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 |
|---|---|---|---|---|---|---|---|---|---|
1999_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | $1 - 2 + 3 -4 + \cdots - 98 + 99 =$
$\mathrm{\textbf{(A)} \ -50 } \qquad \mathrm{\textbf{(B)} \ -49 } \qquad \mathrm{\textbf{(C)} \ 0 } \qquad \mathrm{\textbf{(D)} \ 49 } \qquad \mathrm{\textbf{(E)} \ 50 }$
| [
"If we group consecutive terms together, we get $(-1) + (-1) + \\cdots + 99$, and since there are 49 pairs of terms the answer is $-49 + 99 = 50 \\Rightarrow \\mathrm{\\textbf{(E)}}$.\n\n",
"( Similar to Solution 1 )\nIf we rearranged the terms, we get $1+3-2+5-4 \\cdots + 99-98$ then $1 + 1 + \\cdots + 1$, and s... | 4 | ./CreativeMath/AHSME/1999_AHSME_Problems/1.json | AHSME |
1999_AHSME_Problems | 11 | 0 | Arithmetic | Multiple Choice | The student locker numbers at Olympic High are numbered consecutively beginning with locker number $1$. The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number $9$ and four centers to label locker number $10$. If it costs $137.94 to label all the lockers, ho... | [
"its A\n\n",
"Since all answers are over $2000$, work backwards and find the cost of the first $1999$ lockers. The first $9$ lockers cost $0.18$ dollars, while the next $90$ lockers cost $0.04\\cdot 90 = 3.60$. Lockers $100$ through $999$ cost $0.06\\cdot 900 = 54.00$, and lockers $1000$ through $1999$ inclusiv... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/11.json | AHSME |
1999_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | Which of the following statements is false?
$\mathrm{(A) \ All\ equilateral\ triangles\ are\ congruent\ to\ each\ other.}$
$\mathrm{(B) \ All\ equilateral\ triangles\ are\ convex.}$
$\mathrm{(C) \ All\ equilateral\ triangles\ are\ equianguilar.}$
$\mathrm{(D) \ All\ equilateral\ triangles\ are\ regular\ poly... | [
"An equilateral triangle is isosceles, and we find that $\\angle A=\\angle B=\\angle C$ if we use the property of isosceles triangles that if two sides of a triangle are equal then the opposite angles are equal. Thus equilateral triangles are equiangular, and $C$ is true.\n\n\nRegular polygons are both equilateral ... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/2.json | AHSME |
1999_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$
(ii) $x_1 + \cdots + x_n = 19$; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$.
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$, respectively. Then $\frac Mm =$
$\mat... | [
"Clearly, we can ignore the possibility that some $x_i$ are zero, as adding/removing such variables does not change the truth value of any condition, nor does it change the value of the sum of cubes. Thus we'll only consider $x_i\\in\\{-1,1,2\\}$. \n\n\nAlso, order of the $x_i$ does not matter, so we are only inter... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/28.json | AHSME |
1999_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions $y=p(x)$ and $y=q(x)$, each with leading coefficient 1?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
| [
"The intersections of the two polynomials, $p(x)$ and $q(x)$, are precisely the roots of the equation $p(x)=q(x) \\rightarrow p(x) - q(x) = 0$. Since the leading coefficients of both polynomials are $1$, the degree of $p(x) - q(x) = 0$ is at most three, and the maximum point of intersection is three, because a thir... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/12.json | AHSME |
1999_AHSME_Problems | 25 | 0 | Number Theory | Multiple Choice | There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$. Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$.
... | [
"Multiply out the $7!$ to get \n\n\n\\[5 \\cdot 6! = (3 \\cdot 4 \\cdots 7)a_2 + (4 \\cdots 7)a_3 + (5 \\cdot 6 \\cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .\\]\n\n\nBy Wilson's Theorem (or by straightforward division), $a_7 + 7(a_6 + 6a_5 + \\cdots) \\equiv 5 \\cdot 6! \\equiv -5 \\equiv 2 \\pmod{7}$, so $a_7 = 2$. Then we ... | 4 | ./CreativeMath/AHSME/1999_AHSME_Problems/25.json | AHSME |
1999_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | Define a sequence of real numbers $a_1$, $a_2$, $a_3$, $\dots$ by $a_1 = 1$ and $a_{n + 1}^3 = 99a_n^3$ for all $n \geq 1$. Then $a_{100}$ equals
$\textbf{(A)}\ 33^{33} \qquad \textbf{(B)}\ 33^{99} \qquad \textbf{(C)}\ 99^{33} \qquad \textbf{(D)}\ 99^{99} \qquad \textbf{(E)}\ \text{none of these}$
| [
"We rearrange to get $\\dfrac{a_{n+1}}{a_n} = \\sqrt[3]{99}$. Thus we get $\\dfrac{a_{n+1}}{a_n} = \\sqrt[3]{99}$, $\\dfrac{a_{n}}{a_{n-1}} = \\sqrt[3]{99}$, and so on. Multiplying them all gives $\\dfrac{a_{n+1}}{a_1} = (\\sqrt[3]{99})^{n}$. Plugging in $n = 99$ and $a_1 = 1$, $a_{100} = (\\sqrt[3]{99})^{99} = 99^... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/13.json | AHSME |
1999_AHSME_Problems | 29 | 0 | Probability | Multiple Choice | A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point $P$ is selected at random inside the circumscribed sphere. The pr... | [
"Let the radius of the large sphere be $R$, and of the inner sphere $r$. \nLabel the vertices of the tetrahedron $ABCD$, and let $O$ be the center. Then pyramid $[OABC] + [OABD] + [OACD] + [OBCD] = [ABCD]$, where $[\\ldots]$ denotes volume; thus $[OABC] = \\frac{[ABCD]}{4}$. Since $OABC$ and $ABCD$ are both pyramid... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/29.json | AHSME |
1999_AHSME_Problems | 3 | 0 | Arithmetic | Multiple Choice | The number halfway between $1/8$ and $1/10$ is
$\mathrm{(A) \ } \frac 1{80} \qquad \mathrm{(B) \ } \frac 1{40} \qquad \mathrm{(C) \ } \frac 1{18} \qquad \mathrm{(D) \ } \frac 1{9} \qquad \mathrm{(E) \ } \frac 9{80}$
| [
"To find the number halfway between $\\frac{1}{8}$ and $\\frac{1}{10}$, simply take the arithmetic mean, which is\n\n\n$\\frac{\\frac{1}{8}+\\frac{1}{10}}{2}=\\frac{\\frac{9}{40}}{2}=\\frac{9}{80}.$\n\n\nThus the answer is choice $\\boxed{E}.$\n\n",
"Note that $\\frac{1}{10} = 0.1$ and $\\frac{1}{8} = 0.125$. Th... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/3.json | AHSME |
1999_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | At the end of $1994$, Walter was half as old as his grandmother. The sum of the years in which they were born was $3838$. How old will Walter be at the end of $1999$?
$\textbf{(A)}\ 48 \qquad \textbf{(B)}\ 49\qquad \textbf{(C)}\ 53\qquad \textbf{(D)}\ 55\qquad \textbf{(E)}\ 101$
| [
"In $1994$, if Water is $x$ years old, then Walter's grandmother is $2x$ years old.\n\n\nThis means that Walter was born in $1994 - x$, and Walter's grandmother was born in $1994 - 2x$.\n\n\nThe sum of those years is $3838$, so we have:\n\n\n$1994 - x + 1994 - 2x = 3838$\n\n\n$3988 - 3x = 3838$\n\n\n$x = 50$\n\n\nI... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/8.json | AHSME |
1999_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | The graphs of $y = -|x-a| + b$ and $y = |x-c| + d$ intersect at points $(2,5)$ and $(8,3)$. Find $a+c$.
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18$
| [
"Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a $45^\\circ$ angle, in the second graph they point upwards. One can easily find out that the only way how to get these graphs to intersect in two points is the one depicted below:\n\n\n\\begin{center}\n[asy] units... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/22.json | AHSME |
1999_AHSME_Problems | 18 | 0 | Other | Multiple Choice | How many zeros does $f(x) = \cos(\log x)$ have on the interval $0 < x < 1$?
$\mathrm{(A) \ } 0 \qquad \mathrm{(B) \ } 1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } \text{infinitely\ many}$
| [
"For $0 < x < 1$ we have $-\\infty < \\log x < 0$, and the logarithm is a strictly increasing function on this interval.\n\n\n$\\cos(t)$ is zero for all $t$ of the form $\\frac{\\pi}2 + k\\pi$, where $k\\in\\mathbb{Z}$. There are $\\boxed{\\text{infinitely\\ many}}$ such $t$ in $(-\\infty,0)$.\n\n\nHere's the graph... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/18.json | AHSME |
1999_AHSME_Problems | 4 | 0 | Number Theory | Multiple Choice | Find the sum of all prime numbers between $1$ and $100$ that are simultaneously $1$ greater than a multiple of $4$ and $1$ less than a multiple of $5$.
