The triangle $ABC$ has $CA = CB$. $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$). $D$ is the foot of the perpendicular from $C$ to $PB$. Show that $PA + PB = 2 \cdot PD$.

The cyclic octagon $ABCDEFGH$ has sides $a,a,a,a,b,b,b,b$ respectively. Find the radius of the circle that circumscribes $ABCDEFGH.$

Given a quadrilateral $ABCD$ such that $AB^2 + CD^2 = AD^2 + BC^2$, prove that $AC \perp BD$.

Let $O$ be the center of a circle of radius $26$, and let $A$, $B$ be two distinct points on the circle, with $M$ being the midpoint of $AB$. Consider point $C$ for which $CO=34$ and $\angle COM=15^\circ$. Let $N$ be the midpoint of $CO$. Suppose that $\angle ACB=90^\circ$. Find $MN$.

Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $B$ at a speed of $7$ meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie? [asy] defaultpen(linewidth(0.8)); draw((100,0)--origin--60*dir(60), EndArrow(5)); label("$A$", origin, SW); label("$B$", (100,0), SE); label("$100$", (50,0), S); label("$60^\circ$", (15,0), N);[/asy]

In $\triangle PQR$, $PQ=8$, $QR=13$, and $RP=15$. Prove that there is a point $S$ on line segment $\overline{PR}$, but not at its endpoints, such that $PS$ and $QS$ are also integers. [asy] size(200); defaultpen(linewidth(0.8)); pair P=origin,Q=(8,0),R=(7,10),S=(3/2,15/7); draw(P--Q--R--cycle); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,NE); draw(Q--S,linetype("4 4")); label("$S$",S,NW); [/asy]

Given a square cardboard of area $\frac{1}{4}$, and a paper triangle of area $\frac{1}{2}$ such that the square of its sidelength is a positive integer. Prove that the triangle can be folded in some ways such that the squace can be placed inside the folded figure so that both of its faces are completely covered with paper. [i]Proposed by N.Beluhov, Bulgaria[/i]

Triangle $ABC$ has right angle at $B$, and contains a point $P$ for which $PA = 10$, $PB = 6$, and $\angle APB = \angle BPC = \angle CPA$. Find $PC$. [asy] pair A=(0,5), B=origin, C=(12,0), D=rotate(-60)*C, F=rotate(60)*A, P=intersectionpoint(A--D, C--F); draw(A--P--B--A--C--B^^C--P); dot(A^^B^^C^^P); pair point=P; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$P$", P, NE);[/asy]

In the 2009 Stanford Olympics, Willy and Sammy are two bikers. The circular race track has two lanes, the inner lane with radius 11, and the outer with radius 12. Willy will start on the inner lane, and Sammy on the outer. They will race for one complete lap, measured by the inner track. What is the square of the distance between Willy and Sammy's starting positions so that they will both race the same distance? Assume that they are of point size and ride perfectly along their respective lanes

An equilateral triangle with side length $6$ has a square of side length $6$ attached to each of its edges as shown. The distance between the two farthest vertices of this figure (marked $A$ and $B$ in the figure) can be written as $m + \sqrt{n}$ where $m$ and $n$ are positive integers. Find $m + n$. [asy] draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle); draw((1,0)--(1+sqrt(3)/2,1/2)--(1/2+sqrt(3)/2,1/2+sqrt(3)/2)--(1/2,sqrt(3)/2)); draw((0,0)--(-sqrt(3)/2,1/2)--(-sqrt(3)/2+1/2,1/2+sqrt(3)/2)--(1/2,sqrt(3)/2)); dot((-sqrt(3)/2+1/2,1/2+sqrt(3)/2)); label("A", (-sqrt(3)/2+1/2,1/2+sqrt(3)/2), N); draw((1,0)--(1,-1)--(0,-1)--(0,0)); dot((1,-1)); label("B", (1,-1), SE); [/asy]

Triangle $ ABC$ has side lengths $ AB \equal{} 5$, $ BC \equal{} 6$, and $ AC \equal{} 7$. Two bugs start simultaneously from $ A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $ D$. What is $ BD$? $ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

An equilateral triangle is given. A point lies on the incircle of this triangle. If the smallest two distances from the point to the sides of the triangle is $1$ and $4$, the sidelength of this equilateral triangle can be expressed as $\tfrac{a\sqrt b}c$ where $(a,c)=1$ and $b$ is not divisible by the square of an integer greater than $1$. Find $a+b+c$.

Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a+b\sqrt{c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime. What is $a^{2}+b^{2}+c^{2}$?

Two congruent $ 30^{\circ}$-$ 60^{\circ}$-$ 90^{\circ}$ are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area common to both triangles is $ \textbf{(A)}\ 6\sqrt3 \qquad \textbf{(B)}\ 8\sqrt3 \qquad \textbf{(C)}\ 9\sqrt3 \qquad \textbf{(D)}\ 12\sqrt3 \qquad \textbf{(E)}\ 24$

For a prime $q$, let $\Phi_q(x)=x^{q-1}+x^{q-2}+\cdots+x+1$. Find the sum of all primes $p$ such that $3 \le p \le 100$ and there exists an odd prime $q$ and a positive integer $N$ satisfying \[\dbinom{N}{\Phi_q(p)}\equiv \dbinom{2\Phi_q(p)}{N} \not \equiv 0 \pmod p. \][i]Proposed by Sammy Luo[/i]

For a convex hexagon $ ABCDEF$ with an area $ S$, prove that: \[ AC\cdot(BD\plus{}BF\minus{}DF)\plus{}CE\cdot(BD\plus{}DF\minus{}BF)\plus{}AE\cdot(BF\plus{}DF\minus{}BD)\geq 2\sqrt{3}S \]

Equilateral $\triangle ABC$ has side length $\sqrt{111}$. There are four distinct triangles $AD_1E_1$, $AD_1E_2$, $AD_2E_3$, and $AD_2E_4$, each congruent to $\triangle ABC$, with $BD_1 = BD_2=\sqrt{11}$. Find $\sum^4_{k=1}(CE_k)^2$.

