The figure $ABCD$ is bounded by a semicircle $CDA$ and a quarter circle $ABC$. Given that the distance from $A$ to $C$ is $18$, find the area of the figure. [asy] size(200); defaultpen(linewidth(0.8)); pair A=(-9,0),B=(0,9*sqrt(2)-9),C=(9,0),D=(0,9); dot(A^^B^^C^^D); draw(arc(origin,9,0,180)^^arc((0,-9),9*sqrt(2),45,135)); label("$A$",A,S); label("$B$",B,N); label("$C$",C,S); label("$D$",D,N); [/asy]

Consider a sequence $x_1,x_2,\cdots x_{12}$ of real numbers such that $x_1=1$ and for $n=1,2,\dots,10$ let \[ x_{n+2}=\frac{(x_{n+1}+1)(x_{n+1}-1)}{x_n}. \] Suppose $x_n>0$ for $n=1,2,\dots,11$ and $x_{12}=0$. Then the value of $x_2$ can be written as $\frac{\sqrt{a}+\sqrt{b}}{c}$ for positive integers $a,b,c$ with $a>b$ and no square dividing $a$ or $b$. Find $100a+10b+c$. [i]Proposed by Michael Kural[/i]

The incircle of triangle $ABC$ touches sides $AB$;$BC$;$CA$ at $M$;$N$;$K$, respectively. The line through $A$ parallel to $NK$ meets $MN$ at $D$. The line through $A$ parallel to $MN$ meets $NK$ at $E$. Show that the line $DE$ bisects sides $AB$ and $AC$ of triangle $ABC$. [i]M. Sonkin[/i]

Let $a,b,c$ be the answers to problems $4$, $5$, and $6$, respectively. In $\triangle ABC$, the measures of $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, $c$ in degrees, respectively. Let $D$ and $E$ be points on segments $AB$ and $AC$ with $\frac{AD}{BD} = \frac{AE}{CE} = 2013$. A point $P$ is selected in the interior of $\triangle ADE$, with barycentric coordinates $(x,y,z)$ with respect to $\triangle ABC$ (here $x+y+z=1$). Lines $BP$ and $CP$ meet line $DE$ at $B_1$ and $C_1$, respectively. Suppose that the radical axis of the circumcircles of $\triangle PDC_1$ and $\triangle PEB_1$ pass through point $A$. Find $100x$. [i]Proposed by Evan Chen[/i]

Let $\omega_1$ and $\omega_2$ be two orthogonal circles, and let the center of $\omega_1$ be $O$. Diameter $AB$ of $\omega_1$ is selected so that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$, passing through $O$ and $A$, touch $\omega_2$ at $F$ and $G$. Prove that $FGOB$ is cyclic. [i]Proposed by Eric Chen[/i]

Point $F$ is taken in side $AD$ of square $ABCD$. At $C$ a perpendicular is drawn to $CF$, meeting $AB$ extended at $E$. The area of $ABCD$ is $256$ square inches and the area of triangle $CEF$ is $200$ square inches. Then the number of inches in $BE$ is: [asy] size(6cm); pair A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1), E = (1.3, 0), F = (0, 0.7); draw(A--B--C--D--cycle); draw(F--C--E--B); label("$A$", A, SW); label("$B$", B, S); label("$C$", C, N); label("$D$", D, NW); label("$E$", E, SE); label("$F$", F, W); //Credit to MSTang for the asymptote [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$

A rectangular board of 8 columns has squared numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result? [asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy] $\text{(A)}\ 36 \qquad \text{(B)}\ 64 \qquad \text{(C)}\ 78 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 120$

Diagonals $AC$ and $BD$ of a trapezoid $ABCD$ meet at $P$. The circumcircles of triangles $ABP$ and $CDP$ intersect the line $AD$ for the second time at points $X$ and $Y$ respectively. Let $M$ be the midpoint of segment $XY$. Prove that $BM = CM$.

In a plane we are given two distinct points $A,B$ and two lines $a, b$ passing through $B$ and $A$ respectively $(a \ni B, b \ni A)$ such that the line $AB$ is equally inclined to a and b. Find the locus of points $M$ in the plane such that the product of distances from $M$ to $A$ and a equals the product of distances from $M$ to $B$ and $b$ (i.e., $MA \cdot MA' = MB \cdot MB'$, where $A'$ and $B'$ are the feet of the perpendiculars from $M$ to $a$ and $b$ respectively).

