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 |
|---|---|---|---|---|---|---|---|---|---|
1994_AHSME_Problems
| 24
| 0
|
Other
|
Multiple Choice
|
A sample consisting of five observations has an arithmetic mean of $10$ and a median of $12$. The smallest value that the range (largest observation minus smallest) can assume for such a sample is
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 10$
|
[
"The minimum range occurs in the set $\\{7,7,12,12,12\\}$, so the answer is $\\boxed{\\textbf{(C)}\\ 5}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/24.json
|
AHSME
|
1994_AHSME_Problems
| 25
| 0
|
Algebra
|
Multiple Choice
|
If $x$ and $y$ are non-zero real numbers such that
\[|x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0,\]
then the integer nearest to $x-y$ is
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5$
|
[
"We have two cases to consider: $x$ is positive or $x$ is negative. If $x$ is positive, we have $x+y=3$ and $xy+x^3=0$\n\n\nSolving for $y$ in the top equation gives us $3-x$. Plugging this in gives us: $x^3-x^2+3x=0$. Since we're told $x$ is not zero, we can divide by $x$, giving us: $x^2-x+3=0$\n\n\nThe discriminant of this is $(-1)^2-4(1)(3)=-11$, which means the equation has no real solutions. \n\n\nWe conclude that $x$ is negative. In this case $-x+y=3$ and $-xy+x^3=0$. Negating the top equation gives us $x-y=-3$. We seek $x-y$, so the answer is $\\boxed{(A) -3}$\n\n\n-solution by jmania\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/25.json
|
AHSME
|
1994_AHSME_Problems
| 29
| 0
|
Geometry
|
Multiple Choice
|
Points $A, B$ and $C$ on a circle of radius $r$ are situated so that $AB=AC, AB>r$, and the length of minor arc $BC$ is $r$. If angles are measured in radians, then $AB/BC=$
[asy] draw(Circle((0,0), 13)); draw((-13,0)--(12,5)--(12,-5)--cycle); dot((-13,0)); dot((12,5)); dot((12,-5)); label("A", (-13,0), W); label("B", (12,5), NE); label("C", (12,-5), SE); [/asy]
$\textbf{(A)}\ \frac{1}{2}\csc{\frac{1}{4}} \qquad\textbf{(B)}\ 2\cos{\frac{1}{2}} \qquad\textbf{(C)}\ 4\sin{\frac{1}{2}} \qquad\textbf{(D)}\ \csc{\frac{1}{2}} \qquad\textbf{(E)}\ 2\sec{\frac{1}{2}}$
|
[
"First note that arc length equals $r\\theta$, where $\\theta$ is the central angle in radians. Call the center of the circle $O$. Then $\\angle{BOC} = 1$ radian because the minor arc $BC$ has length $r$. Since $ABC$ is isosceles, $\\angle{AOB} = \\pi - \\tfrac{1}{2}$. We use the Law of Cosines to find that \\[\\frac{AB}{BC} = \\frac{\\sqrt{2r^2 - 2r^2\\cos{(\\pi - \\frac{1}{2})}}}{\\sqrt{2r^2 - 2r^2\\cos1}} = \\frac{\\sqrt{1 + \\cos{(\\frac{1}{2})}}}{\\sqrt{1 - \\cos1}}.\\] \nUsing half-angle formulas, we have that this ratio simplifies to \\[\\frac{\\cos\\frac{1}{4}}{\\sin{\\frac{1}{2}}} = \\frac{\\cos\\frac{1}{4}}{\\sqrt{1 - \\cos^2{\\frac{1}{2}}}} = \\frac{\\cos\\frac{1}{4}}{\\sqrt{(1 + \\cos{\\frac{1}{2}})(1 - \\cos{\\frac{1}{2}})}} = \\frac{\\cos{\\frac{1}{4}}}{2\\cos{\\frac{1}{4}}\\sin{\\frac{1}{4}}}\\] \\[= \\boxed{\\frac{1}{2}\\csc{\\frac{1}{4}}.}\\]\n\n\n",
"[asy] draw(Circle((0,0), 13)); draw((-13,0)--(12,5)--(12,-5)--cycle); dot((-13,0)); dot((12,5)); dot((12,-5)); dot((12,0)); dot((0,0)); draw((-13,0)--(12,0)--cycle); label(\"A\", (-13,0), W); label(\"B\", (12,5), NE); label(\"C\", (12,-5), SE); label(\"D\", (12,0), NW); label(\"O\", (0,0), NE); [/asy]\n\n\nLet the center of this circle be $O$, $\\angle BOC = \\theta$, the radius of $\\odot O$ be $r$.\n\n\nBy the definition of radian, $\\theta =$\n$\\overarc{BC}$\n$/r=1$\n\n\n\n\n$\\angle BAC=\\frac{\\angle BOC}{2} = \\frac12$\n\n\n$\\sin \\angle BAD = \\frac{BD}{AB}$, $\\sin \\frac{\\angle BAC}{2} = \\frac{\\frac{BC}{2}}{AB}$\n\n\n$\\frac{BC}{AB}=2 \\sin \\frac{\\frac12}{2} = 2\\sin \\frac14$\n\n\n$\\frac{AB}{BC}= \\frac{1}{2\\sin \\frac14} = \\boxed{\\textbf{(A) } \\frac{1}{2} \\csc \\frac{1}{4} }$\n\n\n~isabelchen\n\n\n"
] | 2
|
./CreativeMath/AHSME/1994_AHSME_Problems/29.json
|
AHSME
|
1994_AHSME_Problems
| 3
| 0
|
Algebra
|
Multiple Choice
|
How many of the following are equal to $x^x+x^x$ for all $x>0$?
$\textbf{I:}\ 2x^x \qquad\textbf{II:}\ x^{2x} \qquad\textbf{III:}\ (2x)^x \qquad\textbf{IV:}\ (2x)^{2x}$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$
|
[
"We look at each statement individually.\n\n\n$\\textbf{I:}\\ 2x^x$. We note that $x^x+x^x=x^x(1+1)=2x^x$. So statement $\\textbf{I}$ is true.\n\n\n$\\textbf{II:}\\ x^{2x}$. We find a counter example which is $x=1$. $2\\neq 1$. So statement $\\textbf{II}$ is false.\n\n\n$\\textbf{III:}\\ (2x)^x$. We see that this statement is equal to $2^xx^x$. $x=2$ is a counter example. $8\\neq 16$. So statement $\\textbf{III}$ is false.\n\n\n$\\textbf{IV:}\\ (2x)^{2x}$. We see that $x=1$ is again a counter example. $2\\neq 4$. So statement $\\textbf{IV}$ is false.\n\n\nTherefore, our answer is $\\boxed{\\textbf{(B) }1}$.\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/3.json
|
AHSME
|
1994_AHSME_Problems
| 8
| 0
|
Geometry
|
Multiple Choice
|
In the polygon shown, each side is perpendicular to its adjacent sides, and all 28 of the sides are congruent. The perimeter of the polygon is $56$. The area of the region bounded by the polygon is
[asy] draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); [/asy]
$\textbf{(A)}\ 84 \qquad\textbf{(B)}\ 96 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 112 \qquad\textbf{(E)}\ 196$
|
[
"Since the perimeter is $56$ and all of the sides are congruent, the length of each side is $2$. We break the figure into squares as shown below.\n\n\n[asy] unitsize(0.8cm); draw(shift(3,4)*((0,-1)--(1,-1))); draw(shift(3,4)*((-1,-2)--(2,-2))); draw(shift(3,4)*((-2,-3)--(3,-3))); draw(shift(3,4)*((-2,-4)--(3,-4))); draw(shift(3,4)*((-1,-5)--(2,-5))); draw(shift(3,4)*((0,-6)--(1,-6))); draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); draw((1,0)--(1,1)); draw((2,-1)--(2,2)); draw((3,-2)--(3,3)); draw((4,-2)--(4,3)); draw((5,-1)--(5,2)); draw((6,0)--(6,1)); [/asy]\n\n\nWe see that there are a total of $2(1+3+5)+7=25$ squares with side length $2$. Therefore, the total area is $4\\cdot 25=\\boxed{\\textbf{(C) }100.}$\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/8.json
|
AHSME
|
1994_AHSME_Problems
| 22
| 0
|
Counting
|
Multiple Choice
|
Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the six students and decide to choose their chairs so that each professor will be between two students. In how many ways can Professors Alpha, Beta and Gamma choose their chairs?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 36 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 84 \qquad\textbf{(E)}\ 630$
|
[
"Since each professor must sit between two students, they cannot be seated in seats $1$ or $9$ (the seats at either end of the row). Hence, each professor has $7$ seats they can choose from and must be at least $1$ seat apart.\n\n\nThis question is equivalent to choosing $3$ seats from a row of $5$ seats with no restrictions because we can simply generate a valid arrangement by inserting a seat right after the first and second seat chosen.\n\n\nHence, the answer is $\\binom{5}{3} \\cdot 3! =$ $60$ $\\textbf{(C)}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/22.json
|
AHSME
|
1994_AHSME_Problems
| 18
| 0
|
Geometry
|
Multiple Choice
|
Triangle $ABC$ is inscribed in a circle, and $\angle B = \angle C = 4\angle A$. If $B$ and $C$ are adjacent vertices of a regular polygon of $n$ sides inscribed in this circle, then $n=$
[asy] draw(Circle((0,0), 5)); draw((0,5)--(3,-4)--(-3,-4)--cycle); label("A", (0,5), N); label("B", (-3,-4), SW); label("C", (3,-4), SE); dot((0,5)); dot((3,-4)); dot((-3,-4)); [/asy]
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18$
|
[
"We solve for $\\angle A$ as follows: \\[4\\angle A+4\\angle A+\\angle A=180\\implies 9\\angle A=180\\implies \\angle A=20.\\] That means that minor arc $\\widehat{BC}$ has measure $40^\\circ$. We can fit a maximum of $\\frac{360}{40}=\\boxed{\\textbf{(C) }9}$ of these arcs in the circle.\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/18.json
|
AHSME
|
1994_AHSME_Problems
| 4
| 0
|
Geometry
|
Multiple Choice
|
In the $xy$-plane, the segment with endpoints $(-5,0)$ and $(25,0)$ is the diameter of a circle. If the point $(x,15)$ is on the circle, then $x=$
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20$
|
[
"We see that the center of this circle is at $\\left(\\frac{-5+25}{2},0\\right)=(10,0)$. The radius is $\\frac{30}{2}=15$. So the equation of this circle is \\[(x-10)^2+y^2=225.\\] Substituting $y=15$ yields $(x-10)^2=0$ so $x=\\boxed{\\textbf{(A) }10}$.\n\n\n--Solution by TheMaskedMagician\n\n\n",
"The diameter of the circle is $25- (-5)=30$ so the radius is $15$ and the center of the circle is on the $x$-axis at $x=10$. The only point with $y=15$ must be the point exactly above the center.\n\n\n"
] | 2
|
./CreativeMath/AHSME/1994_AHSME_Problems/4.json
|
AHSME
|
1994_AHSME_Problems
| 14
| 0
|
Algebra
|
Multiple Choice
|
Find the sum of the arithmetic series
\[20+20\frac{1}{5}+20\frac{2}{5}+\cdots+40\]
$\textbf{(A)}\ 3000 \qquad\textbf{(B)}\ 3030 \qquad\textbf{(C)}\ 3150 \qquad\textbf{(D)}\ 4100 \qquad\textbf{(E)}\ 6000$
|
[
"\\subsubsection{Brief Introduction}\nFor those that do not know the formula, the sum of an arithmetic series with first term $a_1$, last term $a_n$ as $n$ terms, is \\[S = \\frac{n(a_1+a_n)}{2}.\\] We can prove this as follows:\n\n\nLet $d$ be the common difference between terms of our series and let $n$ be the number of terms in our series. Let $a_1$ be the first term. Our series is \\[a_1,~a_1+d,~a_1+2d,\\dots,a_1+(n-1)d.\\] Note that we have $n-1$ in the last term because $a_1$ is a term. Let $S$ be our sum such that \\[S=a_1+(a_1+d)+\\dots+(a_1+(n-1)d).\\] We can rewrite our sums as \\[S=(a_1+(n-1)d)+(a_1+(n-2)d)+\\dots+(a_1+d)+a_1.\\] Adding these two sums together essentially creates $n$ pairs of $a_1+(a_1+(n-1)d)$ as shown below: \\[2S=n(a_1+(a_1+(n-1)d))\\implies S=\\frac{n(a_1+a_n)}{2}\\] We use $a_n$ in place of $a_1+(n-1)d$ to represent the last term.\n\n\n\n\n\n\n\\hrule\n\\subsubsection{Solving}\nOur first term is $20$ and our last term is $40$. To find the number of terms, $n$, we note that the common difference between each term is $\\frac{1}{5}$. So we have \\[20+\\frac{1}{5}(n-1)=40\\implies n-1=100\\implies n=101.\\] Using our formula, our sum is \\[101\\left(\\frac{20+40}{2}\\right)=101\\times 30=\\boxed{\\textbf{(B) }3030.}\\]\n\n\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/14.json
|
AHSME
|
1994_AHSME_Problems
| 15
| 0
|
Number Theory
|
Multiple Choice
|
For how many $n$ in $\{1, 2, 3, ..., 100 \}$ is the tens digit of $n^2$ odd?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50$
|
[
"Let $n=10a+b$. So $n^2=(10a+b)^2=100a^2+20ab+b^2$. The term $100a^2$ only contributes digits starting at the hundreds place, so this does not affect whether the tens digit is odd. The term $20ab$ only contributes digits starting at the tens place, and the tens digit contributed will be the ones digit of $2ab$ which is even. So we see this term also does not affect whether the tens digit is odd. This means only $b^2$ can affect whether the tens digit is odd. We can quickly check $1^2=1, \\dots, 9^2=81$ and discover only for $b=4$ or $b=6$ does $b^2$ have an odd tens digit. The total number of positive integers less than or equal to $100$ that have $4$ or $6$ as the units digit is \\[10\\times 2=\\boxed{\\textbf{(B) }20.}\\]\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/15.json
|
AHSME
|
1994_AHSME_Problems
| 5
| 0
|
Algebra
|
Multiple Choice
|
Pat intended to multiply a number by $6$ but instead divided by $6$. Pat then meant to add $14$ but instead subtracted $14$. After these mistakes, the result was $16$. If the correct operations had been used, the value produced would have been
$\textbf{(A)}\ \text{less than 400} \qquad\textbf{(B)}\ \text{between 400 and 600} \qquad\textbf{(C)}\ \text{between 600 and 800} \\ \textbf{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}$
|
[
"We reverse the operations that he did and then use the correct operations. His end result is $16$. Before that, he subtracted $14$ which means that his number after the first operation was $30$. He divided by $6$ so his number was $180$. \n\n\nNow, we multiply $180$ by $6$ to get $1080$. Finally, $1080+14=1094$. Since $1094>1000$, our answer is $\\boxed{\\textbf{(E)}\\ \\text{greater than 1000}}$.\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/5.json
|
AHSME
|
1994_AHSME_Problems
| 19
| 0
|
Other
|
Multiple Choice
|
Label one disk "$1$", two disks "$2$", three disks "$3$"$, ...,$ fifty disks "$50$". Put these $1+2+3+ \cdots+50=1275$ labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 51 \qquad\textbf{(C)}\ 415 \qquad\textbf{(D)}\ 451 \qquad\textbf{(E)}\ 501$
|
[
"We can solve this problem by thinking of the worst case scenario, essentially an adaptation of the Pigeon-hole principle. \nWe can start by picking up all the disks numbered 1 to 9 since even if we have all those disks we won't have 10 of any one disk. This gives us 45 disks. \n\n\nFrom disks numbered from 10 to 50, we can pick up at most 9 disks to prevent picking up 10. There are 50-10+1 = 41 different numbers from 10 to 50. We pick up 9 from each number, therefore, we multiply $41 \\cdot 9 = 369$. In total, the maximum number we can pick up without picking up 10 of the same kind is $369+45=414$. We need one more disk to guarantee a complete set of 10. Therefore, the answer is $\\boxed{415}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/19.json
|
AHSME
|
1994_AHSME_Problems
| 23
| 0
|
Geometry
|
Multiple Choice
|
In the $xy$-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at $(0,0), (0,3), (3,3), (3,1), (5,1)$ and $(5,0)$. The slope of the line through the origin that divides the area of this region exactly in half is
[asy] Label l; l.p=fontsize(6); xaxis("$x$",0,6,Ticks(l,1.0,0.5),EndArrow); yaxis("$y$",0,4,Ticks(l,1.0,0.5),EndArrow); draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));[/asy]
$\textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9}$
|
[
"Let the vertices be $A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)$. It is easy to see that the line must pass through $CD$. Let the line intersect $CD$ at the point $G=(3,3-x)$ (i.e. the point $x$ units below $C$). Since the quadrilateral $ABCG$ and pentagon $GDEFA$ must have the same area, we have the equation $3\\times\\frac{1}{2}\\times(x+3)=\\frac{1}{2}\\times3\\times(3-x)+2$. This simplifies into $3x=2$, or $x=\\frac{2}{3}$, so $G=(3,\\frac{7}{3})$. Therefore the slope of the line is $\\boxed{\\textbf{(E)}\\ \\frac{7}{9}}$\n\n\n",
"Consider the small rectangle between $x=3$ and $x=5$ with area $2$. If we exclude that area then the shape becomes a much simpler, its just a rectangle. In order to offset the exclusion of that area from the shape below the line, we must also exclude a shape with the same area above the line. We want the remainder after excluding both areas to be simple, so exclude a rectangle between $x=0$ and $x=3$ and $y=3$ and $y=c$ for some unknown $c$. For the area to be the same we need $(3-c) \\cdot 3 = 2$, or $c=7/3$. After excluding our two offsetting areas, we're left with a rectangle from $x=0$ to $x=3$ and $y=0$ to $y=c$. The area of this region is clearly bisected by its diagonal line. The line passes through $(0,0)$ and $(3,c)$ so its slope is $c/3 = 7/9$ and the answer is $\\fbox{E}$\n\n\n"
] | 2
|
./CreativeMath/AHSME/1994_AHSME_Problems/23.json
|
AHSME
|
1994_AHSME_Problems
| 9
| 0
|
Geometry
|
Multiple Choice
|
If $\angle A$ is four times $\angle B$, and the complement of $\angle B$ is four times the complement of $\angle A$, then $\angle B=$
$\textbf{(A)}\ 10^{\circ} \qquad\textbf{(B)}\ 12^{\circ} \qquad\textbf{(C)}\ 15^{\circ} \qquad\textbf{(D)}\ 18^{\circ} \qquad\textbf{(E)}\ 22.5^{\circ}$
|
[
"Let $\\angle A=x$ and $\\angle B=y$. From the first condition, we have $x=4y$. From the second condition, we have \\[90-y=4(90-x).\\] Substituting $x=4y$ into the previous equation and solving yields \\begin{align*}90-y=4(90-4y)&\\implies 90-y=360-16y\\\\&\\implies 15y=270\\\\&\\implies y=\\boxed{\\textbf{(D) }18^\\circ.}\\end{align*}\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1
|
./CreativeMath/AHSME/1994_AHSME_Problems/9.json
|
AHSME
|
1981_AHSME_Problems
| 20
| 0
|
Geometry
|
Multiple Choice
|
A ray of light originates from point $A$ and travels in a plane, being reflected $n$ times between lines $AD$ and $CD$ before striking a point $B$ (which may be on $AD$ or $CD$) perpendicularly and retracing its path back to $A$ (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for $n=3$). If $\measuredangle CDA=8^\circ$, what is the largest value $n$ can have?
