{(B)}\ 94 \qquad \text{(C)}\ 96 \qquad \text{(D)}\ 102 \qquad \text{(E)}\ 140$
|
[
"Solution by e_power_pi_times_i\n\n\nDenote Al's age, Bob's age, and Carl's age by $a$, $b$, and $c$, respectively. Then, $a = 16 + b + c$ and $a^2 = 1632 + b^2 + c^2$. Substituting the first equation into the second, $(16 + b + c)^2 = b^2 + c^2 + 2bc + 32b + 32c + 256 = b^2 + c^2 + 1632$. Thus, $bc + 16b + 16c = 688$, and $(b+16)(c+16) = 944$. Since $944 = 2^4\\cdot59$, $(b,c) = (0,43)$ or $(43,0)$. Then $a + b + c = 2b + 2c + 16 = \\boxed{\\textbf{(D)}\\ 102}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/12.json
|
AHSME
|
1977_AHSME_Problems
| 24 | 0 |
Algebra
|
Multiple Choice
|
Find the sum $\frac{1}{1(3)}+\frac{1}{3(5)}+\dots+\frac{1}{(2n-1)(2n+1)}+\dots+\frac{1}{255(257)}$.
$\textbf{(A) }\frac{127}{255}\qquad \textbf{(B) }\frac{128}{255}\qquad \textbf{(C) }\frac{1}{2}\qquad \textbf{(D) }\frac{128}{257}\qquad \textbf{(E) }\frac{129}{257}$
|
[
"Note that $\\frac{1}{(2n-1)(2n+1)} = \\frac{1/(2n-1)-1/(2n+1)}{2}$. Indeed, we find the series telescopes and is equal to $\\frac{1-\\frac{1}{257}}{2}$, which is evidently $\\boxed {\\frac{128}{257}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/24.json
|
AHSME
|
1977_AHSME_Problems
| 25 | 0 |
Number Theory
|
Multiple Choice
|
Determine the largest positive integer $n$ such that $1005!$ is divisible by $10^n$.
$\textbf{(A) }102\qquad \textbf{(B) }112\qquad \textbf{(C) }249\qquad \textbf{(D) }502\qquad \textbf{(E) }\text{none of the above}\qquad$
\subsubsection{Solution}
We first observe that since there will be more 2s than 5s in $1005!$, we are looking for the largest $n$ such that $5^n$ divides $1005!$. We will use the fact that:
\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor \cdots\]
(This is an application of Legendre's formula).
From $k=5$ and onwards, $\left \lfloor {\frac{1005}{5^k}}\right \rfloor = 0$. Thus, our calculation becomes
\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor + \left \lfloor {\frac{1005}{5^4}}\right \rfloor\]
\[n = \left \lfloor {\frac{1005}{5}}\right \rfloor + \left \lfloor {\frac{1005}{25}}\right \rfloor + \left \lfloor {\frac{1005}{125}}\right \rfloor + \left \lfloor {\frac{1005}{625}}\right \rfloor\]
\[n = 201 + 40 + 8 + 1 = \boxed{250}\]
Since none of the answer choices equal 250, the answer is $\boxed{\textbf{(E)}}$.
- mako17
|
[
"We first observe that since there will be more 2s than 5s in $1005!$, we are looking for the largest $n$ such that $5^n$ divides $1005!$. We will use the fact that:\n\n\n\\[n = \\left \\lfloor {\\frac{1005}{5^1}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^2}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^3}}\\right \\rfloor \\cdots\\]\n\n\n(This is an application of Legendre's formula).\n\n\nFrom $k=5$ and onwards, $\\left \\lfloor {\\frac{1005}{5^k}}\\right \\rfloor = 0$. Thus, our calculation becomes\n\n\n\\[n = \\left \\lfloor {\\frac{1005}{5^1}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^2}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^3}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^4}}\\right \\rfloor\\]\n\n\n\\[n = \\left \\lfloor {\\frac{1005}{5}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{25}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{125}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{625}}\\right \\rfloor\\]\n\n\n\\[n = 201 + 40 + 8 + 1 = \\boxed{250}\\]\n\n\nSince none of the answer choices equal 250, the answer is $\\boxed{\\textbf{(E)}}$.\n\n\n- mako17\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/25.json
|
AHSME
|
1977_AHSME_Problems
| 13 | 0 |
Algebra
|
Multiple Choice
|
If $a_1,a_2,a_3,\dots$ is a sequence of positive numbers such that $a_{n+2}=a_na_{n+1}$ for all positive integers $n$,
then the sequence $a_1,a_2,a_3,\dots$ is a geometric progression
$\textbf{(A) }\text{for all positive values of }a_1\text{ and }a_2\qquad\\ \textbf{(B) }\text{if and only if }a_1=a_2\qquad\\ \textbf{(C) }\text{if and only if }a_1=1\qquad\\ \textbf{(D) }\text{if and only if }a_2=1\qquad\\ \textbf{(E) }\text{if and only if }a_1=a_2=1$
|
[
"Solution by e_power_pi_times_i\n\n\nThe first few terms are $a_1,a_2,a_1a_2,a_1a_2^2,a_1^2a_2^3,\\dots$ . If this is a geometric progression, $\\dfrac{a_2}{a_1} = a_1 = a_2 = a_1a_2$. $a_1=0,1$, $a_2=0,1$. Since this is a sequence of positive integers, then the answer must be $\\boxed{\\textbf{(E) }\\text{if and only if }a_1=a_2=1 }$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/13.json
|
AHSME
|
1977_AHSME_Problems
| 29 | 0 |
Algebra
|
Multiple Choice
|
Find the smallest integer $n$ such that $(x^2+y^2+z^2)^2\le n(x^4+y^4+z^4)$ for all real numbers $x,y$, and $z$.
$\textbf{(A) }2\qquad \textbf{(B) }3\qquad \textbf{(C) }4\qquad \textbf{(D) }6\qquad \textbf{(E) }\text{There is no such integer n}$
Solution
|
[
"We see squares and one number. And we see an inequality. This calls for Cauchy's inequality. EEEEWWW.\n\n\nAnyways, look at which side is which. The squared side is smaller-- so that's good. It's in the right format.\n\n\nCauchy's states that $(a_1b_1+a_2b_2+a_3b_3+......)^2 \\le (a_1^2+a_2^2+a_3^2+....)(b_1^2+b_2^2+b_3^2+.....)$\n\n\nTherefore, we see that, if we equate $a_1 = x^2, a_2 = y^2, a_3 = z^2$ we get the equality right away. What's the final step? Figuring out this n. Now, note that the equation is basically complete; all we need is for $b_1+b_2+b_3 = n$. So each of them is just 1, and $n = 3$-- answer choice $\\boxed{B}$!\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/29.json
|
AHSME
|
1977_AHSME_Problems
| 3 | 0 |
Algebra
|
Multiple Choice
|
A man has $2.73 in pennies, nickels, dimes, quarters and half dollars. If he has an equal number of coins of each kind, then the total number of coins he has is
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 15$
|
[
"Solution by e_power_pi_times_i\n\n\nDenote the number of pennies, nickels, dimes, quarters, and half-dollars by $x$. Then $x+5x+10x+25x+50x = 273$, and $x = 3$. The total amount of coins is $5x$, which is $\\textbf{\\text{(E)}}\\ 15$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/3.json
|
AHSME
|
1977_AHSME_Problems
| 8 | 0 |
Algebra
|
Multiple Choice
|
For every triple $(a,b,c)$ of non-zero real numbers, form the number $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$.
The set of all numbers formed is
$\textbf{(A)}\ {0} \qquad \textbf{(B)}\ \{-4,0,4\} \qquad \textbf{(C)}\ \{-4,-2,0,2,4\} \qquad \textbf{(D)}\ \{-4,-2,2,4\}\qquad \textbf{(E)}\ \text{none of these}$
|
[
"Solution by e_power_pi_times_i\n\n\n$\\dfrac{x}{|x|} = 1$ or $-1$ depending whether $x$ is positive or negative. If $a$, $b$, and $c$ are positive, then the entire thing amounts to $4$. If one of the three is negative and the other two positive, the answer is $0$. If two of the three is negative and one is positive, the answer is $0$. If all three are negative, the answer is $-4$. Therefore the set is $\\boxed{\\textbf{(B)}\\ \\{-4,0,4\\}}$.\n\n\nThis is very similar to https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_21\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/8.json
|
AHSME
|
1977_AHSME_Problems
| 22 | 0 |
Algebra
|
Multiple Choice
|
If $f(x)$ is a real valued function of the real variable $x$, and $f(x)$ is not identically zero,
and for all $a$ and $b$ $f(a+b)+f(a-b)=2f(a)+2f(b)$, then for all $x$ and $y$
$\textbf{(A) }f(0)=1\qquad \textbf{(B) }f(-x)=-f(x)\qquad \textbf{(C) }f(-x)=f(x)\qquad \\ \textbf{(D) }f(x+y)=f(x)+f(y) \qquad \\ \textbf{(E) }\text{there is a positive real number }T\text{ such that }f(x+T)=f(x)$
|
[
"We can start by finding the value of $f(0)$. \nLet $a = b = 0$\n\\[f(0) + f(0) = 2f(0) + 2f(0) \\Rrightarrow 2f(0) = 4f(0) \\Rrightarrow f(0) = 0\\]\nThus, $A$ is not true.\nTo check $B$, we let $a = 0$. We have\n\\[f(0+b) + f(0-b) = 2f(0) + 2f(b)\\]\n\\[f(b) + f(-b) = 0 + 2f(b)\\]\n\\[f(-b) = f(b)\\]\nThus, $B$ is not true, but $C$ is.\nThus, the answer is $\\boxed{C}$\n\n\n~~JustinLee2017\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/22.json
|
AHSME
|
1977_AHSME_Problems
| 18 | 0 |
Algebra
|
Multiple Choice
|
If $y=(\log_23)(\log_34)\cdots(\log_n[n+1])\cdots(\log_{31}32)$ then
$\textbf{(A) }4<y<5\qquad \textbf{(B) }y=5\qquad \textbf{(C) }5<y<6\qquad \textbf{(D) }y=6\qquad \\ \textbf{(E) }6<y<7$
|
[
"Solution by e_power_pi_times_i\n\n\n\n\nNote that $\\log_{a}b = \\dfrac{\\log{b}}{\\log{a}}$. Then $y=(\\dfrac{\\log3}{\\log2})(\\dfrac{\\log4}{\\log3})\\cdots(\\dfrac{\\log32}{\\log31}) = \\dfrac{\\log32}{\\log2} = \\log_232 = \\boxed{\\text{(B) }y=5}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/18.json
|
AHSME
|
1977_AHSME_Problems
| 4 | 0 |
Geometry
|
Multiple Choice
|
[asy] size(130); pair A = (2, 2.4), C = (0, 0), B = (4.3, 0), E = 0.7*A, F = 0.57*A + 0.43*B, D = (2.4, 0); draw(A--B--C--cycle); draw(E--D--F); label("$A$", A, N); label("$B$", B, E); label("$C$", C, W); label("$D$", D, S); label("$E$", E, NW); label("$F$", F, NE); //Credit to MSTang for the diagram [/asy]
In triangle $ABC, AB=AC$ and $\measuredangle A=80^\circ$. If points $D, E$, and $F$ lie on sides $BC, AC$ and $AB$, respectively, and $CE=CD$ and $BF=BD$, then $\measuredangle EDF$ equals
$\textbf{(A) }30^\circ\qquad \textbf{(B) }40^\circ\qquad \textbf{(C) }50^\circ\qquad \textbf{(D) }65^\circ\qquad \textbf{(E) }\text{none of these}$
|
[
"Solution by e_power_pi_times_i\n\n\nBecause $\\measuredangle A=80^\\circ$, $\\measuredangle B=\\measuredangle C=50^\\circ$. Then, because triangles $CDE$ and $BDF$ are isosceles, $\\measuredangle CDE=\\measuredangle BDF=65^\\circ$. $\\measuredangle EDF=180^\\circ - 2(65^\\circ)=\\textbf{(C) }50^\\circ$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/4.json
|
AHSME
|
1977_AHSME_Problems
| 14 | 0 |
Algebra
|
Multiple Choice
|
How many pairs $(m,n)$ of integers satisfy the equation $m+n=mn$?
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }\text{more than }4$
|
[
"Solution by e_power_pi_times_i\n\n\nIf $m+n=mn$, $mn-m-n = (m-1)(n-1)-1 = 0$. Then $(m-1)(n-1) = 1$, and $(m,n) = (2,2) or (0,0)$. The answer is $\\boxed{\\textbf{(B) }2}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/14.json
|
AHSME
|
1977_AHSME_Problems
| 15 | 0 |
Geometry
|
Multiple Choice
|
Three circles are drawn, so that each circle is externally tangent to the other two circles. Each circle has a radius of $3.$ A triangle is then constructed, such that each side of the triangle is tangent to two circles, as shown below. Find the perimeter of the triangle.
\subsection{Diagram}
[asy] unitsize(1 cm); draw(Circle(dir(90),sqrt(3)/2)); draw(Circle(dir(90 + 120),sqrt(3)/2)); draw(Circle(dir(90 + 240),sqrt(3)/2)); draw((1 + sqrt(3))*dir(90)--(1 + sqrt(3))*dir(90 + 120)--(1 + sqrt(3))*dir(90 + 240)--cycle); [/asy]
Or visit here: https://latex.artofproblemsolving.com/5/6/c/56c2e4a15c4c4654523f9b7ac15535647d8bce47.png
|
[
"Solution by e_power_pi_times_i\n\n\nDraw perpendicular lines from the radii of the circles to the sides of the triangle, and lines from the radii of the circles to the vertices of the triangle. Because the triangle is equilateral, the lines divide the big triangle into a small triangle, three rectangles, and six small $30-60-90$ triangles. The Longer length of the rectangles is 6 because it is the radius of one circle + the radius of another ($3+3$). The length of one side is $2(3\\sqrt{3})+6 = 6\\sqrt{3}+6$. The perimeter is $\\boxed{\\textbf{(D) }18\\sqrt{3}+18}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/15.json
|
AHSME
|
1977_AHSME_Problems
| 5 | 0 |
Geometry
|
Multiple Choice
|
The set of all points $P$ such that the sum of the (undirected) distances from $P$ to two fixed points $A$ and $B$ equals the distance between $A$ and $B$ is
$\textbf{(A) }\text{the line segment from }A\text{ to }B\qquad \textbf{(B) }\text{the line passing through }A\text{ and }B\qquad\\ \textbf{(C) }\text{the perpendicular bisector of the line segment from }A\text{ to }B\qquad\\ \textbf{(D) }\text{an ellipse having positive area}\qquad \textbf{(E) }\text{a parabola}$
|
[
"Solution by e_power_pi_times_i\n\n\nThe answer is $\\textbf{(A)}$ because $P$ has to be on the line segment $AB$ in order to satisfy $PA+PB=AB$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/5.json
|
AHSME
|
1977_AHSME_Problems
| 19 | 0 |
Geometry
|
Multiple Choice
|
Let $E$ be the point of intersection of the diagonals of convex quadrilateral $ABCD$, and let $P,Q,R$, and $S$ be the centers of the circles
circumscribing triangles $ABE, BCE, CDE$, and $ADE$, respectively. Then
$\textbf{(A) }PQRS\text{ is a parallelogram}\\ \textbf{(B) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rhombus}\\ \textbf{(C) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rectangle}\\ \textbf{(D) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a parallelogram}\\ \textbf{(E) }\text{none of the above are true}$
|
[
"Let $L$, $M$, $N$, and $O$ be the intersections of $AC$ and $PS$, $BD$ and $PQ$, $CA$ and $QR$, and $DB$ and $RS$ respectively. Since the circumcenter of a triangle is determined by the intersection of its perpendicular bisectors, each of the pairs mentioned earlier are perpendicular to each other. Let $\\angle SPQ=\\alpha$. Then $\\angle LEM=180^\\circ-\\alpha$ since $\\angle PLC=PMD=90^\\circ$. This means $\\angle DEC=180^\\circ-\\alpha$, which similarly implies $\\angle QRS=\\alpha$. Through a similar method, we find that $\\angle PSR=\\angle PQR=180^\\circ-\\alpha$.\n\n\nSince the opposite angles of $PQRS$ are congruent, $PQRS$ is indeed a parallelogram - and we showed this without any assumption about the type of quadrilateral $ABCD$ was. Therefore, $PQRS$ is always a parallelogram, and so our answer is $\\fbox{A}$. \n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/19.json
|
AHSME
|
1977_AHSME_Problems
| 9 | 0 |
Geometry
|
Multiple Choice
|
[asy] size(120); path c = Circle((0, 0), 1); pair A = dir(20), B = dir(130), C = dir(240), D = dir(330); draw(c); pair F = 3(A-B) + B; pair G = 3(D-C) + C; pair E = intersectionpoints(B--F, C--G)[0]; draw(B--E--C--A); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, E); //Credit to MSTang for the diagram [/asy]
In the adjoining figure $\measuredangle E=40^\circ$ and arc $AB$, arc $BC$, and arc $CD$ all have equal length. Find the measure of $\measuredangle ACD$.
