Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair $A,\,B$ of cities, they are connected with a road along the line segment $AB$ if and only if the following condition holds: [center]For every city $C$ distinct from $A$ and $B$, there exists $R\in\mathcal{S}$ such[/center] [center]that $\triangle PQR$ is directly similar to either $\triangle ABC$ or $\triangle BAC$.[/center] Alice wins the game if (i) the resulting roads allow for travel between any pair of cities via a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine, with proof, which player has a winning strategy. [i]Note:[/i] $\triangle UVW$ is [i]directly similar[/i] to $\triangle XYZ$ if there exists a sequence of rotations, translations, and dilations sending $U$ to $X$, $V$ to $Y$, and $W$ to $Z$.

Three wooden equilateral triangles of side length $18$ inches are placed on axles as shown in the diagram to the right. Each axle is $30$ inches from the other two axles. A $144$-inch leather band is wrapped around the wooden triangles, and a dot at the top corner is painted as shown. The three triangles are then rotated at the same speed and the band rotates without slipping or stretching. Compute the length of the path that the dot travels before it returns to its initial position at the top corner. [asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); pair A=origin,B=(48,0),C=rotate(60,A)*B; path equi=(0,0)--(18,0)--(9,9*sqrt(3))--cycle,circ=circle(centroid(A,B,C)*18/48,1/3); picture a; fill(a,equi,grey); fill(a,circ,white); add(a); add(shift(15,15*sqrt(3))*a); add(shift(30,0)*a); draw(A--B--C--cycle,linewidth(1)); path top = circle(C,2/3); unfill(top); draw(top); real r=-5/2; draw((9,r+1)--(9,r-1)^^(9,r)--(39,r)^^(39,r-1)--(39,r+1)); label("$30$",(24,r),S); [/asy]

Let $H$ be the orthocenter of triangle $\mathrm T$. The sidelines of triangle $\mathrm T_1$ pass through the midpoints of $\mathrm T$ and are perpendicular to the corresponding bisectors of $\mathrm T$. The vertices of triangle $\mathrm T_2$ bisect the bisectors of $\mathrm T$. Prove that the lines joining $H$ with the vertices of $\mathrm T_1$ are perpendicular to the sidelines of $\mathrm T_2$.

Let $D, E$, and $F$ respectively be the feet of the altitudes from $A, B$, and $C$ of acute triangle $\vartriangle ABC$ such that $AF = 28, FB = 35$ and $BD = 45$. Let $P$ be the point on segment $BE$ such that $AP = 42$. Find the length of $CP$.

Circles $ A$, $ B$ and $ C$ are externally tangent to each other and internally tangent to circle $ D$. Circles $ B$ and $ C$ are congruent. Circle $ A$ has radius $ 1$ and passes through the center of $ D$. What is the radius of circle $ B$? [asy] unitsize(15mm); pair A=(-1,0),B=(2/3,8/9),C=(2/3,-8/9),D=(0,0); draw(Circle(D,2)); draw(Circle(A,1)); draw(Circle(B,8/9)); draw(Circle(C,8/9)); label("\(A\)", A); label("\(B\)", B); label("\(C\)", C); label("D", (-1.2,1.8));[/asy] $ \textbf{(A)}\ \frac23 \qquad \textbf{(B)}\ \frac {\sqrt3}{2} \qquad \textbf{(C)}\ \frac78 \qquad \textbf{(D)}\ \frac89 \qquad \textbf{(E)}\ \frac {1 \plus{} \sqrt3}{3}$

Let $\mathcal{B}$ be the set of rectangular boxes that have volume $23$ and surface area $54$. Suppose $r$ is the least possible radius of a sphere that can fit any element of $\mathcal{B}$ inside it. Then $r^{2}$ can be expressed as $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Suppose that $|x+y|+|x-y|=2$. What is the maximum possible value of $x^2-6x+y^2$? $ \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9 $

Prove that there exists a positive constant $c$ such that the following statement is true: Consider an integer $n > 1$, and a set $\mathcal S$ of $n$ points in the plane such that the distance between any two different points in $\mathcal S$ is at least 1. It follows that there is a line $\ell$ separating $\mathcal S$ such that the distance from any point of $\mathcal S$ to $\ell$ is at least $cn^{-1/3}$. (A line $\ell$ separates a set of points S if some segment joining two points in $\mathcal S$ crosses $\ell$.) [i]Note. Weaker results with $cn^{-1/3}$ replaced by $cn^{-\alpha}$ may be awarded points depending on the value of the constant $\alpha > 1/3$.[/i] [i]Proposed by Ting-Feng Lin and Hung-Hsun Hans Yu, Taiwan[/i]

Jesse cuts a circular paper disk of radius $12$ along two radii to form two sectors, the smaller having a central angle of $120$ degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger? $ \textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{1}{4} \qquad\textbf{(C)}\ \frac{\sqrt{10}}{10} \qquad\textbf{(D)}\ \frac{\sqrt{5}}{6} \qquad\textbf{(E)}\ \frac{\sqrt{10}}{5} $

In quadrilateral $ABCD$, $BC=8$, $CD=12$, $AD=10$, and $m\angle A= m\angle B = 60^\circ$. Given that $AB=p + \sqrt{q}$, where $p$ and $q$ are positive integers, find $p+q$.

Let $K$ be a point on the side $[AB]$, and $L$ be a point on the side $[BC]$ of the square $ABCD$. If $|AK|=3$, $|KB|=2$, and the distance of $K$ to the line $DL$ is $3$, what is $|BL|:|LC|$? $ \textbf{(A)}\ \frac78 \qquad\textbf{(B)}\ \frac{\sqrt 3}2 \qquad\textbf{(C)}\ \frac 87 \qquad\textbf{(D)}\ \frac 38 \qquad\textbf{(E)}\ \frac{\sqrt 2}2 $

Triangle $\triangle ABC$ has $\angle{A}=90^\circ$ with $BC=12$. Square $BCDE$ is drawn such that $A$ is in its interior. The line through $A$ tangent to the circumcircle of $\triangle ABC$ intersects $CD$ and $BE$ at $P$ and $Q$, respectively. If $PA=4\cdot QA$, and the area of $\triangle ABC$ can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$, then compute $m+n$. [i]Proposed by Andy Xu[/i]

Show that there does not exist a right triangle with all integer side lengths such that exactly one of the side lengths is odd.

Triangle $ABC$ has a right angle at $C$, and $D$ is the foot of the altitude from $C$ to $AB$. Points $L, M,$ and $N$ are the midpoints of segments $AD, DC,$ and $CA,$ respectively. If $CL = 7$ and $BM = 12,$ compute $BN^2$.

Triangle $\triangle ABC$ has side lengths $\overline{AB} = 13, \overline{BC} = 14,$ and $\overline{AC} = 15$. Suppose $M$ and $N$ are the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively. Let $P$ be a point on $\overline{MN}$, such that if the circumcircles of triangles $\triangle BMP$ and $\triangle CNP$ intersect at a second point $Q$ distinct from $P$, then $PQ$ is parallel to $AB$. The value of $AP^2$ can be expressed as a common fraction of the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.