[asy] unitsize(1.5cm); pair D=origin, A=(-6,0), C=6*dir(160), E=3.2*dir(160), F=(-2.1,0), G=1.5*dir(160), B=(-1.4095,0); draw((-6.5,0)--D--C,black); draw(A--E--F--G--B,black); dotfactor=4; dot("$A$",A,S); dot("$C$",C,N); dot("$R_1$",E,N); dot("$R_2$",F,S); dot("$R_3$",G,N); dot("$B$",B,S); markscalefactor=0.015; draw(rightanglemark(G,B,D)); draw(anglemark(C,E,A,12)); draw(anglemark(F,E,G,12)); draw(anglemark(E,F,A)); draw(anglemark(E,F,A,12)); draw(anglemark(B,F,G)); draw(anglemark(B,F,G,12)); draw(anglemark(E,G,F)); draw(anglemark(E,G,F,12)); draw(anglemark(E,G,F,16)); draw(anglemark(B,G,D)); draw(anglemark(B,G,D,12)); draw(anglemark(B,G,D,16)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.}$
|
[
"Notice that when we start, we want the smallest angle possible of reflection. The ideal reflection would be $0$, but that would be impossible. Therefore we start by working backwards. Since angle $CDA$ is $8$, the reflection would give us a triangle with angles $16, 90$, and $74$. Then, when we reflect again, we will have $180 - 74 - 74$ = $32$. Since the other side of the reflection when we had the $82$ degrees had carried over to the other side, we have a $32-82-66$ triangle. \n\n\nNotice that we keep decreasing by increments of $8$. This is because the starting angle was $8$ and since we always have to decrease $8$ every time and that every triangle has every increasing angles of $8$, we must decrease by $8$ every time. This is the most optimal path of the light beam.\n\n\nThe pattern of light will be $82-74-66-58-50-42-34-26-18-10$. When we get to the angle of $2$ degrees, we have reached angle $A$. Therefore, we don't count the $2$, so our total number of reflections between $CD$ and $AD$ is $\\boxed {(B) 10}$\n\n\n~Arcticturn\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/20.json
|
AHSME
|
1981_AHSME_Problems
| 16
| 0
|
Number Theory
|
Multiple Choice
|
The base three representation of $x$ is
\[12112211122211112222\]
The first digit (on the left) of the base nine representation of $x$ is
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
|
[
"Convert $x$ to base 10 then convert the result to base 9.\n\\[12112211122211112222_{3} = 2150029898\\]\n\n\n\\[2150029898 = 5484584488_{9}\\]\n\n\nTherefore, the answer is $\\textbf{(E)}\\ 5.$\n\n\n-edited by coolmath34\n\n\n",
"Every 2 numbers in base 3 represents 1 number in base 9. \nThe first 2 numbers on the left,12 = 1(3) + 2(1) = 5. \n\n\nSo the answer is $\\textbf{(E)}\\ 5.$\n\n\n"
] | 2
|
./CreativeMath/AHSME/1981_AHSME_Problems/16.json
|
AHSME
|
1981_AHSME_Problems
| 6
| 0
|
Algebra
|
Multiple Choice
|
If $\frac{x}{x-1} = \frac{y^2 + 2y - 1}{y^2 + 2y - 2},$ then $x$ equals
$\text{(A)} \quad y^2 + 2y - 1$
$\text{(B)} \quad y^2 + 2y - 2$
$\text{(C)} \quad y^2 + 2y + 2$
$\text{(D)} \quad y^2 + 2y + 1$
$\text{(E)} \quad -y^2 - 2y + 1$
|
[
"We can cross multiply both sides of the equation and simplify.\n\\[x(y^2 + 2y - 2) = (x-1)(y^2 + 2y - 1)\\]\n\\[xy^2 + 2xy - 2x = xy^2 + 2xy - x - y^2 - 2y + 1\\]\n\\[-x = -y^2 - 2y + 1\\]\n\\[x = y^2 + 2y - 1\\]\n\n\nThe answer is $\\text{A.}$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/6.json
|
AHSME
|
1981_AHSME_Problems
| 7
| 0
|
Number Theory
|
Multiple Choice
|
How many of the first one hundred positive integers are divisible by all of the numbers $2$, $3$, $4$, and $5$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$
|
[
"The least common multiple of 2, 3, 4, and 5 is $2^2 \\cdot 3 \\cdot 5 = 60.$ There is only one multiple of 60 between 0 and 100, so the answer is $\\textbf{(B)}\\ 1.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/7.json
|
AHSME
|
1981_AHSME_Problems
| 17
| 0
|
Algebra
|
Multiple Choice
|
The function $f$ is not defined for $x=0$, but, for all non-zero real numbers $x$, $f(x)+2f\left(\dfrac{1}x\right)=3x$. The equation $f(x)=f(-x)$ is satisfied by
$\textbf{(A)}\ \text{exactly one real number} \qquad \textbf{(B)}\ \text{exactly two real numbers} \qquad\textbf{(C)}\ \text{no real numbers}\qquad \\ \textbf{(D)}\ \text{infinitely many, but not all, non-zero real numbers} \qquad\textbf{(E)}\ \text{all non-zero real numbers}$
|
[
"Substitute $x$ with $\\frac{1}{x}$:\n\n\n$f(\\frac{1}{x})+2f(x)=\\frac{3}{x}$.\n\n\nAdding this to $f(x)+2f\\left(\\dfrac{1}x\\right)=3x$, we get\n\n\n$3f(x)+3f\\left(\\dfrac{1}x\\right)=3x+\\frac{3}{x}$, or\n\n\n$f(x)+f\\left(\\dfrac{1}x\\right)=x+\\frac{1}{x}$.\n\n\nSubtracting this from $f(\\frac{1}{x})+2f(x)=\\frac{3}{x}$, we have\n\n\n$f(x)=\\frac{2}{x}-x$.\n\n\nThen, $f(x)=f(-x)$ when $\\frac{2}{x}-x=\\frac{2}{-x}+x$ or\n\n\n$\\frac{2}{x}-x=0$, so $x=\\pm{\\sqrt2}$ are the two real solutions and the answer is $\\boxed{(B)}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/17.json
|
AHSME
|
1981_AHSME_Problems
| 21
| 0
|
Geometry
|
Multiple Choice
|
In a triangle with sides of lengths $a$, $b$, and $c$, $(a+b+c)(a+b-c) = 3ab$. The measure of the angle opposite the side length $c$ is
$\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ$
|
[
"We will try to solve for a possible value of the variables. First notice that exchanging $a$ for $b$ in the original equation must also work. Therefore, $a=b$ works. Replacing $b$ for $a$ and expanding/simplifying in the original equation yields $4a^2-c^2=3a^2$, or $a^2=c^2$. Since $a$ and $c$ are positive, $a=c$. Therefore, we have an equilateral triangle and the angle opposite $c$ is just $\\textbf{(D)}\\ 60^\\circ\\qquad$.\n\n\n",
"\\[(a+b+c)(a+b-c)=3ab\\]\n\\[a^2+2ab+b^2-c^2=3ab\\]\n\\[a^2+b^2-c^2=ab\\]\n\\[c^2=a^2+b^2-ab\\]\nThis looks a lot like Law of Cosines, which is $c^2=a^2+b^2-2ab\\cos{c}$.\n\\[c^2=a^2+b^2-ab=a^2+b^2-2ab\\cos{c}\\]\n\\[ab=2ab\\cos{c}\\]\n\\[\\frac{1}{2}=\\cos{c}\\]\n$\\cos{c}$ is $\\frac{1}{2}$, so the angle opposite side $c$ is $\\boxed{60^\\circ}$.\n\n\n-aopspandy\n\n\n"
] | 2
|
./CreativeMath/AHSME/1981_AHSME_Problems/21.json
|
AHSME
|
1981_AHSME_Problems
| 26
| 0
|
Probability
|
Multiple Choice
|
Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is $\frac{1}{6}$, independent of the outcome of any other toss.)
$\textbf{(A) } \frac{1}{3} \textbf{(B) } \frac{2}{9} \textbf{(C) } \frac{5}{18} \textbf{(D) } \frac{25}{91} \textbf{(E) } \frac{36}{91}$
|
[
"The probability that Carol wins during the first cycle through is $\\frac{5}{6}*\\frac{5}{6}*\\frac{1}{6}$, and the probability that Carol wins on the second cycle through is $\\frac{5}{6}*\\frac{5}{6}*\\frac{5}{6}*\\frac{5}{6}*\\frac{5}{6}*\\frac{1}{6}$. It is clear that this is an infinite geometric sequence, and we must find the sum of it in order to find the answer to this question. Thus we set up the equation: $\\frac{\\frac{25}{216}}{1-\\frac{125}{216}}$, or $\\frac{\\frac{25}{216}}{\\frac{91}{216}}$, which simplifies into $\\boxed{\\textbf{(D) } \\frac{25}{91}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/26.json
|
AHSME
|
1981_AHSME_Problems
| 30
| 0
|
Algebra
|
Multiple Choice
|
If $a$, $b$, $c$, and $d$ are the solutions of the equation $x^4 - bx - 3 = 0$, then an equation whose solutions are
\[\dfrac {a + b + c}{d^2}, \dfrac {a + b + d}{c^2}, \dfrac {a + c + d}{b^2}, \dfrac {b + c + d}{a^2}\]is
$\textbf{(A)}\ 3x^4 + bx + 1 = 0\qquad \textbf{(B)}\ 3x^4 - bx + 1 = 0\qquad \textbf{(C)}\ 3x^4 + bx^3 - 1 = 0\qquad \\\textbf{(D)}\ 3x^4 - bx^3 - 1 = 0\qquad \textbf{(E)}\ \text{none of these}$
|
[
"Using Vieta's formula, we know the sum of the roots is equal to the negative coefficient of the $x^3$ term. Since the coefficient is 0, $a+b+c+d=0$. Thus, $\\frac{a+b+c}{d^2}$ can be rewritten as $\\frac{-d}{d^2}=\\frac{1}{-d}$. Similarly, the other three new roots can be written as $\\frac{1}{-c}$, $\\frac{1}{-b}$, and $\\frac{1}{-a}$.\n\n\nNow, we need to find a way to transform the function $f(x)=x^4-bx-3$ such that all the roots are its negative reciprocal. We can create this new function by taking the negative reciprocal of the argument. In other words, $f(\\frac{1}{-x})$ satisfies this criteria.\n\n\nThe new equation, $f(\\frac{1}{-x})=0$ has the required roots and can be simplified to $\\frac{1}{x^4}+\\frac{b}{x}-3=0$. Since this is not a polynomial, we can multiply both sides by $x^4$ to become $1+bx^3-3x^4=0$. After rearranging and multiplying by negative one, we arrive at $3x^4-bx^3-1$ so the answer is $\\boxed{\\textbf{(D)} 3x^4-bx^3-1}$\n\n\n",
"As in solution 1, the roots of the new equation are $-\\frac{1}{a}, -\\frac{1}{b}, -\\frac{1}{c},-\\frac{1}{d}$. Furthermore, applying Vieta’s formula to the original equation yields $abcd=-3$ and $abc+abd+acd+bcd=-3(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d})=b$. Therefore, the product of the zeros of the new equation is $\\frac{1}{abcd}=-\\frac{1}{3}$. This limits our choices to options C and D, and we need to look for the sum of the roots of the new equation. This sum equals to $-(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d})$\nThe function whose roots are the reciprocals of the original equation is $-3x^4-bx^3+1$ therefore $-(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d})=-(-\\frac{-b}{-3})=\\frac{b}{3}$. The second term of the chosen equation $a_{2}$ should satisfy that $-\\frac{a_{2}}{3}=\\frac{b}{3}$, hence $a_{2}=-b$. The answer is $D$. \n(Option E looks ridiculous~)\n\n\n"
] | 2
|
./CreativeMath/AHSME/1981_AHSME_Problems/30.json
|
AHSME
|
1981_AHSME_Problems
| 1
| 0
|
Algebra
|
Multiple Choice
|
If $\sqrt{x+2}=2$, then $(x+2)^2$ equals:
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16$
|
[
"If we square both sides of the $\\sqrt{x+2} = 2$, we will get $x+2 = 4$, if we square that again, we get $(x+2)^2 = \\boxed{\\textbf{(E) }16}$\n\n\n",
"We can immediately get that $x = 2$, after we square $(2+2)$, we get $\\boxed{\\textbf{(E) }16}$\n\n\n"
] | 2
|
./CreativeMath/AHSME/1981_AHSME_Problems/1.json
|
AHSME
|
1981_AHSME_Problems
| 11
| 0
|
Geometry
|
Multiple Choice
|
The three sides of a right triangle have integral lengths which form an arithmetic progression. One of the sides could have length
$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 81\qquad\textbf{(D)}\ 91\qquad\textbf{(E)}\ 361$
|
[
"Let the three sides be $a-d,$ $a,$ and $a+d.$ Because of the Pythagorean Theorem,\n\\[(a-d)^2 + a^2 = (a+d)^2\\]\nThis can be simplified to\n\\[a(a-4d) = 0.\\]\n\n\nSo, $a$ is a multiple of $4d$ and the triangle has sides $3d, 4d, 5d.$ We check the answer choices for anything divisible by 3, 4, or 5. The only one that works is 81, which is divisible by 3.\n\n\nThe answer is $\\textbf{(C)}\\ 81.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/11.json
|
AHSME
|
1981_AHSME_Problems
| 2
| 0
|
Geometry
|
Multiple Choice
|
Point $E$ is on side $AB$ of square $ABCD$. If $EB$ has length one and $EC$ has length two, then the area of the square is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5$
|
[
"Note that $\\triangle BCE$ is a right triangle. Thus, we do Pythagorean theorem to find that side $BC=\\sqrt{3}$. Since this is the side length of the square, the area of $ABCD$ is $\\boxed{\\textbf{(C)}\\ 3}$. \n\n\n~superagh\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/2.json
|
AHSME
|
1981_AHSME_Problems
| 28
| 0
|
Algebra
|
Multiple Choice
|
Consider the set of all equations $x^3 + a_2x^2 + a_1x + a_0 = 0$, where $a_2$, $a_1$, $a_0$ are real constants and $|a_i| < 2$ for $i = 0,1,2$. Let $r$ be the largest positive real number which satisfies at least one of these equations. Then
$\textbf{(A)}\ 1 < r < \dfrac{3}{2}\qquad \textbf{(B)}\ \dfrac{3}{2} < r < 2\qquad \textbf{(C)}\ 2 < r < \dfrac{5}{2}\qquad \textbf{(D)}\ \dfrac{5}{2} < r < 3\qquad \\ \textbf{(E)}\ 3 < r < \dfrac{7}{2}$
|
[
"Since $x^3 = -(a_2x^2 + a_1x + a_0)$ and $x$ will be as big as possible, we need $x^3$ to be as big as possible, which means $a_2x^2 + a_1x + a_0$ is as small as possible. Since $x$ is positive (according to the options), it makes sense for all of the coefficients to be $-2$.\n\n\nEvaluating $f(\\frac{5}{2})$ gives a negative number, $f(3)$ 1, and $f(\\frac{7}{2})$ a number greater than 1, so the answer is $\\boxed{D}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/28.json
|
AHSME
|
1981_AHSME_Problems
| 12
| 0
|
Algebra
|
Multiple Choice
|
If $p$, $q$, and $M$ are positive numbers and $q<100$, then the number obtained by increasing $M$ by $p\%$ and decreasing the result by $q\%$ exceeds $M$ if and only if
$\textbf{(A)}\ p>q \qquad\textbf{(B)}\ p>\dfrac{q}{100-q}\qquad\textbf{(C)}\ p>\dfrac{q}{1-q}\qquad \textbf{(D)}\ p>\dfrac{100q}{100+q}\qquad\textbf{(E)}\ p>\dfrac{100q}{100-q}$
