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$

A ship sails $ 10$ miles in a straight line from $ A$ to $ B$, turns through an angle between $ 45^{\circ}$ and $ 60^{\circ}$, and then sails another $ 20$ miles to $ C$. Let $ AC$ be measured in miles. Which of the following intervals contains $ AC^2$? [asy]unitsize(2mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C=20*dir(50); draw(A--B--C); draw(A--C,linetype("4 4")); dot(A); dot(B); dot(C); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C),SE); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE);[/asy]$ \textbf{(A)}\ [400,500] \qquad \textbf{(B)}\ [500,600] \qquad \textbf{(C)}\ [600,700] \qquad \textbf{(D)}\ [700,800]$ $ \textbf{(E)}\ [800,900]$

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 $PA + PB = 2 \cdot PD$.

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]

Let $\triangle ABC$ be an arbitary triangle, and construct $P,Q,R$ so that each of the angles marked is $30^\circ$. Prove that $\triangle PQR$ is an equilateral triangle. [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair ext30(pair pt1, pair pt2) { pair r1 = pt1+rotate(-30)*(pt2-pt1), r2 = pt2+rotate(30)*(pt1-pt2); draw(anglemark(r1,pt1,pt2,25)); draw(anglemark(pt1,pt2,r2,25)); return intersectionpoints(pt1--r1, pt2--r2)[0]; } pair A = (0,0), B=(10,0), C=(3,7), P=ext30(B,C), Q=ext30(C,A), R=ext30(A,B); draw(A--B--C--A--R--B--P--C--Q--A); draw(P--Q--R--cycle, linetype("8 8")); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$P$", P, NE); label("$Q$", Q, NW); label("$R$", R, S);[/asy]

Let $ AB$ be the diameter of semicircle $O$ , $C, D $ be points on the arc $AB$, $P, Q$ be respectively the circumcenter of $\triangle OAC $ and $\triangle OBD $ . Prove that:$CP\cdot CQ=DP \cdot DQ$.[asy] import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black; real h=sqrt(55/64); pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B); D(arc(O,1,0,180),darkgreen); D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue); D(O); [/asy]

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

In the adjoining figure triangle $ ABC$ is such that $ AB \equal{} 4$ and $ AC \equal{} 8$. If $ M$ is the midpoint of $ BC$ and $ AM \equal{} 3$, what is the length of $ BC$? $ \textbf{(A)}\ 2\sqrt{26} \qquad \textbf{(B)}\ 2\sqrt{31} \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 4\plus{}2\sqrt{13} \qquad$ $ \textbf{(E)}\ \text{not enough information given to solve the problem}$ [asy]draw((0,0)--(2.8284,2)--(8,0)--cycle); draw((2.8284,2)--(4,0)); label("A",(2.8284,2),N); label("B",(0,0),S); label("C",(8,0),S); label("M",(4,0),S);[/asy]

The angles in a particular triangle are in arithmetic progression, and the side lengths are $4,5,x$. The sum of the possible values of $x$ equals $a+\sqrt{b}+\sqrt{c}$ where $a, b$, and $c$ are positive integers. What is $a+b+c$? $ \textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44$

Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?

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

In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$ and $\theta$ is the measure of $\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals [asy] size(230); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=14*dir(36), C=intersectionpoint(B--(9001,0), 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, S); label("$N$", N, dir(30)); label("$19$", (A+C)/2, dir(A--C)*dir(90)); label("$14$", (A+B)/2, dir(A--B)*dir(270)); [/asy] $\displaystyle \text{(A)} \ 2 \qquad \text{(B)} \ \frac{5}{2} \qquad \text{(C)} \ \frac{5}{2} - \sin \theta \qquad \text{(D)} \ \frac{5}{2} - \frac{1}{2} \sin \theta \qquad \text{(E)} \ \frac{5}{2} - \frac{1}{2} \sin \left(\frac{1}{2} \theta\right)$

Let $ABC$ be an acute triangle. Denote by $B_0$ and $C_0$ the feet of the altitudes from vertices $B$ and $C$, respectively. Let $X$ be a point inside the triangle $ABC$ such that the line $BX$ is tangent to the circumcircle of the triangle $AXC_0$ and the line $CX$ is tangent to the circumcircle of the triangle $AXB_0$. Show that the line $AX$ is perpendicular to $BC$.

Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \dfrac{1+\sqrt{3}}{\sqrt{2}} \qquad \textbf{(C)}\ \sqrt{3} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \dfrac{4}{\sqrt{3}}$

Point $G$ is where the medians of the triangle $ABC$ intersect and point $D$ is the midpoint of side $BC$. The triangle $BDG$ is equilateral with side length 1. Determine the lengths, $AB$, $BC$, and $CA$, of the sides of triangle $ABC$. [asy] size(200); defaultpen(fontsize(10)); real r=100.8933946; pair A=sqrt(7)*dir(r), B=origin, C=(2,0), D=midpoint(B--C), E=midpoint(A--C), F=midpoint(A--B), G=centroid(A,B,C); draw(A--B--C--A--D^^B--E^^C--F); pair point=G; 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("$G$", G, dir(20)); label("1", B--G, dir(150)); label("1", D--G, dir(30)); label("1", B--D, dir(270));[/asy]

Let $ABC$ be a triangle with circumcenter $O$, incenter $I$, and circumcircle $\Gamma$. It is known that $AB = 7$, $BC = 8$, $CA = 9$. Let $M$ denote the midpoint of major arc $\widehat{BAC}$ of $\Gamma$, and let $D$ denote the intersection of $\Gamma$ with the circumcircle of $\triangle IMO$ (other than $M$). Let $E$ denote the reflection of $D$ over line $IO$. Find the integer closest to $1000 \cdot \frac{BE}{CE}$. [i]Proposed by Evan Chen[/i]

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 $

To the nearest degree, find the measure of the largest angle in a triangle with side lengths $3$, $5$, and $7$.

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]

In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?

$AB$ 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: $\text{(A)} \ 3 \qquad \text{(B)} \ \frac{5\sqrt{2}}{2} \qquad \text{(C)} \frac{5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}$

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 $M$ be the intersection of the diagonals $AC$ and $BD$ of cyclic quadrilateral $ABCD$. If $|AB|=5$, $|CD|=3$, and $m(\widehat{AMB}) = 60^\circ$, what is the circumradius of the quadrilateral? $ \textbf{(A)}\ 5\sqrt 3 \qquad\textbf{(B)}\ \dfrac {7\sqrt 3}{3} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{34} $