$ABCD$ 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 triangle and let $X$ be on $BC$ such that $AX=AB$. let $AX$ meet circumcircle $\omega$ of triangle $ABC$ again at $D$. prove that circumcentre of triangle $BDX$ lies on $\omega$.

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 $\triangle BAC$, $\angle BAC=40^\circ$, $AB=10$, and $AC=6$. Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$? $\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$

Let $ABC$ be an acute triangle. The points $M$ and $N$ are midpoints of $AB$ and $BC$ respectively, and $BH$ is an altitude of $ABC$. The circumcircles of $AHN$ and $CHM$ meet in $P$ where $P\ne H$. Prove that $PH$ passes through the midpoint of $MN$. [i]V. Filimonov[/i]

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]

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]

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]

Let $M$, $N$ be the midpoints of diagonals $AC$, $BD$ of a right-angled trapezoid $ABCD$ ($\measuredangle A=\measuredangle D = 90^\circ$). The circumcircles of triangles $ABN$, $CDM$ meet the line $BC$ in the points $Q$, $R$. Prove that the distances from $Q$, $R$ to the midpoint of $MN$ are equal.

A plane contains points $ A$ and $ B$ with $ AB \equal{} 1$. Let $ S$ be the union of all disks of radius $ 1$ in the plane that cover $ \overline{AB}$. What is the area of $ S$? $ \textbf{(A)}\ 2\pi \plus{} \sqrt3 \qquad \textbf{(B)}\ \frac {8\pi}{3} \qquad \textbf{(C)}\ 3\pi \minus{} \frac {\sqrt3}{2} \qquad \textbf{(D)}\ \frac {10\pi}{3} \minus{} \sqrt3 \qquad \textbf{(E)}\ 4\pi \minus{} 2\sqrt3$

A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$. How many edges does the remaining solid have? $\textbf{(A) }36\qquad \textbf{(B) }60\qquad \textbf{(C) }72\qquad \textbf{(D) }84\qquad \textbf{(E) }108\qquad$

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 $\Gamma_1$ 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]

$[AB]$ and $[CD]$ are not parallel in the convex quadrilateral $ABCD$. Let $E$ and $F$ be the midpoints of $[AD]$ and $[BC]$, respectively. If $|CD|=12$, $|AB|=22$, and $|EF|=x$, what is the sum of integer values of $x$? $ \textbf{(A)}\ 110 \qquad\textbf{(B)}\ 114 \qquad\textbf{(C)}\ 118 \qquad\textbf{(D)}\ 121 \qquad\textbf{(E)}\ \text{None of the above} $

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]

$ABCD$ 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]

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]

A function $ f:\mathbb{R}_{\ge 0}\longrightarrow\mathbb{R} $ has the property that $ \lim_{x\to\infty } \frac{1}{x^2}\int_0^x f(t)dt=1. $ [b]a)[/b] Give an example of what $ f $ could be if it's continuous and $ f/\text{id.} $ doesn't have a limit at $ \infty . $ [b]b)[/b] Prove that if $ f $ is nondecreasing then $ f/\text{id.} $ has a limit at $ \infty , $ and determine it.

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 $

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]

Let $A$, $B$, $C$, and $D$ be points in the coordinate plane on the hyperbola $\tfrac{x^{2}}{20}-\tfrac{y^{2}}{24}=1$ such that $ABCD$ is a rhombus whose diagonals intersect at the origin. Find the greatest real number that is less than $BD^{2}$ for all such rhombi.

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 $XY$.)

The dissection of the $3-4-5$ triangle shown below has diameter $5/2$. [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(23cm); 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 = 1.42, xmax = 24.42, ymin = 3.8, ymax = 15.54; /* image dimensions */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax,defaultpen+black, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); yaxis(ymin, ymax,defaultpen+black, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((9.44,8.52)--(12.44,8.52)); draw((9.44,12.52)--(9.44,8.52)); draw((9.44,12.52)--(12.44,8.52)); draw((9.44,10.52)--(10.94,10.52)); draw((10.94,10.52)--(10.94,8.52)); draw((9.44,8.52)--(10.94,10.52)); /* dots and labels */ dot((9.44,8.52),dotstyle); label("$A$", (9.52,8.64), NE * labelscalefactor); dot((12.44,8.52),dotstyle); label("$B$", (12.52,8.64), NE * labelscalefactor); dot((9.44,12.52),dotstyle); label("$C$", (9.52,12.64), NE * labelscalefactor); dot((10.94,8.52),dotstyle); label("$D$", (11.02,8.64), NE * labelscalefactor); dot((9.44,10.52),dotstyle); label("$E$", (9.52,10.64), NE * labelscalefactor); dot((10.94,10.52),dotstyle); label("$F$", (11.02,10.64), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy] Find the least diameter of this triangle into four parts. (The diameter of a dissection is the least upper bound of the distances between pairs of points belonging to the same part.)