|
[
"Answer Choice $A$: It is obviously incorrect because if $M$ is $50$ and we increase by $50$% and then decrease $49$%, $M$ will be around $37$. \n\n\nAnswer Choice $B$: If $p$ is $100$ and $q$ is $50$, it should be equal but instead we get $100$ is more than $1$. This is therefore also incorrect.\n\n\nAnswer Choice $C$: Obviously incorrect since if $q$ is larger than $1$, this is always valid since $\\frac {1}{1-q}$ is less than $0$ which is obviously false.\n\n\nAnswer Choice $D$: If $p$ is $100$ and $q$ is $50$, it should be equal but instead we get $100$ is less than $\\frac {5000}{150}$. Therefore, $D$ is also incorrect.\n\n\nAnswer Choice $E$: If $p$ is $100$ and $q$ is $50$, it should be equal, and if we check our equation, we get $\\frac {5000}{50}$ = $100$. Therefore, our answer is $\\boxed {(E)\\dfrac{100q}{100-q}}$\n\n\n~Arcticturn\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/12.json
|
AHSME
|
1981_AHSME_Problems
| 24
| 0
|
Algebra
|
Multiple Choice
|
If $\theta$ is a constant such that $0 < \theta < \pi$ and $x + \dfrac{1}{x} = 2\cos{\theta}$, then for each positive integer $n$, $x^n + \dfrac{1}{x^n}$ equals
$\textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta$
|
[
"Multiply both sides by $x$ and rearrange to $x^2-2x\\cos(\\theta)+1=0$. Using the quadratic equation, we can solve for $x$. After some simplifying:\n\n\n\\[x=\\cos(\\theta) + \\sqrt{\\cos^2(\\theta)-1}\\]\n\\[x=\\cos(\\theta) + \\sqrt{(-1)(\\sin^2(\\theta))}\\]\n\\[x=\\cos(\\theta) + i\\sin(\\theta)\\]\n\n\nSubstituting this expression in to the desired $x^n + \\dfrac{1}{x^n}$ gives:\n\n\n\\[(\\cos(\\theta) + i\\sin(\\theta))^n + (\\cos(\\theta) + i\\sin(\\theta))^{-n}\\]\n\n\nUsing DeMoivre's Theorem:\n\n\n\\[=\\cos(n\\theta) + i\\sin(n\\theta) + \\cos(-n\\theta) + i\\sin(-n\\theta)\\]\n\n\nBecause $\\cos$ is even and $\\sin$ is odd:\n\n\n\\[=\\cos(n\\theta) + i\\sin(n\\theta) + \\cos(n\\theta) - i\\sin(n\\theta)\\]\n\n\n$=\\boxed{\\textbf{2\\cos(n\\theta)}},$ (Error compiling LaTeX. Unknown error_msg)\n\n\nwhich gives the answer $\\boxed{\\textbf{D}}.$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/24.json
|
AHSME
|
1981_AHSME_Problems
| 25
| 0
|
Geometry
|
Multiple Choice
|
In $\triangle ABC$ in the adjoining figure, $AD$ and $AE$ trisect $\angle BAC$. The lengths of $BD$, $DE$ and $EC$ are $2$, $3$, and $6$, respectively. The length of the shortest side of $\triangle ABC$ is
[asy] defaultpen(linewidth(.8pt)); pair A = (0,11); pair B = (2,0); pair D = (4,0); pair E = (7,0); pair C = (13,0); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,S); label("$2$",midpoint(B--D),N); label("$3$",midpoint(D--E),NW); label("$6$",midpoint(E--C),NW); draw(A--B--C--cycle); draw(A--D); draw(A--E); [/asy]
$\textbf{(A)}\ 2\sqrt{10}\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 6\sqrt{6}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ \text{not uniquely determined by the given information}$
|
[
"Let $AC=b$, $AB=c$, $AD=d$, and $AE=e$. Then, by the Angle Bisector Theorem, $\\frac{c}{e}=\\frac{2}{3}$ and $\\frac{d}{b}=\\frac12$, thus $e=\\frac{3c}2$ and $d=\\frac b2$.\n\n\nAlso, by Stewart’s Theorem, $198+11d^2=2b^2+9c^2$ and $330+11e^2=5b^2+6c^2$. Therefore, we have the following system of equations using our substitution from earlier:\n\n\n\\[\\begin{cases}198=-\\frac{3b^2}4+9c^2\\\\330=5b^2-\\frac{75c^2}{4}\\end{cases}.\\]\n\n\nThus, we have:\n\n\n\\[\\begin{cases}264=-b^2+12c^2\\\\264=4b^2-15c^2\\end{cases}.\\]\n\n\nTherefore, $5b^2=27c^2$, so $b^2=\\frac{27c^2}5$, thus our first equation from earlier gives $264=\\frac{33c^2}{5}$, so $c^2=40$, thus $b^2=216$. So, $c<b$ and the answer to the original problem is $c=\\sqrt{40}=\\boxed{2\\sqrt{10}~\\textbf{(A)}}$.\n\n\n~ Aops-g5-gethsemanea2\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/25.json
|
AHSME
|
1981_AHSME_Problems
| 29
| 0
|
Algebra
|
Multiple Choice
|
If $a > 1$, then the sum of the real solutions of
$\sqrt{a - \sqrt{a + x}} = x$
is equal to
$\textbf{(A)}\ \sqrt{a} - 1\qquad \textbf{(B)}\ \dfrac{\sqrt{a}- 1}{2}\qquad \textbf{(C)}\ \sqrt{a - 1}\qquad \textbf{(D)}\ \dfrac{\sqrt{a - 1}}{2}\qquad \textbf{(E)}\ \dfrac{\sqrt{4a- 3} - 1}{2}$
|
[
"A solution is available here. Pull up find, and put in \"Since x is the principal\", and you will arrive at the solution.\n\n\nIt's not super clear, and there's some black stuff over it, but its legible. \n\n\nThe solution in the above file/pdf is the following. I tried my best to match it verbatim, but I had to guess at some things. I also did not do the entire solution like this, just parts where I had to figure out what the words/math was, so this transcribed solution could be wrong and different from the solution in the aforementioned file/pdf.\n\n\nAnyways:\n\n\n29. (E) Since $x$ is the principal square root of some quantity, $x\\geq0$. For $x\\geq0$, the given equation is equivalent to \\[a-\\sqrt{a+x}=x^2\\] or \\[a=\\sqrt{a+x}+x^2.\\] The left member is a constant, the right member is an increasing function of $x$, and hence the equation has exactly one solution. We write \n\\begin{align*} \\sqrt{a+x}&=a-x^2 \\\\ \\sqrt{a+x}+x&=(a+x)-x^2 \\\\ &=(\\sqrt{a+x}+x)(\\sqrt{a+x}-x). \\end{align*}\n\n\nSince $\\sqrt{a+x}+x>0$, we may divide by it to obtain \\[1=\\sqrt{a+x}-x\\quad\\text{or}\\quad x+1=\\sqrt{a+x},\\]\nso \\[x^2+2x+1=a+x,\\] and \\[x^2+x+1-a=0.\\]\n\n\nTherefore $x=\\frac{-1\\pm\\sqrt{4a-3}}{2}$, and the positive root is $x=\\frac{-1+\\sqrt{4a-3}}{2}$, the only solution of the original equation. Therefore, this is also the sum of the real solutions.\n\\[\\text{OR}\\]\n\n\nAs above, we derive $a-\\sqrt{a+x}=x^2$, and hence $a-x^2=\\sqrt{a+x}$. Squaring both sides, we find that \\[a^2-2x^2a+x^4=a+x.\\]\n\n\nThis is a quartic equation in $x$, and therefore not easy to solve; but it is only quadratic in $a$, namely \\[a^2-(2x^2+a)a+x^4-x=0.\\]\n\n\nSolving this by the quadratic formula, we find that \n\\begin{align*} a&=\\frac{1}{2}[2x^2+1+\\sqrt{4x^4+4x^2+1-4x^4+4x}] \\\\ &=x^2+x+1. \\end{align*}\n[We took the positive square root since $a>x^2$; indeed $a-x^2=\\sqrt{a+x}$.]\n\n\nNow we have a quadratic equation for $x$, namely \\[x^2+x+1-a=0,\\]\nwhich we solve as in the previous solution.\n\n\n\\textit{Note}: One might notice that when $a=3$, the solution of the original equation is $x=1$. This eliminates all choices except (E).\n\n\n-- OliverA\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/29.json
|
AHSME
|
1981_AHSME_Problems
| 3
| 0
|
Number Theory
|
Multiple Choice
|
What is the least common multiple of ${\frac{1}{x}}$, $\frac{1}{2x}$, and $\frac{1}{3x}$ is $\frac{1}{6x}$?
|
[
"The least common multiple of ${\\frac{1}{x}}$, $\\frac{1}{2x}$, and $\\frac{1}{3x}$ is $\\frac{1}{6x}$. \n\n\n$\\frac{1}{x}$ = $\\frac{6}{6x}$, $\\frac{1}{2x}$ = $\\frac{3}{6x}$, $\\frac{1}{3x}$ = $\\frac{2}{6x}$.\n\n\n$\\frac{6}{6x}$ + $\\frac{3}{6x}$ + $\\frac{2}{6x}$ = $\\frac{11}{6x}$\n\n\nThe answer is $\\boxed{\\left(D\\right) \\frac{11}{6x}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/3.json
|
AHSME
|
1981_AHSME_Problems
| 8
| 0
|
Algebra
|
Multiple Choice
|
For all positive numbers $x$, $y$, $z$, the product
\[(x+y+z)^{-1}(x^{-1}+y^{-1}+z^{-1})(xy+yz+xz)^{-1}[(xy)^{-1}+(yz)^{-1}+(xz)^{-1}]\] equals
$\textbf{(A)}\ x^{-2}y^{-2}z^{-2}\qquad\textbf{(B)}\ x^{-2}+y^{-2}+z^{-2}\qquad\textbf{(C)}\ (x+y+z)^{-1}\qquad \textbf{(D)}\ \dfrac{1}{xyz}\qquad \\ \textbf{(E)}\ \dfrac{1}{xy+yz+xz}$
|
[
"Start simplifying:\n\\[(\\frac{1}{x+y+z})(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z})(\\frac{1}{xy+yz+xz})(\\frac{1}{xy}+\\frac{1}{xz}+\\frac{1}{yz})\\]\n\n\n\\[(\\frac{1}{x+y+z})(\\frac{xy+yz+xz}{xyz})(\\frac{1}{xy+yz+xz})(\\frac{x+y+z}{xyz})\\]\n\n\nThe $xy+yz+xz$ and $x+y+z$ cancel out:\n\n\n\\[\\frac{1}{(xyz)^2}\\]\n\n\nThe answer is $\\textbf{(A)}\\ x^{-2}y^{-2}z^{-2}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/8.json
|
AHSME
|
1981_AHSME_Problems
| 22
| 0
|
Counting
|
Multiple Choice
|
How many lines in a three dimensional rectangular coordinate system pass through four distinct points of the form $(i, j, k)$, where $i$, $j$, and $k$ are positive integers not exceeding four?
$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 64\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 100$
|
[
"No solutions yet!\ninputting...\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/22.json
|
AHSME
|
1981_AHSME_Problems
| 18
| 0
|
Algebra
|
Multiple Choice
|
The number of real solutions to the equation \[\dfrac{x}{100}=\sin x\] is
$\textbf{(A)}\ 61\qquad\textbf{(B)}\ 62\qquad\textbf{(C)}\ 63\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 65$
|
[
"The answer to this problem is the number of intersections between the graph of $f(x) = \\sin x$ and $f(x) = \\frac{1}{100}x.$ We can do the right side of the coordinate plane first. Each cycle of the sine wave, consisting of 2π, will have 2 intersections (From the positive part of the sine wave) The line $f(x) = \\frac{1}{100}x$ will consist of 16 cycles plus a little bit extra for $x$ from 1 to 100. However, the extra is not complete enough to have any intersection at all. Thus, the number of intersections is $2 \\cdot 16 = 32.$ Because of symmetry, we can multiply by two to account for the left side, and subtract one because of the origin. So the answer is $32 \\cdot 2 - 1 = \\textbf{(C)}\\ 63.$\n\n\nhttps://www.desmos.com/calculator/z6edqwu1kx - Graph\n\n\n\n\n~Eric X\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/18.json
|
AHSME
|
1981_AHSME_Problems
| 4
| 0
|
Algebra
|
Multiple Choice
|
If three times the larger of two numbers is four times the smaller and the difference between the numbers is 8, the the larger of two numbers is:
$\text{(A)}\quad 16 \qquad \text{(B)}\quad 24 \qquad \text{(C)}\quad 32 \qquad \text{(D)}\quad 44 \qquad \text{(E)} \quad 52$
|
[
"Let the smaller number be $x$ and the larger number be $y.$ We can use the information given in the problem to write two equations:\n\n\n1) Three times the larger of two numbers is four times the smaller.\n\\[3y = 4x\\]\n\n\n2) The difference between the number is 8.\n\n\n\\[y - x = 8\\]\n\n\nSolving this system of equation yields $x = 24$ and $y = 32.$ The answer is $\\text{C}.$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/4.json
|
AHSME
|
1981_AHSME_Problems
| 14
| 0
|
Algebra
|
Multiple Choice
|
In a geometric sequence of real numbers, the sum of the first $2$ terms is $7$, and the sum of the first $6$ terms is $91$. The sum of the first $4$ terms is
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 84$
|
[
"Denote the sum of the first $2$ terms as $x$. Since we know that the sum of the first $6$ terms is $91$ which is $7 \\cdot 13$, we have $x$ + $xy$ + $xy^2$ = $13x$ because it is a geometric series. We can quickly see that $y$ = $3$, and therefore, the sum of the first $4$ terms is $4x = 4 \\cdot 7 = \\boxed {(A) 28}$\n\n\n~Arcticturn\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/14.json
|
AHSME
|
1981_AHSME_Problems
| 15
| 0
|
Algebra
|
Multiple Choice
|
If $b>1$, $x>0$, and $(2x)^{\log_b 2}-(3x)^{\log_b 3}=0$, then $x$ is
$\textbf{(A)}\ \dfrac{1}{216}\qquad\textbf{(B)}\ \dfrac{1}{6}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \text{not uniquely determined}$
|
[
"$\\boxed {B}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/15.json
|
AHSME
|
1981_AHSME_Problems
| 5
| 0
|
Geometry
|
Multiple Choice
|
In trapezoid $ABCD$, sides $AB$ and $CD$ are parallel, and diagonal $BD$ and side $AD$ have equal length. If $m\angle DCB=110^\circ$ and $m\angle CBD=30^\circ$, then $m\angle ADB=$
$\textbf{(A)}\ 80^\circ\qquad\textbf{(B)}\ 90^\circ\qquad\textbf{(C)}\ 100^\circ\qquad\textbf{(D)}\ 110^\circ\qquad\textbf{(E)}\ 120^\circ$
|
[
"Draw the diagram using the information above. In triangle $DCB,$ note that $m\\angle DCB=110^\\circ$ and $m\\angle CBD=30^\\circ$, so $m\\angle CDB=40^\\circ.$\n\n\nBecause $AB \\parallel CD,$ we have $m\\angle CDB= m\\angle DBA = 40^\\circ.$ Triangle $DAB$ is isosceles, so $m\\angle ADB = 180 - 2(40) = 100^\\circ.$\n\n\nThe answer is $\\textbf{(C)}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/5.json
|
AHSME
|
1981_AHSME_Problems
| 19
| 0
|
Geometry
|
Multiple Choice
|
In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, and $BN\perp AN$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then find $MN$.