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }\left(\frac{45}{2}\right)^\circ\qquad \textbf{(E) }30^\circ$
|
[
"Solution by e_power_pi_times_i\n\n\nIf arcs $AB$, $BC$, and $CD$ are congruent, then $\\measuredangle ACB = \\measuredangle BDC = \\measuredangle CBD = \\theta$. Because $ABCD$ is cyclic, $\\measuredangle CAD = \\measuredangle CBD = \\theta$, and $\\measuredangle ADB = \\measuredangle ACB = \\theta$. Then, $\\measuredangle EAD = \\measuredangle EDA = \\dfrac{180^\\circ - 40^\\circ}{2} = 70^\\circ$. $\\theta = 55^\\circ$. $\\measuredangle ACD = 180^\\circ - 55^\\circ - 110^\\circ = \\boxed{\\textbf{(B) }15^\\circ}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1977_AHSME_Problems/9.json
|
AHSME
|
1962_AHSME_Problems
| 20 | 0 |
Algebra
|
Multiple Choice
|
The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 36$
|
[
"If the angles are in an arithmetic progression, they can be expressed as\n$a$, $a+n$, $a+2n$, $a+3n$, and $a+4n$ for some real numbers $a$ and $n$.\nNow we know that the sum of the degree measures of the angles of a pentagon is $180(5-2)=540$.\nAdding our expressions for the five angles together, we get $5a+10n=540$.\nWe now divide by 5 to get $a+2n=108$. It so happens that $a+2n$ is one of the angles we defined earlier, so that angle must have a measure of $\\boxed{108\\textbf{ (A)}}$.\n(In fact, for any arithmetic progression with an odd number of terms,\nthe middle term is equal to the average of all the terms.)\n\n\n",
"If we write the five terms as $a$, $a - n$, $a - 2n$, $a + n$ and $a + 2n$, we can see that adding them up, we get $5a = 540$ through this, we can see that $a = 108$, $\\fbox{\\textbf{(A)}}$\n\n\n"
] | 2 |
./CreativeMath/AHSME/1962_AHSME_Problems/20.json
|
AHSME
|
1962_AHSME_Problems
| 36 | 0 |
Number Theory
|
Multiple Choice
|
If both $x$ and $y$ are both integers, how many pairs of solutions are there of the equation $(x-8)(x-10) = 2^y$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \text{more than 3}$
|
[
"The equality implies $x-8$ and $x-10$ are both powers of two; since they differ by two, it must be the case that $(x-8,x-10) = (4,2)$ or $(x-8,x-10) = (-2,-4)$. (Note that $(1,-1)$ is not allowed because then the product is negative.) These yield $(x,y) = (12,3)$ or $(x,y) = (6,3)$, for a total of $\\boxed{2\\textbf{ (C)}}$ solutions.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/36.json
|
AHSME
|
1962_AHSME_Problems
| 16 | 0 |
Geometry
|
Multiple Choice
|
Given rectangle $R_1$ with one side $2$ inches and area $12$ square inches. Rectangle $R_2$ with diagonal $15$ inches is similar to $R_1$. Expressed in square inches the area of $R_2$ is:
$\textbf{(A)}\ \frac{9}2\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 9\sqrt{10}\qquad\textbf{(E)}\ \frac{27\sqrt{10}}{4}$
|
[
"Clearly, the other side of $R_1$ has length $\\frac{12}2=6$.\nNow call the sides of $R_2$ a and b, with $b>a$. We know that $\\frac{b}a=\\frac31=3$, because the two rectangles are similar. We also know that $a^2+b^2=15^2=225$. But $b=3a$, so substituting gives \n\\[a^2+(3a)^2=225\\]\n\\[10a^2=225\\]\n\\[a^2=\\frac{45}2\\]\n\\[a=\\frac{3\\sqrt{10}}2\\]\n\\[b=\\frac{9\\sqrt{10}}2\\]\n\\[[R_2]=ab=\\boxed{\\frac{135}2\\textbf{ (C)}}\\]\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/16.json
|
AHSME
|
1962_AHSME_Problems
| 6 | 0 |
Geometry
|
Multiple Choice
|
A square and an equilateral triangle have equal perimeters. The area of the triangle is $9 \sqrt{3}$ square inches. Expressed in inches the diagonal of the square is:
$\textbf{(A)}\ \frac{9}{2}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ \frac{9\sqrt{2}}{2}\qquad\textbf{(E)}\ \text{none of these}$
|
[
"To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is $\\dfrac{x^2\\sqrt{3}}{4}$. This has to be equal to $9 \\sqrt{3}$, which means that $x=6$, or the side length of the triangle is $6$. Thus, the triangle (and the square) have a perimeter of $18$. It follows that each side of the square is $\\dfrac{18}{4}=4.5$. If we draw the diagonal, we create a 45-45-90 triangle, whose hypotenuse (also the diagonal of the square) is $\\boxed{\\text{(D)} \\dfrac{9\\sqrt{2}}{2}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/6.json
|
AHSME
|
1962_AHSME_Problems
| 7 | 0 |
Geometry
|
Multiple Choice
|
Let the bisectors of the exterior angles at $B$ and $C$ of triangle $ABC$ meet at D$.$ Then, if all measurements are in degrees, angle $BDC$ equals:
$\textbf{(A)}\ \frac{1}{2}(90-A)\qquad\textbf{(B)}\ 90-A\qquad\textbf{(C)}\ \frac{1}{2}(180-A)\qquad$
$\textbf{(D)}\ 180-A\qquad\textbf{(E)}\ 180-2A$
|
[
"Calculating for angles $\\angle DBC$ and $\\angle DCB$, we get\n\n\n$\\angle DBC$ = $90 - \\frac{B}{2}$ and\n$\\angle DCB$ = $90 - \\frac{C}{2}$.\n\n\nIn triangle $BCD$, we have\n\n\n$\\angle BDC$ = $180 - (90 - \\frac{B}{2}) - (90 - \\frac{C}{2})$ = $\\frac{B+C}{2}$ = $\\frac{1}{2}\\cdot(180 - A)$, so the answer is $\\fbox{C}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/7.json
|
AHSME
|
1962_AHSME_Problems
| 17 | 0 |
Algebra
|
Multiple Choice
|
If $a = \log_8 225$ and $b = \log_2 15$, then $a$, in terms of $b$, is:
$\textbf{(A)}\ \frac{b}{2}\qquad\textbf{(B)}\ \frac{2b}{3}\qquad\textbf{(C)}\ b\qquad\textbf{(D)}\ \frac{3b}{2}\qquad\textbf{(E)}\ 2b$
|
[
"Using the change-of-base rule: $a = \\frac{\\log 225}{\\log 8}$ and $b = \\frac{\\log 15}{\\log 2}$.\n\\[\\frac{a}{b} = \\frac{\\log 225 \\log 2}{\\log 8 \\log 15}\\]\n\\[a = b \\cdot \\frac{\\log 225}{\\log 15} \\cdot \\frac{\\log 2}{\\log 8}\\]\n\\[a = b \\log_{15} 225 \\log_8 2\\]\n\\[a = \\boxed{\\frac{2b}3 \\textbf{ (B)}}\\]\n\n\n\n\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/17.json
|
AHSME
|
1962_AHSME_Problems
| 40 | 0 |
Algebra
|
Multiple Choice
|
The limiting sum of the infinite series, $\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots$ whose $n$th term is $\frac{n}{10^n}$ is:
$\textbf{(A)}\ \frac{1}9\qquad\textbf{(B)}\ \frac{10}{81}\qquad\textbf{(C)}\ \frac{1}8\qquad\textbf{(D)}\ \frac{17}{72}\qquad\textbf{(E)}\ \text{larger than any finite quantity}$
|
[
"The series can be written as the following:\n\n\n$\\frac{1}{10} + \\frac{1}{10^2} + \\frac{1}{10^3} + ...$\n\n\n$+ \\frac{1}{10^2} + \\frac{1}{10^3} + \\frac{1}{10^4} + ...$\n\n\n$+ \\frac{1}{10^3} + \\frac{1}{10^4} + \\frac{1}{10^5} + ...$\n\n\nand so on.\n\n\nby using the formula for infinite geometric series $(\\frac{a}{1-r})$,\n\n\nWe can get $\\frac{\\frac{1}{10}}{1-\\frac{1}{10}}$ $+$ $\\frac{\\frac{1}{10^2}}{1-\\frac{1}{10}}$ $+$ $\\frac{\\frac{1}{10^3}}{1-\\frac{1}{10}}$ $+$ ...\nSince they all have common denominators, we get $\\frac{(\\frac{1}{10} + \\frac{1}{10^2} + \\frac{1}{10^3})}{\\frac{9}{10}}$.\nUsing the infinite series formula again, we get $\\frac{\\frac{\\frac{1}{10}}{1-\\frac{1}{10}}}{\\frac{9}{10}}$ $=$ $\\frac{\\frac{\\frac{1}{10}}{\\frac{9}{10}}}{\\frac{9}{10}}$ $=$ $\\frac{\\frac{1}{9}}{\\frac{9}{10}}$ $=$ $\\boxed{ (B) \\frac{10}{81}}$\n\n\n",
"So.. we have the sum to be $\\frac{1}{10}+\\frac{2}{100}+\\frac{3}{1000}$...\nNotice that this can be written as \n$\\frac{1}{10}+\\frac{0.2}{10}+\\frac{0.03}{10}+\\frac{0.004}{10}$.\nNow, it is trivial that the new fraction we seek is $\\frac{1.234567891011......}{10}$\n\n\nTesting the answer choices, we see that $\\boxed{B}$ is the correct answer.\n\n\n",
"Let\n\\[S = \\frac{1}{10} + \\frac{2}{10^2} + \\frac{3}{10^3} + ...\\]\nThen\n\\[10S = 1 + \\frac{2}{10} + \\frac{3}{10^2} + \\frac{4}{10^3} + ...\\]\nSubtracting $1S$ from $10S$, we got:\n\\begin{align*} 9S &= 1 + \\frac{1}{10} + \\frac{1}{10^2} + \\frac{1}{10^3} + ... \\\\ &= \\frac{1}{1-\\frac{1}{10}} = \\frac{10}{9} \\\\ S &= \\frac{10}{81} \\end{align*}\nTherefore, the answer is $\\boxed{(B) \\frac{10}{81}}$. -nullptr07\n\n\n"
] | 3 |
./CreativeMath/AHSME/1962_AHSME_Problems/40.json
|
AHSME
|
1962_AHSME_Problems
| 37 | 0 |
Geometry
|
Multiple Choice
|
$ABCD$ is a square with side of unit length. Points $E$ and $F$ are taken respectively on sides $AB$ and $AD$ so that $AE = AF$ and the quadrilateral $CDFE$ has maximum area. In square units this maximum area is:
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ \frac{9}{16}\qquad\textbf{(C)}\ \frac{19}{32}\qquad\textbf{(D)}\ \frac{5}{8}\qquad\textbf{(E)}\ \frac{2}3$
|
[
"Let $AE=AF=x$\n$[CDFE]=[ABCD]-[AEF]-[EBC]=1-\\frac{x^2}{2}-\\frac{1-x}{2}$\nOr\n$[CDFE]=\\frac{\\frac{5}{4}-(x-\\frac{1}{2})^2}{2}\\le \\frac{5}{8}$\nAs $(x-\\frac{1}{2})^2\\ge 0$\nSo $[CDFE]\\le \\frac{5}{8}$\nEquality occurs when $AE=AF=x=\\frac{1}{2}$\nSo the maximum value is $\\frac{5}{8}$\n\n\n",
"Let us first draw a unit square.\n[asy] draw((0,0)--(0,1)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((1,0)--(0,0)); label(\"A\",(0,1),NW); label(\"B\",(1,1),NE); label(\"C\",(1,0),SE); label(\"D\",(0,0),SW); label(\"$1$\",(0,0)--(0,1),W); [/asy]\nWe will now pick arbitrary points $E$ and $F$ on $\\overline{AB}$ and $\\overline{AD}$ respectively. We shall say that $AE = AF = x$\n[asy] draw((0,0)--(0,1)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((1,0)--(0,0)); label(\"A\",(0,1),NW); label(\"B\",(1,1),NE); label(\"C\",(1,0),SE); label(\"D\",(0,0),SW); label(\"$1$\",(0,0)--(0,1),W); draw((1,0)--(0.5,1),blue); draw((0.5,1)--(0,0.5),blue); draw((0,0.5)--(0,0),blue); draw((0,0)--(1,0),blue); label(\"E\",(0.5,1),N); label(\"F\",(0,0.5),SE); [/asy]\nThus, our problem has been simplified to maximizing the area of the blue quadrilateral.\nIf we drop an altitude from $E$ to $\\overline{DC}$, and call the foot of the altitude $G$, we can find the area of $EFDC$ by noting that $[EFDC] = [EGDF] + [EGC]$.\n[asy] draw((0,0)--(0,1)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((1,0)--(0,0)); label(\"A\",(0,1),NW); label(\"B\",(1,1),NE); label(\"C\",(1,0),SE); label(\"D\",(0,0),SW); label(\"$1$\",(0,0)--(0,1),W); draw((1,0)--(0.5,1),blue); draw((0.5,1)--(0,0.5),blue); draw((0,0.5)--(0,0),blue); draw((0,0)--(1,0),blue); label(\"E\",(0.5,1),N); label(\"F\",(0,0.5),SE); draw((0.5,1)--(0.5,0),red); draw((0.5,0.1)--(0.4,0.1),red); draw((0.4,0.1)--(0.4,0),red); label(\"G\",(0.5,0),S); [/asy]\nWe can now finish the problem.\n\n\nSince $EG = 1$ and $FD = 1-x$, we have:\n\\[[EFDC] = [EGDF] + [EGC] = DG\\cdot\\frac{EG + FD}{2} + \\frac{(CG)(EG)}{2} = \\frac{x(2-x)}{2} + \\frac{1-x}{2} = \\frac{1+x-x^2}{2}\\]\n\n\nTo maximize this, we compete the square in the numerator to have:\n\\[[EFDC] = \\frac{-(x-\\frac{1}{2})^2 + \\frac{5}{4}}{2} = \\frac{5}{8} - \\frac{(x-\\frac{1}{2})^2}{2}\\]\n\n\nFinally, we see that $\\frac{(x-\\frac{1}{2})^2}{2}\\geq0$, as:\n\\[\\left(x-\\frac{1}{2}\\right)^2\\geq0\\]\nSo,\n\\[\\frac{(x-\\frac{1}{2})^2}{2}\\geq0,\\]\n\n\nwhere the first inequality was from the Trivial Inequality, and the second came from dividing both sides by $2$, which does not change the inequality sign.\nThus, the maximum area is $\\boxed{\\frac{5}{8}}$ or $\\boxed{\\textbf{(D)}},$ when $\\frac{(x-\\frac{1}{2})^2}{2} = 0.$\n\n\n",
"Let $AE = AF = x$. \n\n\nThe area of the quadrilateral $CDFE$ can be expressed as $A = 1 - \\frac{x^2}{2} - \\frac{(1-x)}{2}$.\n\n\nBy taking the derivative of $A$, we get $A' = -x + \\frac{1}{2}$.\n\n\nWe make $A' = 0$ and get the critical point $x = \\frac{1}{2}$.\n\n\nSubstituting $x = \\frac{1}{2}$ , the maximum area is $\\boxed{\\frac{5}{8}}$ or $\\boxed{\\textbf{(D)}}$.\n\n\n~belanotbella\n\n\n",
"Plot the points of the quadrilateral (with the bottom left corner of the square being at the origin) to plug into shoelace formula:\n\\[(0, 0), (0, 1-x), (x, 1), (1, 0)\\]\nSo, \n\\[\\frac{-x^2+x+1}{2}\\]\nTrying $1/2$ first, you get $5/8$. Any larger or smaller value gives an answer less than $5/8$. Therefore, the answer is $\\boxed{D}$.\n\n\n~MC413551\n\n\n"
] | 4 |
./CreativeMath/AHSME/1962_AHSME_Problems/37.json
|
AHSME
|
1962_AHSME_Problems
| 21 | 0 |
Algebra
|
Multiple Choice
|
It is given that one root of $2x^2 + rx + s = 0$, with $r$ and $s$ real numbers, is $3+2i (i = \sqrt{-1})$. The value of $s$ is:
$\textbf{(A)}\ \text{undetermined}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ -13\qquad\textbf{(E)}\ 26$
|
[
"If a quadratic with real coefficients has two non-real roots, the two roots must be complex conjugates of one another.\nThis means the other root of the given quadratic is $\\overline{3+2i}=3-2i$. \nNow Vieta's formulas say that $s/2$ is equal to the product of the two roots, so\n$s = 2(3+2i)(3-2i) = \\boxed{26 \\textbf{ (E)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/21.json
|
AHSME
|
1962_AHSME_Problems
| 10 | 0 |
Algebra
|
Multiple Choice
|
A man drives $150$ miles to the seashore in $3$ hours and $20$ minutes. He returns from the shore to the starting point in $4$ hours and $10$ minutes. Let $r$ be the average rate for the entire trip. Then the average rate for the trip going exceeds $r$ in miles per hour, by:
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 4\frac{1}{2}\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 1$
|
[
"Since the equation for rate is $r=\\frac{d}{t}$, you only need to find $d$ and $t$. The distance from the seashore to the starting point is $150$ miles, and since he makes a round trip, $d=300$. Also, you know the times it took him to go both directions, so when you add them up ($3$ hours and $20$ minutes $+$ $4$ hours and $10$ minutes), you get $7\\frac{1}{2}$ hours. Since $r=\\frac{d}{t}$, after plugging in the values of $d$ and $t$, you get $r=\\frac{300}{7\\frac{1}{2}}$, or $r=40$. But what the question asks is for the positive difference between the speed going to the seashore and the average speed. The equation is $r=\\frac{d}{t}$, so when you plug in $d$ and $t$, you get $r=\\frac{150}{3\\frac{1}{3}}$. Simplifying, you get $r=45$. $45-40=5 \\rightarrow \\boxed{\\text{A}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/10.json
|
AHSME
|
1962_AHSME_Problems
| 26 | 0 |
Algebra
|
Multiple Choice
|
For any real value of $x$ the maximum value of $8x - 3x^2$ is:
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{8}3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \frac{16}{3}$
|
[
"Let $f(x) = 8x-3x^2$ Since $f(x)$ is a quadratic and the quadratic term is negative, the maximum will be $f\\bigg(- \\dfrac{b}{2a}\\bigg)$ when written in the form $ax^2+bx+c$. We see that $a=-3$, and so $- \\dfrac{b}{2a} = -\\bigg( \\dfrac{8}{-6}\\bigg) = \\dfrac{4}{3}$. Plugging in this value yields $f(\\dfrac{4}{3}) = \\dfrac{32}{3}-3 \\cdot \\dfrac{16}{9} = \\dfrac{32}{3} - \\dfrac{16}{3} = \\boxed{\\text{E}\\ \\dfrac{16}{3}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/26.json
|
AHSME
|
1962_AHSME_Problems
| 30 | 0 |
Other
|
Multiple Choice
|
Consider the statements:
$\textbf{(1)}\ \text{p and q are both true}\qquad\textbf{(2)}\ \text{p is true and q is false}\qquad\textbf{(3)}\ \text{p is false and q is true}\qquad\textbf{(4)}\ \text{p is false and q is false.}$
How many of these imply the negative of the statement "p and q are both true?"
