The diagram below shows the first three figures of a sequence of figures. The first figure shows an equilateral triangle $ABC$ with side length $1$. The leading edge of the triangle going in a clockwise direction around $A$ is labeled $\overline{AB}$ and is darkened in on the figure. The second figure shows the same equilateral triangle with a square with side length $1$ attached to the leading clockwise edge of the triangle. The third figure shows the same triangle and square with a regular pentagon with side length $1$ attached to the leading clockwise edge of the square. The fourth figure in the sequence will be formed by attaching a regular hexagon with side length $1$ to the leading clockwise edge of the pentagon. The hexagon will overlap the triangle. Continue this sequence through the eighth figure. After attaching the last regular figure (a regular decagon), its leading clockwise edge will form an angle of less than $180^\circ$ with the side $\overline{AC}$ of the equilateral triangle. The degree measure of that angle can be written in the form $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] size(250); defaultpen(linewidth(0.7)+fontsize(10)); pair x[],y[],z[]; x[0]=origin; x[1]=(5,0); x[2]=rotate(60,x[0])*x[1]; draw(x[0]--x[1]--x[2]--cycle); for(int i=0;i<=2;i=i+1) { y[i]=x[i]+(15,0); } y[3]=rotate(90,y[0])*y[2]; y[4]=rotate(-90,y[2])*y[0]; draw(y[0]--y[1]--y[2]--y[0]--y[3]--y[4]--y[2]); for(int i=0;i<=4;i=i+1) { z[i]=y[i]+(15,0); } z[5]=rotate(108,z[4])*z[2]; z[6]=rotate(108,z[5])*z[4]; z[7]=rotate(108,z[6])*z[5]; draw(z[0]--z[1]--z[2]--z[0]--z[3]--z[4]--z[2]--z[7]--z[6]--z[5]--z[4]); dot(x[2]^^y[2]^^z[2],linewidth(3)); draw(x[2]--x[0]^^y[2]--y[4]^^z[2]--z[7],linewidth(1)); label("A",(x[2].x,x[2].y-.3),S); label("B",origin,S); label("C",x[1],S);[/asy]

$\Delta ABC$ and its mirror reflection $\Delta A^{\prime}B^{\prime}C^{\prime}$ is arbitrarily placed on the plane. Prove the midpoints of $AA^{\prime},BB^{\prime},CC^{\prime}$ are collinear.

Given a triangle $OAB$ with the vetices $O(0,\ 0,\ 0),\ A(1,\ 0,\ 0),\ B(1,\ 1,\ 0)$ in the $xyz$ space. Let $V$ be the cone obtained by rotating the triangle around the $x$-axis. Find the volume of the solid obtained by rotating the cone $V$ around the $y$-axis.

In the plane, there are two circles $\Gamma_1, \Gamma_2$ intersecting each other at two points $A$ and $B$. Tangents of $\Gamma_1$ at $A$ and $B$ meet each other at $K$. Let us consider an arbitrary point $M$ (which is different of $A$ and $B$) on $\Gamma_1$. The line $MA$ meets $\Gamma_2$ again at $P$. The line $MK$ meets $\Gamma_1$ again at $C$. The line $CA$ meets $\Gamma_2 $ again at $Q$. Show that the midpoint of $PQ$ lies on the line $MC$ and the line $PQ$ passes through a fixed point when $M$ moves on $\Gamma_1$. [color=red][Moderator edit: This problem was also discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=21414 .][/color]

It is known that the graph of the function $y = f(x)$ after a rotation of $45^o$ around a certain point turns into the graph of the function $y = x^3 + ax^2 + 19x + 97$. At what $a$ is this possible?

Find the volume $V$ of the solid formed by a rotation of the region enclosed by the curve $y=2^{x}-1$ and two lines $x=0,\ y=1$ around the $y$ axis.

Let the sides of two rectangles be $ \{a,b\}$ and $ \{c,d\},$ respectively, with $ a < c \leq d < b$ and $ ab < cd.$ Prove that the first rectangle can be placed within the second one if and only if \[ \left(b^2 \minus{} a^2\right)^2 \leq \left(bc \minus{} ad \right)^2 \plus{} \left(bd \minus{} ac \right)^2.\]

The plane is covered with network of regular congruent disjoint hexagons. Prove that there cannot exist a square which has its four vertices in the vertices of the hexagons. [i]Gabriel Nagy[/i]

Triangle $ABC$ has sides $AB = 25$, $BC = 30$, and $CA=20$. Let $P,Q$ be the points on segments $AB,AC$, respectively, such that $AP=5$ and $AQ=4$. Suppose lines $BQ$ and $CP$ intersect at $R$ and the circumcircles of $\triangle{BPR}$ and $\triangle{CQR}$ intersect at a second point $S\ne R$. If the length of segment $SA$ can be expressed in the form $\frac{m}{\sqrt{n}}$ for positive integers $m,n$, where $n$ is not divisible by the square of any prime, find $m+n$. [i]Victor Wang[/i]

Find the volume $V$ of the solid formed by a rotation of the region enclosed by the curve $y=2^{x}-1$ and two lines $x=0,\ y=1$ around the $y$ axis.

