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]

Consider a triangle $ABC$ with $AB=AC$, and $D$ the foot of the altitude from the vertex $A$. The point $E$ lies on the side $AB$ such that $\angle ACE= \angle ECB=18^{\circ}$. If $AD=3$, find the length of the segment $CE$.

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] $\displaystyle \text{(A)} \ 2 \qquad \text{(B)} \ \frac{5}{2} \qquad \text{(C)} \ \frac{5}{2} - \sin \theta \qquad \text{(D)} \ \frac{5}{2} - \frac{1}{2} \sin \theta \qquad \text{(E)} \ \frac{5}{2} - \frac{1}{2} \sin \left(\frac{1}{2} \theta\right)$

If, in a triangle of sides $a, b, c$, the incircle has radius $\frac{b+c-a}{2}$, what is the magnitude of $\angle A$?

Let ABCD be a convex quadrilateral such that $AB + CD = \sqrt{2}AC$ and $BC + DA = \sqrt{2}BD$. Prove that ABCD is a parallelogram.

Let $ABC$ be a triangle with $AB = 42$, $AC = 39$, $BC = 45$. Let $E$, $F$ be on the sides $\overline{AC}$ and $\overline{AB}$ such that $AF = 21, AE = 13$. Let $\overline{CF}$ and $\overline{BE}$ intersect at $P$, and let ray $AP$ meet $\overline{BC}$ at $D$. Let $O$ denote the circumcenter of $\triangle DEF$, and $R$ its circumradius. Compute $CO^2-R^2$. [i]Proposed by Yang Liu[/i]

A convex polygon $\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.

Let $ ABCD$ be a cyclic quadrilateral. The side lengths of $ ABCD$ are distinct integers less than $ 15$ such that $ BC\cdot CD\equal{}AB\cdot DA$. What is the largest possible value of $ BD$? $ \textbf{(A)}\ \sqrt{\frac{325}{2}} \qquad \textbf{(B)}\ \sqrt{185} \qquad \textbf{(C)}\ \sqrt{\frac{389}{2}} \qquad \textbf{(D)}\ \sqrt{\frac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\frac{533}{2}}$

Let $ABC$ be an equilateral triangle and let $P$ be a point in its interior. Let the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ at the points $A_1$, $B_1$, $C_1$, respectively. Prove that $A_1B_1 \cdot B_1C_1 \cdot C_1A_1 \ge A_1B \cdot B_1C \cdot C_1A$.

Point $D$ is on side $CB$ of triangle $ABC$. If \[ \angle{CAD} = \angle{DAB} = 60^\circ,\quad AC = 3\quad\mbox{ and }\quad AB = 6, \] then the length of $AD$ is $\text{(A)} \ 2 \qquad \text{(B)} \ 2.5 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 3.5 \qquad \text{(E)} \ 4$

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]

The sides of a triangle are $ 30$, $ 70$, and $ 80$ units. If an altitude is dropped upon the side of length $ 80$, the larger segment cut off on this side is: $ \textbf{(A)}\ 62\qquad \textbf{(B)}\ 63\qquad \textbf{(C)}\ 64\qquad \textbf{(D)}\ 65\qquad \textbf{(E)}\ 66$

Convex quadrilateral $ABCD$ has sides $AB=BC=7$, $CD=5$, and $AD=3$. Given additionally that $m\angle ABC=60^\circ$, find $BD$.

Let $X$ be a point inside convex quadrilateral $ABCD$ with $\angle AXB+\angle CXD=180^{\circ}$. If $AX=14$, $BX=11$, $CX=5$, $DX=10$, and $AB=CD$, find the sum of the areas of $\triangle AXB$ and $\triangle CXD$. [i]Proposed by Michael Kural[/i]

Let $\Delta ABC$ be a triangle. The internal bisectors of angles $\angle CAB$ and $\angle ABC$ intersect segments $BC$, respectively $AC$ in $D$, respectively $E$. Prove that \[DE\leq (3-2\sqrt{2})(AB+BC+CA).\]