Let $E$ be a point on side $[AD]$ of rhombus $ABCD$. Lines $AB$ and $CE$ meet at $F$, lines $BE$ and $DF$ meet at $G$. If $m(\widehat{DAB}) = 60^\circ $, what is$m(\widehat{DGB})$? $ \textbf{a)}\ 45^\circ \qquad\textbf{b)}\ 50^\circ \qquad\textbf{c)}\ 60^\circ \qquad\textbf{d)}\ 65^\circ \qquad\textbf{e)}\ 75^\circ $

$AB$ is the diameter of a circle centered at $O$. $C$ is a point on the circle such that angle $BOC$ is $60^\circ$. If the diameter of the circle is $5$ inches, the length of chord $AC$, expressed in inches, is: $\text{(A)} \ 3 \qquad \text{(B)} \ \frac{5\sqrt{2}}{2} \qquad \text{(C)} \frac{5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}$

The triangle $ABC$ has $CA = CB$. $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$). $D$ is the foot of the perpendicular from $C$ to $PB$. Show that $PA + PB = 2 \cdot PD$.

In $\triangle A B C$, let $D, E$, and $F$ be the midpoints of the sides of the triangle, and let $P, Q,$ and $R$ be the midpoints of the corresponding medians, $AD ,B E,$ and $C F$, respectively, as shown in the figure at the right. Prove that the value of \[\frac{AQ^2 + A R^2 + B P^2 + B R^2 + C P^2+ C Q^2 }{A B^2 + B C^2 + C A^2}\] does not depend on the shape of $\triangle A B C$ and find that value. [asy] defaultpen(linewidth(0.7)+fontsize(10));size(200); pair A=origin, B=(14,0), C=(9,12), D=midpoint(C--B), E=midpoint(C--A), F=midpoint(A--B), R=midpoint(C--F), P=midpoint(D--A), Q=midpoint(E--B); draw(A--B--C--A, linewidth(1)); draw(A--D^^B--E^^C--F); draw(B--R--A--Q--C--P--cycle, dashed); pair point=centroid(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$P$", P, dir(40)*dir(point--P)); label("$Q$", Q, dir(40)*dir(point--Q)); label("$R$", R, dir(40)*dir(point--R)); dot(P^^Q^^R);[/asy]

In circle $\theta_1$ with radius $1$, circles $\phi_1, \phi_2, \dots, \phi_8$, with equal radii, are drawn such that for $1 \le i \le 8$, $\phi_i$ is tangent to $\omega_1$, $\phi_{i-1}$, and $\phi_{i+1}$, where $\phi_0 = \phi_8$ and $\phi_1 = \phi_9$. There exists a circle $\omega_2$ such that $\omega_1 \neq \omega_2$ and $\omega_2$ is tangent to $\phi_i$ for $1 \le i \le 8$. The radius of $\omega_2$ can be expressed in the form $a - b\sqrt{c} -d\sqrt{e - \sqrt{f}} + g \sqrt{h - j \sqrt{k}}$ such that $a, b, \dots, k$ are positive integers and the numbers $e, f, k, \gcd(h, j)$ are squarefree. What is $a+b+c+d+e+f+g+h+j+k$. [i]Proposed by Eugene Chen [/i]

Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles.

Circles $ C_1,C_2$ intersect at $ P$. A line $ \Delta$ is drawn arbitrarily from $ P$ and intersects with $ C_1,C_2$ at $ B,C$. What is locus of $ A$ such that the median of $ AM$ of triangle $ ABC$ has fixed length $ k$.

In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ$. Find the cosine of $\angle BHD$. [asy] size(200); import three;defaultpen(linewidth(0.7)+fontsize(10)); currentprojection=orthographic(1/3+1/10,1-1/10,1/3); real r=sqrt(3); triple A=(0,0,r), B=(0,r,r), C=(1,r,r), D=(1,0,r), E=O, F=(0,r,0), G=(1,0,0), H=(1,r,0); draw(D--G--H--D--A--B--C--D--B--F--H--B^^C--H); draw(A--E^^G--E^^F--E, linetype("4 4")); label("$A$", A, N); label("$B$", B, dir(0)); label("$C$", C, N); label("$D$", D, W); label("$E$", E, NW); label("$F$", F, S); label("$G$", G, W); label("$H$", H, S); triple H45=(1,r-0.15,0.1), H60=(1-0.05, r, 0.07); label("$45^\circ$", H45, dir(125), fontsize(8)); label("$60^\circ$", H60, dir(25), fontsize(8));[/asy] $\textbf {(A) } \frac{\sqrt{3}}{6} \qquad \textbf {(B) } \frac{\sqrt{2}}{6} \qquad \textbf {(C) } \frac{\sqrt{6}}{3} \qquad \textbf {(D) } \frac{\sqrt{6}}{4} \qquad \textbf {(E) } \frac{\sqrt{6}-\sqrt{2}}{4}$