[asy] size(150); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=14*dir(42), C=intersectionpoint(B--(30,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b); draw(N--B--A--N--M--C--A^^B--M); markscalefactor=0.1; draw(rightanglemark(B,N,A)); pair point=N; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(30)); label(rotate(angle(dir(A--C)))*"$19$", A--C, dir(A--C)*dir(90)); label(rotate(angle(dir(A--B)))*"$14$", A--B, dir(A--B)*dir(90)); [/asy]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ \dfrac{5}{2}\qquad\textbf{(C)}\ \dfrac{5}{2}-\sin\theta\qquad\textbf{(D)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\theta\qquad\textbf{(E)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\left(\dfrac{1}{2}\theta\right)$
|
[
"Extend $BN$ to meet $AC$ at $Q$. Then $\\triangle BNM \\sim \\triangle BQC$, so $BN=NQ$ and $QC=19-AQ=2MN$. \n\n\nSince $\\angle ANB=90^\\circ = \\angle ANQ$, $\\angle BAN=\\angle NAQ$ (since $AN$ is an angle bisector) and $\\triangle ANB$ and $\\triangle ANQ$ share side $AN$, $\\triangle ANB \\cong \\triangle ANQ$. Thus $AQ=14$, and so $MN=\\frac{19-AQ}{2}=\\frac{5}{2}$, hence our answer is $\\fbox{B}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1981_AHSME_Problems/19.json
|
AHSME
|
1971_AHSME_Problems
| 20
| 0
|
Algebra
|
Multiple Choice
|
The sum of the squares of the roots of the equation $x^2+2hx=3$ is $10$. The absolute value of $h$ is equal to
$\textbf{(A) }-1\qquad \textbf{(B) }\textstyle\frac{1}{2}\qquad \textbf{(C) }\textstyle\frac{3}{2}\qquad \textbf{(D) }2\qquad \textbf{(E) }\text{None of these}$
|
[
"We can rewrite the equation as $x^2 + 2hx - 3 = 0.$ By Vieta's Formulas, the sum of the roots is $-2h$ and the product of the roots is $-3.$\n\n\nLet the two roots be $r$ and $s.$ Note that\n\\[r^2 + s^2 = (r+s)^2 - 2rs = (-2h)^2 -2(-3)\\]\n\n\nTherefore, $4h^2 + 6 = 10$ and $h = \\pm 1.$ This doesn't match any of the answer choices, so the answer is $\\textbf{(E)}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/20.json
|
AHSME
|
1971_AHSME_Problems
| 16
| 0
|
Arithmetic
|
Multiple Choice
|
After finding the average of $35$ scores, a student carelessly included the average with the $35$ scores and found the
average of these $36$ numbers. The ratio of the second average to the true average was
$\textbf{(A) }1:1\qquad \textbf{(B) }35:36\qquad \textbf{(C) }36:35\qquad \textbf{(D) }2:1\qquad \textbf{(E) }\text{None of these}$
|
[
"Assume the $35$ scores are the first $35$ natural numbers: $1, 2, 3, \\dots 35.$\n\n\nThe average of the scores is $\\frac{\\frac{(35)(36)}{2}}{35} = 18.$\n\n\nIf we add $18$ to the first $35$ numbers, the new average is $\\frac{648}{36} = 18$ still.\n\n\nThe two averages are the same, therefore the answer is $\\textbf{(A) }1:1.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/16.json
|
AHSME
|
1971_AHSME_Problems
| 6
| 0
|
Algebra
|
Multiple Choice
|
Let $\ast$ be the symbol denoting the binary operation on the set $S$ of all non-zero real numbers as follows:
For any two numbers $a$ and $b$ of $S$, $a\ast b=2ab$. Then the one of the following statements which is not true, is
$\textbf{(A) }\ast\text{ is commutative over }S \qquad \textbf{(B) }\ast\text{ is associative over }S\qquad \\ \textbf{(C) }\frac{1}{2}\text{ is an identity element for }\ast\text{ in }S\qquad \textbf{(D) }\text{Every element of }S\text{ has an inverse for }\ast\qquad \textbf{(E) }\dfrac{1}{2a}\text{ is an inverse for }\ast\text{ of the element }a\text{ of }S$
|
[
"$\\textbf{(A) }\\ast\\text{ is commutative over }S$\n\\[a \\ast b = b \\ast a = 2ab\\]\nStatement A is true.\n\n\n\n\n\n\n$\\textbf{(B) }\\ast\\text{ is associative over }S$\n\\[a \\ast (b \\ast c) = a \\ast 2bc = 2abc\\]\n\\[(a \\ast b) \\ast c = 2ab \\ast c = 2abc\\]\nStatement B is true.\n\n\n\n\n$\\textbf{(C) }\\frac{1}{2}\\text{ is an identity element for }\\ast\\text{ in }S\\qquad$\n\\[a \\ast 1/2 = a\\]\nStatement C is true.\n\n\n\n\n$\\textbf{(E) }\\dfrac{1}{2a}\\text{ is an inverse for }\\ast\\text{ of the element }a\\text{ of }S$\n\\[1/2a \\ast a = 1\\]\n\n\nBy process of elimination, we can see that statement D is false.\n\n\nThe answer is $\\textbf{(D) }\\text{Every element of }S\\text{ has an inverse for }\\ast\\qquad .$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/6.json
|
AHSME
|
1971_AHSME_Problems
| 7
| 0
|
Algebra
|
Multiple Choice
|
$2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}$ is equal to
$\textbf{(A) }2^{-2k}\qquad \textbf{(B) }2^{-(2k-1)}\qquad \textbf{(C) }-2^{-(2k+1)}\qquad \textbf{(D) }0\qquad \textbf{(E) }2$
|
[
"$Let\\ x\\ equal\\ 2^{-2k}\\ \\\\* From\\ this\\ we\\ get \\frac{x}{2}-(\\frac{-x}{\\frac{-1}{2}})+x\\ by\\ using\\ power\\ rule.\\ \\\\*Now\\ we\\ can\\ see\\ this\\ simplies\\ to\\ \\frac{-x}{2}\\ \\\\*Looking\\ at\\ \\frac{x}{2} we\\ can\\ clearly\\ see\\ that\\ -(\\frac{x}{2}) is\\ equal\\ to\\ -2^{-(2k+1)}\\ \\\\*Thus\\ our\\ answer\\ is\\ c$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/7.json
|
AHSME
|
1971_AHSME_Problems
| 17
| 0
|
Geometry
|
Multiple Choice
|
A circular disk is divided by $2n$ equally spaced radii($n>0$) and one secant line. The maximum number of non-overlapping
areas into which the disk can be divided is
$\textbf{(A) }2n+1\qquad \textbf{(B) }2n+2\qquad \textbf{(C) }3n-1\qquad \textbf{(D) }3n\qquad \textbf{(E) }3n+1$
|
[
"We can draw the cases for small values of $n.$\n\\[n = 0 \\rightarrow \\text{areas} = 1\\]\n\\[n = 1 \\rightarrow \\text{areas} = 4\\]\n\\[n = 2 \\rightarrow \\text{areas} = 7\\]\n\\[n = 3 \\rightarrow \\text{areas} = 10\\]\nIt seems that for $2n$ radii, there are $3n+1$ distinct areas. The secant line must pass through $n$ radii for this to occur.\n\n\nThe answer is $\\textbf{(E)} = 3n+1.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/17.json
|
AHSME
|
1971_AHSME_Problems
| 21
| 0
|
Algebra
|
Multiple Choice
|
If $\log_2(\log_3(\log_4 x))=\log_3(\log_4(\log_2 y))=\log_4(\log_2(\log_3 z))=0$, then the sum $x+y+z$ is equal to
$\textbf{(A) }50\qquad \textbf{(B) }58\qquad \textbf{(C) }89\qquad \textbf{(D) }111\qquad \textbf{(E) }1296$
|
[
"If $\\log_{x}{n}=0,$ then $n = 1.$ So, we can rewrite this equation:\n\\[(\\log_3(\\log_4 x))=(\\log_4(\\log_2 y))=(\\log_2(\\log_3 z))=1\\]\n\n\nSolve individually for each variable.\n\\[x = 4^3 = 64\\]\n\\[y = 2^4 = 16\\]\n\\[z = 3^2 = 9\\]\nTherefore, $x + y + z = 89.$\n\n\nThe answer is $\\textbf{(C)}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/21.json
|
AHSME
|
1971_AHSME_Problems
| 10
| 0
|
Other
|
Multiple Choice
|
Each of a group of $50$ girls is blonde or brunette and is blue eyed of brown eyed. If $14$ are blue-eyed blondes,
$31$ are brunettes, and $18$ are brown-eyed, then the number of brown-eyed brunettes is
$\textbf{(A) }5\qquad \textbf{(B) }7\qquad \textbf{(C) }9\qquad \textbf{(D) }11\qquad \textbf{(E) }13$
|
[
"There are $31$ brunettes, so there are $50-31=19$ blondes. We know there are $14$ blue-eyed blondes, so there are $19-14=5$ brown-eyed blondes.\n\n\nNext, we know that $18$ people are brown-eyed, so there are $18-5=13$ brown-eyed brunettes.\n\n\nThe answer is $\\textbf{(E)} \\quad 13.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/10.json
|
AHSME
|
1971_AHSME_Problems
| 26
| 0
|
Geometry
|
Multiple Choice
|
[asy] size(2.5inch); pair A, B, C, E, F, G; A = (0,3); B = (-1,0); C = (3,0); E = (0,0); F = (1,2); G = intersectionpoint(B--F,A--E); draw(A--B--C--cycle); draw(A--E); draw(B--F); label("$A$",A,N); label("$B$",B,W); label("$C$",C,dir(0)); label("$E$",E,S); label("$F$",F,NE); label("$G$",G,SE); //Credit to chezbgone2 for the diagram[/asy]
In $\triangle ABC$, point $F$ divides side $AC$ in the ratio $1:2$. Let $E$ be the point of intersection of side $BC$ and $AG$ where $G$ is the
midpoints of $BF$. The point $E$ divides side $BC$ in the ratio
$\textbf{(A) }1:4\qquad \textbf{(B) }1:3\qquad \textbf{(C) }2:5\qquad \textbf{(D) }4:11\qquad \textbf{(E) }3:8$
|
[
"We will use mass points to solve this problem. $AC$ is in the ratio $1:2,$ so we will assign a mass of $2$ to point $A,$ a mass of $1$ to point $C,$ and a mass of $3$ to point $F.$\n\n\nWe also know that $G$ is the midpoint of $BF,$ so $BG:GF=1:1.$ $F$ has a mass of $3,$ so $B$ also has a mass of $3.$\n\n\nIn line $BC,$ $B$ has a mass of $3$ and $C$ has a mass of $1.$ Therefore, $BE:EC = 1:3.$\n\n\nThe answer is $\\textbf{(B)}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/26.json
|
AHSME
|
1971_AHSME_Problems
| 31
| 0
|
Geometry
|
Multiple Choice
|
[asy] size(2.5inch); pair A = (-2,0), B = 2dir(150), D = (2,0), C; draw(A..(0,2)..D--cycle); C = intersectionpoint(A..(0,2)..D,Circle(B,arclength(A--B))); draw(A--B--C--D--cycle); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,N); label("$D$",D,E); label("$4$",A--D,S); label("$1$",A--B,E); label("$1$",B--C,SE); //Credit to chezbgone2 for the diagram[/asy]
Quadrilateral $ABCD$ is inscribed in a circle with side $AD$, a diameter of length $4$. If sides $AB$ and $BC$ each have length $1$,
then side $CD$ has length
$\textbf{(A) }\frac{7}{2}\qquad \textbf{(B) }\frac{5\sqrt{2}}{2}\qquad \textbf{(C) }\sqrt{11}\qquad \textbf{(D) }\sqrt{13}\qquad \textbf{(E) }2\sqrt{3}$
Solution
|
[
"Note that the length 4 forms a semicircle. We can then use the Law of Cosines. Take the center and form a line segment with the other two points.\n\n\nLet's find the cosine of angle $AOB$. The cosine of that angle is 7/8. We use the double cosine angle to find angle $AOC$: \n\n\n$cos(2x) = 2 cos^2 (x)- 1$\n\n\nTherefore, $cos(2x) = 17/32$\n\n\nApplying the Law of Cosines to the triangle $DOC$, we get \n\n\n$CD^2 = 2^2 + 2^2 - 2(4)(-17/32)$ (because $cos(x) = -cos(180-x)$)\n\n\nTherefore, \n$CD^2 = 8 + 17/4$\n\n\nand we find that is equal to $7/2$, or option A!\n\n\n",
"By the properties of a circle, we see that angle $\\text{DCA}$ and angle $\\text{DBA}$ are both right angles. Thus, if we formulate the diagonal lengths of the cyclic quadrilateral using the Pythagorean theorem, we can finish the problem with Ptolemy's Theorem. \n\n\nWith that train of thought, we see that \n\n\n\\[BD^2=AD^2-AB^2\\]\n\n\n\\[BD^2=4^2-1^2\\]\n\n\n\\[BD^2=15 \\Rightarrow BD = \\sqrt{15}\\]\n\n\nTo let us formulate that rest of our lengths, let $CD=x$. Then, similar to the above, \n\n\n\\[AC^2=AD^2-CD^2\\]\n\n\n\\[AC^2=4^2-x^2=16-x^2\\]\n\n\n\\[AC=\\sqrt{16-x^2}\\]\n\n\nNow that we have all of the side lengths and diagonal lengths in one variable, we use Ptolemy's Theorem to finish from here: \n\n\n\\[4\\cdot1 + x\\cdot1 = \\sqrt{15}\\sqrt{16-x^2}\\]\n\n\n\\[x+4=\\sqrt{240-15x^2}\\]\n\n\n\\[x^2+8x+16=240-15x^2\\]\n\n\n\\[16x^2+8x-224=0\\]\n\n\n\\[2x^2+x-28=0\\]\n\n\n\\[x=\\frac{-1 \\pm \\sqrt{1-(4)(2)(-28)}}{2(2)}=\\frac{-1\\pm \\sqrt{225}}{4}=\\frac{-1 \\pm 15}{4}\\]\n\n\nSince the negative solution is an extraneous solution, we see that our solution is $x=\\frac{7}{2}\\Rightarrow\\boxed{A}$\n\n\n"
] | 2
|
./CreativeMath/AHSME/1971_AHSME_Problems/31.json
|
AHSME
|
1971_AHSME_Problems
| 27
| 0
|
Algebra
|
Multiple Choice
|
A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips, and at most one third the number of red chips. The number which are white or blue is at least $55$. The minimum number of red chips is
$\textbf{(A) }24\qquad \textbf{(B) }33\qquad \textbf{(C) }45\qquad \textbf{(D) }54\qquad \textbf{(E) }57$
|
[
"Let the number of white be $2x$. The number of blue is then $x-y$ for some constant $y$. So we want $2x+x-y=55\\rightarrow 3x-y=55$. We take mod 3 to find y. $55=1\\pmod{3}$, so blue is 1 more than half white. The number of whites is then 36, and the number of blues is 19. So $19*3=\\boxed{57}$\n\n\n\n\n~yofro\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/27.json
|
AHSME
|
1971_AHSME_Problems
| 1
| 0
|
Arithmetic
|
Multiple Choice
|
The number of digits in the number $N=2^{12}\times 5^8$ is
$\textbf{(A) }9\qquad \textbf{(B) }10\qquad \textbf{(C) }11\qquad \textbf{(D) }12\qquad \textbf{(E) }20$
|
[
"$N=2^{12}\\cdot5^8=2^4\\cdot2^8\\cdot5^8=2^4\\cdot(2\\cdot5)^8=2^4\\cdot10^8=16\\cdot100000000$, which has $2+8=10$ digits, so the answer is $(\\textbf{B})$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/1.json
|
AHSME
|
1971_AHSME_Problems
| 11
| 0
|
Number Theory
|
Multiple Choice
|
The numeral $47$ in base a represents the same number as $74$ in base $b$. Assuming that both bases are positive
integers, the least possible value of $a+b$ written as a Roman numeral, is
$\textbf{(A) }\mathrm{XIII}\qquad \textbf{(B) }\mathrm{XV}\qquad \textbf{(C) }\mathrm{XXI}\qquad \textbf{(D) }\mathrm{XXIV}\qquad \textbf{(E) }\mathrm{XVI}$
|
[
"$4a+7=7b+4\\implies 7b=4a+3\\implies b=\\frac{4a+3}{7}$. $4a+3\\equiv 0\\implies 4a\\equiv 4\\implies a\\equiv1 \\pmod{7}$. The smallest possible value of $a$ is $8$. Then, $b=\\frac{4\\cdot8+3}{7}=\\frac{35}{7}=5$. However, the digit $7$ is not valid in base $5$, so we have to try a larger value. $a=8+7=15$, gives a value of $\\frac{15*4+3}{7}=9$, for $b$, which is valid. \n\n\n$15+9=24$, which is $\\mathrm{XXIV}$ as a roman numeral, and thus the answer is $(\\textbf{D})$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/11.json
|
AHSME
|
1971_AHSME_Problems
| 2
| 0
|
Algebra
|
Multiple Choice
|
If $b$ men take $c$ days to lay $f$ bricks, then the number of days it will take $c$ men working at the same rate to lay $b$ bricks, is
$\textbf{(A) }fb^2\qquad \textbf{(B) }b/f^2\qquad \textbf{(C) }f^2/b\qquad \textbf{(D) }b^2/f\qquad \textbf{(E) }f/b^2$
|
[
"We can use a modified version of the equation $\\text{Distance} = \\text{Rate} \\times {\\text{Time}}$, which is $\\text{Work Done} = \\text{Rate of Work} \\times{ \\text{Time Worked}}$. In this case, the work done is the number of bricks laid, the rate of work is the number of men working, and the time worked is the number of days. With these definitions, we see from the first equation that $f = b \\cdot c$. If we let $X$ be the number of days it will take $c$ men working at the same rate to $b$ bricks, then we have the equation $b = c \\cdot x$. So, $x = \\frac{b}{c}$. The first equation says that $\\frac{f}{c^2} = \\frac{b}{c}$, which leads to $x = \\frac{f}{c^2}$. This doesn't match any of our answer choices though, so we have to fiddle around a bit before we realize that $c = \\frac{f}{b}$, a substitution we can make to see that $x = \\frac{b^2}{f}$. Thus, the answer is $\\boxed{\\text{D}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/2.json
|
AHSME
|
1971_AHSME_Problems
| 33
| 0
|
Algebra
|
Multiple Choice
|
If $P$ is the product of $n$ quantities in Geometric Progression, $S$ their sum, and $S'$ the sum of their reciprocals,
then $P$ in terms of $S, S'$, and $n$ is
$\textbf{(A) }(SS')^{\frac{1}{2}n}\qquad \textbf{(B) }(S/S')^{\frac{1}{2}n}\qquad \textbf{(C) }(SS')^{n-2}\qquad \textbf{(D) }(S/S')^n\qquad \textbf{(E) }(S/S')^{\frac{1}{2}(n-1)}$
|
[
"We can just look at a very specific case: $1, 2, 4, 8.$ Here, $n=4, P=64, S=15,$ and $S'=\\frac{30}{16}=\\frac{15}{8}.$\n\n\nThen, plug in values of $S, S',$ and $n$ into each of the answer choices and see if it matches the product. \n\n\nAnswer choice $\\textbf{(B)}$ works: $(\\frac{15}{\\frac{15}{8}})^2 = 64.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/33.json
|
AHSME
|
1971_AHSME_Problems
| 29
| 0
|
Algebra
|
Multiple Choice
|
Given the progression $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$.
