Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: If the sum of the first $10$ terms and the sum of the first $100$ terms of a given arithmetic progression are $100$ and $10$ , 
respectively, then the sum of first $110$ terms is:
$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$

Solution: Let $a$ be the first term of the sequence and let $d$ be the common difference of the sequence.
Sum of the first 10 terms: $\frac{10}{2}(2a+9d)=100 \Longleftrightarrow 2a+9d=20$ Sum of the first 100 terms: $\frac{100}{2}(2a+99d)=10 \Longleftrightarrow 2a+99d=\frac{1}{5}$
Solving the system, we get $d=-\frac{11}{50}$ $a=\frac{1099}{100}$ . The sum of the first 110 terms is $\frac{110}{2}(2a+109d)=55(-2)=-110$
Therefore, $\boxed{110}$

Answer: D 110
Problem: The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$ , respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does $L_2$ , and that $L_1$ has 4 times the slope of $L_2$ . If $L_1$ is not horizontal, then $mn$ is
$\text{(A)} \ \frac{\sqrt{2}}{2} \qquad \text{(B)} \ -\frac{\sqrt{2}}{2} \qquad \text{(C)} \ 2 \qquad \text{(D)} \ -2 \qquad \text{(E)} \ \text{not uniquely determined}$

Solution: Solution by e_power_pi_times_i
$4n = m$ , as stated in the question. In the line $L_1$ , draw a triangle with the coordinates $(0,0)$ $(1,0)$ , and $(1,m)$ . Then $m = \tan(\theta_1)$ . Similarly, $n = \tan(\theta_2)$ . Since $4n = m$ and $\theta_1 = 2\theta_2$ $\tan(2\theta_2) = 4\tan(\theta_2)$ . Using the angle addition formula for tangents, $\dfrac{2\tan(\theta_2)}{1-\tan^2(\theta_2)} = 4\tan(\theta_2)$ . Solving, we have $\tan(\theta_2) = 0, \dfrac{\sqrt{2}}{2}$ . But line $L_1$ is not horizontal, so therefore $(m,n) = (2\sqrt{2},\dfrac{\sqrt{2}}{2})$ . Looking at the answer choices, it seems the answer is $(2\sqrt{2})(\dfrac{\sqrt{2}}{2}) = \boxed{2}$

Answer: C 2
Problem: Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$

Solution: We assume the side length of the cube is $1$ . The side length of the tetrahedron is $\sqrt2$ , so the surface area is $4\times\frac{2\sqrt3}{4}=2\sqrt3$ . The surface area of the cube is $6\times1\times1=6$ , so the ratio of the surface area of the cube to the surface area of the tetrahedron is $\frac{6}{2\sqrt3}=\boxed{3}$

Answer:  3
Problem: Given that $i^2=-1$ , for how many integers $n$ is $(n+i)^4$ an integer?
$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$

Solution: $(n+i)^4=n^4+4in^3-6n^2-4in+1$ , and this has to be an integer, so the sum of the imaginary parts must be $0$ \[4in^3-4in=0\] \[4in^3=4in\] \[n^3=n\] Since $n^3=n$ , there are $\boxed{3}$ solutions for $n$ $0$ and $\pm1$

Answer:  3
Problem: For some real number $r$ , the polynomial $8x^3-4x^2-42x+45$ is divisible by $(x-r)^2$ . Which of the following numbers is closest to $r$
$\text{(A)} \ 1.22 \qquad  \text{(B)} \ 1.32 \qquad  \text{(C)} \ 1.42 \qquad  \text{(D)} \ 1.52 \qquad  \text{(E)} \ 1.62$

Solution: Solution by e_power_pi_times_i
Denote $s$ as the third solution. Then, by Vieta's, $2r+s = \dfrac{1}{2}$ $r^2+2rs = -\dfrac{21}{4}$ , and $r^2s = -\dfrac{45}{8}$ . Multiplying the top equation by $2r$ and eliminating, we have $3r^2 = r+\dfrac{21}{4}$ . Combined with the fact that $s = \dfrac{1}{2}-2r$ , the third equation can be written as $(\dfrac{r+\dfrac{21}{4}}{3})(\dfrac{1}{2}-2r) = -\dfrac{45}{8}$ , or $(4r+21)(4r-1) = 135$ . Solving, we get $r = \dfrac{3}{2}, -\dfrac{13}{2}$ . Plugging the solutions back in, we see that $-\dfrac{13}{2}$ is an extraneous solution, and thus the answer is $\boxed{1.52}$

