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 |
|---|---|---|---|---|---|---|---|---|---|
1965_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | The vertex of the parabola $y = x^2 - 8x + c$ will be a point on the $x$-axis if the value of $c$ is:
$\textbf{(A)}\ - 16 \qquad \textbf{(B) }\ - 4 \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 8 \qquad \textbf{(E) }\ 16$
| [
"Notice that if the vertex of a parabola is on the x-axis, then the x-coordinate of the vertex must be a solution to the quadratic. Since the quadratic is strictly increasing on either side of the vertex, the solution must have double multiplicity, or the quadratic is a perfect square trinomial. This means that for... | 1 | ./CreativeMath/AHSME/1965_AHSME_Problems/9.json | AHSME |
1970_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | Lines $HK$ and $BC$ lie in a plane. $M$ is the midpoint of the line segment $BC$, and $BH$ and $CK$ are perpendicular to $HK$.
Then we
$\text{(A) always have } MH=MK\quad\\ \text{(B) always have } MH>BK\quad\\ \text{(C) sometimes have } MH=MK \text{ but not always}\quad\\ \text{(D) always have } MH>MB\quad\\ \text{(... | [
"$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/20.json | AHSME |
1970_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | If $F(n)$ is a function such that $F(1)=F(2)=F(3)=1$, and such that $F(n+1)= \frac{F(n)\cdot F(n-1)+1}{F(n-2)}$ for $n\ge 3,$
then $F(6)=$
$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 7\quad \text{(D) } 11\quad \text{(E) } 26$
\section{Solution}
Plugging in $n=3$ gives $F(4) = \frac{F(3) \cdot F(2) + 1}{F(1)... | [
"Plugging in $n=3$ gives $F(4) = \\frac{F(3) \\cdot F(2) + 1}{F(1)} = \\frac{1 \\cdot 1 + 1}{1} = 2$.\n\n\nPlugging in $n=4$ gives $F(5) = \\frac{F(4) \\cdot F(3) + 1}{F(2)} = \\frac{2 \\cdot 1 + 1}{1} = 3$.\n\n\nPlugging in $n=5$ gives $F(6) = \\frac{F(5) \\cdot F(4) + 1}{F(3)} = \\frac{3 \\cdot 2 + 1}{1} = 7$.\n\... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/16.json | AHSME |
1970_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | The smallest value of $x^2+8x$ for real values of $x$ is
$\text{(A) } -16.25\quad \text{(B) } -16\quad \text{(C) } -15\quad \text{(D) } -8\quad \text{(E) None of these}$
| [
"Let's imagine this as a quadratic equation. To find the minimum or maximum value, we always need to find the vertex of the quadratic equation. The vertex of the quadratic is $\\frac{-b}{2a}$ in $ax^2+bx+c=0$. Then to find the output, or the y value of the quadratic, we plug the vertex \"x\" value back into the equ... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/6.json | AHSME |
1970_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | Inside square $ABCD$ with side $s$, quarter-circle arcs with radii $s$ and centers at $A$ and $B$ are drawn. These arcs intersect at a point $X$ inside the square. How far is $X$ from the side of $CD$?
$\text{(A) } \tfrac{1}{2} s(\sqrt{3}+4)\quad \text{(B) } \tfrac{1}{2} s\sqrt{3}\quad \text{(C) } \tfrac{1}{2} s(1+\s... | [
"All answers are proportional to $s$, so for ease, let $s=1$.\n\n\nLet $ABCD$ be oriented so that $A(0, 0), B(1, 0), C(1, 1), D(0, 1)$.\n\n\nThe circle centered at $A$ with radius $1$ is $x^2 + y^2 = 1$. The circle centered at $B$ with radius $1$ is $(x - 1)^2 + y^2 = 1$. Solving each equation for $1 - y^2$ to fi... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/7.json | AHSME |
1970_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | If $r>0$, then for all $p$ and $q$ such that $pq\ne 0$ and $pr>qr$, we have
$\text{(A) } -p>-q\quad \text{(B) } -p>q\quad \text{(C) } 1>-q/p\quad \text{(D) } 1<q/p\quad \text{(E) None of these}$
| [
"If $r>0$ and $pr>qr$, we can divide by the positive number $r$ and not change the inequality direction to get $p>q$. Multiplying by $-1$ (and flipping the inequality sign because we're multiplying by a negative number) leads to $-p < -q$, which directly contradicts $A$. Thus, $A$ is always false.\n\n\nIf $p < 0$... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/17.json | AHSME |
1970_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | On an auto trip, the distance read from the instrument panel was $450$ miles. With snow tires on for the return trip over the same route, the reading was $440$ miles. Find, to the nearest hundredth of an inch, the increase in radius of the wheels if the original radius was 15 inches.
$\text{(A) } .33\quad \text{(B) }... | [
"$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/21.json | AHSME |
1970_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | Let $F=.48181\cdots$ be an infinite repeating decimal with the digits $8$ and $1$ repeating. When $F$ is written as a fraction in lowest terms, the denominator exceeds the numerator by
$\text{(A) } 13\quad \text{(B) } 14\quad \text{(C) } 29\quad \text{(D) } 57\quad \text{(E) } 126$
| [
"Multiplying by $100$ gives $100F = 48.181818...$. Subtracting the first equation from the second gives $99F = 47.7$, and all the other repeating parts cancel out. This gives $F = \\frac{47.7}{99} = \\frac{477}{990} = \\frac{159}{330} = \\frac{53}{110}$. Subtracting the numerator from the denominator gives $\\fb... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/10.json | AHSME |
1970_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | The number of distinct points in the $xy$-plane common to the graphs of $(x+y-5)(2x-3y+5)=0$ and $(x-y+1)(3x+2y-12)=0$ is
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4\quad \text{(F) } \infty$
| [
"The graph $(x + y - 5)(2x - 3y + 5) = 0$ is the combined graphs of $x + y - 5=0$ and $2x - 3y + 5 = 0$. Likewise, the graph $(x -y + 1)(3x + 2y - 12) = 0$ is the combined graphs of $x-y+1=0$ and $3x+2y-12=0$. All these lines intersect at one point, $(2,3)$.\nTherefore, the answer is $\\fbox{(B) 1}$.\n\n\n",
"We ... | 2 | ./CreativeMath/AHSME/1970_AHSME_Problems/26.json | AHSME |
1970_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | [asy] draw((0,0)--(2,2)--(5/2,1/2)--(2,0)--cycle,dot); MP("A",(0,0),W);MP("B",(2,2),N);MP("C",(5/2,1/2),SE);MP("D",(2,0),S); MP("a",(1,0),N);MP("b",(17/8,1/8),N); [/asy]
In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of angle $D$ is twice that of angle $B$, and the measures of segments $... | [
"With reference to the diagram above, let $E$ be the point on $AB$ such that $DE||BC$. Let $\\angle ABC=\\alpha$. We then have $\\alpha =\\angle AED = \\angle EDC$ since $AB||CD$, so $\\angle ADE=\\angle ADC-\\angle BDC=2\\alpha-\\alpha = \\alpha$, which means $\\triangle AED$ is isosceles. \n\n\nTherefore, $AB=AE+... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/30.json | AHSME |
1970_AHSME_Problems | 31 | 0 | Probability | Multiple Choice | If a number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to 43, what is the probability that this number will be divisible by 11?
$\text{(A) } \frac{2}{5}\quad \text{(B) } \frac{1}{5}\quad \text{(C) } \frac{1}{6}\quad \text{(D) } \frac{1}{11}\quad \text{(E) } \f... | [
"For the sums of the digits of be equal to 43, this means that the number's digits must fall under one of two categories. It either has 4 9's and a 7, or 3 9's and 2 8's. You can use casework from there to find that there are 15 5-digit numbers in which the sum of the digits is equal to 43. To find the number of wa... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/31.json | AHSME |
1970_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | In a triangle, the area is numerically equal to the perimeter. What is the radius of the inscribed circle?