$\mathrm{(A) \ } 118 \qquad \mathrm{(B) \ }137 \qquad \mathrm{(C) \ } 158 \qquad \mathrm{(D) \ } 187 \qquad \mathrm{(E) \ } 245$
| [
"Numbers that are $1$ less than a multiple of $5$ all end in $4$ or $9$.\n\n\nNo prime number ends in $4$, since all numbers that end in $4$ are divisible by $2$. Thus, we are only looking for numbers that end in $9$.\n\n\nWriting down the ten numbers that so far qualify, we get $9, 19, 29, 39, 49, 59, 69, 79, 89,... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/4.json | AHSME |
1999_AHSME_Problems | 14 | 0 | Counting | Multiple Choice | Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang $7$ songs, which was more than any other girl, and Mary sang $4$ songs, which was fewer than any other girl. How many songs did these trios sing?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qqua... | [
"Alina and Tina must sing more than $4$, but less than $7$, songs. Therefore, Alina sang $5$ or $6$ songs, and Tina sang $5$ or $6$ songs, with $4$ possible combinations.\n\n\n\n\nHowever, since every song is a trio, if you add up all the numbers of songs a person sang for all four singers, it must be divisible by... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/14.json | AHSME |
1999_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | Let $x$ be a real number such that $\sec x - \tan x = 2$. Then $\sec x + \tan x =$
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2 \qquad \textbf{(C)}\ 0.3 \qquad \textbf{(D)}\ 0.4 \qquad \textbf{(E)}\ 0.5$
| [
"$(\\sec x - \\tan x)(\\sec x + \\tan x) = \\sec^{2} x - \\tan^{2} x = 1$, so $\\sec x + \\tan x = \\boxed{\\textbf{(E)}\\ 0.5}$.\n\n\n",
"Note that $\\sec x - \\tan x = (1-\\sin x)/\\cos x$ and $\\sec x + \\tan x = (1+\\sin x)/\\cos x$. Let $(1+\\sin x)/\\cos x = y$. Multiplying, we get $(1-\\sin^{2}x)/\\cos^{2}... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/15.json | AHSME |
1999_AHSME_Problems | 5 | 0 | Arithmetic | Multiple Choice | The marked price of a book was 30% less than the suggested retail price. Alice purchased the book for half the marked price at a Fiftieth Anniversary sale. What percent of the suggested retail price did Alice pay?
$\mathrm{(A)\ }25\%\qquad\mathrm{(B)\ }30\%\qquad\mathrm{(C)\ }35\%\qquad\mathrm{(D)\ }60\%\qquad\math... | [
"Without loss of generality, let's assume that the retail price was $100$ USD.\n\n\nThe marked price of the book is $30 \\%$ off of $100$ which is equal to $100-100(0.3)=70.$\n\n\nHalf of that marked price is $0.5(70)=35.$\n\n\nTherefore the percent Alice payed of the suggested retail price is $35/100=\\boxed{\\mat... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/5.json | AHSME |
1999_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | Consider all triangles $ABC$ satisfying in the following conditions: $AB = AC$, $D$ is a point on $\overline{AC}$ for which $\overline{BD} \perp \overline{AC}$, $AC$ and $CD$ are integers, and $BD^{2} = 57$. Among all such triangles, the smallest possible value of $AC$ is
[asy] pair A,B,C,D; A=(5,12); B=origin; C=(... | [
"Thus $AD = AC - CD$ and $AB = AC$ are integers. By the Pythagorean Theorem, \n\n\n\\[AD^2 + 57 = AB^2 \\Longrightarrow 1 \\cdot 57 = 3 \\cdot 19 = (AB - AD)(AB + AD).\\]\n\n\nThus $AC = AB = \\frac {1 + 57}{2} = 29$ or $\\frac {3 + 19}{2} = 11 \\Longrightarrow \\mathrm{(C)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/19.json | AHSME |
1999_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | The equiangular convex hexagon $ABCDEF$ has $AB = 1, BC = 4, CD = 2,$ and $DE = 4.$ The area of the hexagon is
$\mathrm{(A) \ } \frac {15}2\sqrt{3} \qquad \mathrm{(B) \ }9\sqrt{3} \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }\frac{39}4\sqrt{3} \qquad \mathrm{(E) \ } \frac{43}4\sqrt{3}$
\section{Solution}
| [
"Equiangularity means that each internal angle must be exactly $120^\\circ$.\nThe information given by the problem statement looks as follows:\n\n\n\\begin{center}\n[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW)); dot(O);... | 2 | ./CreativeMath/AHSME/1999_AHSME_Problems/23.json | AHSME |
1999_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | Before Ashley started a three-hour drive, her car's odometer reading was 29792, a palindrome. (A palindrome is a number that reads the same way from left to right as it does from right to left). At her destination, the odometer reading was another palindrome. If Ashley never exceeded the speed limit of 75 miles per hou... | [
"Ashley could have traveled at most $3\\times 75 = 225$ miles.\n\n\nEach $5$-digit palindrome is uniquely determined by its first three digits. The next palindromes are 29892, 29992, 30003, and 30103. We may note that $30103-29792 = 311 > 225$, so this number and all larger ones are too large.\n\n\nOn the other han... | 1 | ./CreativeMath/AHSME/1999_AHSME_Problems/9.json | AHSME |
1973_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | A cowboy is 4 miles south of a stream which flows due east. He is also 8 miles west and 7 miles north of his cabin. He wishes to water his horse at the stream and return home. The shortest distance (in miles) he can travel and accomplish this is
$\textbf{(A)}\ 4+\sqrt{185} \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}... | [
"First, you draw a reflection of the cowboy across the river. Then, you draw the straight line from the \"cowboy\" to his cabin. This will be a $8, 15, 17$ Pythagorean triple, so the answer is $17$, which is $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/20.json | AHSME |
1973_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | If the sum of all the angles except one of a convex polygon is $2190^{\circ}$, then the number of sides of the polygon must be
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 19 \qquad \textbf{(E)}\ 21$
| [
"Let $n$ be the number of sides in the polygon. The number of interior angles in the polygon is $180(n-2)$. We know that the sum of all but one of them is $2190^{\\circ}$, so the sum of all the angles is more than that.\n\\[180(n-2) > 2190\\]\n\\[n-2 > 12 \\tfrac{1}{6}\\]\n\\[n > 14 \\tfrac{1}{6}\\]\n\n\nThe sum ... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/16.json | AHSME |
1973_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If 554 is the base $b$ representation of the square of the number whose base $b$ representation is 24, then $b$, when written in base 10, equals
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 16$
| [
"Write out the numbers using the definition of base numbers.\n\\[554_b = 5b^2 + 5b + 4\\]\n\\[24_b = 2b+4\\]\nSince $554_b = (24_b)^2$, we can write an equation.\n\\[5b^2 + 5b + 4 = (2b+4)^2\\]\n\\[5b^2 + 5b + 4 = 4b^2 + 16b + 16\\]\n\\[b^2 - 11b - 12 = 0\\]\n\\[(b-12)(b+1) = 0\\]\nSince base numbers must be positi... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/6.json | AHSME |
1973_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | The sum of all integers between 50 and 350 which end in 1 is
$\textbf{(A)}\ 5880\qquad\textbf{(B)}\ 5539\qquad\textbf{(C)}\ 5208\qquad\textbf{(D)}\ 4877\qquad\textbf{(E)}\ 4566$
| [
"The numbers that we are adding are $51,61,71 \\cdots 341$. The numbers are part of an arithmetic series with first term $51$, last term $341$, common difference $10$, and $30$ terms. Using the arithmetic series formula, the sum of the terms is $\\tfrac{30 \\cdot 392}{2} = \\boxed{\\textbf{(A)}\\ 5880}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/7.json | AHSME |
1973_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | If $\theta$ is an acute angle and $\sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}$, then $\tan \theta$ equals
$\textbf{(A)}\ x \qquad \textbf{(B)}\ \frac1{x} \qquad \textbf{(C)}\ \frac{\sqrt{x-1}}{x+1} \qquad \textbf{(D)}\ \frac{\sqrt{x^2-1}}{x} \qquad \textbf{(E)}\ \sqrt{x^2-1}$
| [
"Since $\\tfrac{\\theta}{2}$ is acute and $\\cos \\tfrac{\\theta}{2} = \\sqrt{1 - \\sin (\\tfrac{\\theta}{2})^2}$,\n\\[\\cos \\frac{\\theta}{2} = \\sqrt{1 - (\\frac{x-1}{2x})^2}\\]\n\\[\\cos \\frac{\\theta}{2} = \\sqrt{\\frac{x+1}{2x}}\\]\nUsing the definition of tangent,\n\\[\\tan \\frac{\\theta}{2} = \\sqrt{\\fra... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/17.json | AHSME |
1973_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | The number of sets of two or more consecutive positive integers whose sum is 100 is
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
| [
"If the first number of a group of $n$ consecutive numbers is $a$, the $n^\\text{th}$ number is $a+n-1$. We know that the sum of the group of numbers is $100$, so\n\\[\\frac{n(2a+n-1)}{2} = 100\\]\n\\[2a+n-1=\\frac{200}{n}\\]\n\\[2a = 1-n + \\frac{200}{n}\\]\nWe know that $n$ and $a$ are positive integers, so we c... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/21.json | AHSME |
1973_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | If $n$ is a real number, then the simultaneous system
$nx+y = 1$
$ny+z = 1$
$x+nz = 1$
has no solution if and only if $n$ is equal to
$\textbf{(A)}\ -1\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 0\text{ or }1\qquad\textbf{(E)}\ \frac{1}2$
| [
"Add up all three equations to get\n\\[(n+1)(x+y+z) = 3\\]\nIf $n = -1$, then equation results in $0 = 3$. That has no solutions, so the answer is $\\boxed{\\textbf{(A)}}$.\n\n",
"From the first equation,\n\\[y = 1 - nx\\]\nSubstitute that in the second equation to get\n\\[n(1-nx) + z = 1\\]\n\\[z = 1 - n + n^2 ... | 2 | ./CreativeMath/AHSME/1973_AHSME_Problems/10.json | AHSME |
1973_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | The number of terms in an A.P. (Arithmetic Progression) is even. The sum of the odd and even-numbered terms are 24 and 30, respectively. If the last term exceeds the first by 10.5, the number of terms in the A.P. is
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \tex... | [
"Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference. That means the last term is $a+r(n-1)$.\n\n\nWe can write an equation on the difference between the last and first term based on the conditions.\n\\[a+r(n-1)-a =10.5\\]\n\\[rn-r=10.5\\]\nAlso, half of the terms add up to $24$... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/26.json | AHSME |
1973_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | Let $[t]$ denote the greatest integer $\leq t$ where $t \geq 0$ and $S = \{(x,y): (x-T)^2 + y^2 \leq T^2 \text{ where } T = t - [t]\}$. Then we have
$\textbf{(A)}\ \text{the point } (0,0) \text{ does not belong to } S \text{ for any } t \qquad$
$\textbf{(B)}\ 0 \leq \text{Area } S \leq \pi \text{ for all } t \qqua... | [
"The region $S$ is a circle radius $T$ and center $(T,0)$. Since $T = t-[t]$, $0 \\le T < 1$. That means the area of the circle is less than $\\pi$, and since the region can also be just a dot (achieved when $t$ is integer), the answer is $\\boxed{\\textbf{(B)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/30.json | AHSME |
1973_AHSME_Problems | 31 | 0 | Number Theory | Multiple Choice | In the following equation, each of the letters represents uniquely a different digit in base ten:
\[(YE) \cdot (ME) = TTT\]
The sum $E+M+T+Y$ equals
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 24$
| [
"The right side of the equation can be rewritten as $111T = 37 \\cdot 3T$. With trial and error and prime factorization as a guide, we can test different digits of $T$ to see if we can find two two-digit numbers that have the same units digit and multiply to $111T$.\n\n\nThe only possibility that works is $37 \\cd... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/31.json | AHSME |
1973_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | Cars A and B travel the same distance. Car A travels half that distance at $u$ miles per hour and half at $v$ miles per hour. Car B travels half the time at $u$ miles per hour and half at $v$ miles per hour. The average speed of Car A is $x$ miles per hour and that of Car B is $y$ miles per hour. Then we always have
... | [
"Let $t$ be the total number of time in hours that Car B took to drive the distance. This means that for have the time Car B traveled $\\tfrac{ut}{2}$ miles and for half the time Car B traveled $\\tfrac{vt}{2}$ miles. That means the total distance traveled is $\\tfrac{ut+vt}{2}$ miles, so\n\\[y = \\frac{\\frac{ut... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/27.json | AHSME |
1973_AHSME_Problems | 1 | 0 | Geometry | Multiple Choice | A chord which is the perpendicular bisector of a radius of length 12 in a circle, has length
$\textbf{(A)}\ 3\sqrt3\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 6\sqrt3\qquad\textbf{(D)}\ 12\sqrt3\qquad\textbf{(E)}\ \text{ none of these}$
| [
"[asy] draw(circle((0,0),12)); draw((0,0)--(0,12)); draw((0,0)--(10.392,6)--(-10.392,6)--(0,0)); label(\"$12$\",(5.196,3),SE); label(\"$12$\",(-5.196,3),SW); label(\"$6$\",(0,3),E); draw((1,6)--(1,5)--(0,5)); [/asy]\n\n\nDraw a diagram as shown. Using the Pythagorean Theorem (or by using 30-60-90 triangles), hal... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/1.json | AHSME |
1973_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | A circle with a circumscribed and an inscribed square centered at the origin of a rectangular coordinate system with positive and axes and is shown in each figure to below.
[asy] size((400)); draw((0,0)--(22,0), EndArrow); draw((10,-10)--(10,12), EndArrow); draw((25,0)--(47,0), EndArrow); draw((35,-10)--(35,12... | [
"First, note that the following inequality \n\\[|x|+|y|\\leq\\sqrt{2(x^{2}+y^{2})}\\leq 2\\mbox{Max}(|x|, |y|)\\]\nrepresents the graphs $|x| + |y| = a$, $\\sqrt{2(x^{2}+y^{2})} = a$, and $2\\mbox{Max}(|x|, |y|) = a$. We don't actually have to worry about the inequality, we just want to find the picture that graphs... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/11.json | AHSME |
1973_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is
$\textbf{(A)}\ 600\qquad\textbf{(B)}\ 520\qquad\textbf{(C)}\ 488\qquad\textbf{(D)}\ 480\qqu... | [
"The total number of cubes is $10^3$ or $1000$. Because each surface of the large cube is one cube deep, the number of the unpainted cubes is $8^3 = 512$, since we subtract two from the side lengths of the cube itself, and cube it to find the volume of that cube. So there are $1000-512=\\boxed{\\textbf{(C) } 488}$ ... | 2 | ./CreativeMath/AHSME/1973_AHSME_Problems/2.json | AHSME |
1973_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | If $a$, $b$, and $c$ are in geometric progression (G.P.) with $1 < a < b < c$ and $n>1$ is an integer, then $\log_an$, $\log_bn$, $\log_cn$ form a sequence
$\textbf{(A)}\ \text{which is a G.P} \qquad$
$\textbf{(B)}\ \text{which is an arithmetic progression (A.P)} \qquad$
$\textbf{(C)}\ \text{in which the recipro... | [
"Using the change of base formula, the three logarithmic terms can be written as\n\\[\\frac{\\log n}{\\log a}, \\frac{\\log n}{\\log b}, \\frac{\\log n}{\\log c}\\]\nSince $a$, $b$, and $c$ are members of a geometric sequence, $b = ar$ and $c = ar^2$. That means the three logarithmic terms can be rewritten as\n\\... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/28.json | AHSME |
1973_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | The average (arithmetic mean) age of a group consisting of doctors and lawyers in 40. If the doctors average 35 and the lawyers 50 years old, then the ratio of the numbers of doctors to the number of lawyers is
$\textbf{(A)}\ 3: 2\qquad\textbf{(B)}\ 3: 1\qquad\textbf{(C)}\ 2: 3\qquad\textbf{(D)}\ 2: 1\qquad\textbf{(... | [
"Let $d$ be the number of doctors and $l$ be the number of lawyers. The average age of the doctors is $35$, so the sum of the doctors' ages is $35d$. The average age of the lawyers is $50$, so the sum of the lawyers' ages is $50l$. With the two information, we can write an equation.\n\n\n\\[\\frac{35d+50l}{d+l} ... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/12.json | AHSME |
1973_AHSME_Problems | 32 | 0 | Geometry | Multiple Choice | The volume of a pyramid whose base is an equilateral triangle of side length 6 and whose other edges are each of length $\sqrt{15}$ is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 9/2 \qquad \textbf{(C)}\ 27/2 \qquad \textbf{(D)}\ \frac{9\sqrt3}{2} \qquad \textbf{(E)}\ \text{none of these}$
| [
"[asy] import three; unitsize(1cm); size(200); draw((0,0,0)--(6,0,0)--(3,5.196,0)--(0,0,0)); draw((3,1.732,1.732)--(0,0,0)); draw((3,1.732,1.732)--(6,0,0)); draw((3,1.732,1.732)--(3,5.196,0)); draw((3,1.732,1.732)--(3,1.732,0)--(0,0,0),dotted); label(\"6\",(4.5,2.598,0),SW); label(\"$\\sqrt{15}$\",(4.5,0.866,0.866)... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/32.json | AHSME |
1973_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | The check for a luncheon of 3 sandwiches, 7 cups of coffee and one piece of pie came to $$3.15$. The check for a luncheon consisting of 4 sandwiches, 10 cups of coffee and one piece of pie came to $$4.20$ at the same place. The cost of a luncheon consisting of one sandwich, one cup of coffee, and one piece of pie at th... | [
"Let $s$ be the cost of one sandwich, $c$ be the cost of one cup of coffee, and $p$ be the price of one piece of pie. With the information,\n\n\n\\[3s+7c+p=3.15\\]\n\\[4s+10c+p=4.20\\]\n\n\nSubtract the first equation from the second to get\n\n\n\\[s+3c=1.05\\]\n\n\nThat means $s=1.05-3c$. Substituting it back in... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/24.json | AHSME |
1973_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | A circular grass plot 12 feet in diameter is cut by a straight gravel path 3 feet wide, one edge of which passes through the center of the plot. The number of square feet in the remaining grass area is
$\textbf{(A)}\ 36\pi-34 \qquad \textbf{(B)}\ 30\pi - 15 \qquad \textbf{(C)}\ 36\pi - 33 \qquad$
$\textbf{(D)}\ 35... | [
"[asy] draw(circle((0,0),6)); draw((-6,0)--(6,0)); pair A=(-5.196,3),B=(5.196,3); draw(A--B--(0,0)--A); draw((0,0)--(0,3)); label(\"6\",(-2.598,1.5),SW); label(\"3\",(0,1.5),E); draw((-0.5,3)--(-0.5,2.5)--(0,2.5)); [/asy]\n\n\nThe wanted area can be divided into a semicircle and a circular segment. The area of... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/25.json | AHSME |
1973_AHSME_Problems | 33 | 0 | Algebra | Multiple Choice | When one ounce of water is added to a mixture of acid and water, the new mixture is $20\%$ acid. When one ounce of acid is added to the new mixture, the result is $33\frac13\%$ acid. The percentage of acid in the original mixture is
$\textbf{(A)}\ 22\% \qquad \textbf{(B)}\ 24\% \qquad \textbf{(C)}\ 25\% \qquad \textb... | [
"Let $a$ be the original number of ounces of acid and $w$ be the original number of ounces of water. We can write two equations since we know the percentage of acid after some water and acid.\n\\[\\frac{a}{a+w+1} = \\frac{1}{5}\\]\n\\[\\frac{a+1}{a+w+2} = \\frac{1}{3}\\]\nCross-multiply to get rid of the fractions... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/33.json | AHSME |
1973_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | The fraction $\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}$ is equal to
$\textbf{(A)}\ \frac{2\sqrt2}{3} \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ \frac{2\sqrt3}3 \qquad \textbf{(D)}\ \frac43 \qquad \textbf{(E)}\ \frac{16}{9}$
| [
"Squaring the expression and taking the positive square root (since numerator and denominator are positive) of the result yields\n\\[\\sqrt{\\left(\\frac{2(\\sqrt2+\\sqrt6)}{3\\sqrt{2+\\sqrt3}}\\right)^2}\\]\n\\[\\sqrt{\\frac{4(2+4\\sqrt{3}+6)}{9(2+\\sqrt{3})}}\\]\n\\[\\sqrt{\\frac{4(8+4\\sqrt{3})}{9(2+\\sqrt{3})}}... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/13.json | AHSME |
