Let $BD$ be a bisector of triangle $ABC$. Points $I_a$, $I_c$ are the incenters of triangles $ABD$, $CBD$ respectively. The line $I_aI_c$ meets $AC$ in point $Q$. Prove that $\angle DBQ = 90^\circ$.

Find the area of the region bounded by the graphs $y=x^2$, $y=x$, and $x=2$.

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]

$ABCD$ 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]

Triangle $ABC$ has $AC = 3$, $BC = 5$, $AB = 7$. A circle is drawn internally tangent to the circumcircle of $ABC$ at $C$, and tangent to $AB$. Let $D$ be its point of tangency with $AB$. Find $BD - DA$. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 2.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.5, xmax = 7.01, ymin = -3, ymax = 8.02; /* image dimensions */ /* draw figures */ draw(circle((1.37,2.54), 5.17)); draw((-2.62,-0.76)--(-3.53,4.2)); draw((-3.53,4.2)--(5.6,-0.44)); draw((5.6,-0.44)--(-2.62,-0.76)); draw(circle((-0.9,0.48), 2.12)); /* dots and labels */ dot((-2.62,-0.76),dotstyle); label("$C$", (-2.46,-0.51), SW * labelscalefactor); dot((-3.53,4.2),dotstyle); label("$A$", (-3.36,4.46), NW * labelscalefactor); dot((5.6,-0.44),dotstyle); label("$B$", (5.77,-0.17), SE * labelscalefactor); dot((0.08,2.37),dotstyle); label("$D$", (0.24,2.61), SW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); label("$7$",(-3.36,4.46)--(5.77,-0.17), NE * labelscalefactor); label("$3$",(-3.36,4.46)--(-2.46,-0.51),SW * labelscalefactor); label("$5$",(-2.46,-0.51)--(5.77,-0.17), SE * labelscalefactor); /* end of picture */ [/asy]

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]

If $P_1P_2P_3P_4P_5P_6$ is a regular hexagon whose apothem (distance from the center to midpoint of a side) is $2$, and $Q_i$ is the midpoint of side $P_iP_{i+1}$ for $i=1,2,3,4$, then the area of quadrilateral $Q_1Q_2Q_3Q_4$ is $\textbf{(A) }6\qquad\textbf{(B) }2\sqrt{6}\qquad\textbf{(C) }\frac{8\sqrt{3}}{3}\qquad\textbf{(D) }3\sqrt{3}\qquad\textbf{(E) }4\sqrt{3}$

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 $O$ be the center of the circumcircle $\omega$ of an acute-angle triangle $ABC$. A circle $\omega_1$ with center $K$ passes through $A$, $O$, $C$ and intersects $AB$ at $M$ and $BC$ at $N$. Point $L$ is symmetric to $K$ with respect to line $NM$. Prove that $BL \perp AC$.

Let $S$ be the square one of whose diagonals has endpoints $(0.1,0.7)$ and $(-0.1,-0.7)$. A point $v=(x,y)$ is chosen uniformly at random over all pairs of real numbers $x$ and $y$ such that $0\le x \le 2012$ and $0 \le y \le 2012$. Let $T(v)$ be a translated copy of $S$ centered at $v$. What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coordinates in its interior? $ \textbf{(A)}\ 0.125\qquad\textbf{(B)}\ 0.14\qquad\textbf{(C)}\ 0.16\qquad\textbf{(D)}\ 0.25\qquad\textbf{(E)}\ 0.32 $

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.

Let $ABCDE$ be a pentagon inscribed in a circle such that $AB=CD=3$, $BC=DE=10$, and $AE=14$. The sum of the lengths of all diagonals of $ABCDE$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? $\textbf{(A) }129\qquad \textbf{(B) }247\qquad \textbf{(C) }353\qquad \textbf{(D) }391\qquad \textbf{(E) }421\qquad$