Equilateral triangle $ABC$ has been creased and folded so that vertex $A$ now rests at $A'$ on $\overline{BC}$ as shown. If $BA' = 1$ and $A'C = 2$ then the length of crease $\overline{PQ}$ is [asy] size(170); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=(1.5,3*sqrt(3)/2), C=(3,0), D=(1,0), P=B+1.6*dir(B--A), Q=C+1.2*dir(C--A); draw(B--P--D--B^^P--Q--D--C--Q); draw(Q--A--P, linetype("4 4")); label("$A$", A, N); label("$B$", B, W); label("$C$", C, E); label("$A'$", D, S); label("$P$", P, W); label("$Q$", Q, E); [/asy] $ \textbf{(A)}\ \frac{8}{5}\qquad\textbf{(B)}\ \frac{7}{20}\sqrt{21}\qquad\textbf{(C)}\ \frac{1+\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{13}{8}\qquad\textbf{(E)}\ \sqrt{3} $

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 $BC$.

$[BD]$ is a median of $\triangle ABC$. $m(\widehat{ABD})=90^\circ$, $|AB|=2$, and $|AC|=6$. $|BC|=?$ $ \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 3\sqrt2 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 4\sqrt2 \qquad\textbf{(E)}\ 2\sqrt6 $

Let $ ABC$ be a triangle, and $ D$ the midpoint of the side $ BC$. On the sides $ AB$ and $ AC$ we consider the points $ M$ and $ N$, respectively, both different from the midpoints of the sides, such that \[ AM^2\plus{}AN^2 \equal{}BM^2 \plus{} CN^2 \textrm{ and } \angle MDN \equal{} \angle BAC.\] Prove that $ \angle BAC \equal{} 90^\circ$.

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]

Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon. A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a+b\sqrt{c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c.$

Points $A$ and $B$ lie on circle $\omega$. Point $P$ lies on the extension of segment $AB$ past $B$. Line $\ell$ passes through $P$ and is tangent to $\omega$. The tangents to $\omega$ at points $A$ and $B$ intersect $\ell$ at points $D$ and $C$ respectively. Given that $AB=7$, $BC=2$, and $AD=3$, compute $BP$.

Let $a$, $b$, $c$ be the three sides of a triangle, and let $\alpha$, $\beta$, $\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find \[ \frac{\cot \gamma}{\cot \alpha+\cot \beta}. \]

A circle of radius $r$ has chords $\overline{AB}$ of length $10$ and $\overline{CD}$ of length $7$. When $\overline{AB}$ and $\overline{CD}$ are extended through $B$ and $C$, respectively, they intersect at $P$, which is outside the circle. If $\angle APD = 60^{\circ}$ and $BP = 8$, then $r^{2} =$ $ \textbf{(A)}\ 70\qquad\textbf{(B)}\ 71\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 73\qquad\textbf{(E)}\ 74 $

The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon? [asy] size(130); pair A, B, C, D, E, F, G, H, I, J, K, L; A = dir(120); B = dir(60); C = dir(0); D = dir(-60); E = dir(-120); F = dir(180); draw(A--B--C--D--E--F--cycle); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); G = midpoint(A--B); H = midpoint(B--C); I = midpoint(C--D); J = midpoint(D--E); K = midpoint(E--F); L = midpoint(F--A); draw(G--H--I--J--K--L--cycle); label("$A$", A, dir(120)); label("$B$", B, dir(60)); label("$C$", C, dir(0)); label("$D$", D, dir(-60)); label("$E$", E, dir(-120)); label("$F$", F, dir(180)); [/asy] $\textbf{(A)}\ \displaystyle \frac{1}{2} \qquad \textbf{(B)}\ \displaystyle \frac{\sqrt 3}{3} \qquad \textbf{(C)}\ \displaystyle \frac{2}{3} \qquad \textbf{(D)}\ \displaystyle \frac{3}{4} \qquad \textbf{(E)}\ \displaystyle \frac{\sqrt 3}{2}$