1973_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | Two boys start moving from the same point A on a circular track but in opposite directions. Their speeds are 5 ft. per second and 9 ft. per second. If they start at the same time and finish when they first meet at the point A again, then the number of times they meet, excluding the start and finish, is
$\textbf{(A)}... | [
"Let $d$ be the length of the track in feet and $x$ be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is $\\tfrac{dx}{5}$. Since the time elapsed for both boys is equal, one boy ran $5(\\tfrac{dx}{5})$ feet while the other boy ran $9(\\tfrac{dx}{5})$ feet. Beca... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/29.json | AHSME |
1973_AHSME_Problems | 3 | 0 | Number Theory | Multiple Choice | The stronger Goldbach conjecture states that any even integer greater than 7 can be written as the sum of two different prime numbers. For such representations of the even number 126, the largest possible difference between the two primes is
$\textbf{(A)}\ 112\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 92\qquad\text... | [
"We can guess and check small primes, subtract it from $126$, and see if the result is a prime because the further away the two numbers are, the greater the difference will be. Since $126 = 2 \\cdot 3^2 \\cdot 7$, we can eliminate $2$, $3$, and $7$ as an option because subtracting these would result in a composite... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/3.json | AHSME |
1973_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | A plane flew straight against a wind between two towns in 84 minutes and returned with that wind in 9 minutes less than it would take in still air. The number of minutes (2 answers) for the return trip was
$\textbf{(A)}\ 54 \text{ or } 18 \qquad \textbf{(B)}\ 60 \text{ or } 15 \qquad \textbf{(C)}\ 63 \text{ or } 12 \... | [
"Let $d$ be the distance between the two towns, $p$ be the speed of the plane, and $w$ be the speed of the wind. Since the time it took to fly against the wind towards the other town is $84$ minutes,\n\\[\\frac{d}{p-w} = 84\\]\n\\[d = 84p-84w\\]\nSince flying with the wind on the return trip takes $9$ minutes less... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/34.json | AHSME |
1973_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | If 1 pint of paint is needed to paint a statue 6 ft. high, then the number of pints it will take to paint (to the same thickness) 540 statues similar to the original but only 1 ft. high is
$\textbf{(A)}\ 90\qquad\textbf{(B)}\ 72\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 15$
| [
"The statues are similar, and since the height if the smaller statue is $\\tfrac{1}{6}$ of the original statue, the surface area is $\\tfrac{1}{36}$ of the original statue. Thus, $\\tfrac{1}{36}$ pints of paint is needed for one 1 ft. statue, so painting 540 of these statues requires $\\tfrac{540}{36} = \\boxed{\\... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/8.json | AHSME |
1973_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | The set of all real solutions of the inequality
\[|x - 1| + |x + 2| < 5\]
is
$\textbf{(A)}\ x \in ( - 3,2) \qquad \textbf{(B)}\ x \in ( - 1,2) \qquad \textbf{(C)}\ x \in ( - 2,1) \qquad$
$\textbf{(D)}\ x \in \left( - \frac32,\frac72\right) \qquad \textbf{(E)}\ \O \text{ (empty})$
| [
"We can do casework upon the value of $x$. First, consider the case where both absolute values are positive, which is when $x \\geq 1$. In this case, the equation becomes $2x + 1 < 5$. This turns into $x < 2$. Combining this with our original assumption, we get the solutions $1 \\leq x < 2$. \n\n\nNext, consider th... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/22.json | AHSME |
1973_AHSME_Problems | 18 | 0 | Number Theory | Multiple Choice | If $p \geq 5$ is a prime number, then $24$ divides $p^2 - 1$ without remainder
$\textbf{(A)}\ \text{never} \qquad \textbf{(B)}\ \text{sometimes only} \qquad \textbf{(C)}\ \text{always} \qquad$
$\textbf{(D)}\ \text{only if } p =5 \qquad \textbf{(E)}\ \text{none of these}$
| [
"Starting with some experimentation, substituting $p=5$ results in $24$, substituting $p=7$ results in $48$, and substituting $p=11$ results in $120$. For these primes, the resulting numbers are multiples of $24$.\n\n\nTo show that all primes we devise the following proof:\n\n\n$Proof:$\n$p \\ge 5$ result in $p^2 ... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/18.json | AHSME |
1973_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | Two congruent 30-60-90 are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area common to both triangles is
$\textbf{(A)}\ 6\sqrt3\qquad\textbf{(B)}\ 8\sqrt3\qquad\textbf{(C)}\ 9\sqrt3\qquad\textbf{(D)}\ 12\sqrt3\qquad\textbf{(E)}\ 24$
| [
"[asy] fill((0,3.464)--(-6,0)--(6,0)--cycle,yellow); draw((-6,0)--(6,0)--(-3,5.196)--(-6,0)); draw((-6,0)--(6,0)--(3,5.196)--(-6,0)); draw((0,3.464)--(0,0)); label(\"$6$\",(-3,0),S); label(\"$6$\",(3,0),S); [/asy]\n\n\nNote that the altitude of the shared region bisects the hypotenuse of the original two right tr... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/4.json | AHSME |
1973_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | Each valve $A$, $B$, and $C$, when open, releases water into a tank at its own constant rate. With all three valves open, the tank fills in 1 hour, with only valves $A$ and $C$ open it takes 1.5 hours, and with only valves $B$ and $C$ open it takes 2 hours. The number of hours required with only valves $A$ and $B$ open... | [
"Let the rate of water flowing through valve $A$ be $a$, the rate of water flowing through valve $B$ be $b$, and the rate of water flowing through valve $C$ be $c$. WLOG, let the volume of the tank be 1 liter, and let the units for the rates be liters per hour. With this information, we can write three equations.... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/14.json | AHSME |
1973_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | A sector with acute central angle $\theta$ is cut from a circle of radius 6. The radius of the circle circumscribed about the sector is
$\textbf{(A)}\ 3\cos\theta \qquad \textbf{(B)}\ 3\sec\theta \qquad \textbf{(C)}\ 3 \cos \frac12 \theta \qquad \textbf{(D)}\ 3 \sec \frac12 \theta \qquad \textbf{(E)}\ 3$
| [
"Let $O$ be the center of the circle and $A,B$ be two points on the circle such that $\\angle AOB = \\theta$. If the circle circumscribes the sector, then the circle must circumscribe $\\triangle AOB$.\n\n\n[asy] draw((-120,-160)--(0,0)--(120,-160)); draw((-60,-80)--(0,-125)--(60,-80),dotted); draw((0,0)--(0,-125)... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/15.json | AHSME |
1973_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | Of the following five statements, I to V, about the binary operation of averaging (arithmetic mean),
$\text{I. Averaging is associative }$
$\text{II. Averaging is commutative }$
$\text{III. Averaging distributes over addition }$
$\text{IV. Addition distributes over averaging }$
$\text{V. Averaging has an ide... | [
"We can consider each statement independently and see which ones are true. The average of two numbers $x$ and $y$ is $avg(x, y) = \\frac{x + y}{2}$.\n\n\n\\textbf{Statement I}\n\\[avg(avg(x, y), z) = avg(x, avg(y, z))\\]\n\\[\\frac{\\frac{x + y}{2} + z}{2} = \\frac{x + \\frac{y + z}{2}}{2}\\]\n\\[\\frac{x + y + 2z}... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/5.json | AHSME |
1973_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Define $n_a!$ for $n$ and $a$ positive to be
\[n_a ! = n (n-a)(n-2a)(n-3a)...(n-ka)\]
where $k$ is the greatest integer for which $n>ka$. Then the quotient $72_8!/18_2!$ is equal to
$\textbf{(A)}\ 4^5 \qquad \textbf{(B)}\ 4^6 \qquad \textbf{(C)}\ 4^8 \qquad \textbf{(D)}\ 4^9 \qquad \textbf{(E)}\ 4^{12}$
| [
"Using the definition of $n_a!$, the quotient can be rewritten as\n\\[\\frac{72 \\cdot 64 \\cdot 56 \\cdots 8}{18 \\cdot 16 \\cdot 14 \\cdots 2}\\]\nNote that for a given integer $x$, $\\tfrac{72-8x}{18-2x} = 4$. Since $0 \\le x \\le 8$, the quotient simplifies to $\\boxed{\\textbf{(D)}\\ 4^9}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/19.json | AHSME |
1973_AHSME_Problems | 23 | 0 | Probability | Multiple Choice | There are two cards; one is red on both sides and the other is red on one side and blue on the other. The cards have the same probability (1/2) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is red, then the probability that the under-side is also red is
$\text... | [
"There are three red faces, and two are on the card that is completely red, so our answer is $\\frac{2}{3}$, which is $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/23.json | AHSME |
1973_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | In $\triangle ABC$ with right angle at $C$, altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K$, then the area of $\triangle ABC$ is
$\textbf{(A)}\ 6K\qquad\textbf{(B)}\ 4\sqrt3\ K\qquad\textbf{(C)}\ 3\sqrt3\ K\qquad\textbf{(D)}\ 3K\qquad\textbf{(E)}\ 4K$
| [
"[asy] pair A=(-6,0),B=(6,0),C=(-3,5.196),M=(0,0),H=(-3,0); draw((-6,0)--(6,0)--(-3,5.196)--(-6,0)); draw((-3,5.196)--(0,0)); draw(C--H); dot(A); label(\"$A$\",A,SW); dot(B); label(\"$B$\",B,SE); dot(C); label(\"$C$\",C,N); dot(M); label(\"$M$\",M,S); dot(H); label(\"$H$\",H,S); markscalefactor=0.1; draw(anglemark... | 1 | ./CreativeMath/AHSME/1973_AHSME_Problems/9.json | AHSME |
1973_AHSME_Problems | 35 | 0 | Geometry | Multiple Choice | In the unit circle shown in the figure, chords $PQ$ and $MN$ are parallel to the unit radius $OR$ of the circle with center at $O$. Chords $MP$, $PQ$, and $NR$ are each $s$ units long and chord $MN$ is $d$ units long.