A “large” circular disk is attached to a vertical wall. It rotates clockwise with one revolution per minute. An insect lands on the disk and immediately starts to climb vertically upward with constant speed $\frac{\pi}{3}$ cm per second (relative to the disk). Describe the path of the insect [i](a)[/i] relative to the disk; [i](b)[/i] relative to the wall.

Let $\omega_P$ and $\omega_Q$ be two circles of radius $1$, intersecting in points $A$ and $B$. Let $P$ and $Q$ be two regular $n$-gons (for some positive integer $n\ge4$) inscribed in $\omega_P$ and $\omega_Q$, respectively, such that $A$ and $B$ are vertices of both $P$ and $Q$. Suppose a third circle $\omega$ of radius $1$ intersects $P$ at two of its vertices $C$, $D$ and intersects $Q$ at two of its vertices $E$, $F$. Further assume that $A$, $B$, $C$, $D$, $E$, $F$ are all distinct points, that $A$ lies outside of $\omega$, and that $B$ lies inside $\omega$. Show that there exists a regular $2n$-gon that contains $C$, $D$, $E$, $F$ as four of its vertices.

Given triangle $ ABC$. Points $ D,E,F$ outside triangle $ ABC$ are chosen such that triangles $ ABD$, $ BCE$, and $ CAF$ are equilateral triangles. Prove that cicumcircles of these three triangles are concurrent.

Consider a charged capacitor made with two square plates of side length $L$, uniformly charged, and separated by a very small distance $d$. The EMF across the capacitor is $\xi$. One of the plates is now rotated by a very small angle $\theta$ to the original axis of the capacitor. Find an expression for the difference in charge between the two plates of the capacitor, in terms of (if necessary) $d$, $\theta$, $\xi$, and $L$. Also, approximate your expression by transforming it to algebraic form: i.e. without any non-algebraic functions. For example, logarithms and trigonometric functions are considered non-algebraic. Assume $ d << L $ and $ \theta \approx 0 $. $\emph{Hint}$: You may assume that $ \frac {\theta L}{d} $ is also very small. [i]Problem proposed by Trung Phan[/i] [hide="Clarification"] There are two possible ways to rotate the capacitor. Both were equally scored but this is what was meant: [asy]size(6cm); real h = 7; real w = 2; draw((-w,0)--(-w,h)); draw((0,0)--(0,h), dashed); draw((0,0)--h*dir(64)); draw(arc((0,0),2,64,90)); label("$\theta$", 2*dir(77), dir(77)); [/asy] [/hide]

Square $ABCD$ in the coordinate plane has vertices at the points $A(1,1), B(-1,1), C(-1,-1),$ and $D(1,-1).$ Consider the following four transformations: [list=] [*]$L,$ a rotation of $90^{\circ}$ counterclockwise around the origin; [*]$R,$ a rotation of $90^{\circ}$ clockwise around the origin; [*]$H,$ a reflection across the $x$-axis; and [*]$V,$ a reflection across the $y$-axis. [/list] Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying $R$ and then $V$ would send the vertex $A$ at $(1,1)$ to $(-1,-1)$ and would send the vertex $B$ at $(-1,1)$ to itself. How many sequences of $20$ transformations chosen from $\{L, R, H, V\}$ will send all of the labeled vertices back to their original positions? (For example, $R, R, V, H$ is one sequence of $4$ transformations that will send the vertices back to their original positions.) $\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}$

Nine distinct positive integers are arranged in a circle such that the product of any two non-adjacent numbers in the circle is a multiple of $n$ and the product of any two adjacent numbers in the circle is not a multiple of $n$, where $n$ is a fixed positive integer. Find the smallest possible value for $n$.