$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 5\quad \text{(E) } 6$
| [
"One of the most common formulas involving the inradius of a triangle is $A = rs$, where $A$ is the area of the triangle, $r$ is the inradius, and $s$ is the semiperimeter.\n\n\nThe problem states that $A = p = 2s$. This means $2s = rs$, or $r = 2$, which is option $\\fbox{A}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/27.json | AHSME |
1970_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | The fourth power of $\sqrt{1+\sqrt{1+\sqrt{1}}}$ is
$\text{(A) } \sqrt{2}+\sqrt{3}\quad \text{(B) } \tfrac{1}{2}(7+3\sqrt{5})\quad \text{(C) } 1+2\sqrt{3}\quad \text{(D) } 3\quad \text{(E) } 3+2\sqrt{2}$
| [
"We can simplify the expression to $\\sqrt{1+\\sqrt{2}}$ Then we square it so it is now $1+\\sqrt{2}$. We still have to square it one more time, getting $\\boxed{(E)\\ 3+2\\sqrt{2}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/1.json | AHSME |
1970_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | If two factors of $2x^3-hx+k$ are $x+2$ and $x-1$, the value of $|2h-3k|$ is
$\text{(A) } 4\quad \text{(B) } 3\quad \text{(C) } 2\quad \text{(D) } 1\quad \text{(E) } 0$
| [
"From the Remainder Theorem, we have $2(-2)^3 - h(-2) + k = 0$ and $2(1)^3 - h(1) + k = 0$. Simplifying both of those equations gives $-16 + 2h + k = 0$ and $2 - h + k = 0$. Since $k = 16 - 2h$ and $k = h - 2$, we set those equal to get:\n\n\n$16 - 2h = h - 2$\n\n\n$3h = 18$\n\n\n$h = 6$\n\n\nThis gives $k = 4$ w... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/11.json | AHSME |
1970_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | A square and a circle have equal perimeters. The ratio of the area of the circle to the are of the square is
$\text{(A) } \frac{4}{\pi}\quad \text{(B) } \frac{\pi}{\sqrt{2}}\quad \text{(C) } \frac{4}{1}\quad \text{(D) } \frac{\sqrt{2}}{\pi}\quad \text{(E) } \frac{\pi}{4}$
| [
"Let's say the circle has a circumference ( or perimeter of $4\\pi$). Since the perimeter of the square is the same as the perimeter of the circle, the side length of the square is $\\pi$. That means that the area of the square is $\\pi^2$. The area of the circle is $4\\pi$. So the area of the circle over the area ... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/2.json | AHSME |
1970_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | In triangle $ABC$, the median from vertex $A$ is perpendicular to the median from vertex $B$. If the lengths of sides $AC$ and $BC$ are $6$ and $7$ respectively, then the length of side $AB$ is
$\text{(A) } \sqrt{17}\quad \text{(B) } 4\quad \text{(C) } 4\tfrac{1}{2}\quad \text{(D) } 2\sqrt{5}\quad \text{(E) } 4\tfrac... | [
"$\\fbox{A}$\nlet the midpoint be M,N ( i.e. AM,BN are the medians); connecting MN we know that AB = 2x and MN = x hence apply stewart's theorem in triangle ABC with median MN first and then apply stewart's in triangle BNC with median MN\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/28.json | AHSME |
1970_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | A circle with radius $r$ is tangent to sides $AB,AD$ and $CD$ of rectangle $ABCD$ and passes through the midpoint of diagonal $AC$. The area of the rectangle, in terms of $r$, is
$\text{(A) } 4r^2\quad \text{(B) } 6r^2\quad \text{(C) } 8r^2\quad \text{(D) } 12r^2\quad \text{(E) } 20r^2$
| [
"$\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/12.json | AHSME |
1970_AHSME_Problems | 32 | 0 | Algebra | Multiple Choice | $A$ and $B$ travel around a circular track at uniform speeds in opposite directions, starting from diametrically opposite points. If they start at the same time, meet first after $B$ has travelled $100$ yards, and meet a second time $60$ yards before $A$ completes one lap, then the circumference of the track in yards i... | [
"$\\fbox{C}$\n\n\nLet $x$ be half the circumference of the track. They first meet after $B$ has run $100$ yards, meaning that in the time $B$ has run $100$ yards, $A$ has run $x-100$ yards. The second time they meet is when $A$ is 60 yards before he completes the lap. This means that in the time that $A$ has run $2... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/32.json | AHSME |
1970_AHSME_Problems | 24 | 0 | Geometry | Multiple Choice | An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is $2$, then the area of the hexagon is
$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 6\quad \text{(E) } 12$
| [
"Let $ABCDEF$ be our regular hexagon, with centre $O$ - and join $AO, BO, CO, DO, EO,$ and $FO$. Note that we form six equilateral triangles with sidelength $\\frac{s}{2}$, where $s$ is the sidelength of the triangle (since the perimeter of the two polygons are equal). If the area of the original equilateral triang... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/24.json | AHSME |
1970_AHSME_Problems | 25 | 0 | Number Theory | Multiple Choice | For every real number $x$, let $[x]$ be the greatest integer which is less than or equal to $x$. If the postal rate for first class mail is six cents for every ounce or portion thereof, then the cost in cents of first-class postage on a letter weighing $W$ ounces is always
$\text{(A) } 6W\quad \text{(B) } 6[W]\quad \... | [
"This question is trying to convert the floor function, which is more commonly notated as $\\lfloor x \\rfloor$, into the ceiling function, which is $\\lceil x \\rceil$. The identity is $\\lceil x \\rceil = -\\lfloor -x \\rfloor$, which can be verified graphically, or proven using the definition of floor and ceili... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/25.json | AHSME |
1970_AHSME_Problems | 33 | 0 | Arithmetic | Multiple Choice | Find the sum of digits of all the numbers in the sequence $1,2,3,4,\cdots ,10000$.
$\text{(A) } 180001\quad \text{(B) } 154756\quad \text{(C) } 45001\quad \text{(D) } 154755\quad \text{(E) } 270001$
| [
"We can find the sum using the following method. We break it down into cases. The first case is the numbers $1$ to $9$. The second case is the numbers $10$ to $99$. The third case is the numbers $100$ to $999$. The fourth case is the numbers $1,000$ to $9,999$. And lastly, the sum of the digits in $10,000$. The fir... | 3 | ./CreativeMath/AHSME/1970_AHSME_Problems/33.json | AHSME |
1970_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | Given the binary operation $\star$ defined by $a\star b=a^b$ for all positive numbers $a$ and $b$. Then for all positive $a,b,c,n$, we have
$\text{(A) } a\star b=b\star a\quad\qquad\qquad\ \text{(B) } a\star (b\star c)=(a\star b) \star c\quad\\ \text{(C) } (a\star b^n)=(a \star n) \star b\quad \text{(D) } (a\star b)^... | [
"Let $a = 2, b = 3, c=n = 4$. If all of them are false, the answer must be $E$. If one does not fail, we will try to prove it.\n\n\nFor option $A$, we have $2^3 = 3^2$, which is clearly false.\n\n\nFor option $B$, we have $2^{81} = 8^{4}$, which is false.\n\n\nFor option $C$, we have $2^{81} = 16^3$, which is fal... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/13.json | AHSME |
1970_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | It is now between 10:00 and 11:00 o'clock, and six minutes from now, the minute hand of a watch will be exactly opposite the place where the hour hand was three minutes ago. What is the exact time now?
$\text{(A) } 10:05\tfrac{5}{11}\quad \text{(B) } 10:07\tfrac{1}{2}\quad \text{(C) } 10:10\quad \text{(D) } 10:15\qua... | [
"$\\fbox{D}$\n50 + (m-3)/12 = 30 + (m+6)\n= 10hours + m = 10: 15\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/29.json | AHSME |
1970_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | If $x=1+2^p$ and $y=1+2^{-p}$, then $y$ in terms of $x$ is
$\text{(A) } \frac{x+1}{x-1}\quad \text{(B) } \frac{x+2}{x-1}\quad \text{(C) } \frac{x}{x-1}\quad \text{(D) } 2-x\quad \text{(E) } \frac{x-1}{x}$
| [
"Since we want the $y$ expression in terms of $x$, let's convert the $y$ expression. We can convert it to $1+ \\frac{1}{2^p} \\Rightarrow \\frac{2^p+1}{2^p} \\Rightarrow \\frac{x}{x-1} \\Rightarrow$ $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/3.json | AHSME |
1970_AHSME_Problems | 34 | 0 | Number Theory | Multiple Choice | The greatest integer that will divide $13511$, $13903$ and $14589$ and leave the same remainder is
$\text{(A) } 28\quad \text{(B) } 49\quad \text{(C) } 98\quad\\ \text{(D) an odd multiple of } 7 \text{ greater than } 49\quad\\ \text{(E) an even multiple of } 7 \text{ greater than } 98$
\section{Solution}
We know th... | [
"We know that 13903 minus 13511 is equivalent to 392. Additionally, 14589 minus 13903 is equivalent to 686. Since we are searching for the greatest integer that divides these three integers and leaves the same remainder, the answer resides in the greatest common factor of 686 and 392. Therefore, the answer is 98, o... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/34.json | AHSME |
1970_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | If $a=\log_8 225$ and $b=\log_2 15$, then
$\text{(A) } a=b/2\quad \text{(B) } a=2b/3\quad \text{(C) } a=b\quad \text{(D) } b=a/2\quad \text{(E) } a=3b/2$
| [
"The solutions imply that finding the ratio $\\frac{a}{b}$ will solve the problem. We compute $\\frac{a}{b}$, use change-of-base to a neutral base, rearrange the terms, and then use the reverse-change-of-base:\n\n\n$\\frac{\\log_8 225}{\\log_2 15}$\n\n\n$\\frac{\\frac{\\ln 225}{\\ln 8}}{\\frac{\\ln 15}{\\ln 2}}$\n... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/8.json | AHSME |