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$
|
[
"By De Morgan's Law, the negation of $p$ and $q$ are both true is that at least one of them is false, with the exception of the statement 1, i.e.;\n\n\n$\\text{p and q are both true}.$\n\n\nThe other three statements state that at least one statement is false. So, 2, 3, and 4 work, yielding an answer of 3, or $\\textbf{(D)}.$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/30.json
|
AHSME
|
1962_AHSME_Problems
| 31 | 0 |
Geometry
|
Multiple Choice
|
The ratio of the interior angles of two regular polygons with sides of unit length is $3: 2$. How many such pairs are there?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{infinitely many}$
|
[
"The formula for the measure of the interior angle of a regular polygon with $n$-sides is $180 - \\frac{360}{n}$. Letting our two polygons have side length $r$ and $k$, we have that the ratio of the interior angles is $\\frac{180 - \\frac{360}{r}}{180 - \\frac{360}{k}} = \\frac{(r-2) \\cdot k}{(k-2) \\cdot r} = \\frac{3}{2}$. Cross multiplying both sides, we have $2rk-4k = 3kr-6r \\Rightarrow -rk-4k+6r = 0$. Using Simon's Favorite Factoring Trick, we have $-k(r+4)+6r+24 = 24 \\Rightarrow (r+4)(6-k)=24$. Because $k$ and $r$ are both more than $2$, we know that $6-k < r+4$. Now, we just set these factors equal to the factors of 24. We can set $6-k$ to $1$, $2$, $3$, or $4$ and $r+4$ to $24$, $12$, $8$, or $6$ respectively to get the following pairs for $(k, r)$: $(5, 20)$, $(4, 8)$, $(3, 4)$, and $(2, 2)$. However, we have to take out the solution with $k = 2$, because $k$ and $r$ are both more than $2$, leaving us with $\\boxed{\\textbf{(C)}\\ 3}$ as the correct answer.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/31.json
|
AHSME
|
1962_AHSME_Problems
| 27 | 0 |
Algebra
|
Multiple Choice
|
Let $a @ b$ represent the operation on two numbers, $a$ and $b$, which selects the larger of the two numbers, with $a@a = a$. Let $a ! b$ represent the operator which selects the smaller of the two numbers, with $a ! a = a$. Which of the following three rules is (are) correct?
$\textbf{(1)}\ a@b = b@a\qquad\textbf{(2)}\ a@(b@c) = (a@b)@c\qquad\textbf{(3)}\ a ! (b@c) = (a ! b) @ (a ! c)$
$\textbf{(A)}\ (1)\text{ only}\qquad\textbf{(B)}\ (2)\text{ only}\qquad\textbf{(C)}\ \text{(1) and (2) only}$
$\textbf{(D)}\ \text{(1) and (3) only}\qquad\textbf{(E)}\ \text{all three}$
|
[
"The first rule must be correct as both sides of the equation pick the larger out of a and b.\n\n\nThe second rule must also be correct as both sides would end up picking the largest out of a, b, and c.\n\n\nWLOG, lets assume b < c.\n\n\nThe third rule is a little more complex. To see if it works, let’s split the possibilities into three cases. a < b < c, b < a < c, and b < c < a. For the first case, the equation simplifies to a = a, which is correct. In the second case, the equation simplifies to a = a, which is correct. For the last case, the equation simplifies to c = c, which is also correct. Since all three cases work, this rule also works.\n\n\nThus, our answer is $\\boxed{\\textbf{E}}$\n\n\n(Solution by someonenumber011)\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/27.json
|
AHSME
|
1962_AHSME_Problems
| 1 | 0 |
Algebra
|
Multiple Choice
|
The expression $\frac{1^{4y-1}}{5^{-1}+3^{-1}}$ is equal to:
$\textbf{(A)}\ \frac{4y-1}{8}\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{15}{2}\qquad\textbf{(D)}\ \frac{15}{8}\qquad\textbf{(E)}\ \frac{1}{8}$
|
[
"We simplify the expression to yield:\n\n\n$\\dfrac{1^{4y-1}}{5^{-1}+3^{-1}}=\\dfrac{1}{5^{-1}+3^{-1}}=\\dfrac{1}{\\dfrac{1}{5}+\\dfrac{1}{3}}=\\dfrac{1}{\\dfrac{8}{15}}=\\dfrac{15}{8}$.\n\n\nThus our answer is $\\boxed{\\textbf{(D)}\\ \\frac{15}{8}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/1.json
|
AHSME
|
1962_AHSME_Problems
| 11 | 0 |
Algebra
|
Multiple Choice
|
The difference between the larger root and the smaller root of $x^2 - px + (p^2 - 1)/4 = 0$ is:
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ p\qquad\textbf{(E)}\ p+1$
|
[
"Call the two roots $r$ and $s$, with $r \\ge s$. \nBy Vieta's formulas, $p=r+s$ and $(p^2-1)/4=rs.$\n(Multiplying both sides of the second equation by 4 gives $p^2-1=4rs$.)\nThe value we need to find, then, is $r-s$.\nSince $p=r+s$, $p^2=r^2+2rs+s^2$.\nSubtracting $p^2-1=4rs$ from both sides gives $1=r^2-2rs+s^2$.\nTaking square roots, $r-s=1 \\Rightarrow \\boxed{\\textbf{(B)}}$.\n(Another solution is to use the quadratic formula and see that the\nroots are $\\frac{p\\pm 1}2$, and their difference is 1.)\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/11.json
|
AHSME
|
1962_AHSME_Problems
| 2 | 0 |
Algebra
|
Multiple Choice
|
The expression $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}}$ is equal to:
$\textbf{(A)}\ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{-\sqrt{3}}{6}\qquad\textbf{(C)}\ \frac{\sqrt{-3}}{6}\qquad\textbf{(D)}\ \frac{5\sqrt{3}}{6}\qquad\textbf{(E)}\ 1$
|
[
"Simplifying $\\sqrt{\\dfrac{4}{3}}$ yields $\\dfrac{2}{\\sqrt{3}}=\\dfrac{2\\sqrt{3}}{3}$.\n\n\n\n\nSimplifying $\\sqrt{\\dfrac{3}{4}}$ yields $\\dfrac{\\sqrt{3}}{2}$.\n\n\n\n\n$\\dfrac{2\\sqrt{3}}{3}-\\dfrac{\\sqrt{3}}{2}=\\dfrac{4\\sqrt{3}}{6}-\\dfrac{3\\sqrt{3}}{6}=\\dfrac{\\sqrt{3}}{6}$.\n\n\n\n\nSince we cannot simplify further, the correct answer is $\\boxed{\\textbf{(A)}\\ \\frac{\\sqrt{3}}{6}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/2.json
|
AHSME
|
1962_AHSME_Problems
| 28 | 0 |
Algebra
|
Multiple Choice
|
The set of $x$-values satisfying the equation $x^{\log_{10} x} = \frac{x^3}{100}$ consists of:
$\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \text{10, only}\qquad\textbf{(C)}\ \text{100, only}\qquad\textbf{(D)}\ \text{10 or 100, only}\qquad\textbf{(E)}\ \text{more than two real numbers.}$
|
[
"Taking the base-$x$ logarithm of both sides gives $\\log_{10}x=\\log_x\\frac{x^3}{100}$.\nThis simplifies to \n\\[\\log_{10}x=\\log_x{x^3} - \\log_x{100}\\]\n\\[\\log_{10}x+\\log_x{100}=3\\]\n\\[\\log_{10}x+2 \\log_x{10}=3\\]\nAt this point, we substitute $u=\\log_{10}x$.\n\\[u+\\frac2u=3\\]\n\\[u^2+2=3u\\]\n\\[u^2-3u+2=0\\]\n\\[(u-2)(u-1)=0\\]\n\\[u\\in\\{1, 2\\}\\]\n\\[x\\in\\{10, 100\\}\\]\nThe answer is $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/28.json
|
AHSME
|
1962_AHSME_Problems
| 12 | 0 |
Algebra
|
Multiple Choice
|
When $\left ( 1 - \frac{1}{a} \right ) ^6$ is expanded the sum of the last three coefficients is:
$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ -10\qquad\textbf{(E)}\ -11$
|
[
"This is equivalent to $\\frac{(a-1)^6}{a^6}.$\nIts expansion has 7 terms, whose coefficients are the same as those of $(a-1)^6$.\nBy the Binomial Theorem, the sum of the last three coefficients is \n$\\binom{6}{2}-\\binom{6}{1}+\\binom{6}{0}=15-6+1=\\boxed{10 \\textbf{ (C)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/12.json
|
AHSME
|
1962_AHSME_Problems
| 32 | 0 |
Algebra
|
Multiple Choice
|
If $x_{k+1} = x_k + \frac12$ for $k=1, 2, \dots, n-1$ and $x_1=1$, find $x_1 + x_2 + \dots + x_n$.
$\textbf{(A)}\ \frac{n+1}{2}\qquad\textbf{(B)}\ \frac{n+3}{2}\qquad\textbf{(C)}\ \frac{n^2-1}{2}\qquad\textbf{(D)}\ \frac{n^2+n}{4}\qquad\textbf{(E)}\ \frac{n^2+3n}{4}$
|
[
"The sequence $x_1, x_2, \\dots, x_n$ is an arithmetic sequence since every term is $\\frac12$ more than the previous term. Letting $a=1$ and $r=\\frac12$, we can rewrite the sequence as $a, a+r, \\dots, a+(n-1)r$.\nRecall that the sum of the first $n$ terms of an arithmetic sequence is $na+\\binom{n}2r$.\nSubstituting our values for $a$ and $r$, we get $n+\\frac{\\binom{n}2}{2}$. Simplifying gives\n$\\boxed{\\frac{n^2+3n}{4}\\textbf{ (E)}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/32.json
|
AHSME
|
1962_AHSME_Problems
| 24 | 0 |
Algebra
|
Multiple Choice
|
Three machines $\text{P, Q, and R,}$ working together, can do a job in $x$ hours. When working alone, $\text{P}$ needs an additional $6$ hours to do the job; $\text{Q}$, one additional hour; and $R$, $x$ additional hours. The value of $x$ is:
$\textbf{(A)}\ \frac{2}3\qquad\textbf{(B)}\ \frac{11}{12}\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$
|
[
"Machine P takes $x+6$ hours, machine Q takes $x+1$ hours, and machine R takes $2x$ hours.\nWe also know that all three working together take $x$ hours. \nNow the time it takes for all three machines to complete the job is the harmonic mean of the times the three machines take individually; that is,\n\\[x=\\frac1{\\frac1{x+6}+\\frac1{x+1}+\\frac1{2x}}\\]\n\\[\\frac{x}{x+6}+\\frac{x}{x+1}+\\frac{x}{2x}=1\\]\n\\[\\frac{x}{x+6}+\\frac{x}{x+1}=\\frac12\\]\n\\[2x(x+1)+2x(x+6)=(x+1)(x+6)\\]\n\\[2x^2+2x+2x^2+12x=x^2+7x+6\\]\n\\[3x^2+7x-6=0\\]\n\\[(3x-2)(x+3)=0\\]\n\\[x\\in\\{\\frac23, -3\\}\\]\nObviously, the number of hours is positive, so the answer is\n$\\boxed{\\frac23 \\textbf{ (A)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/24.json
|
AHSME
|
1962_AHSME_Problems
| 25 | 0 |
Geometry
|
Multiple Choice
|
Given square $ABCD$ with side $8$ feet. A circle is drawn through vertices $A$ and $D$ and tangent to side $BC$. The radius of the circle, in feet, is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4\sqrt{2}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5\sqrt{2}\qquad\textbf{(E)}\ 6$
|
[
"Let $O$ be the center of the circle and $E$ be the point of tangency of the circle and $BC$ and let $F$ be the point of intersection of lines $OE$ and $AD$ Because of the symmetry, $BE=EC=AF=FD=4$ feet. Let the length of $OF$ be $x$. The length of $OE$ is $EF-OF=-x+8$. By Pythagorean Theorem, $OA=OD=\\sqrt{x^2+4^2}=\\sqrt{x^2+16}$. Because $OA$, $OD$, and $OE$ are radii of the same circle, $-x+8=OE=OA=AD=\\sqrt{x^2+16}$. So, $\\sqrt{x^2+16}=-x+8$. Squaring both sides, we obtain $x^2+16=x^2-16x+64$. Subtracting $x^2+16$ from both sides and adding $16x$, our equation becomes $16x=80$, so our answer is $x=\\boxed{C)5}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/25.json
|
AHSME
|
1962_AHSME_Problems
| 33 | 0 |
Algebra
|
Multiple Choice
|
The set of $x$-values satisfying the inequality $2 \leq |x-1| \leq 5$ is:
$\textbf{(A)}\ -4\leq x\leq-1\text{ or }3\leq x\leq 6\qquad$
$\textbf{(B)}\ 3\leq x\leq 6\text{ or }-6\leq x\leq-3\qquad\textbf{(C)}\ x\leq-1\text{ or }x\geq 3\qquad$
$\textbf{(D)}\ -1\leq x\leq 3\qquad\textbf{(E)}\ -4\leq x\leq 6$
|
[
"This inequality can be split into two inequalities:\n$2\\le x-1\\le5$ or $2\\le1-x\\le5$. Solving for x gives\n$3\\le x\\le6$ or $-4\\le x\\le-1$. $\\boxed{\\textbf{(A)}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/33.json
|
AHSME
|
1962_AHSME_Problems
| 13 | 0 |
Algebra
|
Multiple Choice
|
$R$ varies directly as $S$ and inversely as $T$. When $R = \frac{4}{3}$ and $T = \frac {9}{14}$, $S = \frac37$. Find $S$ when $R = \sqrt {48}$ and $T = \sqrt {75}$.