As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$. [asy] size(200); pair A=origin, B=(14,0), C=(9,12), D=foot(A, B,C), E=foot(B, A, C), F=foot(C, A, B), H=orthocenter(A, B, C); draw(F--C--A--B--C^^A--D^^B--E); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("84", centroid(H, C, E), fontsize(9.5)); label("35", centroid(H, B, D), fontsize(9.5)); label("30", centroid(H, F, B), fontsize(9.5)); label("40", centroid(H, A, F), fontsize(9.5));[/asy]

Let $\omega_1$ and $\omega_2$ be two orthogonal circles, and let the center of $\omega_1$ be $O$. Diameter $AB$ of $\omega_1$ is selected so that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$, passing through $O$ and $A$, touch $\omega_2$ at $F$ and $G$. Prove that $FGOB$ is cyclic. [i]Proposed by Eric Chen[/i]

As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$. Point $F$ lies on $\overline{AD}$ so that $DE=DF$, and $ABCD$ is a square. What is the degree measure of $\angle AFE?$ [asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy] $\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$

Jack wants to bike from his house to Jill's house, which is located three blocks east and two blocks north of Jack's house. After biking each block, Jack can continue either east or north, but he needs to avoid a dangerous intersection one block east and one block north of his house. In how many ways can he reach Jill's house by biking a total of five blocks? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad \textbf{(E) }10$

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]

Find $\frac{area(CDF)}{area(CEF)}$ in the figure. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(5.75cm); 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 = -2, xmax = 21, ymin = -2, ymax = 16; /* image dimensions */ /* draw figures */ draw((0,0)--(20,0)); draw((13.48,14.62)--(7,0)); draw((0,0)--(15.93,9.12)); draw((13.48,14.62)--(20,0)); draw((13.48,14.62)--(0,0)); label("6",(15.16,12.72),SE*labelscalefactor); label("10",(18.56,5.1),SE*labelscalefactor); label("7",(3.26,-0.6),SE*labelscalefactor); label("13",(13.18,-0.71),SE*labelscalefactor); label("20",(5.07,8.33),SE*labelscalefactor); /* dots and labels */ dot((0,0),dotstyle); label("$B$", (-1.23,-1.48), NE * labelscalefactor); dot((20,0),dotstyle); label("$C$", (19.71,-1.59), NE * labelscalefactor); dot((7,0),dotstyle); label("$D$", (6.77,-1.64), NE * labelscalefactor); dot((13.48,14.62),dotstyle); label("$A$", (12.36,14.91), NE * labelscalefactor); dot((15.93,9.12),dotstyle); label("$E$", (16.42,9.21), NE * labelscalefactor); dot((9.38,5.37),dotstyle); label("$F$", (9.68,4.5), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$. [asy] size(200); pair A=origin, B=(14,0), C=(9,12), D=foot(A, B,C), E=foot(B, A, C), F=foot(C, A, B), H=orthocenter(A, B, C); draw(F--C--A--B--C^^A--D^^B--E); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("84", centroid(H, C, E), fontsize(9.5)); label("35", centroid(H, B, D), fontsize(9.5)); label("30", centroid(H, F, B), fontsize(9.5)); label("40", centroid(H, A, F), fontsize(9.5));[/asy]

The shaded region below is called a shark's fin falcata, a figure studied by Leonardo da Vinci. It is bounded by the portion of the circle of radius $3$ and center $(0,0)$ that lies in the first quadrant, the portion of the circle with radius $\tfrac{3}{2}$ and center $(0,\tfrac{3}{2})$ that lies in the first quadrant, and the line segment from $(0,0)$ to $(3,0)$. What is the area of the shark's fin falcata? [asy] import cse5;pathpen=black;pointpen=black; size(1.5inch); D(MP("x",(3.5,0),S)--(0,0)--MP("\frac{3}{2}",(0,3/2),W)--MP("y",(0,3.5),W)); path P=(0,0)--MP("3",(3,0),S)..(3*dir(45))..MP("3",(0,3),W)--(0,3)..(3/2,3/2)..cycle; draw(P,linewidth(2)); fill(P,gray); [/asy] $\textbf{(A) } \dfrac{4\pi}{5} \qquad\textbf{(B) } \dfrac{9\pi}{8} \qquad\textbf{(C) } \dfrac{4\pi}{3} \qquad\textbf{(D) } \dfrac{7\pi}{5} \qquad\textbf{(E) } \dfrac{3\pi}{2} $

