In a triangle $A_1A_2A_3$, the excribed circles corresponding to sides $A_2A_3$, $A_3A_1$, $A_1A_2$ touch these sides at $T_1$, $T_2$, $T_3$, respectively. If $H_1$, $H_2$, $H_3$ are the orthocenters of triangles $A_1T_2T_3$, $A_2T_3T_1$, $A_3T_1T_2$, respectively, prove that lines $H_1T_1$, $H_2T_2$, $H_3T_3$ are concurrent.

Triangle $ABC$ is inscribed in circle $\omega$. Circle $\gamma$ is tangent to $AB$ and $AC$ at points $P$ and $Q$ respectively. Also circle $\gamma$ is tangent to circle $\omega$ at point $S$. Let the intesection of $AS$ and $PQ$ be $T$. Prove that $\angle{BTP}=\angle{CTQ}$.

The triangle $ ABC$ is given. On the extension of the side $ AB$ we construct the point $ R$ with $ BR \equal{} BC$, where $ AR > BR$ and on the extension of the side $ AC$ we construct the point $ S$ with $ CS \equal{} CB$, where $ AS > CS$. Let $ A_1$ be the point of intersection of the diagonals of the quadrilateral $ BRSC$. Analogous we construct the point $ T$ on the extension of the side $ BC$, where $ CT \equal{} CA$ and $ BT > CT$ and on the extension of the side $ BA$ we construct the point $ U$ with $ AU \equal{} AC$, where $ BU > AU$. Let $ B_1$ be the point of intersection of the diagonals of the quadrilateral $ CTUA$. Likewise we construct the point $ V$ on the extension of the side $ CA$, where $ AV \equal{} AB$ and $ CV > AV$ and on the extension of the side $ CB$ we construct the point $ W$ with $ BW \equal{} BA$ and $ CW > BW$. Let $ C_1$ be the point of intersection of the diagonals of the quadrilateral $ AVWB$. Show that the area of the hexagon $ AC_1BA_1CB_1$ is equal to the sum of the areas of the triangles $ ABC$ and $ A_1B_1C_1$.

$A, B$ are points on circle $O$ satisfying $\angle AOB < 120^{\circ} $. $C$ is a point on the tangent line of $O$ passing through $A$ satisfying $AB=AC$ and $\angle BAC < 90^{\circ} $. $D$ is the intersection of $O$ and $BC$ not $B$, and $I$ is the incenter of $ABD$. Prove that $AE=AC$ where $E$ is the intersection of $CI$ and $AD$.

In tetrahedron $SABC$, the circumcircles of faces $SAB$, $SBC$, and $SCA$ each have radius $108$. The inscribed sphere of $SABC$, centered at $I$, has radius $35.$ Additionally, $SI = 125$. Let $R$ be the largest possible value of the circumradius of face $ABC$. Given that $R$ can be expressed in the form $\sqrt{\frac{m}{n}}$, where $m$ and $n$ are relatively prime positive integers, find $m+n$. [i]Author: Alex Zhu[/i]

Let $ABC$ be an acute angled triangle satisfying the conditions $AB>BC$ and $AC>BC$. Denote by $O$ and $H$ the circumcentre and orthocentre, respectively, of the triangle $ABC.$ Suppose that the circumcircle of the triangle $AHC$ intersects the line $AB$ at $M$ different from $A$, and the circumcircle of the triangle $AHB$ intersects the line $AC$ at $N$ different from $A.$ Prove that the circumcentre of the triangle $MNH$ lies on the line $OH$.

$A,B,C$ are on circle $\mathcal C$. $I$ is incenter of $ABC$ , $D$ is midpoint of arc $BAC$. $W$ is a circle that is tangent to $AB$ and $AC$ and tangent to $\mathcal C$ at $P$. ($W$ is in $\mathcal C$) Prove that $P$ and $I$ and $D$ are on a line.

Let $ABC$ be a triangle with incentre $I$. The angle bisectors $AI$, $BI$ and $CI$ meet $[BC]$, $[CA]$ and $[AB]$ at $D$, $E$ and $F$, respectively. The perpendicular bisector of $[AD]$ intersects the lines $BI$ and $CI$ at $M$ and $N$, respectively. Show that $A$, $I$, $M$ and $N$ lie on a circle.

An acute-angle non-isosceles triangle $ ABC $ is given. The point $ H $ is its orthocenter, the points $ O $ and $ I $ are the centers of its circumscribed and inscribed circles, respectively. The circumcircle of the triangle $ OIH $ passes through the vertex $ A $. Prove that one of the angles of the triangle is $ 60^\circ $.

A circle concentric with the inscribed circle of $ABC$ intersects the sides of the triangle at six points forming a convex hexagon $A_1A_2B_1B_2C_1C_2$ (points $C_1$ and $C_2$ on the $AB$ side, $A_1$ and $A_2$ on $BC$, $B_1$ and $B_2$ on $AC$). Prove that if line $A_1B_1$ is parallel to the bisector of angle $B$, then line $A_2C_2$ is parallel to the bisector of angle $C$.