The least positive integer $n$ such that the product of the first $n$ terms of the progression exceeds $100,000$ is
$\textbf{(A) }7\qquad \textbf{(B) }8\qquad \textbf{(C) }9\qquad \textbf{(D) }10\qquad \textbf{(E) }11$
|
[
"The product of the sequence $10^{\\dfrac{1}{11}}, 10^{\\dfrac{2}{11}}, 10^{\\dfrac{3}{11}}, 10^{\\dfrac{4}{11}},\\dots , 10^{\\dfrac{n}{11}}$ is equal to $10^{\\dfrac{1}{11}+\\frac{2}{11}\\dots\\frac{n}{11}}$ since we are looking for the smallest value $n$ that will create $100,000$, or $10^5$. From there, we can set up the equation $10^{\\dfrac{1}{11}+\\frac{2}{11}\\dots\\frac{n}{11}}=10^5$, which simplified to $\\dfrac{1}{11}+\\frac{2}{11}\\dots\\frac{n}{11}=5$, or $1+2+3\\dots n=55$ This can be converted to $\\frac{n(1+n)}{2}=55$ This simplified to the quadratic $n^2+n-110=0$ Or $(n+11)(n-10)=0$ So $n=-11$ or $10$ Since only positive values of $n$ work, $n=10$ makes the expression equal $100000$. However, we have to exceed $100000$, so our answer is $\\boxed{\\textbf{(E) }11}.$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/29.json
|
AHSME
|
1971_AHSME_Problems
| 3
| 0
|
Geometry
|
Multiple Choice
|
If the point $(x,-4)$ lies on the straight line joining the points $(0,8)$ and $(-4,0)$ in the $xy$-plane, then $x$ is equal to
$\textbf{(A) }-2\qquad \textbf{(B) }2\qquad \textbf{(C) }-8\qquad \textbf{(D) }6\qquad \textbf{(E) }-6$
|
[
"Since $(x,-4)$ is on the line $(0,8)$ and $(-4,0)$, the slope between the latter two is the same as the slope between the former two, so $\\frac{-4-8}{x-0}=\\frac{8-0}{0-(-4)}$. Solving for $x$, we get $x=-6$, so the answer is $\\textbf{(E)}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/3.json
|
AHSME
|
1971_AHSME_Problems
| 8
| 0
|
Algebra
|
Multiple Choice
|
The solution set of $6x^2+5x<4$ is the set of all values of $x$ such that
$\textbf{(A) }\textstyle -2<x<1\qquad \textbf{(B) }-\frac{4}{3}<x<\frac{1}{2}\qquad \textbf{(C) }-\frac{1}{2}<x<\frac{4}{3}\qquad \\ \textbf{(D) }x<\textstyle\frac{1}{2}\text{ or }x>-\frac{4}{3}\qquad \textbf{(E) }x<-\frac{4}{3}\text{ or }x>\frac{1}{2}$
|
[
"We are solving the inequality $6x^2 + 5x - 4 < 0.$ This can be factored as\n\\[(2x-1)(3x+4) < 0\\]\n\n\nThe graph of this inequality is a parabola facing upwards, so the area between the roots satisfies the equation. The solution is $x \\in [-\\frac{4}{3}, \\frac{1}{2}]$ and the answer is $\\textbf{(C)}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/8.json
|
AHSME
|
1971_AHSME_Problems
| 4
| 0
|
Algebra
|
Multiple Choice
|
After simple interest for two months at $5$% per annum was credited, a Boy Scout Troop had a total
of $\textdollar 255.31$ in the Council Treasury. The interest credited was a number of dollars plus the following number of cents
$\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }21\qquad \textbf{(E) }31$
|
[
"We can use the formula $A = P(1+rt)$ for simple interest, where $P$ is the principal (initial amount), $A$ is the total, $r$ is the interest rate per year, and $t$ is the time in years.\n\n\nWe are solving for $P,$ so we can rearrange the equation:\n\\[P = \\frac{A}{1+rt}\\]\nPlug in all the necessary numbers:\n\\[P = \\frac{255.31}{1 + (0.05)(2/12)} = \\textdollar 253.21\\]\n\n\nThe answer is $\\textbf{(D) }21$\n\n\n-edited by coolmath34\n\n\nIf there are any mistakes, feel free to edit so that it is correct.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/4.json
|
AHSME
|
1971_AHSME_Problems
| 14
| 0
|
Number Theory
|
Multiple Choice
|
The number $(2^{48}-1)$ is exactly divisible by two numbers between $60$ and $70$. These numbers are
$\textbf{(A) }61,63\qquad \textbf{(B) }61,65\qquad \textbf{(C) }63,65\qquad \textbf{(D) }63,67\qquad \textbf{(E) }67,69$
|
[
"Factor.\n\\[2^{48}-1 = (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)\\]\n\n\nWe only care about two terms: $2^{6}+1$ and $(2^{3}+1)(2^{3}-1)$. These simplify to $65$ and $63.$\n\n\nThe answer is $\\textbf{(C)}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/14.json
|
AHSME
|
1971_AHSME_Problems
| 5
| 0
|
Geometry
|
Multiple Choice
|
Points $A,B,Q,D$, and $C$ lie on the circle shown and the measures of arcs $\widehat{BQ}$ and $\widehat{QD}$
are $42^\circ$ and $38^\circ$ respectively. The sum of the measures of angles $P$ and $Q$ is
$\textbf{(A) }80^\circ\qquad \textbf{(B) }62^\circ\qquad \textbf{(C) }40^\circ\qquad \textbf{(D) }46^\circ\qquad \textbf{(E) }\text{None of these}$
[asy] size(3inch); draw(Circle((1,0),1)); pair A, B, C, D, P, Q; P = (-2,0); B=(sqrt(2)/2+1,sqrt(2)/2); D=(sqrt(2)/2+1,-sqrt(2)/2); Q = (2,0); A = intersectionpoints(Circle((1,0),1),B--P)[1]; C = intersectionpoints(Circle((1,0),1),D--P)[0]; draw(B--P--D); draw(A--Q--C); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SW); label("$D$",D,SE); label("$P$",P,W); label("$Q$",Q,E); //Credit to chezbgone2 for the diagram[/asy]
|
[
"We see that the measure of $P$ equals $(\\widehat{BD}-\\widehat{AC})/2$, and that the measure of $Q$ equals $\\widehat{AC}/2$.\nSince $\\widehat{BD} = \\widehat{BQ} + \\widehat{QD} = 42^{\\circ} + 38^{\\circ} = 80^{\\circ}$, the sum of the measures of $P$ and $Q$ is $\\widehat{BD}/2 = 80^{\\circ}/2 = 40^{\\circ} \\Longrightarrow \\textbf{(C) }$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/5.json
|
AHSME
|
1971_AHSME_Problems
| 19
| 0
|
Algebra
|
Multiple Choice
|
If the line $y=mx+1$ intersects the ellipse $x^2+4y^2=1$ exactly once, then the value of $m^2$ is
$\textbf{(A) }\textstyle\frac{1}{2}\qquad \textbf{(B) }\frac{2}{3}\qquad \textbf{(C) }\frac{3}{4}\qquad \textbf{(D) }\frac{4}{5}\qquad \textbf{(E) }\frac{5}{6}$
|
[
"Plug in $y=mx+1$ into the ellipse's equation to find the intersection points:\n\\[x^2 + 4(mx+1)^2 = 1\\]\nAfter simplifying, we have a quadratic in $x$:\n\\[(4m^2 + 1)x^2 + 8mx +3 = 0\\]\n\n\nBecause there is only one intersection point, then the quadratic has only one solution. This can only happen when the discriminant is 0.\n\\[\\Delta = b^2 - 4ac = (8m)^2 - (4)(1+4m^2)(3) = 0\\]\n\n\nSolving, we find $m^2 = \\frac{3}{4}.$ The answer is $\\textbf{(C)}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/19.json
|
AHSME
|
1971_AHSME_Problems
| 9
| 0
|
Geometry
|
Multiple Choice
|
An uncrossed belt is fitted without slack around two circular pulleys with radii of $14$ inches and $4$ inches.
If the distance between the points of contact of the belt with the pulleys is $24$ inches, then the distance
between the centers of the pulleys in inches is
$\textbf{(A) }24\qquad \textbf{(B) }2\sqrt{119}\qquad \textbf{(C) }25\qquad \textbf{(D) }26\qquad \textbf{(E) }4\sqrt{35}$
|
[
"(I don't know how to use Asymptote, so if you are a visual learner, I apologize)\n\n\nLet the center of the smaller circle be $A$ and the center of the larger circle be $B.$ Draw perpendicular radii $AC$ and $BD.$ The length $CD$ should equal 24.\n\n\nFrom $A,$ draw a perpendicular line $AE,$ where $E$ is the foot of the perpendicular on $BD.$ Line $BE = 14 - 4 = 10.$\n\n\nPoints $A, E,$ and $B$ form a right triangle with legs $10$ and $24.$ We are looking for distance $AB,$ which is $\\sqrt{10^2 + 24^2} = 26$\n\n\nThe answer is $\\textbf{(D)} 26.$\n\n\n-edited by coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1971_AHSME_Problems/9.json
|
AHSME
|
1964_AHSME_Problems
| 20
| 0
|
Algebra
|
Multiple Choice
|
The sum of the numerical coefficients of all the terms in the expansion of $(x-2y)^{18}$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ -1\qquad \textbf{(E)}\ -19$
|
[
"For any polynomial, even a polynomial with more than one variable, the sum of all the coefficients (including the constant, which is the coefficient of $x^0y^0$) is found by setting all variables equal to $1$. Note that we don't have to worry about whether a constant is a coefficient of an \"invisible\" $x^0y^0$ term, because there is no such term here.\n\n\nSetting $x=y=1$ gives $(-1)^{18}$, which is equal to $1$, which is answer $\\boxed{\\textbf{(B)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/20.json
|
AHSME
|
1964_AHSME_Problems
| 36
| 0
|
Geometry
|
Multiple Choice
|
In this figure the radius of the circle is equal to the altitude of the equilateral triangle $ABC$. The circle is made to roll along the side $AB$, remaining tangent to it at a variable point $T$ and intersecting lines $AC$ and $BC$ in variable points $M$ and $N$, respectively. Let $n$ be the number of degrees in arc $MTN$. Then $n$, for all permissible positions of the circle:
$\textbf{(A) }\text{varies from }30^{\circ}\text{ to }90^{\circ}$
$\textbf{(B) }\text{varies from }30^{\circ}\text{ to }60^{\circ}$
$\textbf{(C) }\text{varies from }60^{\circ}\text{ to }90^{\circ}$
$\textbf{(D) }\text{remains constant at }30^{\circ}$
$\textbf{(E) }\text{remains constant at }60^{\circ}$
[asy] pair A = (0,0), B = (1,0), C = dir(60), T = (2/3,0); pair M = intersectionpoint(A--C,Circle((2/3,sqrt(3)/2),sqrt(3)/2)), N = intersectionpoint(B--C,Circle((2/3,sqrt(3)/2),sqrt(3)/2)); draw((0,0)--(1,0)--dir(60)--cycle); draw(Circle((2/3,sqrt(3)/2),sqrt(3)/2)); label("$A$",A,dir(210)); label("$B$",B,dir(-30)); label("$C$",C,dir(90)); label("$M$",M,dir(190)); label("$N$",N,dir(75)); label("$T$",T,dir(-90)); [/asy]
|
[
"E\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/36.json
|
AHSME
|
1964_AHSME_Problems
| 16
| 0
|
Number Theory
|
Multiple Choice
|
Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0, 1, 2, \dots , 25 \}$.
The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:
$\textbf{(A)}\ 25\qquad \textbf{(B)}\ 22\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 17$
|
[
"Note that for all polynomials $f(x)$, $f(x + 6) \\equiv f(x) \\pmod 6$. \n\n\nProof:\nIf $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$, then $f(x+6) = a_n(x+6)^n + a_{n-1}(x+6)^{n-1} +...+ a_0$. In the second equation, we can use the binomial expansion to expand every term, and then subtract off all terms that have a factor of $6^1$ or higher, since subtracting a multiple of $6$ will not change congruence $\\pmod 6$. This leaves $a_nx^n + a_{n-1}x^{n-1} + ... + a_0$, which is $f(x)$, so $f(x+6) \\equiv f(x) \\pmod 6$.\n\n\nSo, we only need to test when $f(x)$ has a remainder of $0$ for $0, 1, 2, 3, 4, 5$. The set of numbers $6, 7, 8, 9, 10, 11$ will repeat remainders, as will all other sets. The remainders are $2, 0, 0, 2, 0, 0$.\n\n\nThis means for $s=1, 2, 4, 5$, $f(s)$ is divisible by $6$. Since $f(1)$ is divisible, so is $f(s)$ for $s=7, 13, 19, 25$, which is $5$ values of $s$ that work. Since $f(2)$ is divisible, so is $f(s)$ for $s=8, 14, 20$, which is $4$ more values of $s$ that work. The values of $s=4, 5$ will also generate $4$ solutions each, just like $f(2)$. This is a total of $17$ values of $s$, for an answer of $\\boxed{\\textbf{(E)}}$\n\n\n",
"\\begin{itemize}\n\\item We know that,\n\\end{itemize}\n$f(x)$ = $x^2$ +$3x$ + $2$ is $0$ congruent modulo 6 that implies $x+1$ or/and $x+2$ is $0$ congruent modulo $6$.The numbers are of the form $6k+5$ and $6k+4$ for some integer $k$ and due to restrictions of number of elements in the set $S$ we get the inequality $k<4$.Then we calculate the possible combinations to get $17$ as answer i.e. option $\\boxed{\\textbf{(E)}}$.\n\n\n\\begin{verbatim}\n$Solution$ $by$ $GEOMETRY-WIZARD$\n\\end{verbatim}\n"
] | 2
|
./CreativeMath/AHSME/1964_AHSME_Problems/16.json
|
AHSME
|
1964_AHSME_Problems
| 6
| 0
|
Algebra
|
Multiple Choice
|
If $x, 2x+2, 3x+3, \dots$ are in geometric progression, the fourth term is:
$\textbf{(A)}\ -27 \qquad \textbf{(B)}\ -13\frac{1}{2} \qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 13\frac{1}{2}\qquad \textbf{(E)}\ 27$
|
[
"Since we know the sequence is a geometric sequence, the ratio of consecutive terms is always the same number. Thus, we can set up an equation:\n\n\n$\\frac{2x+2}{x} = \\frac{3x+3}{2x+2}$.\n\n\nSolving it, we get:\n\n\n$\\frac{2x+2}{x} = \\frac{3x+3}{2x+2}$\n\n\n$(2x+2)(2x+2) = (3x+3)(x)$\n\n\n$4x^2+8x+4 = 3x^2+3x$\n\n\n$x^2+5x+4 = 0$\n\n\n$(x + 4)(x+1) = 0$\n\n\n$x = -4$ or $x = -1$\n\n\nIf $x=-1$, the sequence has a $0$ as the second term, which is not allowed in a geometric sequence, so it is an extraneous solution that came about because we cross-multiplied by $(2x + 2)$, which is $0$.\n\n\nIf $x=-4$, we plug into $x, 2x +2, 3x + 3$ to find the sequence starts as $-4, -6, -9$. The common ratio is $\\frac{-6}{-4} = \\frac{3}{2}$. The next term is $\\frac{3}{2} \\cdot -9 = \\frac{-27}{2} = -13\\frac{1}{2}$, which is option $\\textbf{(B)}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/6.json
|
AHSME
|
1964_AHSME_Problems
| 7
| 0
|
Algebra
|
Multiple Choice
|
Let n be the number of real values of $p$ for which the roots of
$x^2-px+p=0$
are equal. Then n equals:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \text{a finite number greater than 2}\qquad \textbf{(E)}\ \infty$
|
[
"If the roots of the quadratic $Ax^2 + Bx + C = 0$ are equal, then $B^2 - 4AC = 0$. Plugging in $A=1, B=-p, C = p$ into the equation gives $p^2 - 4p = 0$. This leads to $p = 0, 4$, so there are two values of $p$ that work, giving answer $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/7.json
|
AHSME
|
1964_AHSME_Problems
| 17
| 0
|
Geometry
|
Multiple Choice
|
Given the distinct points $P(x_1, y_1), Q(x_2, y_2)$ and $R(x_1+x_2, y_1+y_2)$.
Line segments are drawn connecting these points to each other and to the origin $O$.
Of the three possibilities: (1) parallelogram (2) straight line (3) trapezoid, figure $OPRQ$,
depending upon the location of the points $P, Q$, and $R$, can be:
$\textbf{(A)}\ \text{(1) only}\qquad \textbf{(B)}\ \text{(2) only}\qquad \textbf{(C)}\ \text{(3) only}\qquad \textbf{(D)}\ \text{(1) or (2) only}\qquad \textbf{(E)}\ \text{all three}$
|
[
"Using vector addition can help solve this problem quickly. Note that algebraically, adding $\\overrightarrow{OP}$ to $\\overrightarrow{OQ}$ will give $\\overrightarrow{OR}$. One method of vector addition is literally known as the \"parallelogram rule\" - if you are given $\\overrightarrow{OP}$ and $\\overrightarrow{OQ}$, to find $\\overrightarrow{OR}$, you can literally draw a parallelogram, making a line though $P$ parallel to $OQ$, and a line through $Q$ parallel to $OP$. The intersection of those lines will give the fourth point $R$, and that fourth point will form a parallelogram with $O, P, Q$.\n\n\nThus, $1$ is a possibility. Case $2$ is also a possibility, if $O, P, Q$ are collinear, then $R$ is also on that line. \n\n\nSince $OP \\parallel QR$ and $PQ \\parallel RO$, which can be seen from either the prior reasoning or by examining slopes, the figure can never be a trapezoid, which requires exactly one of parallel sides.\n\n\nThus, the answer is $\\boxed{\\textbf{(D)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/17.json
|
AHSME
|
1964_AHSME_Problems
| 40
| 0
|
Arithmetic
|
Multiple Choice
|
A watch loses $2\frac{1}{2}$ minutes per day. It is set right at $1$ P.M. on March 15. Let $n$ be the positive correction, in minutes, to be added to the time shown by the watch at a given time. When the watch shows $9$ A.M. on March 21, $n$ equals:
$\textbf{(A) }14\frac{14}{23}\qquad\textbf{(B) }14\frac{1}{14}\qquad\textbf{(C) }13\frac{101}{115}\qquad\textbf{(D) }13\frac{83}{115}\qquad \textbf{(E) }13\frac{13}{23}$
|
[
"From March 15 $1$ P.M. on the watch to March 21 $9$ A.M. on the watch, the watch passed $20 + 5 \\times 24 = 140$ hours. \n\n\nSince $1$ watch hour equals $\\frac{24}{23 + \\frac{57.5}{60}} = \\frac{576}{575}$ real hour, the difference between the watch time and the actual time passed is $140 \\times \\left( \\frac{576}{575} - 1 \\right) = \\frac{28}{115}$ hour $=14\\frac{14}{23}$ minutes.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/40.json
|
AHSME
|
1964_AHSME_Problems
| 37
| 0
|
Algebra
|
Multiple Choice
|
Given two positive numbers $a$, $b$ such that $a<b$. Let $A.M.$ be their arithmetic mean and let $G.M.$ be their positive geometric mean. Then $A.M.$ minus $G.M.$ is always less than:
$\textbf{(A) }\dfrac{(b+a)^2}{ab}\qquad\textbf{(B) }\dfrac{(b+a)^2}{8b}\qquad\textbf{(C) }\dfrac{(b-a)^2}{ab}$
$\textbf{(D) }\dfrac{(b-a)^2}{8a}\qquad \textbf{(E) }\dfrac{(b-a)^2}{8b}$
|
[
"$\\boxed{ \\textbf{(D)} }$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/37.json
|
AHSME
|
1964_AHSME_Problems
| 21
| 0
|
Algebra
|
Multiple Choice
|
If $\log_{b^2}x+\log_{x^2}b=1, b>0, b \neq 1, x \neq 1$, then $x$ equals:
$\textbf{(A)}\ 1/b^2 \qquad \textbf{(B)}\ 1/b \qquad \textbf{(C)}\ b^2 \qquad \textbf{(D)}\ b \qquad \textbf{(E)}\ \sqrt{b}$
|
[
"Using natural log as a \"neutral base\", and applying the change of base formula to each term, we get:\n\n\n$\\frac{\\ln x}{\\ln b^2} + \\frac{\\ln b}{\\ln x^2} = 1$\n\n\n\n\n$\\frac{\\ln x}{2\\ln b} + \\frac{\\ln b}{2\\ln x} = 1$\n\n\n\n\n$\\frac{\\ln x \\ln x + \\ln b \\ln b}{2\\ln b \\ln x} = 1$\n\n\n\n\n$\\ln x \\ln x + \\ln b \\ln b = 2\\ln b \\ln x$\n\n\nYou could inspect the equation here and see that $x=b$ is one solution. Or, you can substitute $X = \\ln x$ and $B = \\ln b$ to get a quadratic in $X$:\n\n\n$X^2 + B^2 = 2BX$\n\n\n$X^2 - 2BX + B^2 = 0$\n\n\nThe above is a quadratic with coefficients $(1, -2B, B^2)$. Plug into the QF to get:\n\n\n$X = \\frac{2B \\pm \\sqrt{4B^2 - 4B^2}}{2}$\n\n\n$X = B$\n\n\n$\\ln x = \\ln b$\n\n\n$x = b$\n\n\nEither way, the answer is $\\boxed{\\textbf{(D)}}$.\n\n\n",
"All answers are of the form $x = b^n$, so we substitute that into the equation and try to solve for $n$. We get:\n\n\n$\\log_{b^2}x+\\log_{x^2}b=1$\n\n\n$\\log_{b^2}b^n + \\log_{b^{2n}} b = 1$\n\n\nBy the definition of a logarithm, the first term on the left is asking for the exponent $x$ needed to change the number $b^2$ into $(b^2)^x$ to get to $b^n$. That exponent is $\\frac{n}{2}$.\n\n\nThe second term is asking for a similar exponent needed to change $b^{2n}$ into $b$. That exponent is $\\frac{1}{2n}$.\n\n\nThe equation becomes $\\frac{n}{2} + \\frac{1}{2n} = 1$. Multiplying by $2n$ gives the quadratic $n^2 + 1 = 2n$, which has the solution $n=1$. Thus, $x = b^n = b^1$, and the answer is $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 2
|
./CreativeMath/AHSME/1964_AHSME_Problems/21.json
|
AHSME
|
1964_AHSME_Problems
| 10
| 0
|
Geometry
|
Multiple Choice
|
Given a square side of length $s$. On a diagonal as base a triangle with three unequal sides is constructed so that its area
equals that of the square. The length of the altitude drawn to the base is:
$\textbf{(A)}\ s\sqrt{2} \qquad \textbf{(B)}\ s/\sqrt{2} \qquad \textbf{(C)}\ 2s \qquad \textbf{(D)}\ 2\sqrt{s} \qquad \textbf{(E)}\ 2/\sqrt{s}$
|
[
"The area of the square is $s^2$. The diagonal of a square with side $s$ bisects the square into two $45-45-90$ right triangles, so the diagonal has length $s\\sqrt{2}$.\n\n\nThe area of the triangle is $\\frac{1}{2}bh$. The base $b$ of the triangle is the diagonal of the square, which is $b = s\\sqrt{2}$. If the area of the triangle is equal to the area of the square, we have:\n\n\n$s^2 = \\frac{1}{2}bh$\n\n\n$s^2 = \\frac{1}{2}s\\sqrt{2}\\cdot h$\n\n\n$s = \\frac{\\sqrt{2}}{2}h$\n\n\n$h = \\frac{2}{\\sqrt{2}}s$\n\n\n$h = s\\sqrt{2}$\n\n\nThis is option $\\boxed{\\textbf{(A)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/10.json
|
AHSME
|
1964_AHSME_Problems
| 26
| 0
|
Algebra
|
Multiple Choice
|
In a ten-mile race $\textit{First}$ beats $\textit{Second}$ by $2$ miles and
$\textit{First}$ beats $\textit{Third}$ by $4$ miles.
