A line, which passes through the incentre $I$ of the triangle $ABC$, meets its sides $AB$ and $BC$ at the points $M$ and $N$ respectively. The triangle $BMN$ is acute. The points $K,L$ are chosen on the side $AC$ such that $\angle ILA=\angle IMB$ and $\angle KC=\angle INB$. Prove that $AM+KL+CN=AC$. [i]S. Berlov[/i]

Let $I$ denote the center of the circle inscribed in the right triangle $ABC$ with right angle at the vertex $A$. Next, denote by $M$ and $N$ the midpoints of the lines $AB$ and $BI$. Prove that the line $CI$ is tangent to the circumscribed circle of triangle $BMN$. (Patrik Bak, Josef Tkadlec)

$ABCD$ is a trapezoid with $AB || CD$. There are two circles $\omega_1$ and $\omega_2$ is the trapezoid such that $\omega_1$ is tangent to $DA$, $AB$, $BC$ and $\omega_2$ is tangent to $BC$, $CD$, $DA$. Let $l_1$ be a line passing through $A$ and tangent to $\omega_2$(other than $AD$), Let $l_2$ be a line passing through $C$ and tangent to $\omega_1$ (other than $CB$). Prove that $l_1 || l_2$.

Let $ABC$ be a triangle inscribed in a circle of center $O$ and radius $R$. Let $I$ be the incenter of $ABC$, and let $r$ be the inradius of the same triangle, $O\neq I$, and let $G$ be its centroid. Prove that $IG\perp BC$ if and only if $b=c$ or $b+c=3a$.

Given that circle $ \omega$ is tangent internally to circle $ \Gamma$ at $ S.$ $ \omega$ touches the chord $ AB$ of $ \Gamma$ at $ T$. Let $ O$ be the center of $ \omega.$ Point $ P$ lies on the line $ AO.$ Show that $ PB\perp AB$ if and only if $ PS\perp TS.$

Let $ABC$ to be a triangle with incenter $I$. $\omega_{A}$, $\omega_{B}$ and $\omega_{C}$ are the incircles of the triangles $BIC$, $CIA$ and $AIB$, repectively. After all, $T$ is the tangent point between $\omega_{A}$ and $BC$. Prove that the other internal common tangent to $\omega_{B}$ and $\omega_{C}$ passes through the point $T$.

Let $ M$ be a point on $ BC$ and $ N$ be a point on $ AB$ such that $ AM$ and $ CN$ are angle bisectors of the triangle $ ABC$. Given that $ \frac {\angle BNM}{\angle MNC} \equal{} \frac {\angle BMN}{\angle NMA}$, prove that the triangle $ ABC$ is isosceles.

Let $O$ and $H$ be the circumcenter and the orthocenter of a triangle $ABC$. The line passing through the midpoint of $OH$ and parallel to $BC$ meets $AB$ and $AC$ at points $D$ and $E$. It is known that $O$ is the incenter of triangle $ADE$. Find the angles of $ABC$.

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

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 the triangle $ ABC,$ with $ \angle A \equal{} 60 ^{\circ},$ a parallel $ IF$ to $ AC$ is drawn through the incenter $ I$ of the triangle, where $ F$ lies on the side $ AB.$ The point $ P$ on the side $ BC$ is such that $ 3BP \equal{} BC.$ Show that $ \angle BFP \equal{} \frac{\angle B}{2}.$

Let $ABC$ be a triangle, $I$ its incenter, and $D$ a point on the arc $BC$ of the circumcircle of $ABC$ not containing $A$. The bisector of the angle $\angle ADB$ intesects the segment $AB$ at $E$. The bisector of the angle $\angle CDA$ intesects the segment $AC$ at $F$. Prove that the points $E, F,I$ are collinear. (Malik Talbi)

The points $A_1,B_1,C_1$ lie on the sides $BC,CA$ and $AB$ of the triangle $ABC$ respectively. Suppose that $AB_1-AC_1=CA_1-CB_1=BC_1-BA_1$. Let $O_A,O_B$ and $O_C$ be the circumcentres of triangles $AB_1C_1,A_1BC_1$ and $A_1B_1C$ respectively. Prove that the incentre of triangle $O_AO_BO_C$ is the incentre of triangle $ABC$ too.

Given a cyclic quadrilateral $ABCD$ with the circumcenter $O$, with $BC$ and $AD$ not parallel. Let $P$ be the intersection of $AC$ and $BD$. Let $E$ be the intersection of the rays $AB$ and $DC$. Let $I$ be the incenter of $EBC$ and the incircle of $EBC$ touches $BC$ at $T_1$. Let $J$ be the excenter of $EAD$ that touches $AD$ and the excircle of $EAD$ that touches $AD$ touches $AD$ at $T_2$. Let $Q$ be the intersection between $IT_1$ and $JT_2$. Prove that $O,P,Q$ are collinear.