Given an isosceles triangle $ABC$ with base $AB$, cirumcenter $O$, incenter $S$ and $\angle C<60^\circ$. The circumcircle of $AOS$ intersects $AC$ at $D$. Prove that $SD\parallel BC$ and $AS\perp OD$.

Let $D$, $E$ and $F$ be points in which incircle of triangle $ABC$ touches sides $BC$, $CA$ and $AB$, respectively, and let $I$ be a center of that circle.Furthermore, let $P$ be a foot of perpendicular from point $I$ to line $AD$, and let $M$ be midpoint of $DE$. If $\{N\}=PM\cap{AC}$, prove that $DN \parallel EF$

Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$. [i]Amol Aggarwal.[/i]

Let a cyclic quadrilateral $ ABCD$, $ AC \cap BD \equal{} E$ and let a circle $ \Gamma$ internally tangent to the arch $ BC$ (that not contain $ D$) in $ T$ and tangent to $ BE$ and $ CE$. Call $ R$ the point where the angle bisector of $ \angle ABC$ meet the angle bisector of $ \angle BCD$ and $ S$ the incenter of $ BCE$. Prove that $ R$, $ S$ and $ T$ are collinear. [i](Gabriel Giorgieri)[/i]

Lines that are drawn perpendicular to the faces of a triangular pyramid through the centers of the inscribed circles intersect at one point. Prove that the sums of the opposite edges of such a pyramid are equal to each other.

Let $ ABC$ be a triangle with $ AB \equal{} AC$ . The angle bisectors of $ \angle C AB$ and $ \angle AB C$ meet the sides $ B C$ and $ C A$ at $ D$ and $ E$ , respectively. Let $ K$ be the incentre of triangle $ ADC$. Suppose that $ \angle B E K \equal{} 45^\circ$ . Find all possible values of $ \angle C AB$ . [i]Jan Vonk, Belgium, Peter Vandendriessche, Belgium and Hojoo Lee, Korea [/i]

Let $O$ and $I$ be circumcenter and incenter of triangle $ABC$. Let incircle of $ABC$ touches sides $BC$, $CA$ and $AB$ in points $D$, $E$ and $F$, respectively. Lines $FD$ and $CA$ intersect in point $P$, and lines $DE$ and $AB$ intersect in point $Q$. Furthermore, let $M$ and $N$ be midpoints of $PE$ and $QF$. Prove that $OI \perp MN$

Given points $A,B,C$ and $D$ lie a circle. $AC\cap BD=K$. $I_1, I_2,I_3$ and $I_4$ incenters of $ABK,BCK,CDK,DKA$. $M_1,M_2,M_3,M_4$ midpoints of arcs $AB,BC,CA,DA$ . Then prove that $M_1I_1,M_2I_2,M_3I_3,M_4I_4$ are concurrent.

In acute triangle $ABC, AB<AC$, $I$ is its incenter, $D$ is the foot of perpendicular from $I$ to $BC$, altitude $AH$ meets $BI,CI$ at $P,Q$ respectively. Let $O$ be the circumcenter of $\triangle IPQ$, extend $AO$ to meet $BC$ at $L$. Circumcircle of $\triangle AIL$ meets $BC$ again at $N$. Prove that $\frac{BD}{CD}=\frac{BN}{CN}$.

Let $\Gamma$ be a circumference and $A$ a point outside it. Let $B$ and $C$ be points in $\Gamma$ such that $AB$ and $AC$ are tangent to $\Gamma$. Let $P$ be a point in $\Gamma$. Let $D$, $E$ and $F$ be points in $BC$, $AC$ and $AB$ respectively, such that $PD \perp BC$, $PE \perp AC$, and $PF \perp AB$. Show that $PD^2 = PE \cdot PF$

Diagonals of a cyclic quadrilateral $ABCD$ intersect at point $O$. The circumscribed circles of triangles $AOB$ and $COD$ intersect at point $M$ on the side $AD$. Prove that the point $O$ is the center of the inscribed circle of the triangle $BMC$.

Let $ABC$ be a triangle and $D$ be a point on $BC$ such that $AB+BD=AC+CD$. The line $AD$ intersects the incircle of triangle $ABC$ at $X$ and $Y$ where $X$ is closer to $A$ than $Y$ i. Suppose $BC$ is tangent to the incircle at $E$, prove that: 1) $EY$ is perpendicular to $AD$; 2) $XD=2IM$ where $I$ is the incentre and $M$ is the midpoint of $BC$.

Let $ABCD$ be a convex quadrilateral with $AB = AD$ and $CB = CD$. The bisector of $\angle BDC$ intersects $BC$ at $L$, and $AL$ intersects $BD$ at $M$, and it is known that $BL = BM$. Determine the value of $2\angle BAD + 3\angle BCD$.

Points $A$, $B$, $C$ and $D$ are chosen on a line in that order, with $AB$ and $CD$ greater than $BC$. Equilateral triangles $APB$, $BCQ$ and $CDR$ are constructed so that $P$, $Q$ and $R$ are on the same side with respect to $AD$. If $\angle PQR=120^\circ$, show that \[\frac{1}{AB}+\frac{1}{CD}=\frac{1}{BC}.\]

Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.