1970_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | If the sum of the first $3n$ positive integers is $150$ more than the sum of the first $n$ positive integers, then the sum of the first $4n$ positive integers is
$\text{(A) } 300\quad \text{(B) } 350\quad \text{(C) } 400\quad \text{(D) } 450\quad \text{(E) } 600$
| [
"We can setup our first equation as\n\n\n$\\frac{3n(3n+1)}{2} = \\frac{n(n+1)}{2} + 150$\n\n\nSimplifying we get\n\n\n$9n^2 + 3n = n^2 + n + 300 \\Rightarrow 8n^2 + 2n - 300 = 0 \\Rightarrow 4n^2 + n - 150 = 0$\n\n\nSo our roots using the quadratic formula are\n\n\n$\\dfrac{-b\\pm\\sqrt{b^2 - 4ac}}{2a} \\Rightarrow... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/22.json | AHSME |
1970_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | $\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}$ is equal to
$\text{(A) } 2\quad \text{(B) } 2\sqrt{3}\quad \text{(C) } 4\sqrt{2}\quad \text{(D) } \sqrt{6}\quad \text{(E) } 2\sqrt{2}$
| [
"Square the expression:\n\n\n$(\\sqrt{3+2\\sqrt{2}}-\\sqrt{3-2\\sqrt{2}})^2=3+\\sqrt{2}-2\\sqrt{(3+\\sqrt{2})(3-\\sqrt{2})}+3-2\\sqrt{2}=6-2\\sqrt{9-8}=6-2\\sqrt{1}=4$\n\n\n$\\Rightarrow\\sqrt{3+2\\sqrt{2}}-\\sqrt{3-2\\sqrt{2}}=\\sqrt{4}=2\\Rightarrow\\boxed{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/18.json | AHSME |
1970_AHSME_Problems | 4 | 0 | Number Theory | Multiple Choice | Let $S$ be the set of all numbers which are the sum of the squares of three consecutive integers. Then we can say that
$\text{(A) No member of S is divisible by } 2\quad\\ \text{(B) No member of S is divisible by } 3 \text{ but some member is divisible by } 11\quad\\ \text{(C) No member of S is divisible by } 3 \text... | [
"Consider $3$ consecutive integers $a, b,$ and $c$. Exactly one of these integers must be divisible by 3; WLOG, suppose $a$ is divisible by 3. Then $a \\equiv 0 \\pmod {3}, b \\equiv 1 \\pmod{3},$ and $c \\equiv 2 \\pmod{3}$. Squaring, we have that $a^{2} \\equiv 0 \\pmod{3}, b^{2} \\equiv 1 \\pmod{3},$ and $c^... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/4.json | AHSME |
1970_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | Consider $x^2+px+q=0$, where $p$ and $q$ are positive numbers. If the roots of this equation differ by 1, then $p$ equals
$\text{(A) } \sqrt{4q+1}\quad \text{(B) } q-1\quad \text{(C) } -\sqrt{4q+1}\quad \text{(D) } q+1\quad \text{(E) } \sqrt{4q-1}$
| [
"From the quadratic equation, the two roots of the equation are $\\frac{-p\\pm\\sqrt{p^2-4q}}{2}$. The positive difference between these roots is $\\sqrt{p^2 - 4q}$. Setting $\\sqrt{p^2-4q}=1$ and isolating $p$ gives $\\sqrt{4q+1}$, or choice $\\boxed{\\text{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/14.json | AHSME |
1970_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | Lines in the $xy$-plane are drawn through the point $(3,4)$ and the trisection points of the line segment joining the points $(-4,5)$ and $(5,-1)$. One of these lines has the equation
$\text{(A) } 3x-2y-1=0\quad \text{(B) } 4x-5y+8=0\quad \text{(C) } 5x+2y-23=0\quad\\ \text{(D) } x+7y-31=0\quad \text{(E) } x-4y+13=0$... | [
"The trisection points of $(-4, 5)$ and $(5, -1)$ can be found by trisecting the x-coordinates and the y-coordinates separately. The difference of the x-coordinates is $9$, so the trisection points happen at $-4 + \\frac{9}{3}$ and $-4 + \\frac{9}{3} + \\frac{9}{3}$, which are $-1$ and $2$. Similarly, the y-coord... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/15.json | AHSME |
1970_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | If $f(x)=\frac{x^4+x^2}{x+1}$, then $f(i)$, where $i=\sqrt{-1}$, is equal to
$\text{(A) } 1+i\quad \text{(B) } 1\quad \text{(C) } -1\quad \text{(D) } 0\quad \text{(E) } -1-i$
| [
"$i^4 = 1$ and $i^2=-1$, so the numerator is $0$. As long as the denominator is not $0$, which it isn't, the answer is $0 \\Rightarrow$ $\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/5.json | AHSME |
1970_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | The sum of an infinite geometric series with common ratio $r$ such that $|r|<1$ is $15$, and the sum of the squares of the terms of this series is $45$. The first term of the series is
$\textbf{(A) } 12\quad \textbf{(B) } 10\quad \textbf{(C) } 5\quad \textbf{(D) } 3\quad \textbf{(E) 2}$
| [
"We know that the formula for the sum of an infinite geometric series is $S = \\frac{a}{1-r}$.\n\n\nSo we can apply this to the conditions given by the problem. \n\n\nWe have two equations:\n\n\n\\begin{align*} 15 &= \\frac{a}{1-r} \\\\ 45 &= \\frac{a^{2}}{1-r^{2}} \\end{align*}\n\n\nWe get\n\n\n\\begin{align*} a &... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/19.json | AHSME |
1970_AHSME_Problems | 23 | 0 | Number Theory | Multiple Choice | The number $10!$ ($10$ is written in base $10$), when written in the base $12$ system, ends with exactly $k$ zeros. The value of $k$ is
$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$
| [
"A number in base $b$ that ends in exactly $k$ zeros will be divisible by $b^k$, but not by $b^{k+1}$. Thus, we want to find the highest $k$ for which $12^k | 10!$.\n\n\nThere are $4$ factors of $3$: $3, 6, 9$, and an extra factor from $9$.\n\n\nThere are $8$ factors of $2$: $2, 4, 6, 8, 10$, an extra factor fro... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/23.json | AHSME |
1970_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | Points $P$ and $Q$ are on line segment $AB$, and both points are on the same side of the midpoint of $AB$. Point $P$ divides $AB$ in the ratio $2:3$, and $Q$ divides $AB$ in the ratio $3:4$. If $PQ$=2, then the length of segment $AB$ is
$\text{(A) } 12\quad \text{(B) } 28\quad \text{(C) } 70\quad \text{(D) } 75\quad ... | [
"In order, the points from left to right are $A, P, Q, B$. Let the lengths between successive points be $x, 2, y$, respectively. \n\n\nSince $\\frac{AP}{PB} = \\frac{2}{3}$, we have $\\frac{x}{2 + y} = \\frac{2}{3}$.\n\n\nSince $\\frac{AQ}{QB} = \\frac{3}{4}$, we have $\\frac{x + 2}{y} = \\frac{3}{4}$.\n\n\nThe f... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/9.json | AHSME |
1970_AHSME_Problems | 35 | 0 | Algebra | Multiple Choice | A retiring employee receives an annual pension proportional to the square root of the number of years of his service. Had he served $a$ years more, his pension would have been $p$ dollars greater, whereas had he served $b$ years more $(b\ne a)$, his pension would have been $q$ dollars greater than the original annual p... | [
"Note the original pension as $k\\sqrt{x}$, where $x$ is the number of years served. Then, based on the problem statement, two equations can be set up. \n\n\n\\[k\\sqrt{x+a} = k\\sqrt{x} + p\\]\n\\[k\\sqrt{x+b} = k\\sqrt{x} + q\\]\n\n\nSquare the first equation to get\n\n\n\\[k^2x + ak^2 = k^2x + 2pk\\sqrt{x} + p^2... | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/35.json | AHSME |
1980_AHSME_Problems | 20 | 0 | Probability | Multiple Choice | A box contains $2$ pennies, $4$ nickels, and $6$ dimes. Six coins are drawn without replacement,
with each coin having an equal probability of being chosen. What is the probability that the value of coins drawn is at least $50$ cents?
$\text{(A)} \ \frac{37}{924} \qquad \text{(B)} \ \frac{91}{924} \qquad \text{(C... | [
"We want the number of Successful Outcomes over the number of Total Outcomes. We want to calculate the total outcomes first. Since we have $12$ coins and we need to choose $6$, we have $\\binom{12}{6}$ = $924$ Total outcomes. For our successful outcomes, we can have $(1) 1$ penny and $5$ dimes, $2$ nickels and $4$ ... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/20.json | AHSME |
1980_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$
... | [
"We assume the side length of the cube is $1$. The side length of the tetrahedron is $\\sqrt2$, so the surface area is $4\\times\\frac{2\\sqrt3}{4}=2\\sqrt3$. The surface area of the cube is $6\\times1\\times1=6$, so the ratio of the surface area of the cube to the surface area of the tetrahedron is $\\frac{6}{2\\s... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/16.json | AHSME |
1980_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | A positive number $x$ satisfies the inequality $\sqrt{x} < 2x$ if and only if
$\text{(A)} \ x > \frac{1}{4} \qquad \text{(B)} \ x > 2 \qquad \text{(C)} \ x > 4 \qquad \text{(D)} \ x < \frac{1}{4}\qquad \text{(E)} \ x < 4$
| [
"$\\sqrt{x}< 2x \\\\ x < 4x^2 \\\\ 0 < x(4x-1) \\\\ 0 < 4x-1 \\\\ 1 < 4x \\\\ x >\\frac{1}{4} \\\\ \\boxed{(A)}$\n\n\nNote: You can also draw a rough sketch.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/6.json | AHSME |
1980_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | Sides $AB,BC,CD$ and $DA$ of convex polygon $ABCD$ have lengths 3, 4, 12, and 13, respectively, and $\angle CBA$ is a right angle. The area of the quadrilateral is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); real r=degrees((12,5)), s=degrees((3,4)); pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s)); ... | [
"Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is $\\frac{3\\cdot4}{2}+\\frac{5\\cdot12}{2}=36\\Rightarrow\\boxed{(B)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/7.json | AHSME |
1980_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | Given that $i^2=-1$, for how many integers $n$ is $(n+i)^4$ an integer?