Answer: D 1.52
Problem: In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$ appears $k$ times. It is a fact that there are integers $b, c$ , and $d$ such that for all positive integers $n$ $a_n=b\lfloor \sqrt{n+c} \rfloor +d$ , 
where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$ . The sum $b+c+d$ equals
$\text{(A)} \ 0 \qquad  \text{(B)} \ 1 \qquad  \text{(C)} \ 2 \qquad  \text{(D)} \ 3 \qquad  \text{(E)} \ 4$

Solution: Solution by e_power_pi_times_i
Because the set consists of odd numbers, and since $\lfloor{}\sqrt{n+c}\rfloor{}$ is an integer and can be odd or even, $b = 2$ and $|a| = 1$ . However, given that $\lfloor{}\sqrt{n+c}\rfloor{}$ can be $0$ $a = 1$ . Then, $a_1 = 1 = 2\lfloor{}\sqrt{1+c}\rfloor{}+1$ , and $\lfloor{}\sqrt{1+c}\rfloor{}$ = 0, and $c = -1$ because $c$ is an integer. $b+c+d = 2+(-1)+1 = \boxed{2}$

Answer: C 2
Problem: [asy] draw((-2,1)--(2,1)--(2,-1)--(-2,-1)--cycle); draw((0,0)--(0,-1)--(-2,-1)--(-2,0)--cycle); label("$F$",(0,0),E); label("$A$",(-2,1),W); label("$B$",(2,1),E); label("$C$", (2,-1),E); label("$D$",(-2,-1),WSW); label("$E$",(-2,0),W); label("$G$",(0,-1),S); //Credit to TheMaskedMagician for the diagram [/asy]
If rectangle ABCD has area 72 square meters and E and G are the midpoints of sides AD and CD, respectively, then the area of rectangle DEFG in square meters is
$\textbf{(A) }8\qquad \textbf{(B) }9\qquad \textbf{(C) }12\qquad \textbf{(D) }18\qquad \textbf{(E) }24$

Solution: Solution by e_power_pi_times_i
Since the dimensions of $DEFG$ are half of the dimensions of $ABCD$ , the area of $DEFG$ is $\dfrac{1}{2}\cdot\dfrac{1}{2}$ of $ABCD$ , so the area of $ABCD$ is $\dfrac{1}{4}\cdot72 = \boxed{18}$

Answer: D 18
Problem: For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals
$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$

Solution: Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into \[(x+1)(y-1) = -1\] Plugging in $-1$ and $1$ as the $x$ and $y$ sides respectively, we get $x = -2$ and $y = 2$ . Plugging this in to $\frac{1}{x}-\frac{1}{y}$ gives us $\boxed{1}$ as our final answer.

Answer:  1
Problem: For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals
$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$

Solution: Notice that we can do $\frac{x-y}{xy} = \frac{xy}{xy}$ . We are left with $\frac{1}{y} - \frac{1}{x} = 1$ . Multiply by $-1$ to achieve $\frac{1}{x} - \frac{1}{y} = \boxed{1}$

Answer:  1
Problem: [asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$ . 
What is the measure of $\measuredangle AED$ in degrees?
$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$

Solution: Solution by e_power_pi_times_i
Notice that $\measuredangle DAE = 90^\circ+60^\circ = 150^\circ$ and that $AD = AE$ . Then triangle $ADE$ is isosceles, so $\measuredangle AED = \dfrac{180^\circ-150^\circ}{2} = \boxed{15}$

Answer: C 15
Problem: [asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$ . 
What is the measure of $\measuredangle AED$ in degrees?
$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$

Solution: WLOG, let the side length of the square and the equilateral triangle be $1$ $\angle{DAE}=90^\circ+60^\circ=150^\circ$ . Apply the law of cosines then the law of sines, we find that $\angle{AED}=15^\circ$ . Select $\boxed{15}$