[asy] draw(Circle((0,0),10)); draw((0,0)--(10,0)--(8.5,5.3)--(-8.5,5.3)--(-3,9.5)--(3,9.5)); dot((0... | [
"[asy] pair O=(0,0), R=(10,0), M=(-8.5,5.3), N=(8.5,5.3), P=(-3,9.5), Q=(3,9.5); draw(Circle(O,10)); draw(O--R--N--M--P--Q); dot(O); dot(R); dot(N); dot(M); dot(P); dot(Q); label(\"1\", (5,0), S); label(\"s\", (10.3,2.6)); label(\"s\", (-5,7)); label(\"s\", (0,8.5)); label(\"O\", O, S); label(\"R\", R, E); label(... | 2 | ./CreativeMath/AHSME/1973_AHSME_Problems/35.json | AHSME |
1966_AHSME_Problems | 20 | 0 | Other | Multiple Choice | The negation of the proposition "For all pairs of real numbers $a,b$, if $a=0$, then $ab=0$" is: There are real numbers $a,b$ such that
$\text{(A) } a\ne 0 \text{ and } ab\ne 0 \qquad \text{(B) } a\ne 0 \text{ and } ab=0 \qquad \text{(C) } a=0 \text{ and } ab\ne 0$
$\text{(D) } ab\ne 0 \text{ and } a\ne 0 \qquad \... | [
"$\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/20.json | AHSME |
1966_AHSME_Problems | 36 | 0 | Algebra | Multiple Choice | Let $(1+x+x^2)^n=a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}$ be an identity in $x$. If we let $s=a_0+a_2+a_4+\cdots +a_{2n}$, then $s$ equals:
$\text{(A) } 2^n \quad \text{(B) } 2^n+1 \quad \text{(C) } \frac{3^n-1}{2} \quad \text{(D) } \frac{3^n}{2} \quad \text{(E) } \frac{3^n+1}{2}$
| [
"Let $f(x)=(1+x+x^2)^n$ then we have \n\\[f(1)=a_0+a_1+a_2+...+a_{2n}=(1+1+1)^n=3^n\\]\n\\[f(-1)=a_0-a_1+a_2-...+a_{2n}=(1-1+1)^n=1\\]\nAdding yields\n\\[f(1)+f(-1)=2(a_0+a_2+a_4+...+a_{2n})=3^n+1\\]\nThus $s=\\frac{3^n+1}{2}$, or $\\boxed{E}$.\n\n\n~ Nafer\n\n\n~ Minor edits by Qinglang, Nafer wrote a great soluti... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/36.json | AHSME |
1966_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | If $\frac{4^x}{2^{x+y}}=8$ and $\frac{9^{x+y}}{3^{5y}}=243$, $x$ and $y$ real numbers, then $xy$ equals:
$\text{(A) } \frac{12}{5} \quad \text{(B) } 4 \quad \text{(C) } 6 \quad \text{(D)} 12 \quad \text{(E) } -4$
| [
"$\\frac{4^x}{2^{x+y}}=8\\implies 4^x = 2^{x+y+3}\\implies 2x=x+y+3 \\implies x = y+3$.\n$\\frac{9^{x+y}}{3^{5y}}=243\\implies 9^{x+y}=3^{5y+5}\\implies 2x+2y=5y+5\\implies 2x = 3y +5$.\nSo, $2y+6=3y+5\\implies y = 1 \\implies x = 4$. Therefore, $xy = 4$ or $\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/16.json | AHSME |
1966_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | $AB$ is the diameter of a circle centered at $O$. $C$ is a point on the circle such that angle $BOC$ is $60^\circ$. If the diameter of the circle is $5$ inches, the length of chord $AC$, expressed in inches, is:
$\text{(A)} \ 3 \qquad \text{(B)} \ \frac {5\sqrt {2}}{2} \qquad \text{(C)} \frac {5\sqrt3}{2} \ \qquad \t... | [
"[asy] draw(unitcircle); draw((-1,0)--(1,0)--(1/2, sqrt(3)/2)--cycle); label( \"A\", (-1,0), W); label( \"B\", (1,0), E); label( \"C\", (1/2, sqrt(3)/2), N); [/asy]\nWe see that $\\angle A$ is half the measure of $\\angle BOC$, so $\\angle A = 30^{\\circ}$. That makes $ABC$ a $30-60-90$ triangle and sidelength $\\o... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/6.json | AHSME |
1966_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | Let $\frac {35x - 29}{x^2 - 3x + 2} = \frac {N_1}{x - 1} + \frac {N_2}{x - 2}$ be an identity in $x$. The numerical value of $N_1N_2$ is:
$\text{(A)} \ - 246 \qquad \text{(B)} \ - 210 \qquad \text{(C)} \ - 29 \qquad \text{(D)} \ 210 \qquad \text{(E)} \ 246$
| [
"\\[\\frac {N_1}{x - 1} + \\frac {N_2}{x - 2} = \\frac{N_1(x-2) + N_2(x-1)}{(x-1)(x-2)} = \\frac{(N_1 + N_2)x - (2N_1 + N_2)}{x^2 - 3x + 2}\\]\n\n\nComparing coefficients, we have the system of two equations\n\n\n\\begin{align*} N_1 + N_2 &= 35\\\\ N_1 + 2N_2 &= 29 \\end{align*}\n\n\nSubtracting the first equation ... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/7.json | AHSME |
1966_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | The number of distinct points common to the curves $x^2+4y^2=1$ and $4x^2+y^2=4$ is:
$\text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 3 \quad \text{(E) } 4$
| [
"Let $a=x^2$ and $b=y^2$. \nWe now have a system of 2 linear equations:\n$a+4b=1\\\\4a+b=4$\nMultiplying the first equation by 4 and then subtracting the second equation from the first one, we get:\n$15b=0\\\\b=0$\nNow, we can substitute b to solve for a:\n$4a+0=4\\\\a=1$\nNow note that $x^2=1 \\rightarrow x=\\pm 1... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/17.json | AHSME |
1966_AHSME_Problems | 40 | 0 | Geometry | Multiple Choice | [asy] draw(circle((0,0),10),black+linewidth(1)); MP("O", (0,0), S);MP("A", (-10,0), W);MP("B", (10,0), E);MP("C", (10,10), E);MP("D", (6,8), N); MP("a", (-5,0), S);MP("E", (-6,3), N); dot((0,0));dot((-6,2)); draw((-10,0)--(10,0),black+linewidth(1)); draw((-10,0)--(10,10),black+linewidth(1)); draw((-10,-12)--(-10,12),bl... | [
"$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/40.json | AHSME |
1966_AHSME_Problems | 37 | 0 | Algebra | Multiple Choice | Three men, Alpha, Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let $h$ be the number of hours needed by Alpha and Beta, working together, to do the job. Then $h$ equals:
$\text{(A) }... | [
"$\\fbox{C}$\n\n\n",
"Let $A$,$B$,$C$ denote the number of hours needed by Alpha, Beta, Gamma, respectively. We also have their respective efficiency $\\frac{1}{A}$, $\\frac{1}{B}$, and $\\frac{1}{C}$. Thus we get the equations\n\\[\\frac{1}{A}+\\frac{1}{B}+\\frac{1}{C}=\\frac{1}{A-6}\\]\n\\[\\frac{1}{A}+\\frac{1... | 2 | ./CreativeMath/AHSME/1966_AHSME_Problems/37.json | AHSME |
1966_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | [asy] draw((0,-5)--(-6,10),black+dashed+linewidth(1)); draw((-6,10)--(10,0),black+dashed+linewidth(1)); draw((10,0)--(-10,0),black+dashed+linewidth(1)); draw((-10,-0)--(10,10),black+dashed+linewidth(1)); draw((10,10)--(0,-5),black+dashed+linewidth(1)); draw((-2,0)--(-10/3,10/3),black+linewidth(2)); draw((-10/3,10/3)--(... | [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/21.json | AHSME |
1966_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | If the sum of two numbers is $1$ and their product is $1$, then the sum of their cubes is:
$\text{(A)} \ 2 \qquad \text{(B)} \ - 2 - \frac {3i\sqrt {3}}{4} \qquad \text{(C)} \ 0 \qquad \text{(D)} \ - \frac {3i\sqrt {3}}{4} \qquad \text{(E)} \ - 2$
| [
"Let the two numbers be $a,b$; then $a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)[(a+b)^2 -3ab]$ $= (1)[1^2 - 3 \\cdot 1] = -2 \\Rightarrow \\mathrm{(E)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/10.json | AHSME |
1966_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | Let $m$ be a positive integer and let the lines $13x+11y=700$ and $y=mx-1$ intersect in a point whose coordinates are integers. Then m can be:
$\text{(A) 4 only} \quad \text{(B) 5 only} \quad \text{(C) 6 only} \quad \text{(D) 7 only} \\ \text{(E) one of the integers 4,5,6,7 and one other positive integer}$
| [
"Substitute the second equation into the first one, we have $13x+11mx-11=700$.\n\n\nSo $(13+11m)x=711$. So $13+11m$ is a factor of $711$. $711=3*3*79$, so the factors of $711$ are: $1, 3, 9, 79, 237, 711$.\n\n\nClearly, because $m\\ge1$, so $13+11m\\ge24$. So we only need to check whether $m$ is an integer when $11... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/26.json | AHSME |