$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$
| [
"$(n+i)^4=n^4+4in^3-6n^2-4in+1$, and this has to be an integer, so the sum of the imaginary parts must be $0$. \\[4in^3-4in=0\\] \\[4in^3=4in\\] \\[n^3=n\\]\nSince $n^3=n$, there are $\\boxed{3}$ solutions for $n$: $0$ and $\\pm1$.\n\n\n-aopspandy\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/17.json | AHSME |
1980_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | [asy] defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E); draw(E--B--A--C--B^^A--D); label("$A$", A, dir(D--A)); label("$B$", B, dir(E--B)); label("$C$", C, dir(0)); label("$D$", D, SE); label("$E$", E, N); label("$F$", F, ... | [
"We can use the principle of same height same area to solve this problem. \n$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/21.json | AHSME |
1980_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | The number of teeth in three meshed gears $A$, $B$, and $C$ are $x$, $y$, and $z$, respectively. (The teeth on all gears are the same size and regularly spaced.) The angular speeds, in revolutions per minutes of $A$, $B$, and $C$ are in the proportion
$\text{(A)} \ x: y: z ~~\text{(B)} \ z: y: x ~~ \text{(C)} \ y: z:... | [
"The distance that each of the gears rotate is constant. Let us have the number of teeth per minute equal to $k$. The revolutions per minute are in ratio of:\n\\[\\frac{k}{x}:\\frac{k}{y}:\\frac{k}{z}\\]\n\\[yz:xz:xy.\\]\nTherefore, the answer is $\\fbox{D: yz:xz:xy}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/10.json | AHSME |
1980_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | Four balls of radius $1$ are mutually tangent, three resting on the floor and the fourth resting on the others.
A tetrahedron, each of whose edges have length $s$, is circumscribed around the balls. Then $s$ equals
$\text{(A)} \ 4\sqrt 2 \qquad \text{(B)} \ 4\sqrt 3 \qquad \text{(C)} \ 2\sqrt 6 \qquad \text{(D)}... | [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/26.json | AHSME |
1980_AHSME_Problems | 30 | 0 | Number Theory | Multiple Choice | A six digit number (base 10) is squarish if it satisfies the following conditions:
(i) none of its digits are zero;
(ii) it is a perfect square; and
(iii) the first of two digits, the middle two digits and the last two digits of the number are all perfect squares when considered as two digit numbers.
How ma... | [
"N = a^2*10000 + b^2*100 + c^2*1\n\n\n\\begin{verbatim}\n = (a*100 + c)^2\n\\end{verbatim}\nwe get \n\n\n\\begin{verbatim}\n b^2 = 2*a*c\n\\end{verbatim}\nwhere \n4<=a,b,c<=9\n\n\n\n\n\n\n\\begin{verbatim}\n b^2=2*a*c, so \n\\end{verbatim}\n\\begin{verbatim}\n a=2*2, c=2*2*2, b=8\n a=2*3, c=3 (NO)\n a=2*2*... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/30.json | AHSME |
1980_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | The sum $\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ equals
$\text{(A)} \ \frac 32 \qquad \text{(B)} \ \frac{\sqrt[3]{65}}{4} \qquad \text{(C)} \ \frac{1+\sqrt[6]{13}}{2} \qquad \text{(D)}\ \sqrt[3]{2}\qquad \text{(E)}\ \text{none of these}$
| [
"Lets set our original expression equal to $x$. So $\\sqrt[3] {5+2\\sqrt{13}}+\\sqrt[3]{5-2\\sqrt{13}} = x$. Cubing this gives us $x^3 = \\left(\\sqrt[3] {5+2\\sqrt{13}}+\\sqrt[3]{5-2\\sqrt{13}}\\right)^3 = 5 + 2\\sqrt{13} + 5 - 2\\sqrt{13} + 3\\left(\\sqrt[3] {5+2\\sqrt{13}}*\\sqrt[3]{5-2\\sqrt{13}}\\right)\\left(... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/27.json | AHSME |
1980_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | The largest whole number such that seven times the number is less than 100 is
$\text{(A)} \ 12 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ 16$
| [
"We want to find the smallest integer $x$ so that $7x < 100$. Dividing by 7 gets $x < 14\\dfrac{2}{7}$, so the answer is 14. $\\boxed{(C)}$\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/1.json | AHSME |
1980_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | If the sum of the first $10$ terms and the sum of the first $100$ terms of a given arithmetic progression are $100$ and $10$,
respectively, then the sum of first $110$ terms is:
$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$
| [
"Let $a$ be the first term of the sequence and let $d$ be the common difference of the sequence.\n\n\nSum of the first 10 terms: $\\frac{10}{2}(2a+9d)=100 \\Longleftrightarrow 2a+9d=20$\nSum of the first 100 terms: $\\frac{100}{2}(2a+99d)=10 \\Longleftrightarrow 2a+99d=\\frac{1}{5}$\n\n\nSolving the system, we get ... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/11.json | AHSME |
1980_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is
$\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72$
| [
"It becomes $(x^{8}+...)(x^{9}+...)$ with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or $\\boxed{(D)}$\n\n\n",
"First note that given a polynomial $P(x)$ and a polynomial $Q(x)$:\n\n\n\n\n$deg(P(x))^n = ndeg(P(x))$ and $deg(P(x)Q(x)... | 2 | ./CreativeMath/AHSME/1980_AHSME_Problems/2.json | AHSME |
1980_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | The polynomial $x^{2n}+1+(x+1)^{2n}$ is not divisible by $x^2+x+1$ if $n$ equals
$\text{(A)} \ 17 \qquad \text{(B)} \ 20 \qquad \text{(C)} \ 21 \qquad \text{(D)} \ 64 \qquad \text{(E)} \ 65$
| [
"Let $h(x)=x^2+x+1$.\n\n\nThen we have\n\\[(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \\cdot h(x) + x^n,\\]\nwhere $g(x)$ is $h^{n-1}(x) + nh^{n-2}(x) \\cdot x + ... + x^{n-1}$ (after expanding $(h(x)+x)^n$ according to the Binomial Theorem).\n\n\nNotice that\n\\[x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x -x^{2... | 3 | ./CreativeMath/AHSME/1980_AHSME_Problems/28.json | AHSME |
1980_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$, respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does $L_2$, and that $L_1$ has 4 times the slope of $L_2$. If $L_1$ is not horizontal, then $mn$ is
$\text{(A)} \ \frac{\sqr... | [
"Solution by e_power_pi_times_i\n\n\n$4n = m$, as stated in the question. In the line $L_1$, draw a triangle with the coordinates $(0,0)$, $(1,0)$, and $(1,m)$. Then $m = \\tan(\\theta_1)$. Similarly, $n = \\tan(\\theta_2)$. Since $4n = m$ and $\\theta_1 = 2\\theta_2$, $\\tan(2\\theta_2) = 4\\tan(\\theta_2)$. Using... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/12.json | AHSME |
1980_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | For some real number $r$, the polynomial $8x^3-4x^2-42x+45$ is divisible by $(x-r)^2$. Which of the following numbers is closest to $r$?
