Let there be a convex quadrilateral $ABCD$. The angle bisector of $\angle A$ meets the angle bisector of $\angle B$, the angle bisector of $\angle D$ at $P, Q$ respectively. The angle bisector of $\angle C$ meets the angle bisector of $\angle D$, the angle bisector of $\angle B$ at $R, S$ respectively. $P, Q, R, S$ are all distinct points. $PR$ and $QS$ meets perpendicularly at point $Z$. Denote $l_A, l_B, l_C, l_D$ as the exterior angle bisectors of $\angle A, \angle B, \angle C, \angle D$. Denote $E = l_A \cap l_B$, $F= l_B \cap l_C$, $G = l_C \cap l_D$, and $H= l_D \cap l_A$. Let $K, L, M, N$ be the midpoints of $FG, GH, HE, EF$ respectively. Prove that the area of quadrilateral $KLMN$ is equal to $ZM \cdot ZK + ZL \cdot ZN$.

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 $S$ be the sum of the interior angles of a polygon $P$ for which each interior angle is $7\frac{1}{2}$ times the exterior angle at the same vertex. Then $ \textbf{(A)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may be regular}\qquad$ $\textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad$ $\textbf{(C)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is regular}\qquad$ $\textbf{(D)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad$ $\textbf{(E)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may or may not be regular} $

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]

Given a convex quadrilateral $ABCD$.Let $\ell_A,\ell_B,\ell_C,\ell_D$ be exterior angle bisectors of quadrilateral $ABCD$. Let $\ell_A \cap \ell_B=K,\ell_B \cap \ell_C=L,\ell_C \cap \ell_D=M,\ell_D \cap \ell_A=N$.Prove that if circumcircles of triangles $ABK$ and $CDM$ be externally tangent to each other then circumcircles of the triangles $BCL$ and $DAN$ are externally tangent to each other.(L.Emelyanov)

An angle is given in a plane. Using only a compass, one must find out $(a)$ if this angle is acute. Find the minimal number of circles one must draw to be sure. $(b)$ if this angle equals $31^{\circ}$.(One may draw as many circles as one needs).

Cirlce $\Omega$ is inscribed in triangle $ABC$ with $\angle BAC=40$. Point $D$ is inside the angle $BAC$ and is the intersection of exterior bisectors of angles $B$ and $C$ with the common side $BC$. Tangent form $D$ touches $\Omega$ in $E$. FInd $\angle BEC$.

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]

In triangle $ABC$, $K$ and $L$ are points on the side $BC$ ($K$ being closer to $B$ than $L$) such that $BC \cdot KL = BK \cdot CL$ and $AL$ bisects $\angle KAC$. Show that $AL \perp AB.$

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 $ 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 the bisectors of the exterior angles at $ B$ and $ C$ of triangle $ ABC$ meet at $ D.$ Then, if all measurements are in degrees, angle $ BDC$ equals: $ \textbf{(A)}\ \frac {1}{2} (90 \minus{} A) \qquad \textbf{(B)}\ 90 \minus{} A \qquad \textbf{(C)}\ \frac {1}{2} (180 \minus{} A) \qquad \textbf{(D)}\ 180 \minus{} A \qquad \textbf{(E)}\ 180 \minus{} 2A$

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 $IG$.

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]

In a convex quadrilateral $ABCD$ ,$\angle ABC$ and $\angle BCD$ are $\geq 120^o$. Prove that $|AC|$ + $|BD| \geq |AB|+|BC|+|CD|$. (With $|XY|$ we understand the length of the segment $XY$).

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 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]

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.

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]

Let $ k$ be the inscribed circle of non-isosceles triangle $ \triangle ABC$, which center is $ S$. Circle $ k$ touches sides $ BC,CA,AB$ in points $ P,Q,R$ respectively. Line $ QR$ intersects $ BC$ in point $ M$. Let a circle which contains points $ B$ and $ C$ touch $ k$ in point $ N$. Circumscribed circle of $ \triangle MNP$ intersects line $ AP$ in point $ L$, different from $ P$. Prove that points $ S,L$ and $ M$ are collinear.

In the figure $ \overline{AB} \equal{} \overline{AC}$, angle $ BAD \equal{} 30^{\circ}$, and $ \overline{AE} \equal{} \overline{AD}$. [asy]unitsize(20); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair A=(3,3),B=(0,0),C=(6,0),D=(2,0),E=(5,1); draw(A--B--C--cycle); draw(A--D--E); label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",D,S); label("$E$",E,NE);[/asy]Then angle $ CDE$ equals: $ \textbf{(A)}\ 7\frac {1}{2}^{\circ} \qquad\textbf{(B)}\ 10^{\circ} \qquad\textbf{(C)}\ 12\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 15^{\circ} \qquad\textbf{(E)}\ 20^{\circ}$

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]

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$

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$

In a triangle ABC, D is midpoint of BC . If $\angle ADB = 45 ^{\circ}$ and $\angle ACD = 30^{\circ}$, determine $\angle BAD.$