In the diagram $ABCDEFG$ is a regular heptagon (a $7$ sided polygon). Shown is the star $AEBFCGD$. The degree measure of the obtuse angle formed by $AE$ and $CG$ is $\dfrac{m}{n}$ where m and n are relatively prime positive integers. Find $m + n$. [asy] size(150); defaultpen(linewidth(1)); string lab[]={"A","B","C","D","E","F","G"}; real r = 360/7; pair A=dir(90-r),B=dir(90),C=dir(90+r),D=dir(90+2*r),E=dir(90+3*r),F=dir(90+4*r),G=dir(90+5*r); draw(A--E--B--F--C--G--D--cycle); for(int k = -1;k <= 5;++k) { label("$"+lab[k+1]+"$",dir(90+k*r),dir(90+k*r)); } [/asy]

How many rectangles are in this figure? [asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(20,0); C=(20,20); D=(0,20); draw(A--B--C--D--cycle); E=(-10,-5); F=(13,-5); G=(13,5); H=(-10,5); draw(E--F--G--H--cycle); I=(10,-20); J=(18,-20); K=(18,13); L=(10,13); draw(I--J--K--L--cycle);[/asy] $ \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 $

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]

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

Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]

In the triangle $ABC$, $\angle A$ is biggest. On the circumcircle of $\triangle ABC$, let $D$ be the midpoint of $\widehat{ABC}$ and $E$ be the midpoint of $\widehat{ACB}$. The circle $c_1$ passes through $A,B$ and is tangent to $AC$ at $A$, the circle $c_2$ passes through $A,E$ and is tangent $AD$ at $A$. $c_1$ and $c_2$ intersect at $A$ and $P$. Prove that $AP$ bisects $\angle BAC$. [hide="Diagram"][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.4cm); 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 = -5.23, xmax = 9.18, ymin = -2.97, ymax = 4.82; /* image dimensions */ /* draw figures */ draw(circle((-1.32,1.36), 2.98)); draw(circle((3.56,1.53), 3.18)); draw((0.92,3.31)--(-2.72,-1.27)); draw(circle((0.08,0.25), 3.18)); draw((-2.72,-1.27)--(3.13,-0.65)); draw((3.13,-0.65)--(0.92,3.31)); draw((0.92,3.31)--(2.71,-1.54)); draw((-2.41,-1.74)--(0.92,3.31)); draw((0.92,3.31)--(1.05,-0.43)); /* dots and labels */ dot((-1.32,1.36),dotstyle); dot((0.92,3.31),dotstyle); label("$A$", (0.81,3.72), NE * labelscalefactor); label("$c_1$", (-2.81,3.53), NE * labelscalefactor); dot((3.56,1.53),dotstyle); label("$c_2$", (3.43,3.98), NE * labelscalefactor); dot((1.05,-0.43),dotstyle); label("$P$", (0.5,-0.43), NE * labelscalefactor); dot((-2.72,-1.27),dotstyle); label("$B$", (-3.02,-1.57), NE * labelscalefactor); dot((2.71,-1.54),dotstyle); label("$E$", (2.71,-1.86), NE * labelscalefactor); dot((3.13,-0.65),dotstyle); label("$C$", (3.39,-0.9), NE * labelscalefactor); dot((-2.41,-1.74),dotstyle); label("$D$", (-2.78,-2.07), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

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 $m+n$? $\textbf{(A) }129\qquad \textbf{(B) }247\qquad \textbf{(C) }353\qquad \textbf{(D) }391\qquad \textbf{(E) }421\qquad$

Let $ABC$ be a triangle. Let $A_1$, $A_2$ be points on $AB$ and $AC$ respectively such that $A_1A_2 \parallel BC$ and the circumcircle of $\triangle AA_1A_2$ is tangent to $BC$ at $A_3$. Define $B_3$, $C_3$ similarly. Prove that $AA_3$, $BB_3$, and $CC_3$ are concurrent. [i]Carl Lian.[/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]

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]

If each of two intersecting lines intersects a hyperbola and neither line is tangent to the hyperbola, then the possible number of points of intersection with the hyperbola is: $ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 2\text{ or }3 \qquad\textbf{(C)}\ 2\text{ or }4 \qquad\textbf{(D)}\ 3\text{ or }4 \qquad\textbf{(E)}\ 2,3,\text{ or }4$

Let $a_0,a_1,a_2,\dots$ be a sequence of nonnegative integers such that $a_2=5$, $a_{2014}=2015$, and $a_n=a_{a_{n-1}}$ for all positive integers $n$. Find all possible values of $a_{2015}$.

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 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] $\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$