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60$
|
[
"\\[R=c\\cdot\\frac{S}T\\]\n\n\nfor some constant $c$. \n\n\nYou know that\n\n\n\\[\\frac43=c\\cdot\\frac{3/7}{9/14}=c\\cdot\\frac37\\cdot\\frac{14}9=c\\cdot\\frac23\\,,\\]\n\n\nso\n\n\n\\[c=\\frac{4/3}{2/3}=2\\,.\\]\n\n\nWhen $R=\\sqrt{48}$ and $T=\\sqrt{75}$ we have\n\n\n\\[\\sqrt{48}=\\frac{2S}{\\sqrt{75}}\\,,\\]\n\n\nso\n\n\n\\[S=\\frac12\\sqrt{48\\cdot75}=30\\,.\\] $\\boxed{B}$\n\n\n\\begin{verbatim}\n-- zixuan 12\n\\end{verbatim}\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/13.json
|
AHSME
|
1962_AHSME_Problems
| 29 | 0 |
Algebra
|
Multiple Choice
|
Which of the following sets of $x$-values satisfy the inequality $2x^2 + x < 6$?
$\textbf{(A)}\ -2 < x <\frac{3}{2}\qquad\textbf{(B)}\ x >\frac{3}2\text{ or }x <-2\qquad\textbf{(C)}\ x <\frac{3}2\qquad$
$\textbf{(D)}\ \frac{3}2 < x < 2\qquad\textbf{(E)}\ x <-2$
|
[
"First, subtract 6 from both sides of the inequality, \n$2x^2 + x - 6 < 0$.\nThis is a parabola that opens upward when graphed, it has a positive leading coefficient. So any negative x values must be between its x-axis intersections, namely $x = -2, 1.5$. The answer is A.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/29.json
|
AHSME
|
1962_AHSME_Problems
| 3 | 0 |
Algebra
|
Multiple Choice
|
The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$, in the order shown. The value of $x$ is:
$\textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined}$
|
[
"Let $y$ represent the common difference between the terms. We have $(x+1)-y=(x-1)\\implies y=2$.\n\n\nSubstituting gives us $(2x+3)-2=(x+1)\\implies 2x+1=x+1\\implies x=0$.\n\n\nTherefore, our answer is $\\boxed{\\textbf{(B)}\\ 0}$;\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/3.json
|
AHSME
|
1962_AHSME_Problems
| 34 | 0 |
Algebra
|
Multiple Choice
|
For what real values of $K$ does $x = K^2 (x-1)(x-2)$ have real roots?
$\textbf{(A)}\ \text{none}\qquad\textbf{(B)}\ -2<K<1\qquad\textbf{(C)}\ -2\sqrt{2}< K < 2\sqrt{2}\qquad$
$\textbf{(D)}\ K>1\text{ or }K<-2\qquad\textbf{(E)}\ \text{all}$
|
[
"Factoring, we get\n\\[x = K^2x^2 - 3K^2x + 2K^2\\]\n\\[K^2x^2 - (3K^2+1)x + 2K^2 = 0\\]\nAt this point, we could use the quadratic formula, but we're only interested in whether the roots are real, which occurs if and only if the discriminant ($b^2-4ac$) is non-negative.\n\\[(-3K^2-1)^2-8K^4\\ge0\\]\n\\[9K^4+6K^2+1-8K^4\\ge0\\]\n\\[K^4+6K^2+1\\ge0\\]\nThis function of $K$ is symmetric with respect to the y-axis. Plugging in $0$ for $K$ gives us $1$, and it is clear that the function is strictly increasing over $(0,\\infty)$, and so it is positive for all real values of $K$. $\\boxed{\\textbf{(E)}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/34.json
|
AHSME
|
1962_AHSME_Problems
| 8 | 0 |
Arithmetic
|
Multiple Choice
|
Given the set of $n$ numbers; $n > 1$, of which one is $1 - \frac {1}{n}$ and all the others are $1$. The arithmetic mean of the $n$ numbers is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ n-\frac{1}{n}\qquad\textbf{(C)}\ n-\frac{1}{n^2}\qquad\textbf{(D)}\ 1-\frac{1}{n^2}\qquad\textbf{(E)}\ 1-\frac{1}{n}-\frac{1}{n^2}$
|
[
"Just take $\\frac{1(n-1)+(1-\\frac{1}{n})}{n}$. You get $\\frac{n-1+1-\\frac{1}{n}}{n}$, which is just $\\frac{n-\\frac{1}{n}}{n}$, which is just $\\boxed{D}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/8.json
|
AHSME
|
1962_AHSME_Problems
| 22 | 0 |
Number Theory
|
Multiple Choice
|
The number $121_b$, written in the integral base $b$, is the square of an integer, for
$\textbf{(A)}\ b = 10,\text{ only}\qquad\textbf{(B)}\ b = 10\text{ and }b = 5,\text{ only}\qquad$
$\textbf{(C)}\ 2\leq b\leq 10\qquad\textbf{(D)}\ b > 2\qquad\textbf{(E)}\ \text{no value of }b$
|
[
"$121_b$ can be represented in base 10 as $b^2+2b+1$, which factors as $(b+1)^2$.\nNote that $b>2$ because 2 is a digit in the base-b representation, but for any \n$b>2$, $121_b$ is the square of $b+1$. $\\boxed{\\textbf{(D)}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/22.json
|
AHSME
|
1962_AHSME_Problems
| 18 | 0 |
Geometry
|
Multiple Choice
|
A regular dodecagon ($12$ sides) is inscribed in a circle with radius $r$ inches. The area of the dodecagon, in square inches, is:
$\textbf{(A)}\ 3r^2\qquad\textbf{(B)}\ 2r^2\qquad\textbf{(C)}\ \frac{3r^2\sqrt{3}}{4}\qquad\textbf{(D)}\ r^2\sqrt{3}\qquad\textbf{(E)}\ 3r^2\sqrt{3}$
|
[
"The formula for the area of a regular dodecagon is $3r^2$. The answer is $\\boxed{\\textbf{(A)}}$.\n(If you don't know this formula, it's pretty easy to figure out that the area of a square inscribed in a circle is $2r^2$, and all the choices except $3r^2$ are less than $2r^2$. Remember, the more sides a regular polygon has, the closer its area gets to $\\pi r^2$.)\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/18.json
|
AHSME
|
1962_AHSME_Problems
| 38 | 0 |
Number Theory
|
Multiple Choice
|
The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$, the population was one more than a perfect square. Now, with an additional increase of $100$, the population is again a perfect square.
The original population is a multiple of:
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 17$
|
[
"Let $a^2$ $=$ original population count, $b^2+1$ $=$ the second population count, and $c^2$ $=$ the third population count \nWe first see that $a^2 + 100 = b^2 + 1$ or $99$ $=$ $b^2-a^2$.\nWe then factor the right side getting $99$ $=$ $(b-a)(b+a)$. \nSince we can only have an nonnegative integral population, clearly $b+a$ $>$ $b-a$ and both factor $99$.\nWe factor $99$ into $3^2 \\cdot 11$ $=$ $(b-a)(b+a)$\nThere are a few cases to look at:\n$1)$ $b+a$ $=$ $11$ and $b-a$ $=$ $9$.\nAdding the two equations we get $2b$ $=$ $20$ or $b$ $=$ $10$, which means $a$ $=$ $1$.\nBut looking at the restriction that the second population + $100$ $=$ third population...\n$10^2$ $+$ $1$ $+$ $100$ $=$ $201$ $\\neq$ a perfect square.\n\n\n$2)$ $b+a$ $=$ $33$ and $b-a$ $=$ $3$.\nAdding the two equations we get $2b$ $=$ $36$ or $b$ $=$ $18$, which means $a$ $=$ $15$.\nLooking at the same restriction, we get $18^2$ + $1$ + $100$ $=$ $324$ + $101$ $=$ $425$, which is NOT a perfect square.\n\n\nFinally, $b+a$ $=$ $99$ and $b-a$ $=$ $1$.\n$2b$ $=$ $100$ or $b$ $=$ $50$, which means $a$ $=$ $49$.\nLooking at the same restriction, we get $50^2$ + $1$ + $100$ $=$ $2500$ + $101$ $=$ $2601$ $=$ $51^2$. Thus we find that the original population is $a^2$ $=$ $49^2$ $=$ $7^4$. Or $a^2$ is a multiple of $\\boxed{ (B) 7}$\n\n\n",
"Let $P$ $=$ original population. Translating the word problem into a system of equations, we got:\n\\begin{align} P &= x^2 \\\\ P + 100 &= y^2 + 1 \\\\ P + 200 &= z^2 \\end{align}\nfor some positive integers $x$, $y$ and $z$.\nNow, by subtracting $(2)$ from $(3)$ (i.e. $(3) - (2)$), we got:\n\\begin{align*} 100 &= z^2 - y^2 - 1 \\\\ 101 &= z^2 - y^2 \\\\ 101 &= (z - y)(z + y) \\end{align*}\nSince y and z are both positive integers and 101 is a prime, by factoring, the only working solution for us is $y = 50$ and $z = 51$.\nPlugging that back to $(2)$ and simplify, we got $P = 2401 = (49)^2 = x^2$, a multiple of $7$.\nTherefore, the answer is $\\boxed{(B) 7}$. -nullptr07\n\n\n"
] | 2 |
./CreativeMath/AHSME/1962_AHSME_Problems/38.json
|
AHSME
|
1962_AHSME_Problems
| 4 | 0 |
Algebra
|
Multiple Choice
|
If $8^x = 32$, then $x$ equals:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ \frac{5}{3}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{1}{4}$
|
[
"Recognizing that $8=2^3$, we know that $2^{3x}=32$. Since $2^5=32$, we have $2^{3x}=2^5$. Therefore, $x=\\dfrac{5}{3}$.\n\n\nSo our answer is $\\boxed{\\textbf{(B)}\\ \\frac{5}{3}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/4.json
|
AHSME
|
1962_AHSME_Problems
| 14 | 0 |
Algebra
|
Multiple Choice
|
Let $s$ be the limiting sum of the geometric series $4- \frac83 + \frac{16}{9} - \dots$, as the number of terms increases without bound. Then $s$ equals:
$\textbf{(A)}\ \text{a number between 0 and 1}\qquad\textbf{(B)}\ 2.4\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.6\qquad\textbf{(E)}\ 12$
|
[
"The infinite sum of a geometric series with first term $a$ and common ratio $r$ ($-1<r<1$) is $\\frac{a}{1-r}$.\nNow, in this geometric series, $a=4$, and $r=-\\frac23$. Plugging these into the formula, we get \n$\\frac4{1-(-\\frac23)}$, which simplifies to $\\frac{12}5$, or $\\boxed{2.4\\textbf{ (B)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/14.json
|
AHSME
|
1962_AHSME_Problems
| 15 | 0 |
Geometry
|
Multiple Choice
|
Given triangle $ABC$ with base $AB$ fixed in length and position. As the vertex $C$ moves on a straight line, the intersection point of the three medians moves on:
$\textbf{(A)}\ \text{a circle}\qquad\textbf{(B)}\ \text{a parabola}\qquad\textbf{(C)}\ \text{an ellipse}\qquad\textbf{(D)}\ \text{a straight line}\qquad\textbf{(E)}\ \text{a curve here not listed}$
|
[
"Let $CM$ be the median through vertex $C$, and let $G$ be the point of intersection of the triangle's medians. \n\n\nLet $CH$ be the altitude of the triangle through vertex $C$ and $GP$ be the distance from $G$ to $AB$, with the point $P$ laying on $AB$. \n\n\nUsing Thales' intercept theorem, we derive the proportion:\n\n\n\n\n$\\frac{GP}{CH} = \\frac{GM}{CM}$\n\n\n\n\nThe fraction $\\frac{GM}{CM}$ in any triangle is equal to $\\frac{1}{3}$ . Therefore $GP = \\frac{CH}{3}$ . \n\n\n\n\nSince the problem states that the vertex $C$ is moving on a straight line, the length of $CH$ is a constant value. That means that the length of $GP$ is also a constant. Therefore the point $G$ is moving on a straight line.\n\n\nAnswer: D\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/15.json
|
AHSME
|
1962_AHSME_Problems
| 5 | 0 |
Geometry
|
Multiple Choice
|
If the radius of a circle is increased by $1$ unit, the ratio of the new circumference to the new diameter is:
$\textbf{(A)}\ \pi+2\qquad\textbf{(B)}\ \frac{2\pi+1}{2}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ \frac{2\pi-1}{2}\qquad\textbf{(E)}\ \pi-2$
|
[
"The ratio of a circumference to a diameter always is the same so the answer is obviously C.\n\n\n",
"Let us say that the radius of a circle is $r$. When the radius is increased by $1$, the new radius is $r+1$ so the diameter is $2r+2$. We know that the circumference of a circle is $2\\pi r$ so $2 \\cdot \\pi \\cdot (r+1) = \\pi \\cdot (2r+2)$. Finally, the problem asked for the ratio of the new circumference to the new diameter is $\\frac{\\pi \\cdot (2r+2)}{2r+2}=\\boxed{\\pi}$.\n\n\n~Mathfun1000\n\n\n"
] | 2 |
./CreativeMath/AHSME/1962_AHSME_Problems/5.json
|
AHSME
|
1962_AHSME_Problems
| 39 | 0 |
Geometry
|
Multiple Choice
|
Two medians of a triangle with unequal sides are $3$ inches and $6$ inches. Its area is $3 \sqrt{15}$ square inches. The length of the third median in inches, is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{6}\qquad\textbf{(D)}\ 6\sqrt{3}\qquad\textbf{(E)}\ 6\sqrt{6}$
|
[
"By the area formula:\n\\[A = \\frac43\\sqrt{s(s-m_1)(s-m_2)(s-m_3)}\\]\nWhere $s = \\frac{m_1+m_2+m_3}{2}$.\nPlugging in the numbers:\n\\[3\\sqrt{15} = \\frac43\\sqrt{\\frac{9+m}2\\cdot\\frac{9-m}2\\cdot\\frac{m+3}2\\cdot\\frac{m-3}2}\\]\nSimplifying and squaring both sides:\n\\[1215 = (9-m)(9+m)(m+3)(m-3)\\]\nNow, we can just plug in the answer choices and find that $\\boxed{3\\sqrt6}$ works.\n\n\n",
"We connect all the medians of the triangle. We know that the ratio of the vertice to orthocenter / median is in a $2:3$ ratio. For convenience, call the orthocenter $O$, and label the triangle $\\triangle ABC$ such that the median from $A$ to $BC$ is 3 and the median from $B$ to $AC$ is 6. Then, $BO$ is 4 and $AO$ is 2. Let us call the portion of the third median that goes from $O$ to $AB$ have length $x$, and and $OC$ have length $2x$. Note that the median from $O$ to $AB$ in $\\triangle AOB$ is equal to $x$. \n\n\nNote that the medians split the triangle into 6 triangles of equal area, so $\\triangle AOB$ has area equal to $\\frac{1}{3}$ of $\\triangle ABC = \\sqrt{15}$. Let $AB=2s$. Using Herons*, we get:\n\n\n\\[15=(3+s)(s-1)(s+1)(3-2)\\]\n\\[=(9-s^2)(s^2-1)\\]\n\n\nWe can see that $s^2=4$, meaning that $s$ is 2 and $AB=4$. We can then use Steward's* to find the length of the median from $O$, since we know the median cuts $AB$ into segments each of length $2$. We get:\n\n\n\\[(2^2)(4)+4x^2=(4^2)(2)+(2^2)(2)\\]\n\\[16+4x^2=32+8\\]\n\\[4x^2=24\\]\n\\[x^2=6\\]\n\\[x=\\sqrt{6}\\]\n\n\nSince the length of the actual median from $C$ is equal to $3$, we have that the answer is $\\boxed{3\\sqrt{6}}$.\n\n\n\\begin{itemize}\n\\item If anyone has a better method of either finding $AB$ or the median of $AOB$ from $O$, please feel free to edit\n\\end{itemize}\n~williamxiao\n\n\n"
] | 2 |
./CreativeMath/AHSME/1962_AHSME_Problems/39.json
|
AHSME
|
1962_AHSME_Problems
| 19 | 0 |
Algebra
|
Multiple Choice
|
If the parabola $y = ax^2 + bx + c$ passes through the points $( - 1, 12)$, $(0, 5)$, and $(2, - 3)$, the value of $a + b + c$ is:
$\textbf{(A)}\ -4\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$
|
[
"Substituting in the $(x, y)$ pairs gives the following system of equations:\n\\[a-b+c=12\\]\n\\[c=5\\]\n\\[4a+2b+c=-3\\]\nWe know $c=5$, so plugging this in reduces the system to two variables:\n\\[a-b=7\\]\n\\[4a+2b=-8\\]\nDividing the second equation by 2 gives $2a+b=-4$, which can be added to the first equation to get $3a=3$, or $a=1$. So the solution set is $(1, -6, 5)$, and the sum is $\\boxed{0\\textbf{ (C)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/19.json
|
AHSME
|
1962_AHSME_Problems
| 23 | 0 |
Geometry
|
Multiple Choice
|
In triangle $ABC$, $CD$ is the altitude to $AB$ and $AE$ is the altitude to $BC$. If the lengths of $AB$, $CD$, and $AE$ are known, the length of $DB$ is:
