Let $ABCD$ be a convex quadrilateral, $I_1$ and $I_2$ be the incenters of triangles $ABC$ and $DBC$ respectively. The line $I_1I_2$ intersects the lines $AB$ and $DC$ at points $E$ and $F$ respectively. Given that $AB$ and $CD$ intersect in $P$, and $PE=PF$, prove that the points $A$, $B$, $C$, $D$ lie on a circle.

In a triangle $\triangle ABC$ with orthocenter $H$, let $BH$ and $CH$ intersect $AC$ and $AB$ at $E$ and $F$, respectively. If the tangent line to the circumcircle of $\triangle ABC$ passing through $A$ intersects $BC$ at $P$, $M$ is the midpoint of $AH$, and $EF$ intersects $BC$ at $G$, then prove that $PM$ is parallel to $GH$. [i]Proposed by Sreejato Bhattacharya[/i]

Triangle $ ABC$ has $ AB \equal{} 2 \cdot AC$. Let $ D$ and $ E$ be on $ \overline{AB}$ and $ \overline{BC}$, respectively, such that $ \angle{BAE} \equal{} \angle{ACD}.$ Let $ F$ be the intersection of segments $ AE$ and $ CD$, and suppose that $ \triangle{CFE}$ is equilateral. What is $ \angle{ACB}$? $ \textbf{(A)}\ 60^{\circ}\qquad \textbf{(B)}\ 75^{\circ}\qquad \textbf{(C)}\ 90^{\circ}\qquad \textbf{(D)}\ 105^{\circ}\qquad \textbf{(E)}\ 120^{\circ}$

Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]

Triangle $ ABC$ has $ \angle C \equal{} 60^{\circ}$ and $ BC \equal{} 4$. Point $ D$ is the midpoint of $ BC$. What is the largest possible value of $ \tan{\angle BAD}$? $ \textbf{(A)} \ \frac {\sqrt {3}}{6} \qquad \textbf{(B)} \ \frac {\sqrt {3}}{3} \qquad \textbf{(C)} \ \frac {\sqrt {3}}{2\sqrt {2}} \qquad \textbf{(D)} \ \frac {\sqrt {3}}{4\sqrt {2} \minus{} 3} \qquad \textbf{(E)}\ 1$

Let $\omega_{1}$ be a circle with centre $O$. $P$ is a point on $\omega_{1}$. $\omega_{2}$ is a circle with centre $P$, with radius smaller than $\omega_{1}$. $\omega_{1}$ meets $\omega_{2}$ at points $T$ and $Q$. Let $TR$ be a diameter of $\omega_{2}$. Draw another two circles with $RQ$ as the radius, $R$ and $P$ as the centres. These two circles meet at point $M$, with $M$ and $Q$ lie on the same side of $PR$. A circle with centre $M$ and radius $MR$ intersects $\omega_{2}$ at $R$ and $N$. Prove that a circle with centre $T$ and radius $TN$ passes through $O$.

Let $ABCD$ be a circumscribed quadrilateral (i. e. a quadrilateral which has an incircle). The exterior angle bisectors of the angles $DAB$ and $ABC$ intersect each other at $K$; the exterior angle bisectors of the angles $ABC$ and $BCD$ intersect each other at $L$; the exterior angle bisectors of the angles $BCD$ and $CDA$ intersect each other at $M$; the exterior angle bisectors of the angles $CDA$ and $DAB$ intersect each other at $N$. Let $K_{1}$, $L_{1}$, $M_{1}$ and $N_{1}$ be the orthocenters of the triangles $ABK$, $BCL$, $CDM$ and $DAN$, respectively. Show that the quadrilateral $K_{1}L_{1}M_{1}N_{1}$ is a parallelogram.

Triangle $ABC$ is isosceles with base $AC$. Points $P$ and $Q$ are respectively in $CB$ and $AB$ and such that $AC=AP=PQ=QB$. The number of degrees in angle $B$ is: ${{ \textbf{(A)}\ 25 \frac{5}{7} \qquad\textbf{(B)}\ 26 \frac{1}{3} \qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 40}\qquad\textbf{(E)}\ \text{Not determined by the information given} } $

Let $\triangle ABC$ have incenter $I$ and centroid $G$. Suppose that $P_A$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$, and $Q_A$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$. Define $P_B$, $P_C$, $Q_B$, and $Q_C$ similarly. Show that $P_A, P_B, P_C, Q_A, Q_B,$ and $Q_C$ lie on a circle whose center is on line $IG$.

