The magnitudes of the sides of triangle $ABC$ are $a$, $b$, and $c$, as shown, with $c\le b\le a$. Through interior point $P$ and the vertices $A$, $B$, $C$, lines are drawn meeting the opposite sides in $A'$, $B'$, $C'$, respectively. Let $s=AA'+BB'+CC'$. Then, for all positions of point $P$, $s$ is less than: $\textbf{(A) }2a+b\qquad\textbf{(B) }2a+c\qquad\textbf{(C) }2b+c\qquad\textbf{(D) }a+2b\qquad \textbf{(E) }$ $a+b+c$ [asy] import math; defaultpen(fontsize(11pt)); pair A = (0,0), B = (1,3), C = (5,0), P = (1.5,1); pair X = extension(B,C,A,P), Y = extension(A,C,B,P), Z = extension(A,B,C,P); draw(A--B--C--cycle); draw(A--X); draw(B--Y); draw(C--Z); dot(P); dot(A); dot(B); dot(C); label("$A$",A,dir(210)); label("$B$",B,dir(90)); label("$C$",C,dir(-30)); label("$A'$",X,dir(-100)); label("$B'$",Y,dir(65)); label("$C'$",Z,dir(20)); label("$P$",P,dir(70)); label("$a$",X,dir(80)); label("$b$",Y,dir(-90)); label("$c$",Z,dir(110)); //Credit to bobthesmartypants for the diagram [/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 $\textbf{(A) }24\qquad\textbf{(B) }2\sqrt{119}\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad \textbf{(E) }4\sqrt{35}$

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

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]

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]

$ABCD$ 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]

Let $ABC$ be a right angled triangle at $A$. Denote $D$ the foot of the altitude through $A$ and $O_1, O_2$ the incentres of triangles $ADB$ and $ADC$. The circle with centre $A$ and radius $AD$ cuts $AB$ in $K$ and $AC$ in $L$. Show that $O_1, O_2, K$ and $L$ are on a line.

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 $a,b,c$ 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 $ 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]

On square $ABCD,$ points $E,F,G,$ and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34.$ Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P,$ and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$. [asy] size(200); defaultpen(linewidth(0.8)+fontsize(10.6)); pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(120))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", (H+E)/2,fontsize(15)); label("$x$", (E+F)/2,fontsize(15)); label("$y$", (G+F)/2,fontsize(15)); label("$z$", (H+G)/2,fontsize(15)); label("$w:x:y:z=269:275:405:411$",(sqrt(850)/2,-4.5),fontsize(11)); [/asy]

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 $

The diagram below shows four regular hexagons each with side length $1$ meter attached to the sides of a square. This figure is drawn onto a thin sheet of metal and cut out. The hexagons are then bent upward along the sides of the square so that $A_1$ meets $A_2$, $B_1$ meets $B_2$, $C_1$ meets $C_2$, and $D_1$ meets $D_2$. If the resulting dish is filled with water, the water will rise to the height of the corner where the $A_1$ and $A_2$ meet. there are relatively prime positive integers $m$ and $n$ so that the number of cubic meters of water the dish will hold is $\sqrt{\frac{m}{n}}$. Find $m+n$. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7cm); 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 = -4.3, xmax = 14.52, ymin = -8.3, ymax = 6.3; /* image dimensions */ draw((0,1)--(0,0)--(1,0)--(1,1)--cycle); draw((1,1)--(1,0)--(1.87,-0.5)--(2.73,0)--(2.73,1)--(1.87,1.5)--cycle); draw((0,1)--(1,1)--(1.5,1.87)--(1,2.73)--(0,2.73)--(-0.5,1.87)--cycle); draw((0,0)--(1,0)--(1.5,-0.87)--(1,-1.73)--(0,-1.73)--(-0.5,-0.87)--cycle); draw((0,1)--(0,0)--(-0.87,-0.5)--(-1.73,0)--(-1.73,1)--(-0.87,1.5)--cycle); /* draw figures */ draw((0,1)--(0,0)); draw((0,0)--(1,0)); draw((1,0)--(1,1)); draw((1,1)--(0,1)); draw((1,1)--(1,0)); draw((1,0)--(1.87,-0.5)); draw((1.87,-0.5)--(2.73,0)); draw((2.73,0)--(2.73,1)); draw((2.73,1)--(1.87,1.5)); draw((1.87,1.5)--(1,1)); draw((0,1)--(1,1)); draw((1,1)--(1.5,1.87)); draw((1.5,1.87)--(1,2.73)); draw((1,2.73)--(0,2.73)); draw((0,2.73)--(-0.5,1.87)); draw((-0.5,1.87)--(0,1)); /* dots and labels */ dot((1.87,-0.5),dotstyle); label("$C_1$", (1.72,-0.1), NE * labelscalefactor); dot((1.87,1.5),dotstyle); label("$B_2$", (1.76,1.04), NE * labelscalefactor); dot((1.5,1.87),dotstyle); label("$B_1$", (0.96,1.8), NE * labelscalefactor); dot((-0.5,1.87),dotstyle); label("$A_2$", (-0.26,1.78), NE * labelscalefactor); dot((-0.87,1.5),dotstyle); label("$A_1$", (-0.96,1.08), NE * labelscalefactor); dot((-0.87,-0.5),dotstyle); label("$D_2$", (-1.02,-0.18), NE * labelscalefactor); dot((-0.5,-0.87),dotstyle); label("$D_1$", (-0.22,-0.96), NE * labelscalefactor); dot((1.5,-0.87),dotstyle); label("$C_2$", (0.9,-0.94), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

Let $X$ be an arbitrary point inside the circumcircle of a triangle $ABC$. The lines $BX$ and $CX$ meet the circumcircle in points $K$ and $L$ respectively. The line $LK$ intersects $BA$ and $AC$ at points $E$ and $F$ respectively. Find the locus of points $X$ such that the circumcircles of triangles $AFK$ and $AEL$ touch.

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]

In triangle $ABC$, $AD$ and $AH$ are the angle bisector and the altitude of vertex $A$, respectively. The perpendicular bisector of $AD$, intersects the semicircles with diameters $AB$ and $AC$ which are drawn outside triangle $ABC$ in $X$ and $Y$, respectively. Prove that the quadrilateral $XYDH$ is concyclic. [i]Proposed by Mahan Malihi[/i]

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]

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 $AB+AC=PQ+PR$.

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.

Points $E$ and $F$ lie inside rectangle $ABCD$ with $AE=DE=BF=CF=EF$. If $AB=11$ and $BC=8$, find the area of the quadrilateral $AEFB$.

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]

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} $

$BB_{1}$ is a bisector of an acute triangle $ABC$. A perpendicular from $B_{1}$ to $BC$ meets a smaller arc $BC$ of a circumcircle of $ABC$ in a point $K$. A perpendicular from $B$ to $AK$ meets $AC$ in a point $L$. $BB_{1}$ meets arc $AC$ in $T$. Prove that $K$, $L$, $T$ are collinear. [i]V. Astakhov[/i]

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]