If the runners maintain constant speeds throughout the race,
by how many miles does $\textit{Second}$ beat $\textit{Third}$?
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2\frac{1}{4}\qquad \textbf{(C)}\ 2\frac{1}{2}\qquad \textbf{(D)}\ 2\frac{3}{4}\qquad \textbf{(E)}\ 3$
|
[
"Let the speeds of the runners in miles per hour be $a, b, c$, with $a>b>c$. If person $a$ reaches the finish line in $h$ hours, then we have $10 = ah, 8 = bh, 6 = bh$.\n\n\nThus, the speeds of the second and third place people are $\\frac{8}{h}$ and $\\frac{6}{h}$. The ratio of those speeds is $\\frac{\\frac{8}{h}}{\\frac{6}{h}} = \\frac{4}{3}$, meaning person $b$ runs $\\frac{4}{3}$ times as fast as person $c$, or conversely that person $c$ is $\\frac{3}{4}$ times slower than person $b$.\n\n\nIf $c$ runs $0.75$ times as fast as $b$, then when $b$ has finished the race at $10$ miles, $c$ will run $10 \\cdot 0.75 = 7.5$ miles. This is a difference of $10 - 7.5 = 2.5$ miles, which is answer $\\boxed{\\textbf{(C)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/26.json
|
AHSME
|
1964_AHSME_Problems
| 30
| 0
|
Algebra
|
Multiple Choice
|
The larger root minus the smaller root of the equation \[(7+4\sqrt{3})x^2+(2+\sqrt{3})x-2=0\] is
$\textbf{(A) }-2+3\sqrt{3}\qquad\textbf{(B) }2-\sqrt{3}\qquad\textbf{(C) }6+3\sqrt{3}\qquad\textbf{(D) }6-3\sqrt{3}\qquad \textbf{(E) }3\sqrt{3}+2$
|
[
"Dividing the quadratic by $7 + 4\\sqrt{3}$ to obtain a monic polynomial will give a linear coefficient of $\\frac{2 + \\sqrt{3}}{7 + 4\\sqrt{3}}$. Rationalizing the denominator gives:\n\n\n$\\frac{(2 + \\sqrt{3})(7 - 4\\sqrt{3})}{7^2 - 4^2 \\cdot 3}$\n\n\n$=\\frac{14 - 12 - \\sqrt{3}}{49-48}$\n\n\n$=2 - \\sqrt{3}$\n\n\nDividing the constant term by $7 + 4\\sqrt{3}$ (and using the same radical conjugate as above) gives:\n\n\n$\\frac{-2}{7 + 4\\sqrt{3}}$\n\n\n$=-2(7 - 4\\sqrt{3})$\n\n\n$=8\\sqrt{3} - 14$\n\n\nSo, dividing the original quadratic by the coefficient of $x^2$ gives $x^2 + (2 - \\sqrt{3})x + 8\\sqrt{3} - 14 = 0$\n\n\nFrom the quadratic formula, the positive difference of the roots is $\\frac{\\sqrt{b^2 - 4ac}}{a}$. Plugging in gives:\n\n\n$\\sqrt{(2 - \\sqrt{3})^2 - 4(8\\sqrt{3} - 14)(1)}$\n\n\n$=\\sqrt{7 - 4\\sqrt{3} - 32\\sqrt{3} + 56}$\n\n\n$=\\sqrt{63 - 36\\sqrt{3}}$\n\n\n$=3\\sqrt{7 - 4\\sqrt{3}}$\n\n\nNote that if we take $\\frac{1}{3}$ of one of the answer choices and square it, we should get $7 - 4\\sqrt{3}$.\nThe only answers that are (sort of) divisible by $3$ are $6 \\pm 3\\sqrt{3}$, so those would make a good first guess. And given that there is a negative sign underneath the radical, $6 - 3\\sqrt{3}$ is the most logical place to start.\n\n\nSince $\\frac{1}{3}$ of the answer is $2 - \\sqrt{3}$, and $(2 - \\sqrt{3})^2 = 7 - 4\\sqrt{3}$, the answer is indeed $\\boxed{\\textbf{(D)}}$.\n\n\n",
"\\textbf{Submitted by BinouTheGuineaPig} | \\textit{A step-by-step solution}\n\n\nThe original equation can be manipulated as follows.\n\n\n$(4+4\\sqrt{3}+3)x^2+(2+\\sqrt{3})x-2=0$\n\n\n$(2+\\sqrt{3})^2x^2+(2+\\sqrt{3})x-2=0$\n\n\nSubstituting $u = (2+\\sqrt{3})x$,\n\n\n$\\quad u^2+u-2=0$\n\n\n$\\quad (u-1)(u+2)=0$\n\n\n$\\quad u=1$ or $u=-2$\n\n\nFirst root of $x$:\n\n\n$\\quad (2+\\sqrt{3})x_1=1$\n\n\n$\\quad x_1=\\frac{1}{2+\\sqrt{3}}$\n\n\n$\\quad x_1=2-\\sqrt{3}$\n\n\nSecond root of $x$:\n\n\n$\\quad (2+\\sqrt{3})x_2=-2$\n\n\n$\\quad x_2=\\frac{-2}{2+\\sqrt{3}}$\n\n\n$\\quad x_2=-2(2-\\sqrt{3})$\n\n\n$\\quad x_2=-4+2\\sqrt{3}$\n\n\nNow, to find which root of $x$ is larger:\n\n\nAssume that\n\n\n$\\quad 2-\\sqrt{3}>-4+2\\sqrt{3}$, and so\n\n\n$\\quad 6>3\\sqrt{3}$\n\n\n$\\quad 2>\\sqrt{3}$\n\n\n$\\quad \\sqrt{4}>\\sqrt{3}$ which is true. Hence, $x_1>x_2$.\n\n\nFinally, finding the difference between the larger and smaller roots of $x$:\n\n\n$\\quad x_1-x_2$\n\n\n$\\quad =(2-\\sqrt{3})-(-4+2\\sqrt{3})$\n\n\n$\\quad =6-3\\sqrt{3}$\n\n\nTherefore, the answer is $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 2
|
./CreativeMath/AHSME/1964_AHSME_Problems/30.json
|
AHSME
|
1964_AHSME_Problems
| 31
| 0
|
Algebra
|
Multiple Choice
|
Let \[f(n)=\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\left(\dfrac{1-\sqrt{5}}{2}\right)^n.\]
Then $f(n+1)-f(n-1)$, expressed in terms of $f(n)$, equals:
$\textbf{(A) }\frac{1}{2}f(n)\qquad\textbf{(B) }f(n)\qquad\textbf{(C) }2f(n)+1\qquad\textbf{(D) }f^2(n)\qquad \textbf{(E) }$
$\frac{1}{2}(f^2(n)-1)$
|
[
"We compute $f(n+1)$ and $f(n-1)$, while pulling one copy of the exponential part outside:\n\n\n$f(n+1) = \\dfrac{5+3\\sqrt{5}}{10}\\cdot \\frac{1 + \\sqrt{5}}{2}\\cdot\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n+\\dfrac{5-3\\sqrt{5}}{10}\\cdot \\frac{1 - \\sqrt{5}}{2}\\cdot\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n$\n\n\n$f(n+1) = \\dfrac{20+8\\sqrt{5}}{20}\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n+\\dfrac{20-8\\sqrt{5}}{20}\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n$\n\n\n\n\n\n\n$f(n-1) = \\dfrac{5+3\\sqrt{5}}{10}\\cdot \\frac{2}{1 + \\sqrt{5}}\\cdot\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n+\\dfrac{5-3\\sqrt{5}}{10}\\cdot \\frac{2}{1 - \\sqrt{5}}\\cdot\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n$\n\n\n$f(n-1) = \\dfrac{(5+3\\sqrt{5})(1 - \\sqrt{5})}{5(1 + \\sqrt{5})(1 - \\sqrt{5})}\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n+\\dfrac{(5-3\\sqrt{5})(1 + \\sqrt{5})}{5(1 - \\sqrt{5})(1 + \\sqrt{5})}\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n$\n\n\n$f(n-1) = \\dfrac{-10-2\\sqrt{5}}{5(-4)}\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n+\\dfrac{-10 + 2\\sqrt{5}}{5(-4)}\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n$\n\n\n$f(n-1) = \\dfrac{10+2\\sqrt{5}}{20}\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n+\\dfrac{10 - 2\\sqrt{5}}{20}\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n$\n\n\n\n\nComputing $f(n+1) - f(n-1)$ gives:\n\n\n$f(n+1) - f(n-1) = \\dfrac{10+6\\sqrt{5}}{20}\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n+\\dfrac{10-6\\sqrt{5}}{20}\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n$\n\n\n$f(n+1) - f(n-1) = \\dfrac{5+3\\sqrt{5}}{10}\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n+\\dfrac{5-3\\sqrt{5}}{10}\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n$\n\n\n$f(n+1) - f(n-1) = f(n)$\n\n\nThus, the answer is $\\boxed{\\textbf{(B)}}$.\n\n\n",
"Notice that $\\left(\\frac{1+\\sqrt{5}}{2}\\right)^n$ and $\\left(\\frac{1-\\sqrt{5}}{2}\\right)^n$ are the characteristics roots for the recurrence relation $F_n = F_{n-1} + F_{n-2}$ (think about Binet's formula). And $f(n)$ is the solution (i.e. $a_n$) to the recurrence relation with constants $a = \\frac{5+3\\sqrt{5}}{10}$ and $b = \\frac{5-3\\sqrt{5}}{10}$. Thus, $f(n+1) - f(n-1) = f(n)$, and the answer is $\\boxed{\\textbf{(B)}}$. -nullptr07\n\n\n"
] | 2
|
./CreativeMath/AHSME/1964_AHSME_Problems/31.json
|
AHSME
|
1964_AHSME_Problems
| 27
| 0
|
Geometry
|
Multiple Choice
|
If $x$ is a real number and $|x-4|+|x-3|<a$ where $a>0$, then:
$\textbf{(A)}\ 0<a<.01\qquad \textbf{(B)}\ .01<a<1 \qquad \textbf{(C)}\ 0<a<1\qquad \\ \textbf{(D)}\ 0<a \le 1\qquad \textbf{(E)}\ a>1$
|
[
"Let $A$ be point $3$ on a number line, and let $B$ be point $4$. Let $X$ be a mobile point at $x$. Geometrically, $|x-4| + |x-3|$ represents $AX + BX$. If $X$ is between $A$ and $B$, then $AX + BX = AB = 1$. Otherwise, if $X$ is to the left of $A$, then $AX + BX = AX + (AX + AB) = 2AX + 1$, which is greater than $1$. If $X$ is to the right of $B$, we have $AX + BX = (AB + BX) + BX = 2BX + 1$.\n\n\nIn all $3$ cases, the minimum value of $AX + BX$ is $1$. Thus, $|x-4|+|x-3| < a$ will always be true if $a >1$. If $a=1$, it can be false for $3 \\le x \\le 4$. If $a < 1$, then $|x-4|+|x-3| < a$ is always false because the LHS is too big.\n\n\nThus, the answer is $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/27.json
|
AHSME
|
1964_AHSME_Problems
| 1
| 0
|
Algebra
|
Multiple Choice
|
What is the value of $[\log_{10}(5\log_{10}100)]^2$?
$\textbf{(A)}\ \log_{10}50 \qquad \textbf{(B)}\ 25\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 1$
|
[
"$[\\log_{10}(5\\log_{10}100)]^2$\n\n\nSince $10^2 = 100$, we have $\\log_{10} 100 = 2$, so:\n\n\n$[\\log_{10}(5\\cdot 2)]^2$\n\n\nMultiply:\n\n\n$[\\log_{10}(10)]^2$\n\n\nSince $10^1 = 10$, we have $\\log_{10} 10 = 1$, so:\n\n\n$[1]^2$\n\n\n$1$\n\n\n$\\fbox{E}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/1.json
|
AHSME
|
1964_AHSME_Problems
| 11
| 0
|
Algebra
|
Multiple Choice
|
Given $2^x=8^{y+1}$ and $9^y=3^{x-9}$, find the value of $x+y$
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$
|
[
"Since $8^{y + 1} = 2^{3(y+1)}$ and $9^y = 3^{2y}$, we have:\n\n\n$2^x = 2^{3(y+1)}$ and $3^{2y} = 3^{x - 9}$\n\n\n\n\nNote that if $a^b = a^c$, then $b=c$. Setting the exponents equal gives $x = 3y + 3$ and $2y = x - 9$. Plugging the first equation into the second equation gives:\n\n\n$2y = (3y + 3) - 9$\n\n\n$2y = 3y - 6$\n\n\n$0 = y - 6$\n\n\n$y = 6$\n\n\nPlugging that back in to $x = 3y + 3$ gives $x = 3(6) + 3$, or $x = 21$. Thus, $x+y = 21 + 6$, or $x+y=27$, which is option $\\boxed{\\textbf{(D)}}$\n\n\n\n\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/11.json
|
AHSME
|
1964_AHSME_Problems
| 2
| 0
|
Algebra
|
Multiple Choice
|
The graph of $x^2-4y^2=0$ is:
$\textbf{(A)}\ \text{a parabola} \qquad \textbf{(B)}\ \text{an ellipse} \qquad \textbf{(C)}\ \text{a pair of straight lines}\qquad \\ \textbf{(D)}\ \text{a point}\qquad \textbf{(E)}\ \text{None of these}$
|
[
"In the equation $x^2 - 4y^2 = k$, because the coefficients of $x^2$ and $y^2$ are of opposite sign, the graph is typically a hyperbola for most real values of $k$. However, there is one exception. When $k=0$, the equation can be factored as $(x - 2y)(x+2y) = 0$. This gives the graph of two lines passing though the origin: $x=2y$ and $x=-2y$. Thus, the answer is $\\boxed{\\textbf{(C)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/2.json
|
AHSME
|
1964_AHSME_Problems
| 28
| 0
|
Algebra
|
Multiple Choice
|
The sum of $n$ terms of an arithmetic progression is $153$, and the common difference is $2$.
If the first term is an integer, and $n>1$, then the number of possible values for $n$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 6$
|
[
"Let the progression start at $a$, have common difference $2$, and end at $a + 2(n-1)$.\n\n\n\n\nThe average term is $\\frac{a + (a + 2(n-1))}{2}$, or $a + n - 1$. Since the number of terms is $n$, and the sum of the terms is $153$, we have:\n\n\n$n(a+n-1) = 153$\n\n\nSince $n$ is a positive integer, it must be a factor of $153$. This means $n = 1, 3, 9, 17, 51, 153$ are the only possibilities. We are given $n>1$, leaving the other five factors.\n\n\nWe now must check if $a$ is an integer. We have $a = \\frac{153}{n} + 1 - n$. If $n$ is a factor of $153$, then $\\frac{153}{n}$ will be an integer. Adding $1-n$ wil keep it an integer.\n\n\nThus, there are $5$ possible values for $n$, which is answer $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/28.json
|
AHSME
|
1964_AHSME_Problems
| 12
| 0
|
Other
|
Multiple Choice
|
Which of the following is the negation of the statement: For all $x$ of a certain set, $x^2>0$?