1966_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | If three of the roots of $x^4+ax^2+bx+c=0$ are $1$, $2$, and $3$ then the value of $a+c$ is:
$\text{(A) } 35 \quad \text{(B) } 24 \quad \text{(C) } -12 \quad \text{(D) } -61 \quad \text{(E) } -63$
| [
"Since this is a quartic equation, there are going to be $4$ solutions. \nBy Vieta's formulas, since the $x^3$ term is $0$, the sum of the roots is also $0$. Therefore, the 4th root of this polynomial is $-6$. \nLastly, by Vieta's Formulas, $a+c=-(1\\cdot2+1\\cdot3+1\\cdot-6+2\\cdot3+2\\cdot-6+3\\cdot-6)+1\\cdot2\\... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/30.json | AHSME |
1966_AHSME_Problems | 31 | 0 | Geometry | Multiple Choice | [asy] draw(circle((0,0),10),black+linewidth(1)); draw(circle((-1.25,2.5),4.5),black+linewidth(1)); dot((0,0)); dot((-1.25,2.5)); draw((-sqrt(96),-2)--(-2,sqrt(96)),black+linewidth(.5)); draw((-2,sqrt(96))--(sqrt(96),-2),black+linewidth(.5)); draw((-sqrt(96),-2)--(sqrt(96)-2.5,7),black+linewidth(.5)); draw((-sqrt(96),-2... | [
"We will prove that $\\triangle DOB$ and $\\triangle COD$ is isosceles, meaning that $CD=OD=BD$ and hence $\\fbox{D}$. \n\n\nLet $\\angle A=2\\alpha$ and $\\angle B=2\\beta$. Since the incentre of a triangle is the intersection of its angle bisectors, $\\angle OAB=\\alpha$ and $\\angle ABO=\\beta$. Hence $\\angle D... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/31.json | AHSME |
1966_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | At his usual rate a man rows 15 miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream's current is:
$\text{(A) } 2 \quad \text{(B) } \frac{5}{2} \quad \text{(C) } 3... | [
"Let the speed of rowing in still water and water speed be $r, w$. \n\n\nThen: \n\n\n$\\frac{15}{r-w}-\\frac{15}{r+w}=5$.\n\n\n$\\frac{15}{2r-w}-\\frac{15}{2r+w}=1$.\n\n\nSolving (remember: both variables must be positive!), we have $r=4, w=2$. Select $\\boxed{A}$.\n\n\n~hastapasta\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/27.json | AHSME |
1966_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | Given that the ratio of $3x - 4$ to $y + 15$ is constant, and $y = 3$ when $x = 2$, then, when $y = 12$, $x$ equals:
$\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)} \ \frac72 \qquad \text{(E)} \ 8$
| [
"Let $k$ be the constant ratio. Then $k = \\frac{3(2)-4}{(3)+15} = \\frac{1}{9} = \\frac{3x - 4}{(12) + 15}$. Solving gives $3x - 4 = 3 \\Longrightarrow x = \\frac 73 \\Rightarrow \\mathrm{(B)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/1.json | AHSME |
1966_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | The sides of triangle $BAC$ are in the ratio $2:3:4$. $BD$ is the angle-bisector
drawn to the shortest side $AC$, dividing it into segments $AD$ and $CD$.
If the length of $AC$ is $10$, then the length of the longer segment of $AC$ is:
$\text{(A)} \ 3\frac{1}{2} \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac{5}{7} ... | [
"By the Angle Bisector Theorem, we have $\\frac{BA}{AD}=\\frac{BC}{CD}$ which implies $\\frac{AD}{DC}=\\frac{BA}{BC}=\\frac{3}{4}$. So $AC=10=AD+DC=\\frac{7}{4}DC$, and thus $DC=\\frac{40}{7}=5\\frac{5}{7}$.\nHence\n$\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/11.json | AHSME |
1966_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | Five points $O,A,B,C,D$ are taken in order on a straight line with distances $OA = a$, $OB = b$, $OC = c$, and $OD = d$. $P$ is a point on the line between $B$ and $C$ and such that $AP: PD = BP: PC$. Then $OP$ equals:
$\textbf{(A)} \frac {b^2 - bc}{a - b + c - d} \qquad \textbf{(B)} \frac {ac - bd}{a - b + c - d} ... | [
"$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/28.json | AHSME |
1966_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | The number of real values of $x$ that satisfy the equation \[(2^{6x+3})(4^{3x+6})=8^{4x+5}\] is:
$\text{(A) zero} \qquad \text{(B) one} \qquad \text{(C) two} \qquad \text{(D) three} \qquad \text{(E) greater than 3}$
| [
"We know that\n$2^{6x+3}\\cdot4^{3x+6}=2^{6x+3}\\cdot(2^2)^{3x+6}=2^{6x+3}\\cdot2^{6x+12}=2^{12x+15}$.\nWe also know that\n$8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}$.\nThere are infinite solutions to the equation $2^{12x+15}=2^{12x+15}$, so the answer is $\\boxed{\\text{(E) greater than 3}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/12.json | AHSME |
1966_AHSME_Problems | 32 | 0 | Geometry | Multiple Choice | Let $M$ be the midpoint of side $AB$ of triangle $ABC$. Let $P$ be a point on $AB$ between $A$ and $M$, and let $MD$ be drawn parallel to $PC$ and intersecting $BC$ at $D$. If the ratio of the area of triangle $BPD$ to that of triangle $ABC$ is denoted by $r$, then
$\text{(A) } \frac{1}{2}<r<1 \text{, depending upon... | [
"$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/32.json | AHSME |
1966_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | If $Log_M{N}=Log_N{M},M \ne N,MN>0,M \ne 1, N \ne 1$, then $MN$ equals:
$\text{(A) } \frac{1}{2} \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 10 \\ \text{(E) a number greater than 2 and less than 10}$
| [
"If we change the base of $Log_M{N}$ to base $N$, we get $\\frac{1}{log_N{M}}= Log_N{M}$. Multiplying both sides by $Log_N{M}$, we get $Log_N{M}^2=1$. Since $N\\not = M$, $Log_N{M}=-1$. So $N^{-1}=M \\rightarrow \\frac{1}{N} = M \\rightarrow MN=1$. So the answer is $\\boxed{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/24.json | AHSME |
1966_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | If $F(n+1)=\frac{2F(n)+1}{2}$ for $n=1,2,\cdots$ and $F(1)=2$, then $F(101)$ equals:
$\text{(A) } 49 \quad \text{(B) } 50 \quad \text{(C) } 51 \quad \text{(D) } 52 \quad \text{(E) } 53$
| [
"Notice that $\\frac{2F(n)+1}{2}=F(n)+\\frac{1}{2}.$\n\n\nThis means that for every single increment $n$ goes up from $1$, $F(n)$ will increase by $\\frac{1}{2}.$ Since $101$ is $100$ increments from $1$, $F(n)$ will increase $\\frac{1}{2}\\times100=50.$\n\n\nSince $F(1)=2,$ $F(101)$ will equal $2+50=\\boxed{\\text... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/25.json | AHSME |
1966_AHSME_Problems | 33 | 0 | Algebra | Multiple Choice | If $ab \ne 0$ and $|a| \ne |b|$, the number of distinct values of $x$ satisfying the equation
\[\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b},\]
is:
$\text{(A) zero} \quad \text{(B) one} \quad \text{(C) two} \quad \text{(D) three} \quad \text{(E) four}$
| [
"$\\fbox{D}$\n\n\n",
"Let $m=\\frac{x-a}{b}$ and $n=\\frac{x-b}{a}$ then we have\n\\[m+n=\\frac{1}{m}+\\frac{1}{n}\\]\n\\[m+n=\\frac{m+n}{mn}\\]\nNotice that the equation is possible iff $m+n=0$ or $mn=1$.\n\n\nIf $m+n=0$ then\n\\[\\frac{x-a}{b}+\\frac{x-b}{a}=0\\]\n\\[\\frac{x-a}{b}=\\frac{b-x}{a}\\]\n\\[x=\\fra... | 2 | ./CreativeMath/AHSME/1966_AHSME_Problems/33.json | AHSME |
1966_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | The number of points with positive rational coordinates selected from the set of points in the $xy$-plane such that $x+y \le 5$, is:
$\text{(A)} \ 9 \qquad \text{(B)} \ 10 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E) infinite}$
| [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/13.json | AHSME |
1966_AHSME_Problems | 29 | 0 | Counting | Multiple Choice | The number of positive integers less than $1000$ divisible by neither $5$ nor $7$ is:
$\text{(A) } 688 \quad \text{(B) } 686 \quad \text{(C) } 684 \quad \text{(D) } 658 \quad \text{(E) } 630$
| [
