Circle $C$ with radius $2$ has diameter $\overline{AB}$. Circle $D$ is internally tangent to circle $C$ at $A$. Circle $E$ is internally tangent to circle $C,$ externally tangent to circle $D,$ and tangent to $\overline{AB}$. The radius of circle $D$ is three times the radius of circle $E$ and can be written in the form $\sqrt{m} - n,$ where $m$ and $n$ are positive integers. Find $m+n$.

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]

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]

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]

In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG+JH+CD$? [asy] import cse5;pathpen=black;pointpen=black; size(2inch); pair A=dir(90), B=dir(18), C=dir(306), D=dir(234), E=dir(162); D(MP("A",A,A)--MP("B",B,B)--MP("C",C,C)--MP("D",D,D)--MP("E",E,E)--cycle,linewidth(1.5)); D(A--C--E--B--D--cycle); pair F=IP(A--D,B--E), G=IP(B--E,C--A), H=IP(C--A,B--D), I=IP(D--B,E--C), J=IP(C--E,D--A); D(MP("F",F,dir(126))--MP("I",I,dir(270))--MP("G",G,dir(54))--MP("J",J,dir(198))--MP("H",H,dir(342))--cycle); [/asy] $\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} $

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]

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} $

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]

In triangle $ABC$ lines $CE$ and $AD$ are drawn so that $\dfrac{CD}{DB}=\dfrac{3}{1}$ and $\dfrac{AE}{EB}=\dfrac{3}{2}$. Let $r=\dfrac{CP}{PE}$ where $P$ is the intersection point of $CE$ and $AD$. Then $r$ equals: [asy] size(8cm); pair A = (0, 0), B = (9, 0), C = (3, 6); pair D = (7.5, 1.5), E = (6.5, 0); pair P = intersectionpoints(A--D, C--E)[0]; draw(A--B--C--cycle); draw(A--D); draw(C--E); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, NE); label("$E$", E, S); label("$P$", P, S); //Credit to MSTang for the asymptote [/asy] $\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{3}{2}\qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \dfrac{5}{2}$

A bug travels from $A$ to $B$ along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there? [asy] size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544)); draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767)); draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0)); draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)); draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080)); draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0)); dot((0,0)); dot((22,0)); label("$A$",(0,0),WNW); label("$B$",(22,0),E); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black); filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black); filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black); filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black); filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white); filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black); filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black); filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black); filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black); filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white); filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black); filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black); filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black); filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black); filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);[/asy] $ \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 $

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

Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$. $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$. If $EF = 32$ and $GF = 24$, then $BE$ equals: [asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E = intersectionpoints(A--C, B--F)[0]; draw(A--D--C--B--cycle); draw(A--C); draw(D--F--B); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$F$", F, N); label("$G$", G, NE); label("$E$", E, SE); //Credit to MSTang for the asymptote [/asy] $\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 16$

Points $A$, $B$, and $O$ lie in the plane such that $\measuredangle AOB = 120^\circ$. Circle $\omega_0$ with radius $6$ is constructed tangent to both $\overrightarrow{OA}$ and $\overrightarrow{OB}$. For all $i \ge 1$, circle $\omega_i$ with radius $r_i$ is constructed such that $r_i < r_{i - 1}$ and $\omega_i$ is tangent to $\overrightarrow{OA}$, $\overrightarrow{OB}$, and $\omega_{i - 1}$. If \[ S = \sum_{i = 1}^\infty r_i, \] then $S$ can be expressed as $a\sqrt{b} + c$, where $a, b, c$ are integers and $b$ is not divisible by the square of any prime. Compute $100a + 10b + c$. [i]Proposed by Aaron Lin[/i]

A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$. [asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/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]

An uncrossed belt is fitted without slack around two circular pulleys with radii of $14$ inches and $4$ inches. If the distance between the points of contact of the belt with the pulleys is $24$ inches, then the distance between the centers of the pulleys in inches is $\textbf{(A) }24\qquad\textbf{(B) }2\sqrt{119}\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad \textbf{(E) }4\sqrt{35}$

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

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 $\omega_1$ 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]

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]

[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 $\textbf{(A) }\frac{1}{4}\cos 10^\circ\qquad\textbf{(B) }\frac{\sqrt{3}}{8}\qquad\textbf{(C) }\frac{1}{4}\cos 40^\circ\qquad\textbf{(D) }\frac{1}{4}\cos 50^\circ\qquad\textbf{(E) }\frac{1}{8}$

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 $ABC$ 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]

A variable tangent $t$ to the circle $C_1$, of radius $r_1$, intersects the circle $C_2$, of radius $r_2$ in $A$ and $B$. The tangents to $C_2$ through $A$ and $B$ intersect in $P$. Find, as a function of $r_1$ and $r_2$, the distance between the centers of $C_1$ and $C_2$ such that the locus of $P$ when $t$ varies is contained in an equilateral hyperbola. [b]Note[/b]: A hyperbola is said to be [i]equilateral[/i] if its asymptotes are perpendicular.

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 $\angle DBQ = 90^\circ$.