$\text{(A)} \ 1.22 \qquad \text{(B)} \ 1.32 \qquad \text{(C)} \ 1.42 \qquad \text{(D)} \ 1.52 \qquad \text{(E)} \ 1.62$
| [
"Solution by e_power_pi_times_i\n\n\nDenote $s$ as the third solution. Then, by Vieta's, $2r+s = \\dfrac{1}{2}$, $r^2+2rs = -\\dfrac{21}{4}$, and $r^2s = -\\dfrac{45}{8}$. Multiplying the top equation by $2r$ and eliminating, we have $3r^2 = r+\\dfrac{21}{4}$. Combined with the fact that $s = \\dfrac{1}{2}-2r$, the... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/24.json | AHSME |
1980_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$
appears $k$ times. It is a fact that there are integers $b, c$, and $d$ such that for all positive integers $n$,
$a_n=b\lfloor \sqrt{n+c} \rfloor +d$,
where $\lfloor x \rfloor$ denotes ... | [
"Solution by e_power_pi_times_i\n\n\nBecause the set consists of odd numbers, and since $\\lfloor{}\\sqrt{n+c}\\rfloor{}$ is an integer and can be odd or even, $b = 2$ and $|a| = 1$. However, given that $\\lfloor{}\\sqrt{n+c}\\rfloor{}$ can be $0$, $a = 1$. Then, $a_1 = 1 = 2\\lfloor{}\\sqrt{1+c}\\rfloor{}+1$, and ... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/25.json | AHSME |
1980_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | A bug (of negligible size) starts at the origin on the coordinate plane. First, it moves one unit right to $(1,0)$. Then it makes a $90^\circ$ counterclockwise and travels $\frac 12$ a unit to $\left(1, \frac 12 \right)$. If it continues in this fashion, each time making a $90^\circ$ degree turn counterclockwise and tr... | [
"Writing out the change in $x$ coordinates and then in $y$ coordinates gives the infinite sum $1-\\frac{1}{4}+\\frac{1}{16}-\\dots$ and $\\frac{1}{2}-\\frac{1}{8}+\\dots$ respectively. Using the infinite geometric sum formula, we have $\\frac{1}{1+\\frac{1}{4}}=\\frac{4}{5}$ and $\\frac{\\frac{1}{2}}{1+\\frac{1}{4}... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/13.json | AHSME |
1980_AHSME_Problems | 29 | 0 | Number Theory | Multiple Choice | How many ordered triples (x,y,z) of integers satisfy the system of equations below?
\[\begin{array}{l} x^2-3xy+2y^2-z^2=31 \\ -x^2+6yz+2z^2=44 \\ x^2+xy+8z^2=100\\ \end{array}\]
$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \\ \text{(D)}\ \text{a finite number greater than 2}\qquad\\ \text{(... | [
"Sum of three equations, \n\n\n$x^2-2xy+2y^2+6yz+9z^2 = (x-y)^2+(y+3z)^2 = 175$\n\n\n(x,y,z) are integers, ie. $175 = a^2 + b^2$, \n\n\n$a^2$: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169\n$b^2$: 174, 171, 166, 159, 150, 139, 126, 111, 94, 75, 54, 31, 6\n\n... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/29.json | AHSME |
1980_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | If the ratio of $2x-y$ to $x+y$ is $\frac{2}{3}$, what is the ratio of $x$ to $y$?
$\text{(A)} \ \frac{1}{5} \qquad \text{(B)} \ \frac{4}{5} \qquad \text{(C)} \ 1 \qquad \text{(D)} \ \frac{6}{5} \qquad \text{(E)} \ \frac{5}{4}$
| [
"Cross multiplying gets \n\n\n$6x-3y=2x+2y\\\\4x=5y\\\\ \\dfrac{x}{y}=\\dfrac{5}{4}\\\\ \\boxed{(E)}$\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/3.json | AHSME |
1980_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | How many pairs $(a,b)$ of non-zero real numbers satisfy the equation
\[\frac{1}{a} + \frac{1}{b} = \frac{1}{a+b}\]
$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ \text{one pair for each} ~b \neq 0$
$\text{(E)} \ \text{two pairs for each} ~b \neq 0$
| [
"We hope to simplify this expression into a quadratic in order to find the solutions. To do this, we find a common denominator to the LHS by multiplying by $ab$.\n\\[a+b=\\frac{ab}{a+b}\\]\n\\[a^2+2ab+b^2=ab\\]\n\\[a^2+ab+b^2=0.\\]\n\n\nBy the quadratic formula and checking the discriminant (imagining one of the va... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/8.json | AHSME |
1980_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | For each real number $x$, let $f(x)$ be the minimum of the numbers $4x+1, x+2$, and $-2x+4$. Then the maximum value of $f(x)$ is
$\text{(A)} \ \frac{1}{3} \qquad \text{(B)} \ \frac{1}{2} \qquad \text{(C)} \ \frac{2}{3} \qquad \text{(D)} \ \frac{5}{2} \qquad \text{(E)}\ \frac{8}{3}$
| [
"The first two given functions intersect at $\\left(\\frac{1}{3},\\frac{7}{3}\\right)$, and last two at $\\left(\\frac{2}{3},\\frac{8}{3}\\right)$. Therefore \\[f(x)=\\left\\{ \\begin{matrix} 4x+1 & x<\\frac{1}{3} \\\\ x+2 & \\frac{1}{3}>x>\\frac{2}{3} \\\\ ... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/22.json | AHSME |
1980_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | If $b>1$, $\sin x>0$, $\cos x>0$, and $\log_b \sin x = a$, then $\log_b \cos x$ equals
$\text{(A)} \ 2\log_b(1-b^{a/2}) ~~\text{(B)} \ \sqrt{1-a^2} ~~\text{(C)} \ b^{a^2} ~~\text{(D)} \ \frac 12 \log_b(1-b^{2a}) ~~\text{(E)} \ \text{none of these}$
| [
"\\[\\log_b \\sin x = a\\]\n\\[b^a=\\sin x\\]\n\\[\\log_b \\cos x=c\\]\n\\[b^c=\\cos x\\]\nSince $\\sin^2x+\\cos^2x=1$, \n\\[(b^c)^2+(b^a)^2=1\\]\n\\[b^{2c}+b^{2a}=1\\]\n\\[b^{2c}=1-b^{2a}\\]\n\\[\\log_b (1-b^{2a}) = 2c\\]\n\\[c=\\boxed{\\text{(D)} \\ \\frac 12 \\log_b(1-b^{2a})}\\]\n\n\n-aopspandy\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/18.json | AHSME |
1980_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | In the adjoining figure, CDE is an equilateral triangle and ABCD and DEFG are squares. The measure of $\angle GDA$ is
$\text{(A)} \ 90^\circ \qquad \text{(B)} \ 105^\circ \qquad \text{(C)} \ 120^\circ \qquad \text{(D)} \ 135^\circ \qquad \text{(E)} \ 150^\circ$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair D=... | [
"$m\\angle GDA=360^\\circ-90^\\circ-60^\\circ-90^\\circ=120^\\circ\\Rightarrow\\boxed{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/4.json | AHSME |
1980_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | If the function $f$ is defined by \[f(x)=\frac{cx}{2x+3} ,\quad x\neq -\frac{3}{2} ,\] satisfies $x=f(f(x))$ for all real numbers $x$ except $-\frac{3}{2}$, then $c$ is
$\text{(A)} \ -3 \qquad \text{(B)} \ - \frac{3}{2} \qquad \text{(C)} \ \frac{3}{2} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely d... | [
"As $f(x)=cx/2x+3$, we can plug that into $f(f(x))$ and simplify to get $c^2x/2cx+6x+9 = x$\n. However, we have a restriction on x such that if $x=-3/2$ we have an undefined function. We can use this to our advantage. Plugging that value for x into $c^2x/2cx+6x+9 = x$ yields $c/2 = -3/2$, as the left hand side simp... | 2 | ./CreativeMath/AHSME/1980_AHSME_Problems/14.json | AHSME |
1980_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | A store prices an item in dollars and cents so that when 4% sales tax is added, no rounding is necessary because the result is exactly $n$ dollars where $n$ is a positive integer. The smallest value of $n$ is
$\text{(A)} \ 1 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 25 \qquad \text{(D)} \ 26 \qquad \text{(E)} \ 100$... | [
"Say that the price of the item in cents is $x$ (so $x$ is a positive integer as well). The sales tax would then be $\\frac{x}{25}$, so $n=\\frac{1}{100}\\left( x+\\frac{x}{25}\\right)=\\frac{26x}{2500}=\\frac{13x}{1250}$. \n\n\nSince $x$ is positive integer, the smallest possible integer value for $n=\\frac{13x}{1... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/15.json | AHSME |
1980_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | If $AB$ and $CD$ are perpendicular diameters of circle $Q$, $P$ in $\overline{AQ}$, and $\measuredangle QPC = 60^\circ$, then the length of $PQ$ divided by the length of $AQ$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0); draw(P--C--D^^A--B^^Circle(... | [
"We find that $m\\angle PCQ=30^\\circ$. Because it is a $30^\\circ-60^\\circ-90^\\circ$ right triangle, we can let $PQ=x$, so $CQ=AQ=x\\sqrt{3}$. Thus, $\\frac{PQ}{AQ}=\\frac{x}{x\\sqrt{3}}=\\frac{\\sqrt{3}}{3}\\Rightarrow\\boxed{(B)}$.\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/5.json | AHSME |
1980_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | Let $C_1, C_2$ and $C_3$ be three parallel chords of a circle on the same side of the center.
The distance between $C_1$ and $C_2$ is the same as the distance between $C_2$ and $C_3$.
