Let $ABC$ and $PQR$ be two triangles such that [list] [b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$. [b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$ [/list] Prove that $AB+AC=PQ+PR$.

Let $\Gamma_1$ and $\Gamma_2$ be circles in the plane with centers $O_1$ and $O_2$ and radii $13$ and $10$, respectively. Assume $O_1O_2=2$. Fix a circle $\Omega$ with radius $2$, internally tangent to $\Gamma_1$ at $P$ and externally tangent to $\Gamma_2$ at $Q$ . Let $\omega$ be a second variable circle internally tangent to $\Gamma_1$ at $X$ and externally tangent to $\Gamma_2$ at $Y$. Line $PQ$ meets $\Gamma_2$ again at $R$, line $XY$ meets $\Gamma_2$ again at $Z$, and lines $PZ$ and $XR$ meet at $M$. As $\omega$ varies, the locus of point $M$ encloses a region of area $\tfrac{p}{q} \pi$, where $p$ and $q$ are relatively prime positive integers. Compute $p+q$. [i]Proposed by Michael Kural[/i]

Triangle $ABC$ is equilateral with $AB=1$. Points $E$ and $G$ are on $\overline{AC}$ and points $D$ and $F$ are on $\overline{AB}$ such that both $\overline{DE}$ and $\overline{FG}$ are parallel to $\overline{BC}$. Furthermore, triangle $ADE$ and trapezoids $DFGE$ and $FBCG$ all have the same perimeter. What is $DE+FG$? [asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real s=1/2,m=5/6,l=1; pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m; draw(A--B--C--cycle^^D--E^^F--G); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NW); label("$F$",F,S); label("$G$",G,NW); [/asy] $\textbf{(A) }1\qquad \textbf{(B) }\dfrac{3}{2}\qquad \textbf{(C) }\dfrac{21}{13}\qquad \textbf{(D) }\dfrac{13}{8}\qquad \textbf{(E) }\dfrac{5}{3}\qquad$

Circles $\omega_1$ and $\omega_2$ meet at $P$ and $Q$. Segments $AC$ and $BD$ are chords of $\omega_1$ and $\omega_2$ respectively, such that segment $AB$ and ray $CD$ meet at $P$. Ray $BD$ and segment $AC$ meet at $X$. Point $Y$ lies on $\omega_1$ such that $P Y \parallel BD$. Point $Z$ lies on $\omega_2$ such that $P Z \parallel AC$. Prove that points $Q,~ X,~ Y,~ Z$ are collinear.

The dissection of the $3-4-5$ triangle shown below has diameter $5/2$. [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(23cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 1.42, xmax = 24.42, ymin = 3.8, ymax = 15.54; /* image dimensions */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax,defaultpen+black, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); yaxis(ymin, ymax,defaultpen+black, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((9.44,8.52)--(12.44,8.52)); draw((9.44,12.52)--(9.44,8.52)); draw((9.44,12.52)--(12.44,8.52)); draw((9.44,10.52)--(10.94,10.52)); draw((10.94,10.52)--(10.94,8.52)); draw((9.44,8.52)--(10.94,10.52)); /* dots and labels */ dot((9.44,8.52),dotstyle); label("$A$", (9.52,8.64), NE * labelscalefactor); dot((12.44,8.52),dotstyle); label("$B$", (12.52,8.64), NE * labelscalefactor); dot((9.44,12.52),dotstyle); label("$C$", (9.52,12.64), NE * labelscalefactor); dot((10.94,8.52),dotstyle); label("$D$", (11.02,8.64), NE * labelscalefactor); dot((9.44,10.52),dotstyle); label("$E$", (9.52,10.64), NE * labelscalefactor); dot((10.94,10.52),dotstyle); label("$F$", (11.02,10.64), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy] Find the least diameter of this triangle into four parts. (The diameter of a dissection is the least upper bound of the distances between pairs of points belonging to the same part.)

In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$. [i]Proposed by Ray Li[/i]

Let $ ABC $ be an acute triangle. Denote by $ D $ the foot of the perpendicular line drawn from the point $ A $ to the side $ BC $, by $M$ the midpoint of $ BC $, and by $ H $ the orthocenter of $ ABC $. Let $ E $ be the point of intersection of the circumcircle $ \Gamma $ of the triangle $ ABC $ and the half line $ MH $, and $ F $ be the point of intersection (other than $E$) of the line $ ED $ and the circle $ \Gamma $. Prove that $ \tfrac{BF}{CF} = \tfrac{AB}{AC} $ must hold. (Here we denote $XY$ the length of the line segment $XY$.)

Let $M$, $N$ be the midpoints of diagonals $AC$, $BD$ of a right-angled trapezoid $ABCD$ ($\measuredangle A=\measuredangle D = 90^\circ$). The circumcircles of triangles $ABN$, $CDM$ meet the line $BC$ in the points $Q$, $R$. Prove that the distances from $Q$, $R$ to the midpoint of $MN$ are equal.