The incircle of a right-angled triangle $ABC$ touches hypotenus $AB$ at $P$, $BC$ and $AC$ at $R$ and $Q$ respectively. $C_1$ and $C_2$ are reflections of $C$ in $PQ$ and $PR$. Find the angle $C_1IC_2$, where $I$ is the incenter of $ABC$.

Let $D$ be the point on the side $BC$ of the triangle $ABC$ such that $AD$ is the bisector of $\angle CAB$. Let $I$ be the incenter of$ ABC.$ [i](a)[/i] Construct the points $P$ and $Q$ on the sides $AB$ and $AC$, respectively, such that $PQ$ is parallel to $BC$ and the perimeter of the triangle $APQ$ is equal to $k \cdot BC$, where $k$ is a given rational number. [i](b) [/i]Let $R$ be the intersection point of $PQ$ and $AD$. For what value of $k$ does the equality $AR = RI$ hold? [i](c)[/i] In which case do the equalities $AR = RI = ID$ hold?

The triangle $ABC$ is given. Let $A'$ be the midpoint of the side $BC$, $B_c{}$ be the projection of $B{}$ onto the bisector of the angle $ACB{}$ and $C_b$ be the projection of the point $C{}$ onto the bisector of the angle $ABC$. Let $A_0$ be the center of the circle passing through $A', B_c, C_b$. The points $B_0$ and $C_0$ are defined similarly. Prove that the incenter of the triangle $ABC$ coincides with the orthocenter of the triangle $A_0B_0C_0$.

Let $ABC$, $AC \not= BC$, be an acute triangle with orthocenter $H$ and incenter $I$. The lines $CH$ and $CI$ meet the circumcircle of $\bigtriangleup ABC$ at points $D$ and $L$, respectively. Prove that $\angle CIH = 90^{\circ}$ if and only if $\angle IDL = 90^{\circ}$

Point $C_{1}$ of hypothenuse $AC$ of a right-angled triangle $ABC$ is such that $BC = CC_{1}$. Point $C_{2}$ on cathetus $AB$ is such that $AC_{2} = AC_{1}$; point $A_{2}$ is defined similarly. Find angle $AMC$, where $M$ is the midpoint of $A_{2}C_{2}$.

In scalene $\triangle ABC$, $I$ is the incenter, $I_a$ is the $A$-excenter, $D$ is the midpoint of arc $BC$ of the circumcircle of $ABC$ not containing $A$, and $M$ is the midpoint of side $BC$. Extend ray $IM$ past $M$ to point $P$ such that $IM = MP$. Let $Q$ be the intersection of $DP$ and $MI_a$, and $R$ be the point on the line $MI_a$ such that $AR\parallel DP$. Given that $\frac{AI_a}{AI}=9$, the ratio $\frac{QM} {RI_a}$ can be expressed in the form $\frac{m}{n}$ for two relatively prime positive integers $m,n$. Compute $m+n$. [i]Ray Li.[/i] [hide="Clarifications"][list=1][*]"Arc $BC$ of the circumcircle" means "the arc with endpoints $B$ and $C$ not containing $A$".[/list][/hide]

A circle $\omega$ not passing through any vertex of $\triangle ABC$ intersects each of the segments $AB$, $BC$, $CA$ in 2 distinct points. Prove that the incenter of $\triangle ABC$ lies inside $\omega$. [i]Evan O' Dorney.[/i]

In a triangle $ABC$, point $P$ is the incenter and $A'$, $B'$, $C'$ its orthogonal projections on $BC$, $CA$, $AB$, respectively. Show that $\angle B'A'C'$ is acute.

Let $ I$ be the incenter of triangle $ ABC.$ Let $ M,N$ be the midpoints of $ AB,AC,$ respectively. Points $ D,E$ lie on $ AB,AC$ respectively such that $ BD\equal{}CE\equal{}BC.$ The line perpendicular to $ IM$ through $ D$ intersects the line perpendicular to $ IN$ through $ E$ at $ P.$ Prove that $ AP\perp BC.$

Let $ABCD$ be a tetrahedron with $AB=CD=1300$, $BC=AD=1400$, and $CA=BD=1500$. Let $O$ and $I$ be the centers of the circumscribed sphere and inscribed sphere of $ABCD$, respectively. Compute the smallest integer greater than the length of $OI$. [i] Proposed by Michael Ren [/i]

In isosceles triangle $ABC$ ($AB = BC$) one draws the angle bisector $CD$. The perpendicular to $CD$ through the center of the circumcircle of $ABC$ intersects $BC$ at $E$. The parallel to $CD$ through $E$ meets $AB$ at $F$. Show that $BE$ = $FD$. [i]M. Sonkin[/i]