Answer: C 15
Problem: Find the sum of the digits of the largest even three digit number (in base ten representation) 
which is not changed when its units and hundreds digits are interchanged.
$\textbf{(A) }22\qquad \textbf{(B) }23\qquad \textbf{(C) }24\qquad \textbf{(D) }25\qquad \textbf{(E) }26$

Solution: Solution by e_power_pi_times_i
Since the number doesn't change when the units and hundreds digits are switched, the number must be of the form $aba$ . We want to create the largest even $3$ -digit number, so $a = 8$ and $b = 9$ . The sum of the digits is $8+9+8 = \boxed{25}$

Answer: D 25
Problem: Find a positive integral solution to the equation $\frac{1+3+5+\dots+(2n-1)}{2+4+6+\dots+2n}=\frac{115}{116}$
$\textbf{(A) }110\qquad \textbf{(B) }115\qquad \textbf{(C) }116\qquad \textbf{(D) }231\qquad\\ \textbf{(E) }\text{The equation has no positive integral solutions.}$

Solution: Solution by e_power_pi_times_i
Notice that the numerator and denominator are the sum of the first $n$ odd and even numbers, respectively. Then the numerator is $n^2$ , and the denominator is $n(n+1)$ .  Then $\frac{n}{n+1} = \frac{115}{116}$ , so $n = \boxed{115}$

Answer: B 115
Problem: [asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\circ$",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$ . Point $A$ lies on the extension of $DC$ past $C$ ; 
point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle. 
If length $AB$ equals length $OD$ , and the measure of $\measuredangle EOD$ is $45^\circ$ , then the
measure of $\measuredangle BAO$ is
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$

Solution: Solution by e_power_pi_times_i
Because $AB = OD$ , triangles $ABO$ and $BOE$ are isosceles. Denote $\measuredangle BAO = \measuredangle AOB = \theta$ . Then $\measuredangle ABO = 180^\circ-2\theta$ , and $\measuredangle EBO = \measuredangle OEB = 2\theta$ , so $\measuredangle BOE = 180^\circ-4\theta$ . Notice that $\measuredangle AOB + \measuredangle BOE + 45^\circ = 180^\circ$ . Therefore $\theta+180-4\theta = 135^\circ$ , and $\theta = \boxed{15}$

Answer: B 15
Problem: [asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\circ$",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$ . Point $A$ lies on the extension of $DC$ past $C$ ; 
point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle. 
If length $AB$ equals length $OD$ , and the measure of $\measuredangle EOD$ is $45^\circ$ , then the
measure of $\measuredangle BAO$ is
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$

Solution: Draw $BO$ . Let $y = \angle BAO$ . Since $AB = OD = BO$ , triangle $ABO$ is isosceles, so $\angle BOA = \angle BAO = y$ . Angle $\angle EBO$ is exterior to triangle $ABO$ , so $\angle EBO = \angle BAO + \angle BOA = y + y = 2y$
Triangle $BEO$ is isosceles, so $\angle BEO = \angle EBO = 2y$ . Then $\angle EOD$ is external to triangle $AEO$ , so $\angle EOD = \angle EAO + \angle AEO = y + 2y = 3y$ . But $\angle EOD = 45^\circ$ , so $\angle BAO = y = 45^\circ/3 = \boxed{15}$

Answer: B 15
Problem: A circle with area $A_1$ is contained in the interior of a larger circle with area $A_1+A_2$ . If the radius of the larger circle is $3$ , 
and if $A_1 , A_2, A_1 + A_2$ is an arithmetic progression, then the radius of the smaller circle is
$\textbf{(A) }\frac{\sqrt{3}}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }\frac{2}{\sqrt{3}}\qquad \textbf{(D) }\frac{3}{2}\qquad \textbf{(E) }\sqrt{3}$

Solution: Solution by e_power_pi_times_i
The area of the larger circle is $A_1 + A_2 = 9\pi$ . Then $A_1 , 9\pi-A_1 , 9\pi$ are in an arithmetic progression. Thus $9\pi-(9\pi-A_1) = 9\pi-A_1-A_1$ . This simplifies to $3A_1 = 9\pi$ , or $A_1 = 3\pi$ . The radius of the smaller circle is $\boxed{3}$

Answer: E 3
Problem: Find the sum of the squares of all real numbers satisfying the equation $x^{256}-256^{32}=0$
$\textbf{(A) }8\qquad \textbf{(B) }128\qquad \textbf{(C) }512\qquad \textbf{(D) }65,536\qquad \textbf{(E) }2(256^{32})$