$\textbf{(A)}\ \text{For all x}, x^2 < 0\qquad \textbf{(B)}\ \text{For all x}, x^2 \le 0\qquad \textbf{(C)}\ \text{For no x}, x^2>0\qquad \\ \textbf{(D)}\ \text{For some x}, x^2>0\qquad \textbf{(E)}\ \text{For some x}, x^2 \le 0$
|
[
"In general, the negation of a universal (\"for all\") quantifier will use an existential (\"there exists\") quantifier, and negate the statement inside. \n\n\nIn this case, we change the \"For all\" to \"There exists\", and negate the inner statement from $x^2 > 0$ to $x^2 \\le 0$.\n\n\nSo, the negation of the original statement is \"There exists an $x$ such that $x^2 \\le 0$\". Exactly one of the two statements is true, but not both. This is the same as answer $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/12.json
|
AHSME
|
1964_AHSME_Problems
| 32
| 0
|
Algebra
|
Multiple Choice
|
If $\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}$, then:
$\textbf{(A) }a \text{ must equal }c\qquad\textbf{(B) }a+b+c+d\text{ must equal zero}\qquad$
$\textbf{(C) }\text{either }a=c\text{ or }a+b+c+d=0\text{, or both}\qquad$
$\textbf{(D) }a+b+c+d\ne 0\text{ if }a=c\qquad$
$\textbf{(E) }a(b+c+d)=c(a+b+d)$
|
[
"Cross-multiplying gives:\n\n\n$(a+b)(a+d) = (b+c)(c+d)$\n\n\n$a^2 + ad + ab + bd = bc + bd + c^2 + cd$\n\n\n$a^2 + ad + ab - bc - c^2 - cd = 0$\n\n\n$a(a + b + d) - c(b+c+d)= 0$\n\n\nThis looks close to turning into option C, but we don't have a $c$ term in the first parentheses, and we don't have an $a$ term in the second parentheses to allow us to complete the factorization. However, if we both add $ac$ and subtract $ac$ on the LHS, we get:\n\n\n$a(a + b + d) + ac - c(b+c+d) - ca= 0$\n\n\n$a(a+b+d +c) - c(b+c+d+a) = 0$\n\n\n$(a-c)(a+b+c+d) = 0$\n\n\nThis is equivalent to $\\boxed{\\textbf{(C)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/32.json
|
AHSME
|
1964_AHSME_Problems
| 24
| 0
|
Algebra
|
Multiple Choice
|
Let $y=(x-a)^2+(x-b)^2, a, b$ constants. For what value of $x$ is $y$ a minimum?
$\textbf{(A)}\ \frac{a+b}{2} \qquad \textbf{(B)}\ a+b \qquad \textbf{(C)}\ \sqrt{ab} \qquad \textbf{(D)}\ \sqrt{\frac{a^2+b^2}{2}}\qquad \textbf{(E)}\ \frac{a+b}{2ab}$
|
[
"Expanding the quadratic and collecting terms gives $y = 2x^2 - (2a + 2b)x + (a^2+b^2)$. For a quadratic of the form $y = Ax^2 + Bx + C$ with $A>0$, $y$ is minimized when $x = -\\frac{B}{2A}$, which is the average of the roots.\n\n\nThus, the quadratic is minimized when $x = \\frac{2a+2b}{2\\cdot 2} = \\frac{a+b}{2}$, which is answer $\\boxed{\\textbf{(A)}}$.\n\n\n",
"The problem should return real values for $ab = 0$ and $ab < 0$, which eliminates $E$ and $C$. We want to distinguish between options $A, B, D$, and testing $(a, b) = (2, 0)$ should do that, as answers $A, B, D$ will turn into $1, 2, \\sqrt{2}$, respectively.\n\n\nPLugging in $(a, b) = (2, 0)$ gives $y = (x - 2)^2 + x^2$, or $y = 2x^2 - 4x + 4$. This has a minimum at $x = -\\frac{-4}{4}$, or at $x=1$. This is answer $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 2
|
./CreativeMath/AHSME/1964_AHSME_Problems/24.json
|
AHSME
|
1964_AHSME_Problems
| 25
| 0
|
Algebra
|
Multiple Choice
|
The set of values of $m$ for which $x^2+3xy+x+my-m$ has two factors, with integer coefficients, which are linear in $x$ and $y$, is precisely:
$\textbf{(A)}\ 0, 12, -12\qquad \textbf{(B)}\ 0, 12\qquad \textbf{(C)}\ 12, -12\qquad \textbf{(D)}\ 12\qquad \textbf{(E)}\ 0$
|
[
"Since there are only $3$ candidate values for $m$, we test $0, 12, -12$.\n\n\nIf $m=0$, then the expression is $x^2 + 3xy + x$. The term $x$ appears in each monomial, giving $x(x + 3y + 1)$, which has two factors that are linear in $x$ and $y$ with integer coefficients. This eliminates $C$ and $D$.\n\n\nWe next test $m=12$. This gives $x^2 + 3xy + x + 12y - 12$. The linear factors will be of the form $(Ax + By + C)(Dx + Ey + F)$. We will match coefficients from the skeleton form to the expression that is to be factored.\n\n\nMatching the $y^2$ coefficient gives $BE = 0$, and WLOG we can let $B=0$ to give $(Ax + C)(Dx + Ey + F)$. \n\n\nMatching the $x^2$ coefficients gives $AD = 1$. Since $A, D$ are integers, we either have $(A, D) = (1, 1), (-1, -1)$. WLOG we can pick the former, since if $(a)(b)$ is a factorization, so is $(-a)(-b)$. \n\n\nWe now have $(x + C)(x + Ey + F)$. Matching the $xy$ term gives $E =3$. We now have $(x+C)(x + 3y + F)$. Matching the $y$ term gives $3C = 12$, which leads to $C = 4$. Finally, matching constants leads to $4F = -12$ and $F = -3$ for a provisional factorization of $(x + 4)(x + 3y - 3)$. We've matched every coefficient except for $x$, and finding the $x$ term by selectively multiplying leads to $4x + -3x = x$, which matches, so this is a real factorization.\n\n\nWe now must check $m=-12$. The expression is $x^2+3xy+x-12y+12m$, and the skeleton once again is $(Ax + By + C)(Dx + Ey + F)$. The beginning three steps matching $y^2, x^2, xy$ are the same, leading to $(x + C)(x + 3y + F)$. Matching the $y$ term gives $3C = -12$, or $C = -4$. The skeleton becomes $(x -4)(x + 3y + F)$. The constant locks in $F = -3$ for a provisional factorization of $(x-4)(x + 3y - 3)$. However, the $x$ term does not match up, because it is $-4x + -3x = -7x$, when it needs to be $x$.\n\n\nThus, $m=0, 12$ work and $m=-12$ does not, giving answer $\\boxed{\\textbf{(B)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/25.json
|
AHSME
|
1964_AHSME_Problems
| 33
| 0
|
Geometry
|
Multiple Choice
|
$P$ is a point interior to rectangle $ABCD$ and such that $PA=3$ inches, $PD=4$ inches, and $PC=5$ inches. Then $PB$, in inches, equals:
$\textbf{(A) }2\sqrt{3}\qquad\textbf{(B) }3\sqrt{2}\qquad\textbf{(C) }3\sqrt{3}\qquad\textbf{(D) }4\sqrt{2}\qquad \textbf{(E) }2$
[asy] pair A, B, C, D, P; A = (0, 0); B = (6.5, 0); C = (6.5, 4.5); D = (0, 4.5); P = (2.5, 1.5); draw(A--B--C--D--cycle); draw(A--P); draw(C--P); draw(D--P); draw(B--P, dashed); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$P$", P, S); label("$3$", midpoint(A--P), NW); label("$4$", midpoint(D--P), NE); label("$5$", midpoint(C--P), NW); [/asy]
|
[
"From point $P$, create perpendiculars to all four sides, labeling them $a, b, c, d$ starting from going north and continuing clockwise. Label the length $PB$ as $x$:\n\n\n[asy] unitsize(1cm); pair A, B, C, D, P, ABfoot, BCfoot, CDfoot, DAfoot; A = (0, 0); B = (6.5, 0); C = (6.5, 4.5); D = (0, 4.5); P = (2.5, 1.5); ABfoot = (2.5, 0); BCfoot = (6.5, 1.5); CDfoot = (2.5, 4.5); DAfoot = (0, 1.5); draw(A--B--C--D--cycle); draw(A--P); draw(C--P); draw(D--P); draw(B--P, dashed); draw(ABfoot--CDfoot); draw(DAfoot--BCfoot); draw(rightanglemark(P, CDfoot, D)); draw(rightanglemark(P, BCfoot, C)); draw(rightanglemark(P, ABfoot, B)); draw(rightanglemark(P, DAfoot, A)); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$P$\", P, 3*dir(240)); label(\"$3$\", midpoint(A--P), NW); label(\"$4$\", midpoint(D--P), NE); label(\"$5$\", midpoint(C--P), NW); label(\"$x$\", midpoint(B--P), SW); label(\"$a$\", midpoint(P--CDfoot), E); label(\"$b$\", midpoint(P--BCfoot), N); label(\"$c$\", midpoint(P--ABfoot), E); label(\"$d$\", midpoint(P--DAfoot), N); [/asy]\n\n\nWe have $a^2 + b^2 = 5^2$ and $c^2 + d^2 = 3^2$, leading to $a^2 + b^2 + c^2 + d^2 = 34$.\n\n\nWe also have $a^2 + d^2 = 4^2$ and $b^2 + c^2 = x^2$, leading to $a^2 + b^2 + c^2 + d^2 = 16 + x^2$.\n\n\nThus, $34 = 16 + x^2$, or $x = \\sqrt{18} = 3\\sqrt{2}$, which is option $\\boxed{\\textbf{(B)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/33.json
|
AHSME
|
1964_AHSME_Problems
| 13
| 0
|
Geometry
|
Multiple Choice
|
A circle is inscribed in a triangle with side lengths $8, 13$, and $17$. Let the segments of the side of length $8$,
made by a point of tangency, be $r$ and $s$, with $r<s$. What is the ratio $r:s$?
$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 2:5 \qquad \textbf{(C)}\ 1:2 \qquad \textbf{(D)}\ 2:3 \qquad \textbf{(E)}\ 3:4$
|
[
"Label our triangle $ABC$ where that $AB=17$, $AC=13$, and $BC=8$. Let $L$, $J$, and $K$ be the tangency points of $BC$, $AC$, and $AB$ respectively. Let $AK=x$, which implies $AJ=x$. Thus $KB=BL=17-x$ and $JC=LC=13-x$. \n\n\nSince $BL+LC=(17-x)+(13-x)=8$, $x=11$. Thus $r:s=CL:BL=13-x:17-x=2:6=1:3$, hence our answer is $\\fbox{A}$.\n\n\n",
"Let the three sides of $8, 13, 17$ be split into two lengths each, for a total of six segments. Because the vertices of the triangle form two congruent external tangents to the triangle each, there are really only three unique lengths, and two of these three lengths each make up a side. Thus, we can make a system of three equations in there variables: $a+b = 8, b+c=13, c+a=17$. \n\n\nAdding all three equations and dividing by $2$ gives $a+b+c = 19$. Subtracting each equation from that gives $(a, b, c) = (6,2,11)$\n\n\nThus, $a+b$ makes up the shortest side of $8$. The ratio is $2:6$, or $1:3$, which is answer $\\fbox{A}$.\n\n\n"
] | 2
|
./CreativeMath/AHSME/1964_AHSME_Problems/13.json
|
AHSME
|
1964_AHSME_Problems
| 29
| 0
|
Geometry
|
Multiple Choice
|
In this figure $\angle RFS=\angle FDR$, $FD=4$ inches, $DR=6$ inches, $FR=5$ inches, $FS=7\dfrac{1}{2}$ inches. The length of $RS$, in inches, is:
$\textbf{(A) }\text{undetermined}\qquad\textbf{(B) }4\qquad\textbf{(C) }5\dfrac{1}{2}\qquad\textbf{(D) }6\qquad\textbf{(E) }6\dfrac{1}{2}\qquad$
[asy] import olympiad; draw((0,0)--7.5*dir(0)); draw((0,0)--5*dir(55)); draw((0,0)--4*dir(140)); draw(4*dir(140)--5*dir(55)--7.5*dir(0)); markscalefactor=0.1; draw(anglemark((0,0),4*dir(140),5*dir(55))); draw(anglemark(7.5*dir(0),(0,0),5*dir(55))); label("$F$",(0,0),dir(240)); label("$D$",4*dir(140),dir(180)); label("$R$",5*dir(55),dir(80)); label("$S$",7.5*dir(0),dir(-25)); label("$4$",2*dir(140),dir(230)); label("$5$",2.5*dir(55),dir(155)); label("$6$",(4*dir(140)+5*dir(55))/2,dir(100)); label("$7\frac{1}{2}$",3.75,dir(-90)); [/asy]
|
[
"We examine $\\triangle RDF$ and $\\triangle SFR$. We are given $\\angle RDF \\cong \\angle SFR$. Also note that $\\frac{SF}{RD} = \\frac{7.5}{6} = 1.25$ and $\\frac{FR}{DF} = \\frac{5}{4} = 1.25$, so $\\frac{SF}{RD} = \\frac{FR}{DF}$.\n\n\nIf two triangles have two pairs of sides that are proportional, and the included angles are congruent, then the two triangles are similar by SAS congruence. Therefore, the third pair of sides must also be in the same proportion, so \n\n\n$\\frac{SF}{RD} = \\frac{FR}{DF} = \\frac{RS}{FR} = 1.25$\n\n\n$\\frac{RS}{FR} = 1.25$\n\n\n$\\frac{RS}{5} = 1.25$\n\n\n$RS = 6.25$, which is answer $\\boxed{\\textbf{(E) }}$\n\n\n",
"Let $\\angle RFS = \\angle FDR = \\theta$. By Cosine Law, we have:\n\\begin{align*} RF^2 &= DR^2 + DF^2 - 2(DR)(DF)cos(\\theta) \\\\ 5^2 &= 6^2 + 4^2 - 2(6)(4)cos(\\theta) \\\\ cos(\\theta) &= \\frac{6^2+4^2-5^2}{2(6)(4)} = \\frac{9}{16} \\end{align*}\nApplying Cosine Law again in $\\triangle RFS$, we have:\n\\begin{align*} RS^2 &= RF^2 + FS^2 - 2(RF)(FS)cos(\\theta) \\\\ RS^2 &= 25 + (\\frac{15}{2})^2 - 2(5)(\\frac{15}{2})(\\frac{9}{16}) \\\\ RS^2 &= \\frac{625}{16} \\end{align*}\ngiving us $RS = \\frac{25}{4}$, which is the answer $\\boxed{\\textbf{(E)}}$. -nullptr07\n\n\n"
] | 2
|
./CreativeMath/AHSME/1964_AHSME_Problems/29.json
|
AHSME
|
1964_AHSME_Problems
| 3
| 0
|
Number Theory
|
Multiple Choice
|
When a positive integer $x$ is divided by a positive integer $y$, the quotient is $u$ and the remainder is $v$, where $u$ and $v$ are integers.
What is the remainder when $x+2uy$ is divided by $y$?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 2u \qquad \textbf{(C)}\ 3u \qquad \textbf{(D)}\ v \qquad \textbf{(E)}\ 2v$
|
[
"\\begin{itemize}\n\\item We can solve this problem by elemetary modular arthmetic,\n\\end{itemize}\n$x \\equiv v\\ (\\textrm{mod}\\ y)$ $=>$ $x+2uy \\equiv v\\ (\\textrm{mod}\\ y)$\n\n\n~GEOMETRY-WIZARD\n\n\n",
"By the definition of quotient and remainder, problem states that $x = uy + v$.\n\n\nThe problem asks to find the remainder of $x + 2uy$ when divided by $y$. Since $2uy$ is divisible by $y$, adding it to $x$ will not change the remainder. Therefore, the answer is $\\boxed{\\textbf{(D)}}$.\n\n\n",
"If the statement is true for all values of $(x, y, u, v)$, then it must be true for a specific set of $(x, y, u, v)$.\n\n\nIf you let $x=43$ and $y = 8$, then the quotient is $u = 5$ and the remainder is $v = 3$. The problem asks what the remainder is when you divide $x + 2uy = 43 + 2 \\cdot 5 \\cdot 8 = 123$ by $8$. In this case, the remainder is $3$.\n\n\nWhen you plug in $u=5$ and $v = 3$ into the answer choices, they become $0, 5, 10, 3, 6$, respectively. Therefore, the answer is $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 3
|
./CreativeMath/AHSME/1964_AHSME_Problems/3.json
|
AHSME
|
1964_AHSME_Problems
| 34
| 0
|
Algebra
|
Multiple Choice
|
If $n$ is a multiple of $4$, the sum $s=1+2i+3i^2+\cdots+(n+1)i^n$, where $i=\sqrt{-1}$, equals:
$\textbf{(A) }1+i\qquad\textbf{(B) }\frac{1}{2}(n+2)\qquad\textbf{(C) }\frac{1}{2}(n+2-ni)\qquad$
$\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)$
|
[
"The real part is $1 + 3i^2 + 5i^4 + ...$, which is $1 - 3 + 5 - 7 + ...$. If $n$ is a multiple of $4$, then we have an odd number of terms in total: we start with $1$ at $n=0$, then add two more terms to get $1 - 3 + 5$ at $n=4$, etc. With each successive addition, we're really adding a total of $2$, since $-3 + 5 = 2$, and $-7 + 9 = 2$, etc. \n\n\nAt $n = 0$, the sum is $1$, and at $n=4$, the sum is $3$. Since the sum increases linearly, the real part of the sum is $1 + \\frac{1}{2}n$.\n\n\nThe imaginary part is $\\frac{1}{i}(2i + 4i^3 + 6i^5 + ...)$, which is $2 - 4 + 6 - 8 + ...$. This time, we have an even number of terms. We group pairs of terms to get $(2-4) + (6-8) + ...$, and notice that each pair gives $-2$. Again, with $n=0$ the imaginary part is $0$, while with $n=4$ the imaginary part is $-2$. Since again the sum increases linearly, this means the imaginary part is $-\\frac{n}{2}$.\n\n\nCombining the real and imaginary parts gives $\\frac{n}{2} + 1 - \\frac{n}{2}i$, which is equivalent to option $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/34.json
|
AHSME
|
1964_AHSME_Problems
| 8
| 0
|
Algebra
|
Multiple Choice
|
The smaller root of the equation
$\left(x-\frac{3}{4}\right)\left(x-\frac{3}{4}\right)+\left(x-\frac{3}{4}\right)\left(x-\frac{1}{2}\right) =0$ is:
$\textbf{(A)}\ -\frac{3}{4}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ \frac{5}{8}\qquad \textbf{(D)}\ \frac{3}{4}\qquad \textbf{(E)}\ 1$
|
[
"Note that this equation is of the form $a^2 + ab = 0$, which factors to $a(a + b) = 0$. Plugging in $a = x - \\frac{3}{4}$ and $b = x - \\frac{1}{2}$ gives:\n\n\n$(x - \\frac{3}{4})(x - \\frac{3}{4} + x - \\frac{1}{2}) = 0$\n\n\n$(x - \\frac{3}{4})(2x - \\frac{5}{4}) = 0$\n\n\nThe roots are $x = \\frac{3}{4}$ nad $x = \\frac{1}{2} \\cdot \\frac{5}{4}$. The smaller root is $\\frac{5}{8}$, which is option $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/8.json
|
AHSME
|
1964_AHSME_Problems
| 22
| 0
|
Geometry
|
Multiple Choice
|
Given parallelogram $ABCD$ with $E$ the midpoint of diagonal $BD$. Point $E$ is connected to a point $F$ in $DA$ so that
$DF=\frac{1}{3}DA$. What is the ratio of the area of $\triangle DFE$ to the area of quadrilateral $ABEF$?