The lengths of the chords are $20, 16$, and $8$. The radius of the circle is
$\text{(A)} \ 12 \qquad \text{(B)} \ 4\sqrt{7} \qquad... | [
"Let the center of the circle be on the origin with equation $x^2 + y^2 = r^2$. \nAs the chords are bisected by the x-axis their y-coordinates are $10, 8, 4$ respectively. Let the chord of length $10$ have x-coordinate $a$. Let $d$ be the common distance between chords. Thus, the coordinates of the top of the chord... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/19.json | AHSME |
1980_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | Line segments drawn from the vertex opposite the hypotenuse of a right triangle to the points trisecting the hypotenuse have lengths $\sin x$ and $\cos x$, where $x$ is a real number such that $0<x<\frac{\pi}{2}$. The length of the hypotenuse is
$\text{(A)} \ \frac{4}{3} \qquad \text{(B)} \ \frac{3}{2} \qquad \tex... | [
"Consider right triangle $ABC$ with hypotenuse $BC$. Let points $D$ and $E$ trisect $BC$. WLOG, let $AD=cos(x)$ and $AE=sin(x)$ (the proof works the other way around as well). \n\n\nApplying Stewart's theorem on $\\bigtriangleup ABC$ with point $D$, we obtain the equation\n\n\n\\[cos^{2}(x)\\cdot c=a^2 \\cdot \\fra... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/23.json | AHSME |
1980_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | A man walks $x$ miles due west, turns $150^\circ$ to his left and walks 3 miles in the new direction. If he finishes a a point $\sqrt{3}$ from his starting point, then $x$ is
$\text{(A)} \ \sqrt 3 \qquad \text{(B)} \ 2\sqrt{5} \qquad \text{(C)} \ \frac 32 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely d... | [
"Let us think about this. We only know that he ends up $\\sqrt{3}$ away from the origin. However, think about the locus of points $\\sqrt{3}$ away from the origin, a circle. However, his path could end on any part of the circle below the $x-$axis, so therefore, the answer is\n$\\fbox{E: not uniquely determined}.$\n... | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/9.json | AHSME |
1995_AHSME_Problems | 20 | 0 | Probability | Multiple Choice | If $a,b$ and $c$ are three (not necessarily different) numbers chosen randomly and with replacement from the set $\{1,2,3,4,5 \}$, the probability that $ab + c$ is even is
$\mathrm{(A) \ \frac {2}{5} } \qquad \mathrm{(B) \ \frac {59}{125} } \qquad \mathrm{(C) \ \frac {1}{2} } \qquad \mathrm{(D) \ \frac {64}{125} } ... | [
"The probability of $ab$ being odd is $\\left(\\frac 35\\right)^2 = \\frac{9}{25}$, so the probability of $ab$ being even is $1 - \\frac{9}{25} = \\frac {16}{25}$.\n\n\nThe probability of $c$ being odd is $3/5$ and being even is $2/5$\n\n\n$ab+c$ is even if $ab$ and $c$ are both odd, with probability $\\frac{9}{25}... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/20.json | AHSME |
1995_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | Anita attends a baseball game in Atlanta and estimates that there are 50,000 fans in attendance. Bob attends a baseball game in Boston and estimates that there are 60,000 fans in attendance. A league official who knows the actual numbers attending the two games note that:
i. The actual attendance in Atlanta is within... | [
"Since the number of people at the game in Boston is certainly more than the number of fans in Atlanta, we need to compute the maximum of Bob's game minus the minimum of Anita's game. Note however that there is a slight different between conditions (i) and (ii); the attendence is \\textit{within} $\\pm 10 \\%$ from... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/16.json | AHSME |
1995_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | The figure shown can be folded into the shape of a cube. In the resulting cube, which of the lettered faces is opposite the face marked x?
[asy] defaultpen(linewidth(0.7)); path p=origin--(0,1)--(1,1)--(1,2)--(2,2)--(2,3); draw(p^^(2,3)--(4,3)^^shift(2,0)*p^^(2,0)--origin); draw(shift(1,0)*p, dashed); label("$x$", (... | [
"The marked side is the side with the $x.$ We imagine it folding up. First, we fold the $x$ upwards. Now we fold the $A$ upwards, and thus $x$ is touching $B$ on it's left side. We now fold $B$ up, and we realize that $x$ won't be touching $\\boxed{\\mathrm{(C)}}$ at all.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/6.json | AHSME |
1995_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | The radius of Earth at the equator is approximately 4000 miles. Suppose a jet flies once around Earth at a speed of 500 miles per hour relative to Earth. If the flight path is a neglibile height above the equator, then, among the following choices, the best estimate of the number of hours of flight is:
$\mathrm{(A... | [
"We want the number of hours that it takes the jet to fly the length of the circumference. $\\frac{8000\\pi}{500}=16\\pi$. The best estimate of that is $50\\Rightarrow \\mathrm{(C)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/7.json | AHSME |
1995_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | Given regular pentagon $ABCDE$, a circle can be drawn that is tangent to $\overline{DC}$ at $D$ and to $\overline{AB}$ at $A$. The number of degrees in minor arc $AD$ is
[asy]size(100); defaultpen(linewidth(0.7)); draw(rotate(18)*polygon(5)); real x=0.6180339887; draw(Circle((-x,0), 1)); int i; for(i=0; i<5; i=i+1) {... | [
"Define major arc DA as $DA$, and minor arc DA as $da$. Extending DC and AB to meet at F, we see that $\\angle CFB=36=\\frac{DA-da}{2}$. We now have two equations: $DA-da=72$, and $DA+da=360$. Solving, $DA=216$ and $da=144\\Rightarrow \\mathrm{(E)}$.\nhello\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/17.json | AHSME |
1995_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | Two nonadjacent vertices of a rectangle are $(4,3)$ and $(-4,-3)$, and the coordinates of the other two vertices are integers. The number of such rectangles is
$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 }$
| [
"The center of the rectangle is $(0,0)$, and the distance from the center to a corner is $\\sqrt{4^2+3^2}=5$. The remaining two vertices of the rectangle must be another pair of points opposite each other on the circle of radius 5 centered at the origin. Let these points have the form $(\\pm x,\\pm y)$, where $x^2+... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/21.json | AHSME |
1995_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | The area of the triangle bounded by the lines $y = x, y = - x$ and $y = 6$ is
$\mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 12\sqrt{2} } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 24\sqrt{2} } \qquad \mathrm{(E) \ 36 }$
| [
"\\begin{center}\n[asy] defaultpen(fontsize(8)); draw((0,0)--(6,6)--(-6,6)--(0,0)); draw((0,-1)--(0,8)); draw((-7,0)--(7,0)); label(\"$(6,6)$\",(6,6), (1,1));label(\"$(-6,6)$\",(-6,6),(-1,1)); label(\"$y=x$\",(3,3),(1,-1));label(\"$y=-x$\",(-3,3),(-1,-0.5)); label(\"$6$\",(-3,6),(0,1));label(\"$6$\",(3,6),(0,1)); l... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/10.json | AHSME |
1995_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | In the figure, $\overline{AB}$ and $\overline{CD}$ are diameters of the circle with center $O$, $\overline{AB} \perp \overline{CD}$, and chord $\overline{DF}$ intersects $\overline{AB}$ at $E$. If $DE = 6$ and $EF = 2$, then the area of the circle is
[asy]size(120); defaultpen(linewidth(0.7)); pair O=origin, A=(-5,0),... | [
"\\textbf{Solution 1}\n\n\nLet the radius of the circle be $r$ and let $x=\\overline{OE}$. \n\n\nBy the Pythagorean Theorem, $OD^2+OE^2=DE^2 \\Rightarrow r^2+x^2=6^2=36$. \n\n\nBy Power of a point, $AE \\cdot EB = DE \\cdot EF \\Rightarrow (r+x)(r-x)=r^2-x^2=6\\cdot2=12$. \n\n\nAdding these equations yields $2r^2=... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/26.json | AHSME |
1995_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathr... | [
"Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.\n\n\nNow consider the diagonal from $(0,0,0)$ to $(3,3,3)$. The midpoint of this diagonal is at $\\left(\\frac 32,\\frac 32,\\frac 32\\right)$. The plane that passes through this point and is orthogonal ... | 2 | ./CreativeMath/AHSME/1995_AHSME_Problems/30.json | AHSME |
1995_AHSME_Problems | 27 | 0 | Number Theory | Multiple Choice | Consider the triangular array of numbers with 0,1,2,3,... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 1 through 6 are shown.
\[\begin{tabular}{ccccccccccc} & & & & & 0 & & & & & \\ & & & & 1 & & 1 & & & & \\ & & & 2 & & 2 & & 2 & & & \\ & & 3 & & 4 & & 4 ... | [
"Note that if we re-draw the table with an additional diagonal row on each side, the table is actually just two of Pascal's Triangles, except translated and summed. \n\\[\\begin{tabular}{ccccccccccccccc} & & & & & 1 & & 0 & & 1 & & & & \\\\ & & & & 1 & & 1 & & 1 & & 1 & & & \\\\ & & & 1 & & 2 & & 2 & & 2 & & 1 & & ... | 3 | ./CreativeMath/AHSME/1995_AHSME_Problems/27.json | AHSME |
1995_AHSME_Problems | 1 | 0 | Arithmetic | Multiple Choice | Kim earned scores of 87,83, and 88 on her first three mathematics examinations. If Kim receives a score of 90 on the fourth exam, then her average will
$\mathrm{(A) \ \text{remain the same} } \qquad \mathrm{(B) \ \text{increase by 1} } \qquad \mathrm{(C) \ \text{increase by 2} } \qquad \mathrm{(D) \ \text{increase... | [
"The average of the first three test scores is $\\frac{88+83+87}{3}=86$. The average of all four exams is $\\frac{87+83+88+90}{4}=87$. It increased by one point. $\\mathrm{(B)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/1.json | AHSME |
1995_AHSME_Problems | 11 | 0 | Counting | Multiple Choice | How many base 10 four-digit numbers, $N = \underline{a} \underline{b} \underline{c} \underline{d}$, satisfy all three of the following conditions?