Solution: Solution by e_power_pi_times_i
Notice that the solutions to the equation $x^{256}-1=0$ are the $256$ roots of unity. Then the solutions to the equation $x^{256}-256^{32}=0$ are the $256$ roots of unity dilated by $\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2$ . However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are $\pm1$ , and dilating by $2$ gives $\pm2$ . The sum of the squares is $(2)^2+(-2)^2 = \boxed{8}$

Answer: A 8
Problem: If $a=\tfrac{1}{2}$ and $(a+1)(b+1)=2$ then the radian measure of $\arctan a + \arctan b$ equals
$\textbf{(A) }\frac{\pi}{2}\qquad \textbf{(B) }\frac{\pi}{3}\qquad \textbf{(C) }\frac{\pi}{4}\qquad \textbf{(D) }\frac{\pi}{5}\qquad \textbf{(E) }\frac{\pi}{6}$

Solution: Solution by e_power_pi_times_i
Since $a=\frac{1}{2}$ $b=\frac{1}{3}$ . Now we evaluate $\arctan a$ and $\arctan b$ . Denote $x$ and $\theta$ such that $\arctan x = \theta$ . Then $\tan(\arctan(x)) = \tan(\theta)$ , and simplifying gives $x = \tan(\theta)$ . So $a = \tan(\theta_a) = \frac{1}{2}$ and $b = \tan(\theta_b) = \frac{1}{3}$ . The question asks for $\theta_a + \theta_b$ , so we try to find $\tan(\theta_a + \theta_b)$ in terms of $\tan(\theta_a)$ and $\tan(\theta_b)$ . Using the angle addition formula for $\tan(\alpha+\beta)$ , we get that $\tan(\theta_a + \theta_b) = \frac{\tan(\theta_a)+\tan(\theta_b)}{1-\tan(\theta_a)\tan(\theta_b)}$ . Plugging $\tan(\theta_a) = \frac{1}{2}$ and $\tan(\theta_b) = \frac{1}{3}$ in, we have $\tan(\theta_a + \theta_b) = \frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}$ . Simplifying, $\tan(\theta_a + \theta_b) = 1$ , so $\theta_a + \theta_b$ in radians is $\boxed{4}$

Answer: C 4
Problem: Find the number of pairs $(m, n)$ of integers which satisfy the equation $m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1$
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }3\qquad \textbf{(D) }9\qquad \textbf{(E) }\infty$

Solution: The equation is equivalent to $m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1$ . Taking mod 3, we get $m(m+1)(m+2)=1 (\bmod 3)$ . However, $m(m+1)(m+2)$ is always divisible by $3$ for any integer $m$ . Thus, the answer is $\boxed{0}$ Solution by mickyboy789

Answer: A 0
Problem: Sides $AB,~ BC$ , and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5$ , and $20$ , respectively. 
If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5}$ , then side $AD$ has length
$\textbf{(A) }24\qquad \textbf{(B) }24.5\qquad \textbf{(C) }24.6\qquad \textbf{(D) }24.8\qquad \textbf{(E) }25$

Solution: We know that $\sin(C)=-\cos(B)=\frac{3}{5}$ . Since $B$ and $C$ are obtuse, we have $\sin(180-C)=\cos(180-B)=\frac{3}{5}$ . It is known that $\sin(x)=\cos(90-x)$ , so $180-C=90-(180-C)=180-B$ . We simplify this as follows:
\[-90+C=180-B\]
\[B+C=270^{\circ}\]
Since $B+C=270^{\circ}$ , we know that $A+D=360-(B+C)=90^{\circ}$ . Now extend $AB$ and $CD$ as follows:
[asy] size(10cm); label("A",(-1,0)); dot((0,0)); label("B",(-1,4)); dot((0,4)); label("E",(-1,7)); dot((0,7)); label("C",(4,8)); dot((4,7)); label("D",(24,8)); dot((24,7)); draw((0,0)--(0,4)); draw((0,4)--(4,7)); draw((4,7)--(24,7)); draw((24,7)--(0,0)); draw((0,4)--(0,7), dashed); draw((0,7)--(4,7), dashed); //diagram by WannabeCharmander [/asy]
Let $AB$ and $CD$ intersect at $E$ . We know that $\angle AED=90^{\circ}$ because $\angle E = 180 - (A+D)=180-90 = 90^{\circ}$
Since $\sin BCD = \frac{3}{5}$ , we get $\sin ECB=\sin(180-BCD)=\sin BCD = \frac{3}{5}$ . Thus, $EB=3$ and $EC=4$ from simple sin application.
$AD$ is the hypotenuse of right $\triangle AED$ , with leg lengths $AB+BE=7$ and $EC+CD=24$ . Thus, $AD=\boxed{25}$