Suppose circles $\mathit{W}_1$ and $\mathit{W}2$, with centres $\mathit{O}_1$ and $\mathit{O}_2$ respectively, intersect at points $\mathit{M}$ and $\mathit{N}$. Let the tangent on $\mathit{W}_2$ at point $\mathit{N}$ intersect $\mathit{W}_1$ for the second time at $\mathit{B}_1$. Similarly, let the tangent on $\mathit{W}_1$ at point $\mathit{N}$ intersect $\mathit{W}_2$ for the second time at $\mathit{B}_2$. Let $\mathit{A}_1$ be a point on $\mathit{W}_1$ which is on arc $\mathit{B}_1\mathit{N}$ not containing $\mathit{M}$ and suppose line $\mathit{A}_1\mathit{N}$ intersects $\mathit{W}_2$ at point $\mathit{A}_2$. Denote the incentres of triangles $\mathit{B}_1\mathit{A}_1\mathit{N}$ and $\mathit{B}_2\mathit{A}_2\mathit{N}$ by $\mathit{I}_1$ and $\mathit{I}_2$, respectively.* [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10.1cm); 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 = -0.9748626324969808, xmax = 13.38440254515721, ymin = 0.5680051903627492, ymax = 10.99430986899034; /* image dimensions */ pair O_2 = (7.682929606970993,6.084708172218866), O_1 = (2.180000000000002,6.760000000000007), M = (4.560858774883258,8.585242858926296), B_2 = (10.07334553576748,9.291873850408265), A_2 = (11.49301008867042,4.866805580476367), B_1 = (2.113311869970955,9.759258690628950), A_1 = (0.2203184186713625,4.488514120712773); /* draw figures */ draw(circle(O_2, 4.000000000000000)); draw(circle(O_1, 3.000000000000000)); draw((4.048892687647541,4.413249028538064)--B_2); draw(B_2--A_2); draw(A_2--(4.048892687647541,4.413249028538064)); draw((4.048892687647541,4.413249028538064)--B_1); draw(B_1--A_1); draw(A_1--(4.048892687647541,4.413249028538064)); /* dots and labels */ dot(O_2,dotstyle); label("$O_2$", (7.788512439159622,6.243082420501817), NE * labelscalefactor); dot(O_1,dotstyle); label("$O_1$", (2.298205165350667,6.929370829727937), NE * labelscalefactor); dot(M,dotstyle); label("$M$", (4.383466101076183,8.935444641311980), NE * labelscalefactor); dot((4.048892687647541,4.413249028538064),dotstyle); label("$N$", (3.855551940133015,3.761885864068922), NE * labelscalefactor); dot(B_2,dotstyle); label("$B_2$", (10.19052187145104,9.463358802255147), NE * labelscalefactor); dot(A_2,dotstyle); label("$A_2$", (11.80066006232771,4.659339937672310), NE * labelscalefactor); dot(B_1,dotstyle); label("$B_1$", (1.981456668784765,10.09685579538695), NE * labelscalefactor); dot(A_1,dotstyle); label("$A_1$", (0.08096568938935705,3.973051528446190), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy] Show that \[\angle\mathit{I}_1\mathit{MI}_2=\angle\mathit{O}_1\mathit{MO}_2.\] *[size=80]Given a triangle ABC, the incentre of the triangle is defined to be the intersection of the angle bisectors of A, B, and C. To avoid cluttering, the incentre is omitted in the provided diagram. Note also that the diagram serves only as an aid and is not necessarily drawn to scale.[/size]

Circles $\omega_1$ and $\omega_2$ meet at $P$ and $Q$. Segments $AC$ and $BD$ are chords of $\omega_1$ and $\omega_2$ respectively, such that segment $AB$ and ray $CD$ meet at $P$. Ray $BD$ and segment $AC$ meet at $X$. Point $Y$ lies on $\omega_1$ such that $P Y \parallel BD$. Point $Z$ lies on $\omega_2$ such that $P Z \parallel AC$. Prove that points $Q,~ X,~ Y,~ Z$ are collinear.