Let $ABC$ be a triangle with $\angle ABC $ as the largest angle. Let $R$ be its circumcenter. Let the circumcircle of triangle $ARB$ cut $AC$ again at $X$. Prove that $RX$ is perpendicular to $BC$.

Let $ABC$ be an acute triangle, and let $X$ be a variable interior point on the minor arc $BC$ of its circumcircle. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively. Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$. Let $\ell$ be the line through $P$ parallel to $XR$. Prove that as $X$ varies along minor arc $BC$, the line $\ell$ always passes through a fixed point. (Specifically: prove that there is a point $F$, determined by triangle $ABC$, such that no matter where $X$ is on arc $BC$, line $\ell$ passes through $F$.) [i]Robert Simson et al.[/i]

$ABCD$ is a cyclic quadrilateral inscribed in the circle $\omega$. Let $AB \cap CD = E$, $AD \cap BC = F$. Let $\omega_1, \omega_2$ be the circumcircles of $AEF, CEF$, respectively. Let $\omega \cap \omega_1 = G$, $\omega \cap \omega_2 = H$. Show that $AC, BD, GH$ are concurrent. [i]Proposed by Yang Liu[/i]

A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$. How many edges does the remaining solid have? $\textbf{(A) }36\qquad \textbf{(B) }60\qquad \textbf{(C) }72\qquad \textbf{(D) }84\qquad \textbf{(E) }108\qquad$

The letter F shown below is rotated $90^\circ$ clockwise around the origin, then reflected in the $y$-axis, and then rotated a half turn around the origin. What is the final image? [asy] import cse5;pathpen=black;pointpen=black; size(2cm); D((0,-2)--MP("y",(0,7),N)); D((-3,0)--MP("x",(5,0),E)); D((1,0)--(1,2)--(2,2)--(2,3)--(1,3)--(1,4)--(3,4)--(3,5)--(0,5)); [/asy][asy] import cse5;pathpen=black;pointpen=black; unitsize(0.2cm); D((0,-2)--MP("y",(0,7),N)); D(MP("\textbf{(A) }",(-3,0),W)--MP("x",(5,0),E)); D((1,0)--(1,2)--(2,2)--(2,3)--(1,3)--(1,4)--(3,4)--(3,5)--(0,5)); // D((18,-2)--MP("y",(18,7),N)); D(MP("\textbf{(B) }",(13,0),W)--MP("x",(21,0),E)); D((17,0)--(17,2)--(16,2)--(16,3)--(17,3)--(17,4)--(15,4)--(15,5)--(18,5)); // D((36,-2)--MP("y",(36,7),N)); D(MP("\textbf{(C) }",(29,0),W)--MP("x",(38,0),E)); D((31,0)--(31,1)--(33,1)--(33,2)--(34,2)--(34,1)--(35,1)--(35,3)--(36,3)); // D((0,-17)--MP("y",(0,-8),N)); D(MP("\textbf{(D) }",(-3,-15),W)--MP("x",(5,-15),E)); D((3,-15)--(3,-14)--(1,-14)--(1,-13)--(2,-13)--(2,-12)--(1,-12)--(1,-10)--(0,-10)); // D((15,-17)--MP("y",(15,-8),N)); D(MP("\textbf{(E) }",(13,-15),W)--MP("x",(22,-15),E)); D((15,-14)--(17,-14)--(17,-13)--(18,-13)--(18,-14)--(19,-14)--(19,-12)--(20,-12)--(20,-15)); [/asy]

The incircle of triangle $ABC$ touches the side $AB$ at point $C'$; the incircle of triangle $ACC'$ touches the sides $AB$ and $AC$ at points $C_1, B_1$; the incircle of triangle $BCC'$ touches the sides $AB$ and $BC$ at points $C_2$, $A_2$. Prove that the lines $B_1C_1$, $A_2C_2$, and $CC'$ concur.

A square piece of paper has sides of length $ 100$. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at distance $ \sqrt {17}$ from the corner, and they meet on the diagonal at an angle of $ 60^\circ$ (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form $ \sqrt [n]{m}$, where $ m$ and $ n$ are positive integers, $ m < 1000$, and $ m$ is not divisible by the $ n$th power of any prime. Find $ m \plus{} n$. [asy]import math; unitsize(5mm); defaultpen(fontsize(9pt)+Helvetica()+linewidth(0.7)); pair O=(0,0); pair A=(0,sqrt(17)); pair B=(sqrt(17),0); pair C=shift(sqrt(17),0)*(sqrt(34)*dir(75)); pair D=(xpart(C),8); pair E=(8,ypart(C)); draw(O--(0,8)); draw(O--(8,0)); draw(O--C); draw(A--C--B); draw(D--C--E); label("$\sqrt{17}$",(0,2),W); label("$\sqrt{17}$",(2,0),S); label("cut",midpoint(A--C),NNW); label("cut",midpoint(B--C),ESE); label("fold",midpoint(C--D),W); label("fold",midpoint(C--E),S); label("$30^\circ$",shift(-0.6,-0.6)*C,WSW); label("$30^\circ$",shift(-1.2,-1.2)*C,SSE);[/asy]