In acute triangle $ABC$, $AB > AC$. Let $M$ be the midpoint of side $BC$. The exterior angle bisector of $\widehat{BAC}$ meet ray $BC$ at $P$. Point $K$ and $F$ lie on line $PA$ such that $MF \perp BC$ and $MK \perp PA$. Prove that $BC^2 = 4 PF \cdot AK$. [asy] defaultpen(fontsize(10)); size(7cm); pair A = (4.6,4), B = (0,0), C = (5,0), M = midpoint(B--C), I = incenter(A,B,C), P = extension(A, A+dir(I--A)*dir(-90), B,C), K = foot(M,A,P), F = extension(M, (M.x, M.x+1), A,P); draw(K--M--F--P--B--A--C); pair point = I; pair[] p={A,B,C,M,P,F,K}; string s = "A,B,C,M,P,F,K"; int size = p.length; real[] d; real[] mult; for(int i = 0; i<size; ++i) { d[i] = 0; mult[i] = 1;} string[] k= split(s,","); for(int i = 0;i<p.length;++i) { label("$"+k[i]+"$",p[i],mult[i]*dir(point--p[i])*dir(d[i])); }[/asy]

Triangle $ ABC$ has $ AB \equal{} 2 \cdot AC$. Let $ D$ and $ E$ be on $ \overline{AB}$ and $ \overline{BC}$, respectively, such that $ \angle{BAE} \equal{} \angle{ACD}.$ Let $ F$ be the intersection of segments $ AE$ and $ CD$, and suppose that $ \triangle{CFE}$ is equilateral. What is $ \angle{ACB}$? $ \textbf{(A)}\ 60^{\circ}\qquad \textbf{(B)}\ 75^{\circ}\qquad \textbf{(C)}\ 90^{\circ}\qquad \textbf{(D)}\ 105^{\circ}\qquad \textbf{(E)}\ 120^{\circ}$

In circle $ O$ chord $ AB$ is produced so that $ BC$ equals a radius of the circle. $ CO$ is drawn and extended to $ D$. $ AO$ is drawn. Which of the following expresses the relationship between $ x$ and $ y$? [asy]size(200);defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, D=dir(195), A=dir(150), B=dir(30), C=B+1*dir(0); draw(O--A--C--D); dot(A^^B^^C^^D^^O); pair point=O; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$O$", O, dir(285)); label("$x$", O+0.1*dir(172.5), dir(172.5)); label("$y$", C+0.4*dir(187.5), dir(187.5)); draw(Circle(O,1)); [/asy] $ \textbf{(A)}\ x\equal{}3y \\ \textbf{(B)}\ x\equal{}2y \\ \textbf{(C)}\ x\equal{}60^\circ \\ \textbf{(D)}\ \text{there is no special relationship between }x\text{ and }y \\ \textbf{(E)}\ x\equal{}2y \text{ or }x\equal{}3y\text{, depending upon the length of }AB$

Denote by $M$ and $N$ feets of perpendiculars from $A$ to angle bisectors of exterior angles at $B$ and $C,$ in triangle $\triangle ABC.$ Prove that the length of segment $MN$ is equal to semiperimeter of triangle $\triangle ABC.$

Let $ ABC$ be a triangle. The $ A$-excircle of triangle $ ABC$ has center $ O_{a}$ and touches the side $ BC$ at the point $ A_{a}$. The $ B$-excircle of triangle $ ABC$ touches its sidelines $ AB$ and $ BC$ at the points $ C_{b}$ and $ A_{b}$. The $ C$-excircle of triangle $ ABC$ touches its sidelines $ BC$ and $ CA$ at the points $ A_{c}$ and $ B_{c}$. The lines $ C_{b}A_{b}$ and $ A_{c}B_{c}$ intersect each other at some point $ X$. Prove that the quadrilateral $ AO_{a}A_{a}X$ is a parallelogram. [i]Remark.[/i] The $ A$[i]-excircle[/i] of a triangle $ ABC$ is defined as the circle which touches the segment $ BC$ and the extensions of the segments $ CA$ and $ AB$ beyound the points $ C$ and $ B$, respectively. The center of this circle is the point of intersection of the interior angle bisector of the angle $ CAB$ and the exterior angle bisectors of the angles $ ABC$ and $ BCA$. Similarly, the $ B$-excircle and the $ C$-excircle of triangle $ ABC$ are defined. [hide="Source of the problem"][i]Source of the problem:[/i] Theorem (88) in: John Sturgeon Mackay, [i]The Triangle and its Six Scribed Circles[/i], Proceedings of the Edinburgh Mathematical Society 1 (1883), pages 4-128 and drawings at the end of the volume.[/hide]