Answer: E 25
Problem: For each positive number $x$ , let $f(x)=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2} {\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}$ . 
The minimum value of $f(x)$ is
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }6$

Solution: Let $a = \left( x + \frac{1}{x} \right)^3$ and $b = x^3 + \frac{1}{x^3}$ . Then \begin{align*} f(x) &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^6 + \frac{1}{x^6}) - 2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^6 + 2 + \frac{1}{x^6})}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^3 + \frac{1}{x^3})^2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{a^2 - b^2}{a + b}. \end{align*}
By difference of squares, \begin{align*} f(x) &= \frac{(a - b)(a + b)}{a + b} \\ &= a - b \\ &= \left( x + \frac{1}{x} \right)^3 - \left( x^3 + \frac{1}{x^3} \right) \\ &= \left( x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \right) - \left( x^3 + \frac{1}{x^3} \right) \\ &= 3x + \frac{3}{x} \\ &= 3 \left( x + \frac{1}{x} \right). \end{align*}
By the AM-GM inequality, \[x + \frac{1}{x} \ge 2,\] so $f(x) \ge 6$ . Furthermore, when $x = 1$ $f(1) = 6$ , so the minimum value of $f(x)$ is $\boxed{6}$

Answer:  6
Problem: If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad  \textbf{(E) }-1\text{ or }-2$

Solution: By guessing and checking, 2 works. $\frac{2}{x} = \boxed{1}$ ~awin

Answer: B 1
Problem: If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad  \textbf{(E) }-1\text{ or }-2$

Solution: Multiplying each side by $x^2$ , we get $x^2-4x+4 = 0$ . Factoring, we get $(x-2)(x-2) = 0$ . Therefore, $x = 2$ $\frac{2}{x} = \boxed{1}$ ~awin

Answer: B 1
Problem: If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad  \textbf{(E) }-1\text{ or }-2$

Solution: Directly factoring, we get $(1-\frac{2}{x})^2 = 0$ . Thus $\frac{2}{x}$ must equal $\boxed{1}$

Answer: B 1
Problem: If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is
$\textbf{(A) }\frac{1}{\pi^2}\qquad \textbf{(B) }\frac{1}{\pi}\qquad \textbf{(C) }1\qquad \textbf{(D) }\pi\qquad \textbf{(E) }\pi^2$

Solution: Creating equations, we get $4\cdot\frac{1}{2\pi r} = 2r$ . Simplifying, we get $\frac{1}{\pi r} = r$ . Multiplying each side by $r$ , we get $\frac{1}{\pi} = r^2$ . Because the formula of the area of a circle is $\pi r^2$ , we multiply each side by $\pi$ to get $1 = \pi r^2$ .
Therefore, our answer is $\boxed{1}$

Answer: C 1
Problem: If $a = 1,~ b = 10, ~c = 100$ , and $d = 1000$ , then $(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)$ is equal to
$\textbf{(A) }1111\qquad \textbf{(B) }2222\qquad \textbf{(C) }3333\qquad \textbf{(D) }1212\qquad  \textbf{(E) }4242$

Solution: Adding all four of the equations up, we can see that it equals \[3(a+b+c+d)\] This is equal to $3(1111) = \boxed{3333}$ ~awin

Answer: C 3333
Problem: Four boys bought a boat for $\textdollar 60$ . The first boy paid one half of the sum of the amounts paid by the other boys; 
the second boy paid one third of the sum of the amounts paid by the other boys; 
and the third boy paid one fourth of the sum of the amounts paid by the other boys. How much did the fourth boy pay?
$\textbf{(A) }\textdollar 10\qquad \textbf{(B) }\textdollar 12\qquad \textbf{(C) }\textdollar 13\qquad \textbf{(D) }\textdollar 14\qquad \textbf{(E) }\textdollar 15$