"The number of positive integers under $1000$ that are divisible by $5$ is $\\lfloor\\frac{999}{5}\\rfloor=199$. The number of positive integers under $1000$ that are divisible by $7$ is $\\lfloor\\frac{1000}{7}\\rfloor=142$. Adding them together, we get $341$. However, we have over counted the numbers which are di... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/29.json | AHSME |
1966_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | If the arithmetic mean of two numbers is $6$ and their geometric mean is $10$, then an equation with the given two numbers as roots is:
$\text{(A)} \ x^2 + 12x + 100 = 0 ~~ \text{(B)} \ x^2 + 6x + 100 = 0 ~~ \text{(C)} \ x^2 - 12x - 10 = 0$
$\text{(D)} \ x^2 - 12x + 100 = 0 \qquad \text{(E)} \ x^2 - 6x + 100 = 0$
| [
"Let the numbers be $\\eta$ and $\\zeta$.\n\n\n\\[\\dfrac{\\eta+\\zeta}{2}=6\\Rightarrow \\eta+\\zeta=12\\].\n\n\n\\[\\sqrt{\\eta\\zeta}=10\\Rightarrow \\eta\\zeta=100\\].\n\n\nThe monic quadratic with roots $\\eta$ and $\\zeta$ is $x^2-(\\eta+\\zeta)x+\\eta\\zeta$. Therefore, an equation with $\\eta$ and $\\zeta$ ... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/3.json | AHSME |
1966_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | Let $r$ be the speed in miles per hour at which a wheel, $11$ feet in circumference, travels. If the time for a complete rotation of the wheel is shortened by $\frac{1}{4}$ of a second, the speed $r$ is increased by $5$ miles per hour. Then $r$ is:
$\text{(A) } 9 \quad \text{(B) } 10 \quad \text{(C) } 10\frac{1}{2} \... | [
"The circumference of the wheel is $\\frac{11}{5280}$ miles. Let the time for the rotation in seconds be $t$. So $rt=\\frac{11}{5280}*3600$. We also know reducing the time by $\\frac{1}{4}$ of a second makes $r$ increase by $5$. So $(r+5)(t-\\frac{1}{4})=\\frac{11}{5280}*3600$. Solving for $r$ we get $r=10$. So our... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/34.json | AHSME |
1966_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | Consider the statements: (I)$\sqrt{a^2+b^2}=0$, (II) $\sqrt{a^2+b^2}=ab$, (III) $\sqrt{a^2+b^2}=a+b$, (IV) $\sqrt{a^2+b^2}=a - b$, where we allow $a$ and $b$ to be real or complex numbers. Those statements for which there exist solutions other than $a=0$ and $b=0$, are:
$\text{(A) (I),(II),(III),(IV)}\quad \te... | [
"$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/22.json | AHSME |
1966_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | In a given arithmetic sequence the first term is $2$, the last term is $29$, and the sum of all the terms is $155$. The common difference is:
$\text{(A) } 3 \qquad \text{(B) } 2 \qquad \text{(C) } \frac{27}{19} \qquad \text{(D) } \frac{13}{9} \qquad \text{(E) } \frac{23}{38}$
| [
"$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/18.json | AHSME |
1966_AHSME_Problems | 38 | 0 | Geometry | Multiple Choice | In triangle $ABC$ the medians $AM$ and $CN$ to sides $BC$ and $AB$, respectively, intersect in point $O$. $P$ is the midpoint of side $AC$, and $MP$ intersects $CN$ in $Q$. If the area of triangle $OMQ$ is $n$, then the area of triangle $ABC$ is:
$\text{(A) } 16n \quad \text{(B) } 18n \quad \text{(C) } 21n \quad \tex... | [
"Construct triangle $\\triangle ABC$ with points $M,N,P$ being the midpoints of sides $\\overline{CB}, \\overline{AB}, \\overline{AC}$, respectively. Proceed by drawing all medians. Then draw all medians (so draw $\\overline{AM}, \\overline{BP}, \\overline{CN}$). Next, draw line $\\overline{PM}$ and label $\\overli... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/38.json | AHSME |
1966_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | Circle I is circumscribed about a given square and circle II is inscribed in the given square. If $r$ is the ratio of the area of circle I to that of circle II, then $r$ equals:
$\text{(A) } \sqrt{2} \quad \text{(B) } 2 \quad \text{(C) } \sqrt{3} \quad \text{(D) } 2\sqrt{2} \quad \text{(E) } 2\sqrt{3}$
| [
"Make half of the square's side $x$. Now the radius of the smaller circle is $x$, so it's area is $\\pi x^2$.\nNow find the diameter of the bigger circle. Since half of the square's side is $x$, the full side is $2x$. Using the Pythagorean theorem, you get the diagonal to be $2\\sqrt{2}x$. Half of that is the radiu... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/4.json | AHSME |
1966_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | The length of rectangle $ABCD$ is 5 inches and its width is 3 inches. Diagonal $AC$ is divided into three equal segments by points $E$ and $F$. The area of triangle $BEF$, expressed in square inches, is:
$\text{(A)} \frac{3}{2} \qquad \text{(B)} \frac {5}{3} \qquad \text{(C)} \frac{5}{2} \qquad \text{(D)} \frac{1}{3}... | [
"Draw the rectangle $ABCD$ with $AB$ = $5$ and $AD$ = $3$. We created our diagonal, $AC$ and use the Pythagorean Theorem to find the length of $AC$, which is $\\sqrt34$. Since $EF$ breaks the diagonal into $3$ equal parts, the lenght of $EF$ is $\\frac {\\sqrt34}{3}$. The only other thing we need is the height of $... | 2 | ./CreativeMath/AHSME/1966_AHSME_Problems/14.json | AHSME |
1966_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | If $x-y>x$ and $x+y<y$, then
$\text{(A) } y<x \quad \text{(B) } x<y \quad \text{(C) } x<y<0 \quad \text{(D) } x<0,y<0 \quad \text{(E) } x<0,y>0$
| [
"From $x-y>x$, we get that $-y>0\\implies y<0$.\nFrom $x+y<y$, we get that $x<0$. \nSo, our final answer is $\\fbox{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/15.json | AHSME |
1966_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | The number of values of $x$ satisfying the equation
\[\frac {2x^2 - 10x}{x^2 - 5x} = x - 3\]
is:
$\text{(A)} \ \text{zero} \qquad \text{(B)} \ \text{one} \qquad \text{(C)} \ \text{two} \qquad \text{(D)} \ \text{three} \qquad \text{(E)} \ \text{an integer greater than 3}$
| [
"Since $x^2 - 5x$ is in the denominator, $x \\neq 0,5$. Simplifying,\n\n\n\\[2\\left(\\frac{x^2 - 5}{x^2 - 5}\\right) = 2 = x-3\\]\n\n\nThus $x = 5$, which isn't in the domain of the equation. Thus there are no values of $x \\Rightarrow \\mathrm{(A)}$. \n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/5.json | AHSME |
1966_AHSME_Problems | 39 | 0 | Algebra | Multiple Choice | In base $R_1$ the expanded fraction $F_1$ becomes $.373737\cdots$, and the expanded fraction $F_2$ becomes $.737373\cdots$. In base $R_2$ fraction $F_1$, when expanded, becomes $.252525\cdots$, while the fraction $F_2$ becomes $.525252\cdots$. The sum of $R_1$ and $R_2$, each written in the base ten, is:
$\text{(A) }... | [
"First, let's write $F_1$ as a proper fraction in base $R_1$. To do that, note that:\n$F_1=0.373737\\dots$\nMultiplying this equation on both sides $R_1^2$, we get:\n$R_1^2F_1=37.373737\\dots$\nSubtracting the first equation from the second one, we get:\n$R_1^2F_1-F_1=37\\\\F_1(R_1^2-1)=37\\\\F_1=\\frac{3R_1+7}{R_1... | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/39.json | AHSME |
1966_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Let $s_1$ be the sum of the first $n$ terms of the arithmetic sequence $8,12,\cdots$ and let $s_2$ be the sum of the first $n$ terms of the arithmetic sequence $17,19,\cdots$. Assume $n \ne 0$. Then $s_1=s_2$ for:
$\text{(A) no value of } n \quad \text{(B) one value of } n \quad \text{(C) two values of } n \quad \tex... | [
"$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/19.json | AHSME |
