Let $ ABCD$ be a quadrilateral which is a cyclic quadrilateral and a tangent quadrilateral simultaneously. (By a [i]tangent quadrilateral[/i], we mean a quadrilateral that has an incircle.) Let the incircle of the quadrilateral $ ABCD$ touch its sides $ AB$, $ BC$, $ CD$, and $ DA$ in the points $ K$, $ L$, $ M$, and $ N$, respectively. The exterior angle bisectors of the angles $ DAB$ and $ ABC$ intersect each other at a point $ K^{\prime}$. The exterior angle bisectors of the angles $ ABC$ and $ BCD$ intersect each other at a point $ L^{\prime}$. The exterior angle bisectors of the angles $ BCD$ and $ CDA$ intersect each other at a point $ M^{\prime}$. The exterior angle bisectors of the angles $ CDA$ and $ DAB$ intersect each other at a point $ N^{\prime}$. Prove that the straight lines $ KK^{\prime}$, $ LL^{\prime}$, $ MM^{\prime}$, and $ NN^{\prime}$ are concurrent.

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)

Suppose $n>2$ and let $A_1,\dots,A_n$ be points on the plane such that no three are collinear. [b](a)[/b] Suppose $M_1,\dots,M_n$ be points on segments $A_1A_2,A_2A_3,\dots ,A_nA_1$ respectively. Prove that if $B_1,\dots,B_n$ are points in triangles $M_2A_2M_1,M_3A_3M_2,\dots ,M_1A_1M_n$ respectively then \[|B_1B_2|+|B_2B_3|+\dots+|B_nB_1| \leq |A_1A_2|+|A_2A_3|+\dots+|A_nA_1|\] Where $|XY|$ means the length of line segment between $X$ and $Y$. [b](b)[/b] If $X$, $Y$ and $Z$ are three points on the plane then by $H_{XYZ}$ we mean the half-plane that it's boundary is the exterior angle bisector of angle $\hat{XYZ}$ and doesn't contain $X$ and $Z$ ,having $Y$ crossed out. Prove that if $C_1,\dots ,C_n$ are points in ${H_{A_nA_1A_2},H_{A_1A_2A_3},\dots,H_{A_{n-1}A_nA_1}}$ then \[|A_1A_2|+|A_2A_3|+\dots +|A_nA_1| \leq |C_1C_2|+|C_2C_3|+\dots+|C_nC_1|\] Time allowed for this problem was 2 hours.

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 $ZM \cdot ZK + ZL \cdot ZN$.

A circle which is tangent to sides $AB$ and $BC$ of triangle $ABC$ is also tangent to its circumcircle at point $T$. If $I$ in the incenter of triangle $ABC$, show that $\angle ATI=\angle CTI$.

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 $ ABC$ be a right triangle with $ \angle A \equal{} 90^\circ$. Let $ D$ be the midpoint of $ AB$ and let $ E$ be a point on segment $ AC$ such that $ AD \equal{} AE$. Let $ BE$ meet $ CD$ at $ F$. If $ \angle BFC \equal{} 135^\circ$, determine $ BC / 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]

Let [i]O[/i] be the center of circle[i] W[/i]. Two equal chords [i]AB[/i] and [i]CD [/i]of[i] W [/i]intersect at [i]L [/i]such that [i]AL>LB [/i]and [i]DL>LC[/i]. Let [i]M [/i]and[i] N [/i]be points on [i]AL[/i] and [i]DL[/i] respectively such that ([i]ALC[/i])=2*([i]MON[/i]). Prove that the chord of [i]W[/i] passing through [i]M [/i]and [i]N[/i] is equal to [i]AB[/i] and [i]CD[/i].

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).

In triangle $ABC$ the ratio $AC:CB$ is $3:4$. The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$ ($A$ is between $P$ and $B$). The ratio $PA:AB$ is: ${{ \textbf{(A)}\ 1:3 \qquad\textbf{(B)}\ 3:4 \qquad\textbf{(C)}\ 4:3 \qquad\textbf{(D)}\ 3:1 }\qquad\textbf{(E)}\ 7:1 } $

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} $

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 non isosceles triangle $ ABC$ let $ w$ be the angle bisector of the exterior angle at $ C$. Let $ D$ be the point of intersection of $ w$ with the extension of $ AB$. Let $ k_A$ be the circumcircle of the triangle $ ADC$ and analogy $ k_B$ the circumcircle of the triangle $ BDC$. Let $ t_A$ be the tangent line to $ k_A$ in A and $ t_B$ the tangent line to $ k_B$ in B. Let $ P$ be the point of intersection of $ t_A$ and $ t_B$. Given are the points $ A$ and $ B$. Determine the set of points $ P\equal{}P(C )$ over all points $ C$, so that $ ABC$ is a non isosceles, acute-angled triangle.

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 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 $\omega_{1}$ 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.

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 $ 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 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 $AL \perp AB.$

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 $XY$).

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).

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

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}$