$\textbf{(A)}\ \text{not determined by the information given} \qquad$
$\textbf{(B)}\ \text{determined only if A is an acute angle} \qquad$
$\textbf{(C)}\ \text{determined only if B is an acute angle} \qquad$
$\textbf{(D)}\ \text{determined only if ABC is an acute triangle} \qquad$
$\textbf{(E)}\ \text{none of these is correct}$
|
[
"We can actually determine the length of $DB$ no matter what type of angles $A$ and $B$ are. This can be easily proved through considering all possible cases, though for the purposes of this solution, we'll show that we can determine $DB$ if $A$ is an obtuse angle. \n\n\nLet's see what happens when $A$ is an obtuse angle. $\\triangle AEB\\sim \\triangle CDB$ by SSS, so $\\frac{AE}{CD}=\\frac{AB}{DB}$. Hence $DB=\\frac{AE}{CD\\times AB}$. Since we've determined the length of $DB$ even though we have an obtuse angle, $DB$ is not $\\bf{only}$ determined by what type of angle $A$ may be. Hence our answer is $\\fbox{E}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/23.json
|
AHSME
|
1962_AHSME_Problems
| 9 | 0 |
Algebra
|
Multiple Choice
|
When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:
$\textbf{(A)}\ \text{more than 5}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2$
|
[
"Obviously, we can factor out an $x$ first to get $x(x^8-1)$.\nNext, we repeatedly factor differences of squares:\n\\[x(x^4+1)(x^4-1)\\]\n\\[x(x^4+1)(x^2+1)(x^2-1)\\]\n\\[x(x^4+1)(x^2+1)(x+1)(x-1)\\]\nNone of these 5 factors can be factored further, so the answer is \n$\\boxed{\\textbf{(B) } 5}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/9.json
|
AHSME
|
1962_AHSME_Problems
| 35 | 0 |
Algebra
|
Multiple Choice
|
A man on his way to dinner short after $6: 00$ p.m. observes that the hands of his watch form an angle of $110^{\circ}$. Returning before $7: 00$ p.m. he notices that again the hands of his watch form an angle of $110^{\circ}$. The number of minutes that he has been away is:
$\textbf{(A)}\ 36\frac{2}3\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 42.4\qquad\textbf{(E)}\ 45$
|
[
"Let $n$ be the number of minutes after 6:00. Let $h(n)=180+\\frac{n}2$ be the angle, in degrees, of the hour hand (with $0^{\\circ}$ at the top and increasing in the clockwise direction); similarly, let $m(n)=6n$ be the angle of the minute hand. We want $|h(n)-m(n)|=110$. This is equivalent to \n\\[180-\\frac{11n}2=\\pm110\\]\n\\[-\\frac{11n}2\\in\\{-70,-290\\}\\]\n\\[\\frac{11n}2\\in\\{70,290\\}\\]\n\\[11n\\in\\{140,580\\}\\]\n\\[n\\in\\{\\frac{140}{11},\\frac{580}{11}\\}\\]\nThe difference between the two values of $n$ is $\\frac{440}{11}=\\boxed{40\\textbf{ (B)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1962_AHSME_Problems/35.json
|
AHSME
|
1988_AHSME_Problems
| 20 | 0 |
Geometry
|
Multiple Choice
|
In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints
of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown
in the second figure. The ratio of height to base, $XY / YZ$, in this rectangle is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,1), B=(0,-1), C=(2,-1), D=(2,1), E=(1,-1), F=(1,1), G=(.8,.6); pair X=(4,sqrt(5)), Y=(4,-sqrt(5)), Z=(4+2/sqrt(5),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5)); draw(A--B--C--D--A^^B--F^^E--D^^A--G^^rightanglemark(A,G,F)); draw(X--Y--Z--W--X^^T--V--X^^Y--U); label("A", A, NW); label("B", B, SW); label("C", C, SE); label("D", D, NE); label("E", E, S); label("F", F, N); label("G", G, E); label("X", X, NW); label("Y", Y, SW); label("Z", Z, SE); label("W", W, NE); [/asy]
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 1+2\sqrt{3}\qquad \textbf{(C)}\ 2\sqrt{5}\qquad \textbf{(D)}\ \frac{8+4\sqrt{3}}{3}\qquad \textbf{(E)}\ 5$
|
[
"Within $WXYZ$, the parallelogram piece has vertical side $BF = \\sqrt{1^2 + 2^2} = \\sqrt{5}$, and diagonal side $FD = 1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $\\triangle AFG$, so we deduce $YZ = AG = \\frac{\\frac{1}{2} \\times 1 \\times 2}{\\frac{1}{2} \\times \\sqrt{5}} = \\frac{2}{\\sqrt{5}}.$ Now the rectangle must have the same area as the square, as the pieces were put together without gaps or overlaps, so its area is $2^2 = 4$, and thus the vertical side $WZ$ is $\\frac{4}{\\frac{2}{\\sqrt{5}}} = 2\\sqrt{5}$, so the required ratio is $\\frac{2\\sqrt{5}}{\\frac{2}{\\sqrt{5}}} = \\sqrt{5} \\times \\sqrt{5} = 5$, which is $\\boxed{\\text{E}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/20.json
|
AHSME
|
1988_AHSME_Problems
| 16 | 0 |
Geometry
|
Multiple Choice
|
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, B=(1,-(1/sqrt(3))), C=(-1,-(1/sqrt(3))), A=(0,(2/sqrt(3))), E=(2,-(2/sqrt(3))), F=(-2,-(2/sqrt(3))), D=(0,(4/sqrt(3))); draw(A--B--C--A^^D--E--F--D); label("A'", A, N); label("B'", B, SE); label("C'", C, SW); label("A", D, E); label("B", E, E); label("C", F, W); [/asy]
$ABC$ and $A'B'C'$ are equilateral triangles with parallel sides and the same center,
as in the figure. The distance between side $BC$ and side $B'C'$ is $\frac{1}{6}$ the altitude of $\triangle ABC$.
The ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ is
$\textbf{(A)}\ \frac{1}{36}\qquad \textbf{(B)}\ \frac{1}{6}\qquad \textbf{(C)}\ \frac{1}{4}\qquad \textbf{(D)}\ \frac{\sqrt{3}}{4}\qquad \textbf{(E)}\ \frac{9+8\sqrt{3}}{36}$
|
[
"Let $\\triangle ABC$ have side length $s$ and $\\triangle A'B'C'$ have side length $t$. Thus the altitude of $\\triangle ABC$ is $\\frac{s\\sqrt{3}}{2}$. Now observe that this altitude is made up of three parts: the distance from $BC$ to $B'C'$, plus the altitude of $\\triangle A'B'C'$, plus a top part which is equal to the length of the diagonal line from the bottom-left corner of $\\triangle ABC$ to the bottom left corner of $\\triangle A'B'C'$ (as an isosceles trapezium is formed with parallel sides $AB$ and $A'B'$, and legs $AA'$ and $BB'$). We drop a perpendicular from $B'$ to $BC$, which meets $BC$ at $D$. $\\triangle BDB'$ has angles $30^{\\circ}$, $60^{\\circ}$, and $90^{\\circ}$, and the vertical side is that distance from $BC$ to $B'C'$, which is given as $\\frac{1}{6} \\times \\frac{s\\sqrt{3}}{2} = \\frac{s\\sqrt{3}}{12}$, so that by simple trigonometry, the length of the diagonal line is $\\frac{s\\sqrt{3}}{12} \\times \\csc{30^{\\circ}} = \\frac{s\\sqrt{3}}{6}.$ Thus using the \"altitude in three parts\" idea, we get $\\frac{s\\sqrt{3}}{2} = \\frac{s\\sqrt{3}}{12} + \\frac{t\\sqrt{3}}{2} + \\frac{s\\sqrt{3}}{6} \\implies \\frac{s\\sqrt{3}}{4} = \\frac{t\\sqrt{3}}{2} \\implies t = \\frac{1}{2}s.$ Thus the sides of $\\triangle A'B'C'$ are half as long as $\\triangle ABC$, so the area ratio is $(\\frac{1}{2}) ^ {2} = \\frac{1}{4}$, which is $\\boxed{\\text{C}}$.\n\n\n\n\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/16.json
|
AHSME
|
1988_AHSME_Problems
| 6 | 0 |
Geometry
|
Multiple Choice
|
A figure is an equiangular parallelogram if and only if it is a
$\textbf{(A)}\ \text{rectangle}\qquad \textbf{(B)}\ \text{regular polygon}\qquad \textbf{(C)}\ \text{rhombus}\qquad \textbf{(D)}\ \text{square}\qquad \textbf{(E)}\ \text{trapezoid}$
|
[
"The definition of an equiangular parallelogram is that all the angles are equal, and that pairs of sides are parallel. It may be a rectangle, because all the angles are equal and it is a parallelogram. It is not necessarily a regular polygon, because if the polygon is a pentagon, it is not a parallelogram. It is not necessarily a rhombus, because all the angles are not necessarily equal. It may be a square, since it is a parallelogram and all the angles are equal. It is not necessarily a trapezoid, because the angles are not necessarily equal. \nWe have that it could be a square or a rectangle. A square is a rectangle, but a rectangle is not necessarily a square. We want the all-encompassing answer so the answer is a rectange $\\implies \\boxed{\\text{A}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/6.json
|
AHSME
|
1988_AHSME_Problems
| 7 | 0 |
Arithmetic
|
Multiple Choice
|
Estimate the time it takes to send $60$ blocks of data over a communications channel if each block consists of $512$
"chunks" and the channel can transmit $120$ chunks per second.
$\textbf{(A)}\ 0.04 \text{ seconds}\qquad \textbf{(B)}\ 0.4 \text{ seconds}\qquad \textbf{(C)}\ 4 \text{ seconds}\qquad \textbf{(D)}\ 4\text{ minutes}\qquad \textbf{(E)}\ 4\text{ hours}$
|
[
"We want to figure out the number of chunks in $60$ blocks, so we have $60\\cdot 512 \\approx 30000$. We divide this by $120$ to determine the number of seconds necessary to transmit. $30000/120 \\approx 250$, which means that it takes approximately $4$ minutes to transmit. Thus, the answer is $\\boxed{\\text{D}}$.\n\n\n",
"This solution is if you are running out of time and just want to write down an answer. So, this is quite unreliable. You can logic it out. It doesn't make sense for the first three options to be the answer since that is way too quick. The last option is way too long. That just leaves $\\boxed{\\text{D}}$.\n\n\n"
] | 2 |
./CreativeMath/AHSME/1988_AHSME_Problems/7.json
|
AHSME
|
1988_AHSME_Problems
| 17 | 0 |
Algebra
|
Multiple Choice
|
If $|x| + x + y = 10$ and $x + |y| - y = 12$, find $x + y$
$\textbf{(A)}\ -2\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ \frac{18}{5}\qquad \textbf{(D)}\ \frac{22}{3}\qquad \textbf{(E)}\ 22$
|
[
"We proceed by casework:\n\n\nCase 1: x, y >= 0\n\n\nThus we have 2x + y = 10, x = 12. This makes y < 0, a contradiction.\n\n\nCase 2: x >= 0 > y\n\n\n2x + y = 10, x - 2y = 12. By multiplying the first equation by two and adding, we have 4x + x = 2*10 + 12 thus x = 32/5. Since the question only specifies 1 solution, plugging in x = 32/5 into the first equation yields y = -14/5 thus x + y = 18/5 $C$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/17.json
|
AHSME
|
1988_AHSME_Problems
| 21 | 0 |
Algebra
|
Multiple Choice
|
The complex number $z$ satisfies $z + |z| = 2 + 8i$. What is $|z|^{2}$? Note: if $z = a + bi$, then $|z| = \sqrt{a^{2} + b^{2}}$.
$\textbf{(A)}\ 68\qquad \textbf{(B)}\ 100\qquad \textbf{(C)}\ 169\qquad \textbf{(D)}\ 208\qquad \textbf{(E)}\ 289$
|
[
"Let the complex number $z$ equal $a+bi$. Then the preceding equation can be expressed as \\[a+bi+\\sqrt{a^2+b^2} = 2+8i\\] Because $a$ and $b$ must both be real numbers, we immediately have that $bi = 8i$, giving $b = 8$. Plugging this in back to our equation gives us $a+\\sqrt{a^2+64} = 2$. \nRearranging this into $2-a = \\sqrt{a^2+64}$, we can square each side of the equation resulting in \\[4-4a+a^2 = a^2+64\\] Further simplification will yield \n$60 = -4a$ meaning that $-15 = a$. Knowing both $a$ and $b$, we can plug them in into $a^2+b^2$. Our final answer is $\\boxed{289}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/21.json
|
AHSME
|
1988_AHSME_Problems
| 10 | 0 |
Arithmetic
|
Multiple Choice
|
In an experiment, a scientific constant $C$ is determined to be $2.43865$ with an error of at most $\pm 0.00312$.
The experimenter wishes to announce a value for $C$ in which every digit is significant.
That is, whatever $C$ is, the announced value must be the correct result when $C$ is rounded to that number of digits.
The most accurate value the experimenter can announce for $C$ is
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2.4\qquad \textbf{(C)}\ 2.43\qquad \textbf{(D)}\ 2.44\qquad \textbf{(E)}\ 2.439$
|
[
"If added together, we have:\n\\[2.43865+0.00312=2.44177\\]\nThis rounds to $2.44$. If they subtracted, we have:\n\\[2.43865-0.00312=2.43553.\\]\nThis rounds to $2.44$. Therefore, we have the answer to be $\\fbox{\\textbf{(D)} 2.44}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/10.json
|
AHSME
|
1988_AHSME_Problems
| 26 | 0 |
Algebra
|
Multiple Choice
|
Suppose that $p$ and $q$ are positive numbers for which \[\operatorname{log}_{9}(p) = \operatorname{log}_{12}(q) = \operatorname{log}_{16}(p+q).\] What is the value of $\frac{q}{p}$?
$\textbf{(A)}\ \frac{4}{3}\qquad \textbf{(B)}\ \frac{1+\sqrt{3}}{2}\qquad \textbf{(C)}\ \frac{8}{5}\qquad \textbf{(D)}\ \frac{1+\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{16}{9}$
|
[
"We can rewrite the equation as $\\frac{\\log{p}}{\\log{9}} = \\frac{\\log{q}}{\\log{12}} = \\frac{\\log{(p + q)}}{\\log{16}}$. Then, the system can be split into 3 pairs: $\\frac{\\log{p}}{\\log{9}} = \\frac{\\log{q}}{\\log{12}}$, $\\frac{\\log{q}}{\\log{12}} = \\frac{\\log{(p + q)}}{\\log{16}}$, and $\\frac{\\log{p}}{\\log{9}} = \\frac{\\log{(p + q)}}{\\log{16}}$. Cross-multiplying in the first two, we obtain: \\[(\\log{12})(\\log{p}) = (2\\log{3})(\\log{q})\\] and \\[(\\log{12})(\\log{(p + q)}) = (2\\log{4})(\\log{q})\\]\nAdding these equations results in: \\[(\\log{12})(\\log{p(p+q)}) = (2\\log{12})(\\log{q})\\] which simplifies to \\[p(p + q) = q^2\\] Dividing by $pq$ on both sides gives: $\\frac{p+q}{q} = \\frac{q}{p} = \\frac{p}{q} + 1$. We set the desired value, $q/p$ to $x$ and substitute it into our equation: $\\frac{1}{x} + 1 = x \\implies x^2 - x - 1 = 0$ which is solved to get our answer: $\\boxed{\\text{(D) } \\frac{1 + \\sqrt{5}}{2}}$. -lucasxia01\n\n\n",
"For some number t:\n\n\n$p = 9^{t}$\n\n\n$q = 12^{t}$\n\n\n$p + q = 16^{t}$\n\n\nNext we can divide $p + q$ by $p$ to obtain\n$\\frac{p+q}{p} = 1 + \\frac{q}{p}$\n\n\nFurthermore, we know that\n\n\n$\\frac{p+q}{p} = (\\frac{16}{9})^{t}$ and $\\frac{q}{p} = (\\frac{4}{3})^{t}$\n\n\nSubstituting into the previous equation, we get $(\\frac{16}{9})^{t} = 1 + (\\frac{4}{3})^{t}$\n\n\nLet $x = (\\frac{4}{3})^{t}$ and we can observe that $x = \\frac{q}{p}$, then similarly to solution 1: $x^2 = 1 + x$, in which we get: $\\boxed{\\text{(D) } \\frac{1 + \\sqrt{5}}{2}}$ - ehmmaq\n\n\n"
] | 2 |
./CreativeMath/AHSME/1988_AHSME_Problems/26.json
|
AHSME
|
1988_AHSME_Problems
| 30 | 0 |
Algebra
|
Multiple Choice
|
Let $f(x) = 4x - x^{2}$. Give $x_{0}$, consider the sequence defined by $x_{n} = f(x_{n-1})$ for all $n \ge 1$.