Consider the following layouts of nine triangles with the letters $A, B, C, D, E, F, G, H, I$ in its interior. [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(200); 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 = 1.740000000000003, xmax = 8.400000000000013, ymin = 3.500000000000005, ymax = 9.360000000000012; /* image dimensions */ draw((5.020000000000005,8.820000000000011)--(2.560000000000003,4.580000000000005)--(7.461947712046029,4.569577506690286)--cycle); /* draw figures */ draw((5.020000000000005,8.820000000000011)--(2.560000000000003,4.580000000000005)); draw((2.560000000000003,4.580000000000005)--(7.461947712046029,4.569577506690286)); draw((7.461947712046029,4.569577506690286)--(5.020000000000005,8.820000000000011)); draw((3.382989341689345,5.990838871467448)--(4.193333333333338,4.580000000000005)); draw((4.202511849578174,7.405966442513598)--(5.828619600041468,4.573707435672692)); draw((5.841878190157451,7.408513542990484)--(4.193333333333338,4.580000000000005)); draw((6.656214943659867,5.990342259816768)--(5.828619600041468,4.573707435672692)); draw((4.202511849578174,7.405966442513598)--(5.841878190157451,7.408513542990484)); draw((3.382989341689345,5.990838871467448)--(6.656214943659867,5.990342259816768)); label("\textbf{A}",(4.840000000000007,8.020000000000010),SE*labelscalefactor,fontsize(22)); label("\textbf{B}",(3.980000000000006,6.640000000000009),SE*labelscalefactor,fontsize(22)); label("\textbf{C}",(4.820000000000007,7.000000000000010),SE*labelscalefactor,fontsize(22)); label("\textbf{D}",(5.660000000000008,6.580000000000008),SE*labelscalefactor,fontsize(22)); label("\textbf{E}",(3.160000000000005,5.180000000000006),SE*labelscalefactor,fontsize(22)); label("\textbf{F}",(4.020000000000006,5.600000000000008),SE*labelscalefactor,fontsize(22)); label("\textbf{G}",(4.800000000000007,5.200000000000007),SE*labelscalefactor,fontsize(22)); label("\textbf{H}",(5.680000000000009,5.620000000000007),SE*labelscalefactor,fontsize(22)); label("\textbf{I}",(6.460000000000010,5.140000000000006),SE*labelscalefactor,fontsize(22)); /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy] A sequence of letters, each letter chosen from$ A, B, C, D, E, F, G, H, I$ is said to be [i]triangle-friendly[/i] if the first and last letter of the sequence is $C$, and for every letter except the first letter, the triangle containing this letter shares an edge with the triangle containing the previous letter in the sequence. For example, the letter after $C$ must be either $A, B$, or $D$. For example, $CBF BC$ is triangle-friendly, but $CBF GH$ and $CBBHC$ are not. [list] [*] (a) Determine the number of triangle-friendly sequences with $2012$ letters. [*] (b) Determine the number of triangle-friendly sequences with exactly $2013$ letters.[/list]

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 $AB+AC=PQ+PR$.

Triangle $ABC$ has $AC = 3$, $BC = 5$, $AB = 7$. A circle is drawn internally tangent to the circumcircle of $ABC$ at $C$, and tangent to $AB$. Let $D$ be its point of tangency with $AB$. Find $BD - DA$. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 2.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.5, xmax = 7.01, ymin = -3, ymax = 8.02; /* image dimensions */ /* draw figures */ draw(circle((1.37,2.54), 5.17)); draw((-2.62,-0.76)--(-3.53,4.2)); draw((-3.53,4.2)--(5.6,-0.44)); draw((5.6,-0.44)--(-2.62,-0.76)); draw(circle((-0.9,0.48), 2.12)); /* dots and labels */ dot((-2.62,-0.76),dotstyle); label("$C$", (-2.46,-0.51), SW * labelscalefactor); dot((-3.53,4.2),dotstyle); label("$A$", (-3.36,4.46), NW * labelscalefactor); dot((5.6,-0.44),dotstyle); label("$B$", (5.77,-0.17), SE * labelscalefactor); dot((0.08,2.37),dotstyle); label("$D$", (0.24,2.61), SW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); label("$7$",(-3.36,4.46)--(5.77,-0.17), NE * labelscalefactor); label("$3$",(-3.36,4.46)--(-2.46,-0.51),SW * labelscalefactor); label("$5$",(-2.46,-0.51)--(5.77,-0.17), SE * labelscalefactor); /* end of picture */ [/asy]

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 $ABCD$ be a circumscribed quadrilateral and let $P$ be the orthogonal projection of its in center on $AC$. Prove that $\angle {APB}=\angle {APD}$

[asy] import cse5; size(180); real a=4, b=3; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)--(b,a)); D(CP((X+C)/2,C)); D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0)))); //Credit to chezbgone2 for the diagram[/asy] In $\triangle ABC$, $AB = 10~ AC = 8$ and $BC = 6$. Circle $P$ is the circle with smallest radius which passes through $C$ and is tangent to $AB$. Let $Q$ and $R$ be the points of intersection, distinct from $C$ , of circle $P$ with sides $AC$ and $BC$, respectively. The length of segment $QR$ is $\textbf{(A) }4.75\qquad\textbf{(B) }4.8\qquad\textbf{(C) }5\qquad\textbf{(D) }4\sqrt{2}\qquad \textbf{(E) }3\sqrt{3}$

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 $AB+AC=PQ+PR$.