(i) $4,000 \leq N < 6,000;$ (ii) $N$ is a multiple of 5; (iii) $3 \leq b < c \leq 6$.
$\mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 18 } \qquad \mathrm{(C) \ 24 } \qquad \m... | [
"\\begin{itemize}\n\\item For condition (i), the restriction is put on $a$; $N<4000$ if $a<4$, and $N \\ge 6$ if $a \\ge 6$. Therefore, $a=4,5$.\n\n\\item For condition (ii), the restriction is put on $d$; it must be a multiple of $5$. Therefore, $d=0,5$.\n\n\\item For condition (iii), the restriction is put on $b$... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/11.json | AHSME |
1995_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | If $\sqrt {2 + \sqrt {x}} = 3$, then $x =$
$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ \sqrt{7} } \qquad \mathrm{(C) \ 7 } \qquad \mathrm{(D) \ 49 } \qquad \mathrm{(E) \ 121 }$
| [
"$2 + \\sqrt{x} = 9 \\Longrightarrow \\sqrt{x} = 7 \\Longrightarrow x = 49 \\Rightarrow \\mathrm{(D)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/2.json | AHSME |
1995_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | Let $f$ be a linear function with the properties that $f(1) \leq f(2), f(3) \geq f(4),$ and $f(5) = 5$. Which of the following is true?
$\mathrm{(A) \ f(0) < 0 } \qquad \mathrm{(B) \ f(0) = 0 } \qquad \mathrm{(C) \ f(1) < f(0) < f( - 1) } \qquad \mathrm{(D) \ f(0) = 5 } \qquad \mathrm{(E) \ f(0) > 5 }$
| [
"A linear function has the property that $f(a)\\leq f(b)$ either for all $a<b$, or for all $b<a$. Since $f(3)\\geq f(4)$, $f(1)\\geq f(2)$. Since $f(1)\\leq f(2)$, $f(1)=f(2)$. And if $f(a)=f(b)$ for $a\\neq b$, then $f(x)$ is a constant function. Since $f(5)=5$, $f(0)=5\\Rightarrow \\mathrm{(D)}$\n\n\n",
"If $f$... | 3 | ./CreativeMath/AHSME/1995_AHSME_Problems/12.json | AHSME |
1995_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | There exist positive integers $A,B$ and $C$, with no common factor greater than $1$, such that
\[A \log_{200} 5 + B \log_{200} 2 = C.\]
What is $A + B + C$?
$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$
| [
"\\[A \\log_{200} 5 + B \\log_{200} 2 = C\\]\n\n\nSimplifying and taking the logarithms away,\n\n\n\\[5^A \\cdot 2^B=200^C=2^{3C} \\cdot 5^{2C}\\]\n\n\nTherefore, $A=2C$ and $B=3C$. Since $A, B,$ and $C$ are relatively prime, $C=1$, $B=3$, $A=2$. $A+B+C=6 \\Rightarrow \\mathrm{(A)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/24.json | AHSME |
1995_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | A list of five positive integers has mean $12$ and range $18$. The mode and median are both $8$. How many different values are possible for the second largest element of the list?
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$
| [
"Let $a$ be the smallest element, so $a+18$ is the largest element. Since the mode is $8$, at least two of the five numbers must be $8$. The last number we denote as $b$.\n\n\nThen their average is $\\frac{a + (8) + (8) + b + (a+18)}5 = 12 \\Longrightarrow 2a + b = 26$. Clearly $a \\le 8$. Also we have $b \\le a + ... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/25.json | AHSME |
1995_AHSME_Problems | 13 | 0 | Arithmetic | Multiple Choice | The addition below is incorrect. The display can be made correct by changing one digit $d$, wherever it occurs, to another digit $e$. Find the sum of $d$ and $e$.
$\begin{tabular}{ccccccc} & 7 & 4 & 2 & 5 & 8 & 6 \\ + & 8 & 2 & 9 & 4 & 3 & 0 \\ \hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}$
$\mathrm{(A) \ 4 } \qqu... | [
"If we change $0$, the units column would be incorrect. \n\n\nIf we change $1$, then the leading $1$ in the sum would be incorrect.\n\n\nHowever, looking at the $2$ in the hundred-thousands column, it would be possible to change the $2$ to either a $5$ (no carry) or a $6$ (carry) to create a correct statement. \n... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/13.json | AHSME |
1995_AHSME_Problems | 29 | 0 | Counting | Multiple Choice | For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$?
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$
| [
"$2310 = 2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11$. We wish to figure out the number of ways to distribute these prime factors amongst 3 different integers, without over counting triples which are simply permutations of one another.\n\n\nWe can account for permutations by assuming WLOG that $a$ contains the prime fac... | 2 | ./CreativeMath/AHSME/1995_AHSME_Problems/29.json | AHSME |
1995_AHSME_Problems | 3 | 0 | Arithmetic | Multiple Choice | The total in-store price for an appliance is $\textdollar 99.99$. A television commercial advertises the same product for three easy payments of $\textdollar 29.98$ and a one-time shipping and handling charge of $\textdollar 9.98$. How many cents are saved by buying the appliance from the television advertiser?
$\mat... | [
"IMPORTANT NOTICE: The original problem statement had \"how much is saved\". However, because this made little sense when the calculations were done, the problem statement was changed to \"how many cents\".\n\n\nWe see that 3 payments of $\\textdollar 29.98$ will be a total cost of $3\\cdot(30-.02)=90-.06$\n\n\nAdd... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/3.json | AHSME |
1995_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | In $\triangle ABC$, $\angle C = 90^\circ, AC = 6$ and $BC = 8$. Points $D$ and $E$ are on $\overline{AB}$ and $\overline{BC}$, respectively, and $\angle BED = 90^\circ$. If $DE = 4$, then $BD =$
$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ \frac {16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15... | [
"[asy] size(120); defaultpen(0.7); pair A = (0,6), B = (8,0), C= (0,0), D = (8/3,4), E = (8/3,0), F = (0, 3), G = (38/15,1.6); draw(A--B--E--D--E--B--C--A--B--cycle); label(\"\\(A\\)\",A,W); label(\"\\(B\\)\",B,E); label(\"\\(C\\)\",C,SW); label(\"\\(D\\)\",D,NE); label(\"\\(E\\)\",E,S); label(\"\\(6\\)\",F,W); la... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/8.json | AHSME |
1995_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13, 19, 20, 25$ and $31$, although this is not necessarily their order around the pentagon. The area of the pentagon is
$\mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C... | [
"\\begin{center}\n[asy] defaultpen(linewidth(0.7)); draw((0,0)--(31,0)--(31,25)--(12,25)--(0,20)--cycle); draw((0,20)--(0,25)--(12,25)--cycle,linetype(\"4 4\")); [/asy]\\end{center}\nSince the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follo... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/22.json | AHSME |
1995_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is
$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ \frac {1 + \sqrt {3}}{\sqrt 2} } \qquad \mathrm{(C) \ \sqrt{3} } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \... | [
"Triangle $OAB$ has the property that $\\angle O=30^{\\circ}$ and $AB=1$. \n\n\nFrom the Law of Sines, $\\frac{\\sin{\\angle A}}{OB}=\\frac{\\sin{\\angle O}}{AB}$. \n\n\nSince $\\sin 30^\\circ = \\frac{1}{2}$, we have:\n\n\n$\\frac{\\sin{\\angle A}}{OB}=\\frac{\\frac{1}{2}}{1}$\n\n\n$2\\sin{\\angle A}=OB$. \n\n\nTh... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/18.json | AHSME |
1995_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | If $M$ is $30 \%$ of $Q$, $Q$ is $20 \%$ of $P$, and $N$ is $50 \%$ of $P$, then $\frac {M}{N} =$
$\mathrm{(A) \ \frac {3}{250} } \qquad \mathrm{(B) \ \frac {3}{25} } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ \frac {6}{5} } \qquad \mathrm{(E) \ \frac {4}{3} }$
| [
"We are given: $M=\\frac{3Q}{10}$, $Q=\\frac{P}{5}$, $N=\\frac{P}{2}$. We want M in terms of N, so we substitute N into everything:\n\n\n$\\frac{2}{5}N=\\frac{P}{5}=Q$\n\n\n$M=\\frac{3N}{25}$\n\n\n$\\frac{M}{N}=\\frac{3}{25} \\Rightarrow \\mathrm{(B)}$\n\n\n",
"Alternatively, picking an arbitrary value for $Q$ of... | 2 | ./CreativeMath/AHSME/1995_AHSME_Problems/4.json | AHSME |