For how many real numbers $x_{0}$ will the sequence $x_{0}, x_{1}, x_{2}, \ldots$ take on only a finite number of different values?
$\textbf{(A)}\ \text{0}\qquad \textbf{(B)}\ \text{1 or 2}\qquad \textbf{(C)}\ \text{3, 4, 5 or 6}\qquad \textbf{(D)}\ \text{more than 6 but finitely many}\qquad \textbf{(E) }\infty$
|
[
"Note that $x_{0} = 0$ gives the constant sequence $0, 0, ...$, since $f(0) = 4 \\cdot 0 - 0^2 = 0$. Because $f(4)=0, x_{0} = 4$ gives the sequence $4, 0, 0, ...$ with two different values. Similarly, $f(2) = 4$, so $x_{0} = 2$ gives the sequence $2, 4, 0, 0, ...$ with three values. In general, if $x_{0} = a_{n}$ gives the sequence $a_{n}, a_{n-1}, ... , a_{2}, a_{1}, ...$ with $n$ different values, and $f(a_{n+1}) = a_{n}$, then $x_{0} = a_{n+1}$ gives a sequence with $n+1$ different values. (It is easy to see that we could not have $a_{n+1} = a_{i}$ for some $i < n + 1$.) Thus, it follows by induction that there is a sequence with $n$ distinct values for every positive integer $n$, as long as we can verify that there is always a real number $a_{n+1}$ such that $f(a_{n+1}) = a_{n}$. This makes the answer $\\boxed{\\text{E}}$. The verification alluded to above, which completes the proof, follows from the quadratic formula: the solutions to $f(a_{n+1}) = 4a_{n+1} - a_{n+1}^{2} = a_{n}$ are $a_{n+1} = 2 \\pm \\sqrt{4 - a_{n}}$. Hence if $0 \\leq a_{n} \\leq 4$, then $a_{n+1}$ is real, since the part under the square root is non-negative, and in fact $0 \\leq a_{n+1} \\leq 4$, since $4-a_{n}$ will be between $0$ and $4$, so the square root will be between $0$ and $2$, and $2 \\pm$ something between $0$ and $2$ gives something between $0$ and $4$. Finally, since $a_{1} = 0 \\leq 4$, it follows by induction that all terms satisfy $0 \\leq a_{n} \\leq 4$; in particular, they are all real.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/30.json
|
AHSME
|
1988_AHSME_Problems
| 27 | 0 |
Geometry
|
Multiple Choice
|
In the figure, $AB \perp BC, BC \perp CD$, and $BC$ is tangent to the circle with center $O$ and diameter $AD$.
In which one of the following cases is the area of $ABCD$ an integer?
[asy] pair O=origin, A=(-1/sqrt(2),1/sqrt(2)), B=(-1/sqrt(2),-1), C=(1/sqrt(2),-1), D=(1/sqrt(2),-1/sqrt(2)); draw(unitcircle); dot(O); draw(A--B--C--D--A); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$O$",O,dir(45)); [/asy]
$\textbf{(A)}\ AB=3, CD=1\qquad \textbf{(B)}\ AB=5, CD=2\qquad \textbf{(C)}\ AB=7, CD=3\qquad\\ \textbf{(D)}\ AB=9, CD=4\qquad \textbf{(E)}\ AB=11, CD=5$
|
[
"Let $E$ and $F$ be the intersections of lines $AB$ and $BC$ with the circle. One can prove that $BCDE$ is a rectangle, so $BE=CD$.\n\n\nIn order for the area of trapezoid $ABCD$ to be an integer, the expression $\\frac{(AB+CD)BC}2=(AB+CD)BF$ must be an integer, so $BF$ must be rational.\n\n\nBy Power of a Point, $AB\\cdot BE=BF^2\\implies AB\\cdot CD=BF^2$, so $AB\\cdot CD$ must be a perfect square. Among the choices, the only one where $AB\\cdot CD$ is a perfect square is $\\textbf{(D)}\\ AB=9, CD=4$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/27.json
|
AHSME
|
1988_AHSME_Problems
| 1 | 0 |
Algebra
|
Multiple Choice
|
$\sqrt{8}+\sqrt{18}=$
\[\text{(A)}\ \sqrt{20}\qquad\text{(B)}\ 2(\sqrt{2}+\sqrt{3})\qquad\text{(C)}\ 7\qquad\text{(D)}\ 5\sqrt{2}\qquad\text{(E)}\ 2\sqrt{13}\]
|
[
"$\\sqrt{8} = 2 \\sqrt{2}$\n\n\n$\\sqrt{18} = 3 \\sqrt{2}$\n\n\nSo adding the two terms we get $2 \\sqrt{2} + 3 \\sqrt{2} = 5 \\sqrt{2}$ , which corresponds to answer choice $\\boxed{\\textbf{(D)}}$ .\n\n\n\n\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/1.json
|
AHSME
|
1988_AHSME_Problems
| 11 | 0 |
Arithmetic
|
Multiple Choice
|
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(5,0), B=(7,0), C=(10,0), D=(13,0), E=(16,0); pair F=(4,3), G=(5,3), H=(7,3), I=(10,3), J=(12,3); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(I); dot(J); draw((0,0)--(18,0)^^(0,3)--(18,3)); draw((0,0)--(0,.5)^^(5,0)--(5,.5)^^(10,0)--(10,.5)^^(15,0)--(15,.5)); draw((0,3)--(0,2.5)^^(5,3)--(5,2.5)^^(10,3)--(10,2.5)^^(15,3)--(15,2.5)); draw((1,0)--(1,.2)^^(2,0)--(2,.2)^^(3,0)--(3,.2)^^(4,0)--(4,.2)^^(6,0)--(6,.2)^^(7,0)--(7,.2)^^(8,0)--(8,.2)^^(9,0)--(9,.2)^^(10,0)--(10,.2)^^(11,0)--(11,.2)^^(12,0)--(12,.2)^^(13,0)--(13,.2)^^(14,0)--(14,.2)^^(16,0)--(16,.2)^^(17,0)--(17,.2)^^(18,0)--(18,.2)); draw((1,3)--(1,2.8)^^(2,3)--(2,2.8)^^(3,3)--(3,2.8)^^(4,3)--(4,2.8)^^(6,3)--(6,2.8)^^(7,3)--(7,2.8)^^(8,3)--(8,2.8)^^(9,3)--(9,2.8)^^(10,3)--(10,2.8)^^(11,3)--(11,2.8)^^(12,3)--(12,2.8)^^(13,3)--(13,2.8)^^(14,3)--(14,2.8)^^(16,3)--(16,2.8)^^(17,3)--(17,2.8)^^(18,3)--(18,2.8)); label("A", A, S); label("B", B, S); label("C", C, S); label("D", D, S); label("E", E, S); label("A", F, N); label("B", G, N); label("C", H, N); label("D", I, N); label("E", J, N); label("1970", (0,3), W); label("1980", (0,0), W); label("0", (0,1.5)); label("50", (5,1.5)); label("100", (10,1.5)); label("150", (15,1.5)); label("Population in thousands", (9,-3)); [/asy]
On each horizontal line in the figure below, the five large dots indicate the populations of cities $A, B, C, D$ and $E$ in the year indicated.
Which city had the greatest percentage increase in population from $1970$ to $1980$?
$\textbf{(A)}\ A\qquad \textbf{(B)}\ B\qquad \textbf{(C)}\ C\qquad \textbf{(D)}\ D\qquad \textbf{(E)}\ E$
|
[
"We simply have to divide the present population by the original, and whichever one is the largest have the greatest percent change. City $\\text{A}$ is \n$\\frac{50}{40}=1.25$. City $\\text{B}$ is $\\frac{70}{50}=1.4$. City $\\text{C}$ is $\\frac{100}{70} \\approx 1.42$. City $\\text{D}$ is $\\frac{130}{100}=1.3$. City $\\text{E}$ is $\\frac{160}{120} \\approx 1.33$. City $\\text{C}$ has the greatest value, so the answer is $\\boxed{\\text{C}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/11.json
|
AHSME
|
1988_AHSME_Problems
| 2 | 0 |
Geometry
|
Multiple Choice
|
Triangles $ABC$ and $XYZ$ are similar, with $A$ corresponding to $X$ and $B$ to $Y$. If $AB=3, BC=4$, and $XY=5$, then $YZ$ is:
$\text{(A)}\ 3\frac{3}{4} \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 6\frac{1}{4} \qquad \text{(D)}\ 6\frac{2}{3} \qquad \text{(E)}\ 8$
|
[
"Since the triangles are similar, we know that the corresponding sides of the triangle are in ratio to each other. \nWe have $\\frac{\\overline{AB}}{\\overline{XY}}=\\frac{\\overline{BC}}{\\overline{YZ}}$. Plugging in values we have:\n$\\frac{3}{4}=\\frac{5}{\\overline{YZ}}$. Solving for $\\overline{YZ}$, we have $\\overline{YZ}=6\\frac{2}{3}$. So, the answer is $\\boxed{\\text{D}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/2.json
|
AHSME
|
1988_AHSME_Problems
| 28 | 0 |
Probability
|
Multiple Choice
|
An unfair coin has probability $p$ of coming up heads on a single toss.
Let $w$ be the probability that, in $5$ independent toss of this coin,
heads come up exactly $3$ times. If $w = 144 / 625$, then
$\textbf{(A)}\ p\text{ must be }\tfrac{2}{5}\qquad \textbf{(B)}\ p\text{ must be }\tfrac{3}{5}\qquad\\ \textbf{(C)}\ p\text{ must be greater than }\tfrac{3}{5}\qquad \textbf{(D)}\ p\text{ is not uniquely determined}\qquad\\ \textbf{(E)}\ \text{there is no value of } p \text{ for which }w =\tfrac{144}{625}$
|
[
"We have $w = {5\\choose3}p^{3}(1-p)^{2} = 10p^{3}(1-p)^{2}$, so we need to solve $10p^{3}(1-p)^{2} = \\frac{144}{625} \\implies p^{3}(1-p)^{2} = \\frac{72}{3125}$. Now observe that when $p=0$, the left-hand side evaluates to $0$, which is less than $\\frac{72}{3125}$; when $p=\\frac{1}{2}$, it evaluates to $\\frac{1}{32}$, which is more than $\\frac{72}{3125}$, and when $p=1$, it evaluates to $0$ again. Thus, since $p^{3}(1-p)^{2}$ is continuous, the Intermediate Value Theorem tells us there is a solution between $0$ and $\\frac{1}{2}$ and another solution between $\\frac{1}{2}$ and $1$, meaning that there is not a unique value of $p$, so the answer is $\\boxed{\\text{D}}$.\n\n\n\n\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/28.json
|
AHSME
|
1988_AHSME_Problems
| 12 | 0 |
Probability
|
Multiple Choice
|
Each integer $1$ through $9$ is written on a separate slip of paper and all nine slips are put into a hat.
Jack picks one of these slips at random and puts it back. Then Jill picks a slip at random.
Which digit is most likely to be the units digit of the sum of Jack's integer and Jill's integer?
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ \text{each digit is equally likely}$
|
[
"We can draw a sample space diagram, and find that of the $9^2 = 81$ possibilities, $9$ of them give a sum of $0$, and each other sum mod $10$ (from $1$ to $9$) is given by $8$ of the possibilities (and indeed we can check that $9 + 8 \\times 9 = 81$). Thus $0$ is the most likely, so the answer is $\\boxed{\\text{A}}$.\n\n\n",
"The largest sum that can be formed from two slips of these papers is 18, and if we list the composition of each of the numbers greater than 1 and less than 19, we can find that there are 9 combinations in total which make a sum of 10 ($1+9$, $2+8$, etc). Notice that the order in which either number appears doesn't matter since they all count toward the number of possibilities.\n\n\n"
] | 2 |
./CreativeMath/AHSME/1988_AHSME_Problems/12.json
|
AHSME
|
1988_AHSME_Problems
| 24 | 0 |
Geometry
|
Multiple Choice
|
An isosceles trapezoid is circumscribed around a circle. The longer base of the trapezoid is $16$,
and one of the base angles is $\arcsin(.8)$. Find the area of the trapezoid.
$\textbf{(A)}\ 72\qquad \textbf{(B)}\ 75\qquad \textbf{(C)}\ 80\qquad \textbf{(D)}\ 90\qquad \textbf{(E)}\ \text{not uniquely determined}$
|
[
"Let the trapezium have diagonal legs of length $x$ and a shorter base of length $y$. Drop altitudes from the endpoints of the shorter base to the longer base to form two right-angled triangles, which are congruent since the trapezium is isosceles. Thus using the base angle of $\\arcsin(0.8)$ gives the vertical side of these triangles as $0.8x$ and the horizontal side as $0.6x$. Now notice that the sides of the trapezium can be seen as being made up of tangents to the circle, and thus using the fact that \"the tangents from a point to a circle are equal in length\" gives $2y + 0.6x + 0.6x = 2x$. Also, using the given length of the longer base tells us that $y + 0.6x + 0.6x = 16$. Solving these equations simultaneously gives $x=10$ and $y=4$, so the height of the trapezium is $0.8 \\times 10 = 8$. Thus the area is $\\frac{1}{2}(4+16)(8) = 80$, which is $\\boxed{\\text{C}}$.\n\n\n\n\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/24.json
|
AHSME
|
1988_AHSME_Problems
| 25 | 0 |
Algebra
|
Multiple Choice
|
$X, Y$ and $Z$ are pairwise disjoint sets of people. The average ages of people in the sets
$X, Y, Z, X \cup Y, X \cup Z$ and $Y \cup Z$ are $37, 23, 41, 29, 39.5$ and $33$ respectively.
Find the average age of the people in set $X \cup Y \cup Z$.
$\textbf{(A)}\ 33\qquad \textbf{(B)}\ 33.5\qquad \textbf{(C)}\ 33.6\overline{6}\qquad \textbf{(D)}\ 33.83\overline{3}\qquad \textbf{(E)}\ 34$
|
[
"Let the variables $X$, $Y$, and $Z$ represent the sums of the ages of the people in sets $X$, $Y$, and $Z$ respectively. Let $x$, $y$, and $z$ represent the numbers of people who are in sets $X$, $Y$, and $Z$ respectively. Since the sets are disjoint, we know, for example, that the number of people in the set $X \\cup Y$ is $x+y$ and the sum of their ages is $X+Y$, and similar results apply for the other unions of sets. Thus we have $X=37x$, $Y=23y$, $Z=41z$, $X+Y=29(x+y) \\implies 37x + 23y = 29x + 29y \\implies 8x = 6y \\implies y = \\frac{4}{3}x$, and $X+Z = 39.5(x+z) \\implies 37x + 41z = 39.5x + 39.5z \\implies 74x + 82z = 79x + 79z \\implies 3z = 5x \\implies z = \\frac{5}{3}x$. Hence the answer is $\\frac{X+Y+Z}{x+y+z} = \\frac{37x + 23\\times\\frac{4}{3}x + 41\\times\\frac{5}{3}x}{x+\\frac{4}{3}x+\\frac{5}{3}x} = \\frac{136x}{4x} = 34$, which is $\\boxed{\\text{E}}$. Notice that we did not need all three of the means of the unions of the sets - any two of them would have been sufficient to determine the answer.\n\n\n\n\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/25.json
|
AHSME
|
1988_AHSME_Problems
| 13 | 0 |
Other
|
Multiple Choice
|
If $\sin(x) =3 \cos(x)$ then what is $\sin(x) \cdot \cos(x)$?