The sum of the distances from one vertex of a square with sides of length two to the midpoints of each of the sides of the square is $\textbf{(A) }2\sqrt{5}\qquad\textbf{(B) }2+\sqrt{3}\qquad\textbf{(C) }2+2\sqrt{3}\qquad\textbf{(D) }2+\sqrt{5}\qquad \textbf{(E) }2+2\sqrt{5}$

A cylinder radius $12$ and a cylinder radius $36$ are held tangent to each other with a tight band. The length of the band is $m\sqrt{k}+n\pi$ where $m$, $k$, and $n$ are positive integers, and $k$ is not divisible by the square of any prime. Find $m + k + n$. [asy] size(150); real t=0.3; void cyl(pair x, real r, real h) { pair xx=(x.x,t*x.y); path B=ellipse(xx,r,t*r), T=ellipse((x.x,t*x.y+h),r,t*r), S=xx+(r,0)--xx+(r,h)--(xx+(-r,h))--xx-(r,0); unfill(S--cycle); draw(S); unfill(B); draw(B); unfill(T); draw(T); } real h=8, R=3,r=1.2; pair X=(0,0), Y=(R+r)*dir(-50); cyl(X,R,h); draw(shift((0,5))*yscale(t)*arc(X,R,180,360)); cyl(Y,r,h); void str (pair x, pair y, real R, real r, real h, real w) { real u=(angle(y-x)+asin((R-r)/(R+r)))*180/pi+270; path P=yscale(t)*(arc(x,R,180,u)--arc(y,r,u,360)); path Q=shift((0,h))*P--shift((0,h+w))*reverse(P)--cycle; fill(Q,grey);draw(Q); } str(X,Y,R,r,3.5,1.5);[/asy]

Let $ABC$ be a triangle with $AB = 130$, $BC = 140$, $CA = 150$. Let $G$, $H$, $I$, $O$, $N$, $K$, $L$ be the centroid, orthocenter, incenter, circumenter, nine-point center, the symmedian point, and the de Longchamps point. Let $D$, $E$, $F$ be the feet of the altitudes of $A$, $B$, $C$ on the sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$. Let $X$, $Y$, $Z$ be the $A$, $B$, $C$ excenters and let $U$, $V$, $W$ denote the midpoints of $\overline{IX}$, $\overline{IY}$, $\overline{IZ}$ (i.e. the midpoints of the arcs of $(ABC)$.) Let $R$, $S$, $T$ denote the isogonal conjugates of the midpoints of $\overline{AD}$, $\overline{BE}$, $\overline{CF}$. Let $P$ and $Q$ denote the images of $G$ and $H$ under an inversion around the circumcircle of $ABC$ followed by a dilation at $O$ with factor $\frac 12$, and denote by $M$ the midpoint of $\overline{PQ}$. Then let $J$ be a point such that $JKLM$ is a parallelogram. Find the perimeter of the convex hull of the self-intersecting $17$-gon $LETSTRADEBITCOINS$ to the nearest integer. A diagram has been included but may not be to scale. [asy] size(6cm); import olympiad; import cse5; pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A,B,C); pair E = foot(B,C,A); pair F = foot(C,A,B); pair G = centroid(A,B,C); pair H = orthocenter(A,B,C); pair I = incenter(A,B,C); pair isocon(pair targ) { return extension(A,2*foot(targ,I,A)-targ, C,2*foot(targ,I,C)-targ); } pair O = circumcenter(A,B,C); pair K = isocon(G); pair N = midpoint(O--H); pair U = extension(O,midpoint(B--C),A,I); pair V = extension(O,midpoint(C--A),B,I); pair W = extension(O,midpoint(A--B),C,I); pair X = -I + 2*U; pair Y = -I + 2*V; pair Z = -I + 2*W; pair R = isocon(midpoint(A--D)); pair S = isocon(midpoint(B--E)); pair T = isocon(midpoint(C--F)); pair L = 2*H-O; pair P = 0.5/conj(G); pair Q = 0.5/conj(H); pair M = midpoint(P--Q); pair J = K+M-L; draw(A--B--C--cycle); void draw_cevians(pair target) { draw(A--extension(A,target,B,C)); draw(B--extension(B,target,C,A)); draw(C--extension(C,target,A,B)); } draw_cevians(H); draw_cevians(G); draw_cevians(I); draw(unitcircle); draw(circumcircle(D,E,F)); draw(O--P); draw(O--Q); draw(P--Q); draw(CP(X,foot(X,B,C))); draw(CP(Y,foot(Y,C,A))); draw(CP(Z,foot(Z,A,B))); draw(J--K--L--M); draw(X--Y--Z--cycle); draw(A--X); draw(B--Y); draw(C--Z); draw(A--foot(X,A,B)); draw(A--foot(X,A,C)); draw(B--foot(Y,B,C)); draw(B--foot(Y,B,A)); draw(C--foot(Z,C,A)); draw(C--foot(Z,C,B)); pen p = black; dot(A, p); dot(B, p); dot(C, p); dot(D, p); dot(E, p); dot(F, p); dot(G, p); dot(H, p); dot(I, p); dot(J, p); dot(K, p); dot(L, p); dot(M, p); dot(N, p); dot(O, p); dot(P, p); dot(Q, p); dot(R, p); dot(S, p); dot(T, p); dot(U, p); dot(V, p); dot(W, p); dot(X, p); dot(Y, p); dot(Z, p); [/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]

In a circle of radius 42, two chords of length 78 intersect at a point whose distance from the center is 18. The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lenghts, and the area of either of them can be expressed uniquley in the form $m\pi-n\sqrt{d},$ where $m, n,$ and $d$ are positive integers and $d$ is not divisible by the square of any prime number. Find $m+n+d.$