Let $ABCD$ be a square and let $\ell$ be a line. Let $M$ be the centre of the square. The diagonals of the square have length 2 and the distance from $M$ to $\ell$ exceeds 1. Let $A',B',C',D'$ be the orthogonal projections of $A,B,C,D$ onto $\ell$. Suppose that one rotates the square, such that $M$ is invariant. The positions of $A,B,C,D,A',B',C',D'$ change. Prove that the value of $AA'^2 + BB'^2 + CC'^2 + DD'^2$ does not change.

Let $S$ be the graph of $y=x^3$, and $T$ be the graph of $y=\sqrt[3]{y}$. Let $S^*$ be $S$ rotated around the origin $15$ degrees clockwise, and $T^*$ be T rotated around the origin 45 degrees counterclockwise. $S^*$ and $T^*$ will intersect at a point in the first quadrant a distance $M+\sqrt{N}$ from the origin where $M$ and $N$ are positive integers. Find $M+N$.

Cut the triangle shown in the picture into three pieces and rearrange them into a rectangle. [asy] size(200); pen tpen = defaultpen + 1.337; draw((1,0)--(1,8)); draw((2,0)--(2,8)); draw((3,0)--(3,8)); draw((4,0)--(4,8)); draw((5,0)--(5,8)); draw((6,0)--(6,8)); draw((7,0)--(7,8)); draw((8,0)--(8,8)); draw((9,0)--(9,8)); draw((10,0)--(10,8)); draw((11,0)--(11,8)); draw((12,0)--(12,8)); draw((13,0)--(13,8)); draw((0,1)--(13.5,1)); draw((0,2)--(13.5,2)); draw((0,3)--(13.5,3)); draw((0,4)--(13.5,4)); draw((0,5)--(13.5,5)); draw((0,6)--(13.5,6)); draw((0,7)--(13.5,7)); draw((1,1)--(5,7), tpen); draw((1,1)--(13,1),tpen); draw((5,7)--(13,1),tpen); [/asy]

We define a [i]chessboard polygon[/i] to be a polygon whose sides are situated along lines of the form $ x \equal{} a$ or $ y \equal{} b$, where $ a$ and $ b$ are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping $ 1 \times 2$ rectangles. Finally, a [i]tasteful tiling[/i] is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a $ 3 \times 4$ rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner. [asy]size(300); pathpen = linewidth(2.5); void chessboard(int a, int b, pair P){ for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j) if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6)); D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle); } chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12)); /* draw lines */ D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy] a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully. b) Prove that such a tasteful tiling is unique.

If circular arcs $ AC$ and $ BC$ have centers at $ B$ and $ A$, respectively, then there exists a circle tangent to both $ \stackrel{\frown}{AC}$ and $ \stackrel{\frown}{BC}$, and to $ \overline{AB}$. If the length of $ \stackrel{\frown}{BC}$ is $ 12$, then the circumference of the circle is [asy]unitsize(4cm); defaultpen(fontsize(8pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,.375); pair A=(-.5,0); pair B=(.5,0); pair C=shift(-.5,0)*dir(60); draw(Arc(A,1,0,60)); draw(Arc(B,1,120,180)); draw(A--B); draw(Circle(O,.375)); dot(A); dot(B); dot(C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N);[/asy]$ \textbf{(A)}\ 24 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 28$

Five people are sitting at a round table. Let $f \ge 0$ be the number of people sitting next to at least one female and $m \ge 0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is $ \textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11 $

$B$ is the center of unit circle. $A,C$ are points on the circle (the order of $A,B,C$ is clockwise), and $\angle ABC=2\alpha(0<\alpha<\frac{\pi}{3})$. Then we will rotate $\triangle ABC$ anticlockwise. In the first rotation, $A$ is the center of rotation, the result is that $B$ is on the circle. In the second rotation, $B$ is the center of rotation, the result is that $C$ is on the circle. In the third rotation, $C$ is the center of rotation, the result is that $A$ is on the circle. ... After we rotate for $100$ times, the distance $A$ travelled is $\text{(A)}22\pi(1+\sin\alpha)-66\alpha\qquad\text{(B)}\frac{67}{3}\pi\qquad\text{(C)}22\pi+\frac{68}{3}\pi\sin\alpha-66\alpha\qquad\text{(D)}33\pi-66\alpha$

Find the volume of the solid of a circle $x^2+(y-1)^2=4$ generated by a rotation about the $x$-axis.

Let $H$ be a rectangle with angle between two diagonal $\leq 45^{0}$. Rotation $H$ around the its center with angle $0^{0}\leq x\leq 360^{0}$ we have rectangle $H_{x}$. Find $x$ such that $[H\cap H_{x}]$ minimum, where $[S]$ is area of $S$.