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{5}\qquad \textbf{(C)}\ \frac{2}{9}\qquad \textbf{(D)}\ \frac{1}{4}\qquad \textbf{(E)}\ \frac{3}{10}$
|
[
"In the problem we are given that $\\sin{(x)}=3\\cos{(x)}$, and we want to find $\\sin{(x)}\\cos{(x)}$. We can divide both sides of the original equation by $\\cos{(x)}$ to get \\[\\frac{\\sin{(x)}}{\\cos{(x)}}=\\tan{(x)}=3.\\]\nWe can now use right triangle trigonometry to finish the problem.\n[asy] pair A,B,C; A = (0,0); B = (3,0); C = (0,1); draw(A--B--C--A); draw(rightanglemark(B,A,C,8)); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$3$\",B/2,S); label(\"$1$\",C/2,W); label(\"$\\sqrt{10}$\",(C+B)/2,NE); [/asy]\n\n\n(Note that this assumes that $x$ is acute. If $x$ is obtuse, then $\\sin{(x)}$ is positive and $\\cos{(x)}$ is negative, so the equation cannot be satisfied. If $x$ is reflex, then both $\\sin{(x)}$ and $\\cos{(x)}$ are negative, so the equation is satisfied, but when we find $\\sin{(x)}\\cos{(x)}$, the two negatives will cancel out and give the same (positive) answer as in the acute case.)\n\n\nSince the problem asks us to find $\\sin{(x)}\\cos{(x)}$.\n\\[\\sin{(x)}\\cos{(x)}=\\left(\\frac{3}{\\sqrt{10}}\\right)\\left(\\frac{1}{\\sqrt{10}}\\right)=\\frac{3}{10}.\\]\nSo $\\boxed{\\textbf{(E)}\\ \\frac{3}{10}}$ is our answer.\n\n\n",
"Squaring both sides gives ${\\sin}^2 x = 9{\\cos}^2 x$. We can take the Pythagorean identity, ${\\sin}^2 x + {\\cos}^2 x = 1$ and substitute the 1st equation in, giving $10{\\cos}^2 x = 1$. So ${\\cos}^2 x = \\frac{1}{10}$, and ${\\sin}^2 x = \\frac{9}{10}$. \n\n\nMultiplying the 2 together gives ${\\sin}^2 x {\\cos}^2 x = \\frac{9}{100}$, and then taking the square root gives $\\mp \\frac{3}{10}$. However, $-\\frac{3}{10}$ implies one of $\\sin x$ and $\\cos x$ is negative while the other is positive, but $\\sin x = 3 \\cos x$ means they have the same sign, which contradicts the first statement. This means $\\boxed{\\textbf{(E)}\\ \\frac{3}{10}}$ is the only answer.\n\n\n-ThisUsernameIsTaken\n\n\n"
] | 2 |
./CreativeMath/AHSME/1988_AHSME_Problems/13.json
|
AHSME
|
1988_AHSME_Problems
| 29 | 0 |
Algebra
|
Multiple Choice
|
You plot weight $(y)$ against height $(x)$ for three of your friends and obtain the points
$(x_{1},y_{1}), (x_{2},y_{2}), (x_{3},y_{3})$. If $x_{1} < x_{2} < x_{3}$ and $x_{3} - x_{2} = x_{2} - x_{1}$,
which of the following is necessarily the slope of the line which best fits the data?
"Best fits" means that the sum of the squares of the vertical distances from the data points to the line is smaller than for any other line.
$\textbf{(A)}\ \frac{y_{3}-y_{1}}{x_{3}-x_{1}}\qquad \textbf{(B)}\ \frac{(y_{2}-y_{1})-(y_{3}-y_{2})}{x_{3}-x_{1}}\qquad\\ \textbf{(C)}\ \frac{2y_{3}-y_{1}-y_{2}}{2x_{3}-x_{1}-x_{2}}\qquad \textbf{(D)}\ \frac{y_{2}-y_{1}}{x_{2}-x_{1}}+\frac{y_{3}-y_{2}}{x_{3}-x_{2}}\qquad\\ \textbf{(E)}\ \text{none of these}$
|
[
"Apply one of the standard formulae for the gradient of the line of best fit, e.g. $\\frac{\\frac{\\sum {x_i y_i}}{n} - \\bar{x} \\bar{y}}{\\frac{\\sum {x_{i}^2}}{n} - \\bar{x}^2}$, and substitute in the given condition $x_3 - x_2 = x_2 - x_1$. The answer is $\\boxed{\\text{A}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/29.json
|
AHSME
|
1988_AHSME_Problems
| 3 | 0 |
Geometry
|
Multiple Choice
|
[asy] draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)--(0,1)--(-1,1)--(-1,2)); draw((-1,2)--(0,2)--(0,4)--(-1,4)--(-1,5)--(1,5)--(1,6)--(0,6)); draw((0,6)--(0,5)--(3,5)--(3,6)--(4,6)--(4,2)--(5,2)); draw((5,2)--(5,1)--(1,1)--(3,1)--(3,0)--(4,0)--(4,1)); draw((1,4)--(3,4)--(3,2)--(1,2)--(4,2)--(3,2)--(3,6)); draw((3,6)--(4,6)--(4,5)--(5,5)--(5,4)--(4,4)); [/asy]
Four rectangular paper strips of length $10$ and width $1$ are put flat on a table and overlap perpendicularly as shown. How much area of the table is covered?
$\text{(A)}\ 36 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 44 \qquad \text{(D)}\ 98 \qquad \text{(E)}\ 100$
|
[
"We first notice that the paper strips cover up part of the others. Since the width of the overlap is $1$ and the length of the overlap is $1$, the area of \neach of the strips with the overlap is $(10\\cdot 1)-1=9$. Since there are 4 strips, $4\\cdot 9=36 \\implies \\boxed{\\text{A}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/3.json
|
AHSME
|
1988_AHSME_Problems
| 8 | 0 |
Algebra
|
Multiple Choice
|
If $\frac{b}{a} = 2$ and $\frac{c}{b} = 3$, what is the ratio of $a + b$ to $b + c$?
$\textbf{(A)}\ \frac{1}{3}\qquad \textbf{(B)}\ \frac{3}{8}\qquad \textbf{(C)}\ \frac{3}{5}\qquad \textbf{(D)}\ \frac{2}{3}\qquad \textbf{(E)}\ \frac{3}{4}$
|
[
"Since we are finding ratios, it would be helpful to put everything in terms of one variable. Since $b$ is in both equations, that would be a place to start. \nWe manipulate the equations yielding $\\frac{b}{2}=a$ and $c=3b$. Since we are asked to find the ratio of $a+b$ to $b+c$, we need to find $\\frac{a+b}{b+c}$. We found the $a$ and $c$ in terms of $b$ so that means we can plug them in. We have: $\\frac{\\frac{b}{2}+b}{b+3b}=\\frac{\\frac{3}{2}b}{4b}=\\frac{3}{8}$. Thus the answer is $\\frac{3}{8} \\implies \\boxed{\\text{B}}$.\n\n\n",
"WLOG, let $b=4, a=2, c=12.$\nThus, the answer is $\\frac{4+2}{12+4}= \\frac{3}{8}$\n\n\n"
] | 2 |
./CreativeMath/AHSME/1988_AHSME_Problems/8.json
|
AHSME
|
1988_AHSME_Problems
| 22 | 0 |
Geometry
|
Multiple Choice
|
For how many integers $x$ does a triangle with side lengths $10, 24$ and $x$ have all its angles acute?
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{more than } 7$
|
[
"We first notice that the sides $10$ and $24$, can be part of $2$ different right triangles, one with sides $10,24,26$, and the other with a leg\nsomewhere between $21$ and $22$. We now notice that if $x$ is less than or equal to $21$, one of the angles is obtuse, and that the same is the same for \nany value of $x$ above $26$. Thus the only integer values of $x$ that fit the conditions, are $x=22, 23, 24, \\text{and }25.$\nSo, the answer is $\\boxed{\\text{A}}$\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/22.json
|
AHSME
|
1988_AHSME_Problems
| 18 | 0 |
Counting
|
Multiple Choice
|
At the end of a professional bowling tournament, the top 5 bowlers have a playoff.
First #5 bowls #4. The loser receives $5$th prize and the winner bowls #3 in another game.
The loser of this game receives $4$th prize and the winner bowls #2.
The loser of this game receives $3$rd prize and the winner bowls #1.
The winner of this game gets 1st prize and the loser gets 2nd prize.
In how many orders can bowlers #1 through #5 receive the prizes?
$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 120\qquad \textbf{(E)}\ \text{none of these}$
|
[
"We have $2$ choices for who wins the first game, and that uniquely determines $5^{\\text{th}}$ place. Then there are $2$ choices for a next game and that uniquely determines $4^{\\text{th}}$ place, followed by $2$ choices for the next game that uniquely determines $3^{\\text{rd}}$ place. Finally, there are $2$ choices for the last game, which uniquely determines both $1^{\\text{st}}$ and $2^{\\text{nd}}$ places, since the winner is $1^{\\text{st}}$ and the loser is $2^{\\text{nd}}$. Thus the number of possible orders is $2 \\times 2 \\times 2 \\times 2 = 16$, which is $\\boxed{\\text{B}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/18.json
|
AHSME
|
1988_AHSME_Problems
| 4 | 0 |
Algebra
|
Multiple Choice
|
The slope of the line $\frac{x}{3} + \frac{y}{2} = 1$ is
$\textbf{(A)}\ -\frac{3}{2}\qquad \textbf{(B)}\ -\frac{2}{3}\qquad \textbf{(C)}\ \frac{1}{3}\qquad \textbf{(D)}\ \frac{2}{3}\qquad \textbf{(E)}\ \frac{3}{2}$
|
[
"To find the slope, all we have to do is put the equation into slope-intercept form. We subtract $\\frac{x}{3}$ from both sides and then multiple all \nterms by $2$. This yields $y=-\\frac{2}{3}x+1$, so the slope is $-\\frac{2}{3} \\implies \\boxed{\\text{B}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/4.json
|
AHSME
|
1988_AHSME_Problems
| 14 | 0 |
Algebra
|
Multiple Choice
|
For any real number a and positive integer k, define
${a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}$
What is
${-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}$?
$\textbf{(A)}\ -199\qquad \textbf{(B)}\ -197\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 197\qquad \textbf{(E)}\ 199$
|
[
"We expand both the numerator and the denominator.\n\n\n\\begin{align*} \\binom{-\\frac{1}{2}}{100}\\div\\binom{\\frac{1}{2}}{100} &= \\frac{ \\dfrac{ (-\\frac{1}{2}) (-\\frac{1}{2} - 1) (-\\frac{1}{2} - 2) \\cdots (-\\frac{1}{2} - (100 - 1)) }{\\cancel{(100)(99)\\cdots(1)}} }{ \\dfrac{ (\\frac{1}{2}) (\\frac{1}{2} - 1) (\\frac{1}{2} - 2) \\cdots (\\frac{1}{2} - (100 - 1)) }{\\cancel{(100)(99)\\cdots(1)}} } \\\\ &= \\frac{ (-\\frac{1}{2}) (-\\frac{1}{2} - 1) (-\\frac{1}{2} - 2) \\cdots (-\\frac{1}{2} - 99) }{ (\\frac{1}{2}) (\\frac{1}{2} - 1) (\\frac{1}{2} - 2) \\cdots (\\frac{1}{2} - 99) } \\end{align*}\n\n\nNow, note that $-\\frac{1}{2}-1=\\frac{1}{2}-2$, $-\\frac{1}{2}-2=\\frac{1}{2}-3$, etc.; in essence, $-\\frac{1}{2}-n=\\frac{1}{2}-(n+1)$. We can then simplify the numerator and cancel like terms.\n\n\n\\begin{align*} \\frac{ (-\\frac{1}{2}) (-\\frac{1}{2} - 1) (-\\frac{1}{2} - 2) \\cdots (-\\frac{1}{2} - 99) }{ (\\frac{1}{2}) (\\frac{1}{2} - 1) (\\frac{1}{2} - 2) \\cdots (\\frac{1}{2} - 99) } &= \\frac{ \\cancel{(\\frac{1}{2} - 1)} \\cancel{(\\frac{1}{2} - 2)} \\cancel{(\\frac{1}{2} - 3)} \\cdots (\\frac{1}{2} - 100) }{ (\\frac{1}{2}) \\cancel{(\\frac{1}{2} - 1)} \\cancel{(\\frac{1}{2} - 2)} \\cdots \\cancel{(\\frac{1}{2} - 99)} } \\\\ &= \\frac{\\frac{1}{2}-100}{\\frac{1}{2}} \\\\ &= \\frac{-\\frac{199}{2}}{\\frac{1}{2}} \\\\ &= \\boxed{\\textbf{(A) } -199.} \\end{align*}\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/14.json
|
AHSME
|
1988_AHSME_Problems
| 15 | 0 |
Algebra
|
Multiple Choice
|
If $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1$, then $b$ is
$\textbf{(A)}\ -2\qquad \textbf{(B)}\ -1\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2$
|
[
"Using polynomial division, we find that the remainder is $(2a+b)x+(a+b+1)$, so for the condition to hold, we need this remainder to be $0$. This gives $2a+b=0$ and $a+b+1=0$, so $b=-2a$ and $a-2a+1=0 \\implies a=1 \\implies b=-2$, which is $\\boxed{\\text{A}}.$\n\n\n\n\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/15.json
|
AHSME
|
1988_AHSME_Problems
| 5 | 0 |
Algebra
|
Multiple Choice
|
If $b$ and $c$ are constants and $(x + 2)(x + b) = x^2 + cx + 6$, then $c$ is
$\textbf{(A)}\ -5\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 5$
|
[
"We first start out by expanding the left side of the equation, $(x+2)(x+b)=x^{2}+bx+2x+2b=x^2+(2+b)x+2b=x^2+cx+6$. We know the constant \nterms have to be equal so we have $2b=6$, so $b=3$. Plugging $b=3$ back in yields $x^2+(2+3)x+6=x^2+cx+6$. Thus, $c=5 \\implies \\boxed{\\text{E}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/5.json
|
AHSME
|
1988_AHSME_Problems
| 19 | 0 |
Algebra
|
Multiple Choice
|
Simplify
$\frac{bx(a^2x^2 + 2a^2y^2 + b^2y^2) + ay(a^2x^2 + 2b^2x^2 + b^2y^2)}{bx + ay}$
$\textbf{(A)}\ a^2x^2 + b^2y^2\qquad \textbf{(B)}\ (ax + by)^2\qquad \textbf{(C)}\ (ax + by)(bx + ay)\qquad\\ \textbf{(D)}\ 2(a^2x^2+b^2y^2)\qquad \textbf{(E)}\ (bx+ay)^2$
|
[
"We can multiply each answer choice by $bx + ay$ and then compare with the numerator. This gives $\\boxed{\\text{B}}$.\n\n\n",
"Expanding everything in the brackets, we get\n$\\frac{ba^2x^3 + 2ba^2xy^2 + b^3xy^2 + a^3x^2y + 2ab^2x^2y + ab^2y^3}{bx + ay}$. We can then group numbers up in pairs so they equal $n(bx + ay)$:\n\n\n$= \\frac{ba^2x^3 + a^3x^2y + 2ab^2x^2y + 2ba^2xy^2 + b^3xy^2 + ab^2y^3}{bx+ay}$\n\n\n$= \\frac{bx + ay(a^2x^2) + bx + ay(2baxy) + bx + ay(b^2y^2)}{bx+ay}$\n\n\n$= a^2x^2 + 2baxy + b^2y^2$\n\n\n$= (ax + by)^2$\n\n\nWe get $\\boxed{\\text{B}}$.\n\n\n-ThisUsernameIsTaken\n\n\n",
"If you were out of time and your algebra isn't that good, you could just plug in some values for the variables and see which answer choice works.\n\n\n",
"After regrouping, the numerator becomes $(bx+ay)(a^2x^2+b^2y^2)+2bxa^2y^2+2ayb^2x^2$. Factoring further, we get $(bx+ay)(a^2x^2+b^2y^2)+2bxay(bx+ay)$. After dividing, we get $a^2x^2+b^2y^2+2bxay$, which can be factored as $(ax+by)^2$, so the answer is $\\boxed{\\text{B}}$.\n\n\n-Pengu14\n\n\n"
] | 4 |
./CreativeMath/AHSME/1988_AHSME_Problems/19.json
|
AHSME
|
1988_AHSME_Problems
| 23 | 0 |
Geometry
|
Multiple Choice
|
The six edges of a tetrahedron $ABCD$ measure $7, 13, 18, 27, 36$ and $41$ units. If the length of edge $AB$ is $41$, then the length of edge $CD$ is
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$
|
[
"By the triangle inequality in $\\triangle ABC$, we find that $BC$ and $CA$ must sum to greater than $41$, so they must be (in some order) $7$ and $36$, $13$ and $36$, $18$ and $27$, $18$ and $36$, or $27$ and $36$. We try $7$ and $36$, and now by the triangle inequality in $\\triangle ABD$, we must use the remaining numbers $13$, $18$, and $27$ to get a sum greater than $41$, so the only possibility is $18$ and $27$. This works as we can put $BC = 36$, $AC = 7$, $AD = 18$, $BD = 27$, $CD = 13$, so that $\\triangle ADC$ and $\\triangle BDC$ also satisfy the triangle inequality. Hence we have found a solution that works, and it can be verified that the other possibilities don't work, though as this is a multiple-choice competition, you probably wouldn't do that in order to save time. In any case, the answer is $CD = 13$, which is $\\boxed{\\text{B}}$.\n\n\n\n\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/23.json
|
AHSME
|
1988_AHSME_Problems
| 9 | 0 |
Geometry
|
Multiple Choice
|
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,0), B=(16,0), C=(16,16), D=(0,16), E=(32,0), F=(48,0), G=(48,16), H=(32,16), I=(0,8), J=(10,8), K=(10,16), L=(32,6), M=(40,6), N=(40,16); draw(A--B--C--D--A^^E--F--G--H--E^^I--J--K^^L--M--N); label("S", (18,8)); label("S", (50,8)); label("Figure 1", (A+B)/2, S); label("Figure 2", (E+F)/2, S); label("10'", (I+J)/2, S); label("8'", (12,12)); label("8'", (L+M)/2, S); label("10'", (42,11)); label("table", (5,12)); label("table", (36,11)); [/asy]
An $8'\times 10'$ table sits in the corner of a square room, as in Figure $1$ below.