1995_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$, then $f(3) =$
$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$
| [
"$f(-x) = a(-x)^4 - b(-x)^2 - x + 5$\n\n\n$f(-x) = ax^4 - bx^2 - x + 5$\n\n\n$f(-x) = (ax^4 - bx^2 + x + 5) - 2x$\n\n\n$f(-x) = f(x) - 2x$. \n\n\nThus $f(3) = f(-3)-2(-3) = 8 \\Rightarrow \\mathrm{(E)}$.\n\n\n",
"If $f(-3) = 2$, then $81a - 9b + -3 + 5 = 2$. Simplifying, we get $81a - 9b = 0$.\n\n\nGetting an e... | 3 | ./CreativeMath/AHSME/1995_AHSME_Problems/14.json | AHSME |
1995_AHSME_Problems | 15 | 0 | Number Theory | Multiple Choice | Five points on a circle are numbered 1,2,3,4, and 5 in clockwise order. A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points. If the bug begins on point 5, after 1995 jumps it ... | [
"Let us see how the bug moves.\n\n\nFirst, we see that if it starts at point 5, it moves to point 1.\n\n\nAt point 1, it moves to point 2.\n\n\nAt point 2, since 2 is even, it moves to point 4.\n\n\nThen at point 4, it moves to point 1.\n\n\nWe can see that at this point, the bug will cycle between 1, 2, and 4\n\n\... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/15.json | AHSME |
1995_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | A rectangular field is 300 feet wide and 400 feet long. Random sampling indicates that there are, on the average, three ants per square inch through out the field. [12 inches = 1 foot.] Of the following, the number that most closely approximates the number of ants in the field is
$\mathrm{(A) \ \text{500 thousand}... | [
"The rectangular field is $300 \\text{ feet} \\cdot \\frac{12 \\text{ inches}}{1 \\text{ foot}} = 3600 \\text{ inches}$ wide and $400 \\text{ feet} \\cdot \\frac{12 \\text{ inches}}{1 \\text{ foot}} = 4800 \\text{ inches}$ inches long.\n\n\nIt has an area of $3.6\\cdot 10^3 \\cdot 4.8 \\cdot 10^3 = 17.28 \\cdot 1... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/5.json | AHSME |
1995_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | Equilateral triangle $DEF$ is inscribed in equilateral triangle $ABC$ such that $\overline{DE} \perp \overline{BC}$. The ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ is
$\mathrm{(A) \ \frac {1}{6} } \qquad \mathrm{(B) \ \frac {1}{4} } \qquad \mathrm{(C) \ \frac {1}{3} } \qquad \mathrm{(D) \ \fr... | [
"[asy]draw((0,0)--(3,0)); draw((0,0)--(1.5, 2.59807621)); draw((3,0)--(1.5, 2.59807621)); draw((1,0)--(1, 1.73205081)); draw((1,0)--(2.5, 0.866025404)); draw((1, 1.73205081)--(2.5, 0.866025404));[/asy]\n\n\nLet's take one of the smaller right triangles. Without loss of generality, let the smaller leg be $1$. Since... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/19.json | AHSME |
1995_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | The sides of a triangle have lengths $11,15,$ and $k$, where $k$ is an integer. For how many values of $k$ is the triangle obtuse?
$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 13 } \qquad \mathrm{(E) \ 14 }$
| [
"By the Law of Cosines, a triangle is obtuse if the sum of the squares of two of the sides of the triangles is less than the square of the third. The largest angle is either opposite side $15$ or side $k$. If $15$ is the largest side, \n\n\n\\[15^2 >11^2 + k^2 \\Longrightarrow k < \\sqrt{104}\\]\n\n\nBy the Triangl... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/23.json | AHSME |
1995_AHSME_Problems | 9 | 0 | Counting | Multiple Choice | Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is
$\mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 12 } \qquad \mathrm{(C) \ 14 } \qquad \mathrm{(D) \ 16 } \qquad \mathrm{(E) \ 18 }$
| [
"[asy]draw((0,0)--(2,2)); draw((0,2)--(2,0)); draw((0,0)--(0,2)); draw((0,2)--(2,2)); draw((2,2)--(2,0)); draw((0,0)--(2,0)); draw((1,0)--(1,2)); draw((0,1)--(2,1));[/asy]\n\n\nThere are 8 little triangles, 4 triangles with twice the area, and 4 triangles with four times the area of the smaller triangles. $8+4+4=16... | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/9.json | AHSME |
1977_AHSME_Problems | 20 | 0 | Counting | Multiple Choice | $\begin{tabular}{ccccccccccccc}& & & & & & C & & & & & &\\ & & & & & C & O & C & & & & &\\ & & & & C & O & N & O & C & & & &\\ & & & C & O & N & T & N & O & C & & &\\ & & C & O & N & T & E & T & N & O & C & &\\ & C & O & N & T & E & S & E & T & N & O & C &\\ C & O & N & T & E & S & T & S & E & T & N & O & C \end{... | [
"The path will either start on the left side, start on the right side, or go down the middle. Say it starts on the left side. Instead of counting it starting from the $C$, let's start at the T and work backwards. Starting at the $T$, there are $2$ options to choose the $S$. Then, there are $2$ options to chose the ... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/20.json | AHSME |
1977_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | If $i^2 = -1$, then the sum \[\cos{45^\circ} + i\cos{135^\circ} + \cdots + i^n\cos{(45 + 90n)^\circ} + \cdots + i^{40}\cos{3645^\circ}\]
equals
$\text{(A)}\ \frac{\sqrt{2}}{2} \qquad \text{(B)}\ -10i\sqrt{2} \qquad \text{(C)}\ \frac{21\sqrt{2}}{2} \qquad\\ \text{(D)}\ \frac{\sqrt{2}}{2}(21 - 20i) \qquad \text{(E... | [
"Solution by e_power_pi_times_i\n\n\nNotice that the sequence repeats every $4$ terms. These $4$ terms are $\\cos(45^\\circ)$, $i\\cos(135^\\circ)$, $-\\cos(225^\\circ)$, and $-i\\cos(315^\\circ)$. This is because $\\cos(\\theta) = \\cos(360^\\circ - \\theta)$. Also $\\cos(\\theta) = -\\cos(180^\\circ - \\theta)$, ... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/16.json | AHSME |
1977_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $x, y$ and $2x + \frac{y}{2}$ are not zero, then
$\left( 2x + \frac{y}{2} \right)^{-1} \left[(2x)^{-1} + \left( \frac{y}{2} \right)^{-1} \right]$
equals
$\textbf{(A) }1\qquad \textbf{(B) }xy^{-1}\qquad \textbf{(C) }x^{-1}y\qquad \textbf{(D) }(xy)^{-1}\qquad \textbf{(E) }\text{none of these}$
| [
"We can write $\\left( 2x+ \\frac{y}{2} \\right)^{-1} = \\left( \\frac{4x+y}{2} \\right)^{-1} = \\frac{2}{4x+y}$.\n\n\nThen, the expression simplifies to \\[\\left( 2x + \\frac{y}{2} \\right)^{-1} \\left[(2x)^{-1} + \\left( \\frac{y}{2} \\right)^{-1} \\right] \\Rightarrow \\frac{2}{4x+y} \\left( \\frac{1}{2x} + \\f... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/6.json | AHSME |
1977_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | If $t = \frac{1}{1 - \sqrt[4]{2}}$, then $t$ equals
$\text{(A)}\ (1-\sqrt[4]{2})(2-\sqrt{2})\qquad \text{(B)}\ (1-\sqrt[4]{2})(1+\sqrt{2})\qquad \text{(C)}\ (1+\sqrt[4]{2})(1-\sqrt{2}) \qquad \\ \text{(D)}\ (1+\sqrt[4]{2})(1+\sqrt{2})\qquad \text{(E)}-(1+\sqrt[4]{2})(1+\sqrt{2})$
| [
"Solution by e_power_pi_times_i\n\n\n$t = \\dfrac{1}{1-\\sqrt[4]{2}} = (\\dfrac{1}{1-\\sqrt[4]{2}})(\\dfrac{1+\\sqrt[4]{2}}{1+\\sqrt[4]{2}}) = (\\dfrac{1+\\sqrt[4]{2}}{1-\\sqrt{2}})(\\dfrac{1+\\sqrt{2}}{1+\\sqrt{2}}) = \\dfrac{(1+\\sqrt[4]{2})(1+\\sqrt{2})}{-1} = \\text{(E)}-(1+\\sqrt[4]{2})(1+\\sqrt{2})$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/7.json | AHSME |
1977_AHSME_Problems | 17 | 0 | Probability | Multiple Choice | Three fair dice are tossed at random (i.e., all faces have the same probability of coming up).
What is the probability that the three numbers turned up can be arranged to form an arithmetic progression with common difference one?
$\textbf{(A) }\frac{1}{6}\qquad \textbf{(B) }\frac{1}{9}\qquad \textbf{(C) }\frac{1}{27... | [
"Solution by e_power_pi_times_i\n\n\n\n\nThe denominator of the probability is $6^3 = 216$. We are asked to find the probability that the three numbers are consecutive. The three numbers can be arranged in $6$ ways, and there are $4$ triplets of consecutive numbers from $1-6$. The probability is $\\dfrac{6\\cdot4}{... | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/17.json | AHSME |
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