$\textbf{(A)}\ 1:2 \qquad \textbf{(B)}\ 1:3 \qquad \textbf{(C)}\ 1:5 \qquad \textbf{(D)}\ 1:6 \qquad \textbf{(E)}\ 1:7$
|
[
"If it works for a parallelogram $ABCD$, it should also work for a unit square, with $A(0, 0), B(0, 1), C(1, 1), D(1, 0)$. We are given that $E$ is the midpoint of $BD$, so $E(0.5, 0.5)$. If $F$ is on $DA$, then $F(x, 0)$. We note that $DF = 1-x$ and $DA = 1$, so $DF = \\frac{1}{3}DA$ means $1-x = \\frac{1}{3}$, or $x = \\frac{2}{3}$, and hence $F(\\frac{2}{3}, 0)$.\n\n\nWe note that $\\triangle DFE$ has a base $DF$ that is $\\frac{1}{3}$ and an altitude from $E$ to $DF$ that is $\\frac{1}{2}$. Therefore, $[DEF] = \\frac{1}{2}bh = \\frac{1}{2} \\cdot \\frac{1}{3} \\cdot \\frac{1}{2} = \\frac{1}{12}$.\n\n\nQuadrilateral $ABEF$ can be split into $\\triangle ABE$ and $\\triangle AEF$. The first triangle is $\\frac{1}{4}$ of the unit square cut diagonally, so $[ABE] = \\frac{1}{4}$. The second triangle has base $AF$ that is $\\frac{2}{3}$ and height $E$ to $AF$ that is $\\frac{1}{2}$. Therefore, $[AEF] = \\frac{1}{2}bh = \\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{1}{2} = \\frac{1}{6}$.\n\n\nThe entire quadrilateral $ABEF$ has area $\\frac{1}{4} + \\frac{1}{6} = \\frac{5}{12}$. This is $5$ times larger than the area of $\\triangle DFE$, so the ratio is $1:5$, or $\\boxed{\\textbf{(C)}}$.\n\n\n",
"\\begin{align*} \\frac{A_{DFE}}{A_{ADE}} &= \\frac{\\frac{1}{3}DA}{DA} = \\frac{1}{3} \\\\ A_{DFE} &= \\frac{1}{3}A_{ADE} \\\\ A_{ADE} &= \\frac{1}{2}A_{ADB} = \\frac{1}{4}A_{ABCD} \\\\ \\implies A_{DFE} &= \\frac{1}{3} \\cdot \\frac{1}{4} A_{ABCD} = \\frac{1}{12} A_{ABCD} \\\\ A_{ABEF} &= A_{ABD} - A_{DFE} \\\\ &= \\frac{1}{2}A_{ABCD} - \\frac{1}{12}A_{ABCD} \\\\ &= \\frac{5}{12}A_{ABCD} \\end{align*}\nTherefore, $\\frac{A_{DFE}}{A_{ABEF}} = \\frac{\\frac{1}{12}}{\\frac{5}{12}} = \\frac{1}{5}$, giving us the answer $\\boxed{\\textbf{(C)}}$. -nullptr07\n\n\n"
] | 2
|
./CreativeMath/AHSME/1964_AHSME_Problems/22.json
|
AHSME
|
1964_AHSME_Problems
| 18
| 0
|
Algebra
|
Multiple Choice
|
Let $n$ be the number of pairs of values of $b$ and $c$ such that $3x+by+c=0$ and $cx-2y+12=0$ have the same graph. Then $n$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ \text{finite but more than 2}\qquad \textbf{(E)}\ \infty$
|
[
"For two lines to be the same, their slopes must be equal and their intercepts must be equal. This is a necessary and sufficient condition.\n\n\nThe slope of the first line is $\\frac{-3}{b}$, while the slope of the second line is $\\frac{c}{2}$. Thus, $\\frac{-3}{b} = \\frac{c}{2}$, or $bc = -6$.\n\n\nThe intercept of the first line is $\\frac{-c}{b}$, while the intercept of the second line is $6$. Thus, $6 = \\frac{-c}{b}$, or $-6b = c$.\n\n\nPlugging $-6b = c$ into $bc = -6$ gives $b(-6b) = -6$, or $b^2 = 1$. This means $b = \\pm 1$ This in turn gives $c = \\mp 6$. Thus, $(b, c) = (\\pm 1, \\mp 6)$, for two solutions, which is answer $\\boxed{\\textbf{(C)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/18.json
|
AHSME
|
1964_AHSME_Problems
| 38
| 0
|
Geometry
|
Multiple Choice
|
The sides $PQ$ and $PR$ of triangle $PQR$ are respectively of lengths $4$ inches, and $7$ inches. The median $PM$ is $3\frac{1}{2}$ inches. Then $QR$, in inches, is:
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$
|
[
"By the Median Formula, $PM = \\frac12\\sqrt{2PQ^2+2PR^2-QR^2}$\n\n\nPlugging in the numbers given in the problem, we get\n\\[\\frac72=\\frac12\\sqrt{2\\cdot4^2+2\\cdot7^2-QR^2}\\]\n\n\nSolving,\n\\[7=\\sqrt{2(16)+2(49)-QR^2}\\]\n\\[49=32+98-QR^2\\]\n\\[QR^2=81\\]\n\\[QR=9=\\boxed{D}\\]\n\n\n-AOPS81619\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/38.json
|
AHSME
|
1964_AHSME_Problems
| 4
| 0
|
Algebra
|
Multiple Choice
|
The expression
\[\frac{P+Q}{P-Q}-\frac{P-Q}{P+Q}\]
where $P=x+y$ and $Q=x-y$, is equivalent to:
$\textbf{(A)}\ \frac{x^2-y^2}{xy}\qquad \textbf{(B)}\ \frac{x^2-y^2}{2xy}\qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{x^2+y^2}{xy}\qquad \textbf{(E)}\ \frac{x^2+y^2}{2xy}$
|
[
"Simplifying before substituting by obtaining a common denominator gives: \n\n\n$\\frac{(P+Q)^2 - (P - Q)^2}{(P-Q)(P+Q)}$\n\n\n$\\frac{(P^2 + 2PQ + Q^2) - (P^2 - 2PQ + Q^2)}{P^2 - Q^2}$\n\n\n$\\frac{4PQ}{P^2 - Q^2}$\n\n\nSubstituting gives:\n\n\n$\\frac{4(x+y)(x-y)}{(x+y)^2 - (x - y)^2}$\n\n\n$\\frac{4(x^2 - y^2)}{(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2)}$\n\n\n$\\frac{4(x^2 - y^2)}{4xy}$\n\n\n$\\frac{x^2 - y^2}{xy}$\n\n\n$\\boxed{\\textbf{(A)}}$\n\n\n",
"Substituting and then simplifying, noting that $P+Q = 2x$ and $P - Q = 2y$, gives:\n\n\n$\\frac{2x}{2y} - \\frac{2y}{2x}$\n\n\n$\\frac{x^2}{xy} - \\frac{y^2}{xy}$\n\n\n$\\frac{x^2 - y^2}{xy}$\n\n\n$\\boxed{\\textbf{(A)}}$\n\n\n",
"If it's true for all values of $x$ and $y$, then it's true for a specific value of $x$ and $y$. Letting $x=5$ and $y = 4$ gives $P = 9$ and $Q = 1$. The fraction becomes:\n\n\n$\\frac{9 + 1}{9 - 1} - \\frac{9 - 1}{9 + 1}$\n\n\n$\\frac{10}{8} - \\frac{8}{10}$\n\n\n$\\frac{5}{4} - \\frac{4}{5}$\n\n\n$\\frac{25}{20} - \\frac{16}{20}$\n\n\n$\\frac{9}{20}$\n\n\nPlugging in $(x, y) = (5, 4)$ into the answer choices gives $\\frac{9}{20}, \\frac{9}{40}, 1, \\frac{41}{20}, \\frac{41}{40}$. Therefore, the answer is $\\boxed{\\textbf{(A)}}$\n\n\n"
] | 3
|
./CreativeMath/AHSME/1964_AHSME_Problems/4.json
|
AHSME
|
1964_AHSME_Problems
| 14
| 0
|
Algebra
|
Multiple Choice
|
A farmer bought $749$ sheep. He sold $700$ of them for the price paid for the $749$ sheep.
The remaining $49$ sheep were sold at the same price per head as the other $700$.
Based on the cost, the percent gain on the entire transaction is:
$\textbf{(A)}\ 6.5 \qquad \textbf{(B)}\ 6.75 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 7.5 \qquad \textbf{(E)}\ 8$
|
[
"Let us say each sheep cost $x$ dollars. The farmer paid $749x$ for the sheep. He sold $700$ of them for $749x$, so each sheep sold for $\\frac{749}{700} = 1.07x$.\n\n\nSince every sheep sold for the same price per head, and since every sheep cost $x$ and sold for $1.07x$, there is an increase of $\\frac{1.07x - 1x}{1x} = 0.07$, or $7\\%$, which is option $\\boxed{\\textbf{(C)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/14.json
|
AHSME
|
1964_AHSME_Problems
| 15
| 0
|
Geometry
|
Multiple Choice
|
A line through the point $(-a,0)$ cuts from the second quadrant a triangular region with area $T$. The equation of the line is:
$\textbf{(A)}\ 2Tx+a^2y+2aT=0 \qquad \textbf{(B)}\ 2Tx-a^2y+2aT=0 \qquad \textbf{(C)}\ 2Tx+a^2y-2aT=0 \qquad \\ \textbf{(D)}\ 2Tx-a^2y-2aT=0 \qquad \textbf{(E)}\ \text{none of these}$
|
[
"The right triangle has area $T$ and base $a$, so the height $h$ satisfies $\\frac{1}{2}ha = T$. This means $h = \\frac{2T}{a}$. Becuase the triangle is in the second quadrant, the coordinates are the origin, $(-a, 0)$, and $(0, h)$, which means the third coordinate is $(0, \\frac{2T}{a})$.\n\n\nSo, we want the equation of a line though $(-a, 0)$ and $(0, \\frac{2T}{a})$. The slope is $\\frac{\\frac{2T}{a} - 0}{0 - (-a)}$, which simplifies to $\\frac{2T}{a^2}$.\n\n\nThe y-intercept is $(0, \\frac{2T}{a})$, so the line in slope-intercept form is $y = mx + b$, or:\n\n\n\n\n$y = \\frac{2T}{a^2}x + \\frac{2T}{a}$\n\n\nAll the solutions have a positive x-coefficient and no fractions, so we clear the fractions by multiplying by $a^2$ and move the $y$ term to the right to get:\n\n\n$0 = 2Tx + 2aT - a^2y$\n\n\nThis is equivalent to option $\\boxed{\\textbf{(B)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/15.json
|
AHSME
|
1964_AHSME_Problems
| 5
| 0
|
Algebra
|
Multiple Choice
|
If $y$ varies directly as $x$, and if $y=8$ when $x=4$, the value of $y$ when $x=-8$ is:
$\textbf{(A)}\ -16 \qquad \textbf{(B)}\ -4 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 4k, k= \pm1, \pm2, \dots \qquad \\ \textbf{(E)}\ 16k, k=\pm1,\pm2,\dots$
|
[
"If $y$ varies directly as $x$, then $\\frac{x}{y}$ will be a constant for every pair of $(x, y)$. Thus, we have:\n\n\n$\\frac{4}{8} = \\frac{-8}{y}$\n\n\n$y \\cdot \\frac{1}{2} = -8$\n\n\n$y = -16$\n\n\n$\\boxed{\\textbf{(A)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/5.json
|
AHSME
|
1964_AHSME_Problems
| 39
| 0
|
Geometry
|
Multiple Choice
|
The magnitudes of the sides of triangle $ABC$ are $a$, $b$, and $c$, as shown, with $c\le b\le a$. Through interior point $P$ and the vertices $A$, $B$, $C$, lines are drawn meeting the opposite sides in $A'$, $B'$, $C'$, respectively. Let $s=AA'+BB'+CC'$. Then, for all positions of point $P$, $s$ is less than:
$\textbf{(A) }2a+b\qquad\textbf{(B) }2a+c\qquad\textbf{(C) }2b+c\qquad\textbf{(D) }a+2b\qquad \textbf{(E) }$ $a+b+c$
[asy] import math; pair A = (0,0), B = (1,3), C = (5,0), P = (1.5,1); pair X = extension(B,C,A,P), Y = extension(A,C,B,P), Z = extension(A,B,C,P); draw(A--B--C--cycle); draw(A--X); draw(B--Y); draw(C--Z); dot(P); dot(A); dot(B); dot(C); label("$A$",A,dir(210)); label("$B$",B,dir(90)); label("$C$",C,dir(-30)); label("$A'$",X,dir(-100)); label("$B'$",Y,dir(65)); label("$C'$",Z,dir(20)); label("$P$",P,dir(70)); label("$a$",X,dir(80)); label("$b$",Y,dir(-90)); label("$c$",Z,dir(110)); [/asy]
|
[
"We know that in a $\\triangle DEF$, if $\\angle D \\le \\angle E$ then $EF \\le DF$, we can use this fact in the different triangles to form inequalities, and then add the inequalities.\n\n\nIn $\\triangle ABC$, since $c \\le b \\le a$, we have $\\angle C \\le \\angle B \\le \\angle A$ by the above argument.\n\n\n\n\nNow, $\\angle AA'C > \\angle B \\ge \\angle C$, hence we have $AC > AA' \\implies b > AA'$\n\n\nAnd, $\\angle BB'C > \\angle A \\ge \\angle C$, hence we have $BC > BB' \\implies a > BB'$\n\n\nAnd, $\\angle CC'B > \\angle A \\ge \\angle B$, hence we have $BC > CC' \\implies a > CC'$\n\n\nFinally, adding all three inequalities, we have $b + a + a > AA' + BB' + CC' \\implies AA' + BB' + CC' < \\textbf{2a + b}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/39.json
|
AHSME
|
1964_AHSME_Problems
| 19
| 0
|
Algebra
|
Multiple Choice
|
If $2x-3y-z=0$ and $x+3y-14z=0, z \neq 0$, the numerical value of $\frac{x^2+3xy}{y^2+z^2}$ is:
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ -20/17\qquad \textbf{(E)}\ -2$
|
[
"If the value of $\\frac{x^2+3xy}{y^2+z^2}$ is constant, as the answers imply, we can pick a value of $z$, and then solve the two linear equations for the corresponding $(x, y)$. We can then plug in $(x,y, z)$ into the expression to get the answer.\n\n\nIf $z=1$, then $2x - 3y = 1$ and $x + 3y = 14$. We can solve each equation for $3y$ and set them equal, which leads to $2x - 1 = 14 - x$. This leads to $x = 5$. Plugging in $x=5$ into $x + 3y = 14$ gives $y = 3$. Thus, $(5, 3, 1)$ is one solution to the intersection of the two planes given.\n\n\nPlugging $(5, 3, -1)$ into the expression gives $\\frac{x^2+3xy}{y^2+z^2}$ gives $\\frac{25 + 45}{10}$, or $7$, which is answer $\\boxed{\\textbf{(A)}}$\n\n\n",
"If we think of $z$ as a parameter, we get $2x - 3y = z$ and $x + 3y = 14z$. Adding the equations leads to $3x = 15z$, or $x = 5z$. Plugging that into $x + 3y = 14z$ gives $5z + 3y = 14z$, or $y = 3z$. Thus, the intersection of the two planes is given by the parametric line $(5z, 3z, z)$, where $z$ varies along all real numbers.\n\n\nWe plug this in to $\\frac{x^2+3xy}{y^2+z^2}$ to get $\\frac{25z^2 + 45z^2}{10z^2}$, or $7$, which is answer $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 2
|
./CreativeMath/AHSME/1964_AHSME_Problems/19.json
|
AHSME
|
1964_AHSME_Problems
| 23
| 0
|
Algebra
|
Multiple Choice
|
Two numbers are such that their difference, their sum, and their product are to one another as $1:7:24$. The product of the two numbers is:
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96$
|
[
"Set the two numbers as $x$ and $y$. Therefore, \n$x+y=7(x-y), xy=24(x-y)$, and\n$24(x+y)=7xy$. Simplifying the first equation gives $y=\\frac{3}{4}x$. Substituting for $y$ in the second equation gives $\\frac{3}{4}x^2=6x.$ Solving yields $x=8$ or $x=0$. Substituting $x=0$ back into the first equation yields $1=-7$ which is false, so $x=0$ is not valid and $x=8$. Substituting into $y=\\frac{3}{4}x$ gives $y=6$ and $xy=\\boxed{\\textbf{(D) } 48}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/23.json
|
AHSME
|
1964_AHSME_Problems
| 9
| 0
|
Arithmetic
|
Multiple Choice
|
A jobber buys an article at $$24$ less $12\frac{1}{2}\%$. He then wishes to sell the article at a gain of $33\frac{1}{3}\%$ of his cost
after allowing a $20\%$ discount on his marked price. At what price, in dollars, should the article be marked?
$\textbf{(A)}\ 25.20 \qquad \textbf{(B)}\ 30.00 \qquad \textbf{(C)}\ 33.60 \qquad \textbf{(D)}\ 40.00 \qquad \textbf{(E)}\ \text{none of these}$
|
[
"The item is bought for $24$ minus $12\\frac{1}{2}\\%$ of $24$. Since $12\\frac{1}{2}\\% = \\frac{1}{8}$, the item was purchased for $24 - 24 \\cdot \\frac{1}{8}$, or $21$.\n\n\nHe wants to make $33\\frac{1}{3}\\%$ of the purchase price, or $21 \\cdot \\frac{1}{3} = 7$. This means the price must actually sell for $21 + 7 = 28$.\n\n\nIf the article were marked for $x$ dollars, then after the $20\\%$ discount, it would sell for $\\frac{4}{5}x$. Thus, we have $\\frac{4}{5}x = 28$, which leads to $x = 28 \\cdot \\frac{5}{4}$, or $35$ dollars. The correct answer is $\\boxed{\\textbf{(E)}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1964_AHSME_Problems/9.json
|
AHSME
|
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