Solution: If the first boy paid one half of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{3}$ of the total.
If the second boy paid one third of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{4}$ of the total.
If the third boy paid one fourth of the sum of the  amounts paid by the other boys, that means that he paid $\frac{1}{5}$ of the total.
Summing it up, we get $\textdollar 20 + \textdollar 15 + \textdollar 12 = \textdollar 47$ .
Therefore, our answer is $\textdollar 60 - \textdollar 47 = \boxed{13}$ ~awin

Answer: C 13
Problem: The number of distinct pairs $(x,y)$ of real numbers satisfying both of the following equations:
\[x=x^2+y^2 \   \ y=2xy\] is
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad  \textbf{(E) }4$

Solution: If $x=x^2+y^2$ and $y=2xy$ , then we can break this into two cases.
Case 1: $y = 0$
If $y = 0$ , then $x = x^2$ and $0 = 0$
Therefore, $x = 0$ or $x = 1$
This yields 2 solutions
Case 2: $x = \frac{1}{2}$
If $x = \frac{1}{2}$ , this means that $y = y$ , and $\frac{1}{2} = \frac{1}{4} + y^2$
Because y can be negative or positive, this yields $y = \frac{1}{2}$ or $y = -\frac{1}{2}$
This yields another 2 solutions.
$2+2 = \boxed{4}$

Answer: E 4
Problem: If $r$ is positive and the line whose equation is $x + y = r$ is tangent to the circle whose equation is $x^2 + y ^2 = r$ , then $r$ equals
$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }\sqrt{2}\qquad  \textbf{(E) }2\sqrt{2}$

Solution: The circle $x^2 + y^2 = r$ has center $(0,0)$ and radius $\sqrt{r}$ . Therefore, if the line $x + y = r$ is tangent to the circle $x^2 + y^2 = r$ , then the distance between $(0,0)$ and the line $x + y = r$ is $\sqrt{r}$
The distance between $(0,0)$ and the line $x + y = r$ is \[\frac{|0 + 0 - r|}{\sqrt{1^2 + 1^2}} = \frac{r}{\sqrt{2}}.\] Hence, \[\frac{r}{\sqrt{2}} = \sqrt{r}.\] Then $r = \sqrt{r} \cdot \sqrt{2}$ , so $\sqrt{r} = \sqrt{2}$ , which means $r = \boxed{2}$ or (B), $2$

Answer:  2
Problem: If $a,b,c$ , and $d$ are non-zero numbers such that $c$ and $d$ are the solutions of $x^2+ax+b=0$ and $a$ and $b$ are 
the solutions of $x^2+cx+d=0$ , then $a+b+c+d$ equals
$\textbf{(A) }0\qquad \textbf{(B) }-2\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad  \textbf{(E) }(-1+\sqrt{5})/2$

Solution: By Vieta's formulas, $c + d = -a$ $cd = b$ $a + b = -c$ , and $ab = d$ . From the equation $c + d = -a$ $d = -a - c$ , and from the equation $a + b = -c$ $b = -a - c$ , so $b = d$
Then from the equation $cd = b$ $cb = b$ . Since $b$ is nonzero, we can divide both sides of the equation by $b$ to get $c = 1$ . Similarly, from the equation $ab = d$ $ab = b$ , so $a = 1$ . Then $b = d = -a - c = -2$ . Therefore, $a + b + c + d = 1 + (-2) + 1 + (-2) = \boxed{2}$ . The answer is (B).

Answer:  2
Problem: If an integer $n > 8$ is a solution of the equation $x^2 - ax+b=0$ and the representation of $a$ in the base- $n$ number system is $18$ , 
then the base-n representation of $b$ is
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 80 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 280$

Solution: Assuming the solutions to the equation are n and m, by Vieta's formulas, $n_n + m_n = 18_n$
$n_n = 10_n$ , so $10_n + m_n = 18_n$
\[m_n = 8_n\]
Also by Vieta's formulas, $n_n \cdot m_n = b_n$ \[10_n \cdot 8_n = \boxed{80}\]

Answer:  80