The owners desire to move the table to the position shown in Figure $2$.
The side of the room is $S$ feet. What is the smallest integer value of $S$ for which the table can be moved as desired without tilting it or taking it apart?
$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15$
|
[
"Small correction: The writer below has maximized the area of the rectangle (with sides parallel to the walls) that fits around the table, but there is a larger single dimension we can find in the table. The height or width is maximized when the diagonal of the table is horizontal or vertical. By the Pythagorean Theorem, this diagonal is $\\sqrt{8^2+10^2} = \\sqrt{164},$ which is between $\\sqrt{144}$ and $\\sqrt{169},$ so the answer is still $\\textbf{(C)}.$ -hailstone\n\n\nWe begin by thinking about the motion of the table. As it moves, the table will have it's maximum height and width when the rectangle's sides form $45$ degree angles relative to the sides of the square. Therefore, by the Pythagorean Theorem, we have that $S= 4\\sqrt{2}+5\\sqrt{2}$, with $4\\sqrt{2}$ being the length of the leg formed by the side of the square with length $8$ and $5\\sqrt{2}$ being the length of the leg formed by the side of the square with length $10$. Adding these up yields $9\\sqrt{2}$.\n\n\nWe have that $\\sqrt{2}\\approx 1.414\\approx 1.4$. That means that $9\\sqrt{2}\\approx 12.6$, which rounds up to $\\boxed{\\textbf{C)} 13}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1988_AHSME_Problems/9.json
|
AHSME
|
1963_AHSME_Problems
| 20 | 0 |
Algebra
|
Multiple Choice
|
Two men at points $R$ and $S$, $76$ miles apart, set out at the same time to walk towards each other.
The man at $R$ walks uniformly at the rate of $4\tfrac{1}{2}$ miles per hour; the man at $S$ walks at the constant
rate of $3\tfrac{1}{4}$ miles per hour for the first hour, at $3\tfrac{3}{4}$ miles per hour for the second hour,
and so on, in arithmetic progression. If the men meet $x$ miles nearer $R$ than $S$ in an integral number of hours, then $x$ is:
$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 2$
|
[
"First, find the number of hours it takes for the two to meet together. After $h$ hours, the person at $R$ walks $4.5h$ miles. In the same amount of time, the person at $S$ has been walking at $3.25+0.5(h-1)$ mph for the past hour, so the person walks $\\frac{h(6.5+0.5(h-1))}{2}$ miles.\n\n\nIn order for both to meet, the sum of both of the distances walked must total $76$ miles, so\n\\[4.5h + \\frac{h(6+0.5h)}{2} = 76\\]\n\\[4.5h + 3h + 0.25h^2 = 76\\]\n\\[0.25h^2 + 7.5h - 76 = 0\\]\n\\[h^2 + 30h - 304 = 0\\]\n\\[(h + 38)(h - 8) = 0\\]\nSince $h$ must be positive, $h = 8$. Because it takes $8$ hours to meet, the person from $R$ traveled $36$ miles while the person from $S$ traveled $40$ miles. Thus, they are $4$ miles closer to $R$ than $S$, so the answer is $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1963_AHSME_Problems/20.json
|
AHSME
|
1963_AHSME_Problems
| 36 | 0 |
Probability
|
Multiple Choice
|
A person starting with $$64$ and making $6$ bets, wins three times and loses three times,
the wins and losses occurring in random order. The chance for a win is equal to the chance for a loss.
If each wager is for half the money remaining at the time of the bet, then the final result is:
$\textbf{(A)}\text{ a loss of }$ 27 \qquad \textbf{(B)}\text{ a gain of }$ 27 \qquad \textbf{(C)}\text{ a loss of }$ 37 \qquad \\ \textbf{(D)}\text{ neither a gain nor a loss}\qquad \\ \textbf{(E)}\text{ a gain or a loss depending upon the order in which the wins and losses occur}$
|
[
"If the person wins the bet, the person has $\\frac{3}{2}$ of the previous amount. If the person loses the bet, the person only has $\\frac{1}{2}$ of the previous amount.\n\n\nBecause of the Commutative Property, the order of multiplying the multipliers does not matter. Thus, the person walks away with $64 \\cdot \\frac{3}{2} \\cdot \\frac{3}{2} \\cdot \\frac{3}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} = 27$ dollars, so the person loses $$ 37$. The answer is $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1963_AHSME_Problems/36.json
|
AHSME
|
1963_AHSME_Problems
| 16 | 0 |
Algebra
|
Multiple Choice
|
Three numbers $a,b,c$, none zero, form an arithmetic progression. Increasing $a$ by $1$ or increasing $c$ by $2$ results
in a geometric progression. Then $b$ equals:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$
|
[
"Let $d$ be the common difference of the arithmetic sequence, so $a = b-d$ and $c = b+d$.\n\n\nSince increasing $a$ by $1$ or $c$ by $2$ results in a geometric sequence,\n\\[\\frac{b}{b-d+1} = \\frac{b+d}{b}\\]\n\\[\\frac{b}{b-d} = \\frac{b+d+2}{b}\\]\nCross-multiply in both equations to get a system of equations.\n\\[b^2 = b^2 - d^2 + b + d\\]\n\\[b^2 = b^2 - d^2 + 2b - 2d\\]\nRearranging terms results in\n\\[d^2 = b+d\\]\n\\[d^2 = 2b-2d\\]\nSubstitute and solve for $d$.\n\\[b+d = 2b-2d\\]\n\\[d = \\frac{b}{3}\\]\nFinally, substitute $d$ and solve for $b$. Since $b \\ne 0$, dividing by $b$ is allowed.\n\\[(\\frac{b}{3})^2 = b + \\frac{b}{3}\\]\n\\[\\frac{b^2}{9} = \\frac{4b}{3}\\]\n\\[\\frac{b}{9} = \\frac{4}{3}\\]\n\\[b = 12\\]\nThe answer is $\\boxed{\\textbf{(C)}}$.\n\n\n\n\n\n\n"
] | 1 |
./CreativeMath/AHSME/1963_AHSME_Problems/16.json
|
AHSME
|
1963_AHSME_Problems
| 6 | 0 |
Geometry
|
Multiple Choice
|
$\triangle BAD$ is right-angled at $B$. On $AD$ there is a point $C$ for which $AC=CD$ and $AB=BC$. The magnitude of $\angle DAB$ is:
$\textbf{(A)}\ 67\tfrac{1}{2}^{\circ}\qquad \textbf{(B)}\ 60^{\circ}\qquad \textbf{(C)}\ 45^{\circ}\qquad \textbf{(D)}\ 30^{\circ}\qquad \textbf{(E)}\ 22\tfrac{1}{2}^{\circ}$
|
[
"[asy] draw((0,0)--(0,10)--(17.321,0)--(0,0)); draw((0,0)--(8.661,5)); dot((0,0)); label(\"$B$\",(0,0),SW); dot((0,10)); label(\"$A$\",(0,10),NW); dot((17.321,0)); label(\"$D$\",(17.321,0),SE); dot((8.661,5)); label(\"$C$\",(8.661,5),NE); [/asy]\nNote that because $C$ is the midpoint of $AD$, $C$ is the circumcenter of the triangle, so $BC = AC = AB$. Thus, $\\triangle ABC$ is equilateral, so $\\angle DAB = 60^{\\circ}$, which is answer choice $\\boxed{\\textbf{(B)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1963_AHSME_Problems/6.json
|
AHSME
|
1963_AHSME_Problems
| 7 | 0 |
Geometry
|
Multiple Choice
|
Given the four equations:
$\textbf{(1)}\ 3y-2x=12 \qquad\textbf{(2)}\ -2x-3y=10 \qquad\textbf{(3)}\ 3y+2x=12 \qquad\textbf{(4)}\ 2y+3x=10$
The pair representing the perpendicular lines is:
$\textbf{(A)}\ \text{(1) and (4)}\qquad \textbf{(B)}\ \text{(1) and (3)}\qquad \textbf{(C)}\ \text{(1) and (2)}\qquad \textbf{(D)}\ \text{(2) and (4)}\qquad \textbf{(E)}\ \text{(2) and (3)}$
|
[
"Write each equation in slope-intercept from since the slopes are easier to compare.\n\n\nEquation $(1)$ in slope-intercept form is $y = \\frac{2}{3}x+4$.\n\n\nEquation $(2)$ in slope-intercept form is $y = -\\frac{2}{3}x - \\frac{10}{3}$.\n\n\nEquation $(3)$ in slope-intercept form is $y = -\\frac{2}{3}x + 4$.\n\n\nEquation $(4)$ in slope-intercept form is $y = -\\frac{3}{2}x + 5$.\n\n\nRemember that if the two lines are perpendicular, then the product of two slopes equals $-1$. Equations $(1)$ and $(4)$ satisfy the condition, so the answer is $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1963_AHSME_Problems/7.json
|
AHSME
|
1963_AHSME_Problems
| 17 | 0 |
Algebra
|
Multiple Choice
|
The expression $\dfrac{\dfrac{a}{a+y}+\dfrac{y}{a-y}}{\dfrac{y}{a+y}-\dfrac{a}{a-y}}$, $a$ real, $a\neq 0$, has the value $-1$ for:
$\textbf{(A)}\ \text{all but two real values of }y \qquad \\ \textbf{(B)}\ \text{only two real values of }y \qquad \\ \textbf{(C)}\ \text{all real values of }y\qquad \\ \textbf{(D)}\ \text{only one real value of }y\qquad \\ \textbf{(E)}\ \text{no real values of }y$
|
[
"First, note that $y \\ne \\pm a$ because that would make the denominator $0$.\n\n\nCreate common denominators in the complex fraction.\n\\[\\frac{\\frac{a^2-ay}{a^2-y^2} + \\frac{ay+y^2}{a^2-y^2}}{\\frac{ay-y^2}{a^2-y^2} - \\frac{a^2+ay}{a^2-y^2}}\\]\n\\[\\frac{\\frac{a^2+y^2}{a^2-y^2}}{\\frac{-a^2-y^2}{a^2-y^2}}\\]\nSince $a \\ne 0$, $a^2 + y^2 > 0$, so no other restrictions need to be made.\n\\[\\frac{a^2 + y^2}{a^2 - y^2} \\cdot \\frac{a^2 - y^2}{-a^2 - y^2}\\]\n\\[\\frac{a^2 + y^2}{-(a^2 + y^2)}\\]\n\\[-1\\]\nThe expression (with suitable restrictions) simplifies to $-1$, so the answer is $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1963_AHSME_Problems/17.json
|
AHSME
|
1963_AHSME_Problems
| 40 | 0 |
Algebra
|
Multiple Choice
|
If $x$ is a number satisfying the equation $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$, then $x^2$ is between:
$\textbf{(A)}\ 55\text{ and }65\qquad \textbf{(B)}\ 65\text{ and }75\qquad \textbf{(C)}\ 75\text{ and }85\qquad \textbf{(D)}\ 85\text{ and }95\qquad \textbf{(E)}\ 95\text{ and }105$
|
[
"Let $a = \\sqrt[3]{x + 9}$ and $b = \\sqrt[3]{x - 9}$. Cubing these equations, we get $a^3 = x + 9$ and $b^3 = x - 9$, so $a^3 - b^3 = 18$. The left-hand side factors as\n\\[(a - b)(a^2 + ab + b^2) = 18.\\]\n\n\nHowever, from the given equation $\\sqrt[3]{x + 9} - \\sqrt[3]{x - 9} = 3$, we get $a - b = 3$. Then $3(a^2 + ab + b^2) = 18$, so $a^2 + ab + b^2 = 18/3 = 6$.\n\n\nSquaring the equation $a - b = 3$, we get $a^2 - 2ab + b^2 = 9$. Subtracting this equation from the equation $a^2 + ab + b^2 = 6$, we get $3ab = -3$, so $ab = -1$. But $a = \\sqrt[3]{x + 9}$ and $b = \\sqrt[3]{x - 9}$, so $ab = \\sqrt[3]{(x + 9)(x - 9)} = \\sqrt[3]{x^2 - 81}$, so $\\sqrt[3]{x^2 - 81} = -1$. Cubing both sides, we get $x^2 - 81 = -1$, so $x^2 = 80$. The answer is $\\boxed{\\textbf{(C)}}$.\n\n\n\n\n\n\n",
"$\\sqrt[3]{x+9}-\\sqrt[3]{x-9}-3=0$ i.e, $\\sqrt[3]{x+9}+\\sqrt[3]{-x+9}+(-3)=0$\n\n\n\\begin{verbatim}\nif the sum of three numbers is zero, then their sum of cubes is thrice the product of each number.\nthen, $x+9-x+9+27=3(\\sqrt[3]{x+9})(\\sqrt[3]{9-x})(-3)$ . by solving this, we get $-9=-9\\sqrt[3]{9^2-x^2}$\n$x^2=80$. this gives the step what we had done in solution 1.The answer is $\\boxed{\\textbf{(C)}}$.\n\\end{verbatim}\n",
"Cubing both sides, we get\n\\[x+9-3\\sqrt[3]{(x+9)^2(x-9)}+3\\sqrt[3]{(x+9)(x-9)^2}-x+9=27,\\]so \\[18-3\\sqrt[3]{(x-9)(x+9)}(\\sqrt[3]{x+9}-\\sqrt[3]{x-9})=27\\implies 3\\sqrt[3]{x^2-81}=-3.\\] Dividing both sides by 3 and cubing, we find $x^2=80$, which is between $\\boxed{(C)75\\text{ and }85}$.\n\n\n~pfalcon\n\n\n",
"We consider the formula $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$. Factoring out $3ab$, and the commutative property gives $( a^3 - b^3) - 3ab(a-b)$. Cubing both sides gives us\n\\[(\\sqrt[3]{x+9}-\\sqrt[3]{x-9})^3=27\\]\nUsing our formula, we have\n\\[18 - 9 \\sqrt[3]{x^2 - 81} = 27\\]\nSolving this gives $x^2 = 80$, therefore, the answer is $\\boxed{\\textbf{(C)}}$.\n~zixuan12\n\n\n"
] | 4 |
./CreativeMath/AHSME/1963_AHSME_Problems/40.json
|
AHSME
|
1963_AHSME_Problems
| 37 | 0 |
Geometry
|
Multiple Choice
|
Given points $P_1, P_2,\cdots,P_7$ on a straight line, in the order stated (not necessarily evenly spaced).
Let $P$ be an arbitrarily selected point on the line and let $s$ be the sum of the undirected lengths
$PP_1, PP_2, \cdots , PP_7$. Then $s$ is smallest if and only if the point $P$ is:
$\textbf{(A)}\ \text{midway between }P_1\text{ and }P_7\qquad \\ \textbf{(B)}\ \text{midway between }P_2\text{ and }P_6\qquad \\ \textbf{(C)}\ \text{midway between }P_3\text{ and }P_5\qquad \\ \textbf{(D)}\ \text{at }P_4 \qquad \textbf{(E)}\ \text{at }P_1$
|
[
"By the Triangle Inequality, $P_1P + P_7P \\ge P_1P_7$, with equality happening when $P$ is between $P_1$ and $P_7$. Using similar logic, $P$ must be between $P_3$ and $P_5$ in order for the distance to be minimized.\n\n\nThe only point left to deal with is $P_4$ (which is also between $P_3$ and $P_5$). The minimum possible distance is $0$ (when $P$ is on $P_4$), so the answer is $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 1 |
./CreativeMath/AHSME/1963_AHSME_Problems/37.json
|
AHSME
|
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