A non-Hookian spring has force $F = -kx^2$ where $k$ is the spring constant and $x$ is the displacement from its unstretched position. For the system shown of a mass $m$ connected to an unstretched spring initially at rest, how far does the spring extend before the system momentarily comes to rest? Assume that all surfaces are frictionless and that the pulley is frictionless as well. [asy] size(250); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((0,0)--(0,-1)--(2,-1)--(2+sqrt(3),-2)); draw((2.5,-2)--(4.5,-2),dashed); draw(circle((2.2,-0.8),0.2)); draw((2.2,-0.8)--(1.8,-1.2)); draw((0,-0.6)--(0.6,-0.6)--(0.75,-0.4)--(0.9,-0.8)--(1.05,-0.4)--(1.2,-0.8)--(1.35,-0.4)--(1.5,-0.8)--(1.65,-0.4)--(1.8,-0.8)--(1.95,-0.6)--(2.2,-0.6)); draw((2+0.3*sqrt(3),-1.3)--(2+0.3*sqrt(3)+0.6/2,-1.3+sqrt(3)*0.6/2)--(2+0.3*sqrt(3)+0.6/2+0.2*sqrt(3),-1.3+sqrt(3)*0.6/2-0.2)--(2+0.3*sqrt(3)+0.2*sqrt(3),-1.3-0.2)); //super complex Asymptote code gg draw((2+0.3*sqrt(3)+0.3/2,-1.3+sqrt(3)*0.3/2)--(2.35,-0.6677)); draw(anglemark((2,-1),(2+sqrt(3),-2),(2.5,-2))); label("$30^\circ$",(3.5,-2),NW); [/asy] $ \textbf{(A)}\ \left(\frac{3mg}{2k}\right)^{1/2} $ $ \textbf{(B)}\ \left(\frac{mg}{k}\right)^{1/2} $ $ \textbf{(C)}\ \left(\frac{2mg}{k}\right)^{1/2} $ $ \textbf{(D)}\ \left(\frac{\sqrt{3}mg}{k}\right)^{1/3} $ $ \textbf{(E)}\ \left(\frac{3\sqrt{3}mg}{2k}\right)^{1/3} $

Let $ABC$ be a triangle such that $AB < AC$. Let $P$ lie on a line through $A$ parallel to line $BC$ such that $C$ and $P$ are on the same side of line $AB$. Let $M$ be the midpoint of segment $BC$. Define $D$ on segment $BC$ such that $\angle BAD = \angle CAM$, and define $T$ on the extension of ray $CB$ beyond $B$ so that $\angle BAT = \angle CAP$. Given that lines $PC$ and $AD$ intersect at $Q$, that lines $PD$ and $AB$ intersect at $R$, and that $S$ is the midpoint of segment $DT$, prove that if $A$,$P$,$Q$, and $R$ lie on a circle, then $Q$, $R$, and $S$ are collinear. [i]David Rush[/i]

[asy] /*Using regular asymptote, this diagram would take 30 min to make. Using cse5, this takes 5 minutes. Conclusion? CSE5 IS THE BEST PACKAGE EVER CREATED!!!!*/ size(100); import cse5; pathpen=black; anglefontpen=black; pointpen=black; anglepen=black; dotfactor=3; pair A=(0,0),B=(0.5,0.5*sqrt(3)),C=(3,0),D=(1.7,0),EE; EE=(B+C)/2; D(MP("$A$",A,W)--MP("$B$",B,N)--MP("$C$",C,E)--cycle); D(MP("$E$",EE,N)--MP("$D$",D,S)); D(D);D(EE); MA("80^\circ",8,D,EE,C,0.1); MA("20^\circ",8,EE,C,D,0.3,2,shift(1,3)*C); draw(arc(shift(-0.1,0.05)*C,0.25,100,180),arrow =ArcArrow()); MA("100^\circ",8,A,B,C,0.1,0); MA("60^\circ",8,C,A,B,0.1,0); //Credit to TheMaskedMagician for the diagram [/asy] In $\triangle ABC$, $E$ is the midpoint of side $BC$ and $D$ is on side $AC$. If the length of $AC$ is $1$ and $\measuredangle BAC = 60^\circ$, $\measuredangle ABC = 100^\circ$, $\measuredangle ACB = 20^\circ$ and $\measuredangle DEC = 80^\circ$, then the area of $\triangle ABC$ plus twice the area of $\triangle CDE$ equals $\textbf{(A) }\frac{1}{4}\cos 10^\circ\qquad\textbf{(B) }\frac{\sqrt{3}}{8}\qquad\textbf{(C) }\frac{1}{4}\cos 40^\circ\qquad\textbf{(D) }\frac{1}{4}\cos 50^\circ\qquad\textbf{(E) }\frac{1}{8}$

In a triangle $ABC$, let $I$ denote its incenter. Points $D, E, F$ are chosen on the segments $BC, CA, AB$, respectively, such that $BD + BF = AC$ and $CD + CE = AB$. The circumcircles of triangles $AEF, BFD, CDE$ intersect lines $AI, BI, CI$, respectively, at points $K, L, M$ (different from $A, B, C$), respectively. Prove that $K, L, M, I$ are concyclic.

Let $ABC$ be a triangle with symmedian point $K$. Select a point $A_1$ on line $BC$ such that the lines $AB$, $AC$, $A_1K$ and $BC$ are the sides of a cyclic quadrilateral. Define $B_1$ and $C_1$ similarly. Prove that $A_1$, $B_1$, and $C_1$ are collinear. [i]Proposed by Sammy Luo[/i]

Here’s a screenshot of the problem. If someone could LaTEX a diagram, that would be great!

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 $XY$.)

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]