1966_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | If $x$ is real and $4y^2+4xy+x+6=0$, then the complete set of values of $x$ for which $y$ is real, is:
$\text{(A) } x\le-2 \text{ or } x\ge3 \quad \text{(B) } x\le2 \text{ or } x\ge3 \quad \text{(C) } x\le-3 \text{ or } x\ge2 \quad \\ \text{(D) } -3\le x\le2 \quad \text{(E) } -2\le x\le3$
| [
"We treat the equation as a quadratic equation in $y$ for which the discriminant \n\n\n\\[D=16x^2-16(x+6)=16(x^2-x-6)=16(x-3)(x+2)\\] \nFor $y$ to be real $D \\ge 0$. This inequality is satisfied when $x \\le -2$ or $x \\ge3$ or $\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/23.json | AHSME |
1966_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | If $x = (\log_82)^{(\log_28)}$, then $\log_3x$ equals:
$\text{(A)} \ - 3 \qquad \text{(B)} \ - \frac13 \qquad \text{(C)} \ \frac13 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 9$
| [
"By definition, $\\log_8 2 = \\frac 13$ and $\\log_2 8 = 3$, so $\\log_3 \\left(\\frac{1}{3}\\right)^3 = \\log_3 3^{-3} = -3 \\Rightarrow \\mathrm{(A)}$. \n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/9.json | AHSME |
1966_AHSME_Problems | 35 | 0 | Geometry | Multiple Choice | Let $O$ be an interior point of triangle $ABC$, and let $s_1=OA+OB+OC$. If $s_2=AB+BC+CA$, then
$\text{(A) for every triangle } s_2>2s_1,s_1 \le s_2 \\ \text{(B) for every triangle } s_2>2s_1,s_1 < s_2 \\ \text{(C) for every triangle } s_1> \tfrac{1}{2}s_2,s_1 < s_2 \\ \text{(D) for every triangle } s_2\ge 2s_1,s_... | [
"By the Triangle Inequality, we see that $s_1 > \\frac12 s_2$, therefore $\\fbox{C}$. -Dark_Lord\n\n\n"
] | 1 | ./CreativeMath/AHSME/1966_AHSME_Problems/35.json | AHSME |
1984_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | The number of the distinct solutions to the equation
$|x-|2x+1||=3$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$
| [
"We can create a tree of possibilities, progressively eliminating the absolute value signs by creating different cases.\n\n\n[asy] unitsize(2cm); draw((0,0)--(2,-2)); draw((0,0)--(-2,-2)); label(\"$|x-|2x+1||=3$\",(0,0),N); label(\"$x-|2x+1|=3$\",(-2,-2),S); label(\"$x-|2x+1|=-3$\",(2,-2),S); label(\"$|2x+1|=x-3$\"... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/20.json | AHSME |
1984_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | The function $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x$. If the equation $f(x)=0$ has exactly four distinct real roots, then the sum of these roots is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$
| [
"Let one of the roots be $r_1$. Also, define $x$ such that $2+x=r_1$. Thus, we have $f(2+x)=f(r_1)=0$ and $f(2+x)=f(2-x)$. Therefore, we have $f(2-x)=0$, and $2-x$ is also a root. Let this root be $r_2$. The sum $r_1+r_2=2+x+2-x=4$. Similarly, we can let $r_3$ be a root and define $y$ such that $2+y=r_3$, and we wi... | 2 | ./CreativeMath/AHSME/1984_AHSME_Problems/16.json | AHSME |
1984_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | In a certain school, there are $3$ times as many boys as girls and $9$ times as many girls as teachers. Using the letters $b, g, t$ to represent the number of boys, girls, and teachers, respectively, then the total number of boys, girls, and teachers can be represented by the expression
$\mathrm{(A) \ }31b \qquad \ma... | [
"From the given, we have $3g=b$ and $9t=g$, or $t=\\frac{g}{9}$. The sum of these, in terms of $g$, is $3g+g+\\frac{g}{9}$, or, with a common denominator, $\\frac{37g}{9}$. We can see that this isn't one of the choices. So we write it in terms of $b$. We can see from the first equation that $g=\\frac{b}{3}$, so sub... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/6.json | AHSME |
1984_AHSME_Problems | 7 | 0 | Arithmetic | Multiple Choice | When Dave walks to school, he averages $90$ steps per minute, and each of his steps is $75$ cm long. It takes him $16$ minutes to get to school. His brother, Jack, going to the same school by the same route, averages $100$ steps per minute, but his steps are only $60$ cm long. How long does it take Jack to get to schoo... | [
"At $90$ steps per minute and $75 cm$ per step, Dave walks at a rate of $90\\times75$ cm per minute, and with $16$ minutes, the distance Dave walks to school is $90\\times75\\times16$. Also, at $100$ steps per minute and $60$ cm per step, Jack walks at a rate of $100\\times60$ cm per minute. Jack must walk $90\\tim... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/7.json | AHSME |
1984_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | A right triangle $ABC$ with hypotenuse $AB$ has side $AC=15$. Altitude $CH$ divides $AB$ into segments $AH$ and $HB$, with $HB=16$. The area of $\triangle ABC$ is:
$\mathrm{(A) \ }120 \qquad \mathrm{(B) \ }144 \qquad \mathrm{(C) \ } 150 \qquad \mathrm{(D) \ }216 \qquad \mathrm{(E) \ } 144\sqrt{5}$
| [
"[asy] unitsize(.4cm); draw((0,0)--(0,12)); draw((9,0)--(-16,0)); draw((9,0)--(0,12)); draw((-16,0)--(0,12)); label(\"$A$\",(9,0),ENE); label(\"$B$\",(-16,0),WNW); label(\"$C$\",(0,12),ENE); label(\"$H$\",(0,0),S); label(\"$15$\",(4.5,6),NE); label(\"$16$\",(-8,0),S); [/asy]\n$AHC\\sim ACB$ by $AA$, so $\\frac{AH}{... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/17.json | AHSME |
1984_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | The number of triples $(a, b, c)$ of positive integers which satisfy the simultaneous equations
$ab+bc=44$
$ac+bc=23$
is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$
| [
"We can factor the second equation to get $c(a+b)=23$, so we see that $c$ must be a factor of $23$, and since this is prime, $c=1$ or $c=23$. However, if $c=23$, then $a+b=1$, which is impossible for the field of positive integers. Therefore, $c=1$ for all possible solutions. Substituting this into the original equ... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/21.json | AHSME |
1984_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are $1+2i, -2+i$, and $-1-2i$. The fourth number is
$\mathrm{(A) \ }2+i \qquad \mathrm{(B) \ }2-i \qquad \mathrm{(C) \ } 1-2i \qquad \mathrm{(D) \ }-1+2i \qquad \mathrm{(E) \ } -2-i$
| [
"Perhaps the easiest way to attack this is to transfer this to the Cartesian plane. The points then would be $(1, 2), (-2, 1),$ and $(-1, -2)$, assuming the real axis was horizontal. Let these points be $A, B$ and $C$, respectively. The remaining point is then the intersection of the following perpendicular lines:\... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/10.json | AHSME |
1984_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | For any complex number $w=a+bi$, $|w|$ is defined to be the real number $\sqrt{a^2+b^2}$. If $w=\cos40^\circ+i\sin40^\circ$, then $|w+2w^2+3w^3+...+9w^9|^{-1}$ equals
$\text{(A) }\frac{1}{9}\sin40^\circ \qquad \text{(B) }\frac{2}{9}\sin20^\circ \qquad \text{(C) } \frac{1}{9}\cos40^\circ \qquad \text{(D) }\frac{1}{18}... | [
"Let $S=w+2w^2+3w^3+...+9w^9$. Note that\n\n\n\\[S=\\sum_{i=1}^{9}\\sum_{j=i}^{9} w^j\\]\n\n\nNow we multiply $S$ by $1-w$:\n\n\n\\[S(1-w)=\\sum_{i=1}^{9}\\sum_{j=i}^{9} w^j(1-w)\\]\n\n\nBy the geometric series formula, $\\sum_{j=i}^{9} w^j(1-w)$ is simply $w^i-w^{10}$. Therefore\n\\[S(1-w)=\\sum_{i=1}^{9} w^i-w^{1... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/30.json | AHSME |
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