Given $\triangle ABC$ with circumcircle $\ell$. Point $M$ in $\triangle ABC$ such that $AM$ is the angle bisector of $\angle BAC$. Circle with center $M$ and radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively, $(B \not= D, B \not= E)$. Let $P$ be the midpoint of arc $BC$ in $\ell$ that didn't have $A$. Prove that $AP$ angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.

$ ABC$ is a triangle, the bisector of angle $ A$ meets the circumcircle of triangle $ ABC$ in $ A_1$, points $ B_1$ and $ C_1$ are defined similarly. Let $ AA_1$ meet the lines that bisect the two external angles at $ B$ and $ C$ in $ A_0$. Define $ B_0$ and $ C_0$ similarly. Prove that the area of triangle $ A_0B_0C_0 \equal{} 2 \cdot$ area of hexagon $ AC_1BA_1CB_1 \geq 4 \cdot$ area of triangle $ ABC$.

Let $ABC$ be a triangle with $90^\circ \ne \angle A \ne 135^\circ$. Let $D$ and $E$ be external points to the triangle $ABC$ such that $DAB$ and $EAC$ are isoscele triangles with right angles at $D$ and $E$. Let $F = BE \cap CD$, and let $M$ and $N$ be the midpoints of $BC$ and $DE$, respectively. Prove that, if three of the points $A$, $F$, $M$, $N$ are collinear, then all four are collinear.

$ABC$ is an acute triangle with orthocenter $H$. Point $P$ is in triangle $BHC$ that $\angle HPC = 3 \angle HBC $ and $\angle HPB =3 \angle HCB $. Reflection of point $P$ through $BH,CH$ is $X,Y$. if $S$ is the center of circumcircle of $AXY$ , Prove that: $$\angle BAS = \angle CAP$$ [i]Proposed by Pouria Mahmoudkhan Shirazi [/i]

Let $ABC$ be an acuted-angle triangle and let $H$ be it's orthocenter. Let $PQ$ be a segment through $H$ such that $P$ lies on $AB$ and $Q$ lies on $AC$ and such that $ \angle PHB= \angle CHQ$. Finally, in the circumcircle of $\triangle ABC$, consider $M$ such that $M$ is the mid point of the arc $BC$ that doesn't contain $A$. Prove that $MP=MQ$ Proposed by Eduardo Velasco/Marco Figueroa

The incircle of triangle $ABC$ touches its sides $AB$ and $BC$ at points $P$ and $Q.$ The line $PQ$ meets the circumcircle of triangle $ABC$ at points $X$ and $Y.$ Find $\angle XBY$ if $\angle ABC = 90^\circ.$ [i]Proposed by A. Smirnov[/i]

Lines tangent to circle $O$ in points $A$ and $B$, intersect in point $P$. Point $Z$ is the center of $O$. On the minor arc $AB$, point $C$ is chosen not on the midpoint of the arc. Lines $AC$ and $PB$ intersect at point $D$. Lines $BC$ and $AP$ intersect at point $E$. Prove that the circumcentres of triangles $ACE$, $BCD$, and $PCZ$ are collinear.

Points $P$ and $Q$ lie respectively on sides $AB$ and $AC$ of a triangle $ABC$ and $BP=CQ$. Segments $BQ$ and $CP$ cross at $R$. Circumscribed circles of triangles $BPR$ and $CQR$ cross again at point $S$ different from $R$. Prove that point $S$ lies on the bisector of angle $BAC$.

Let $\triangle ABC$ be a triangle with circumcenter $O$ and orthocenter $H$. Let $D$ be a point on the circumcircle of $ABC$ such that $AD \perp BC$. Suppose that $AB = 6, DB = 2$, and the ratio $\tfrac{\text{area}(\triangle ABC)}{\text{area}(\triangle HBC)}=5.$ Then, if $OA$ is the length of the circumradius, then $OA^2$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.

Let $ABC$ be an acute triangle, with $\angle A > 60^\circ$, and let $H$ be it's orthocenter. Let $M$ and $N$ be points on $AB$ and $AC$, respectively, such that $\angle HMB = \angle HNC = 60^\circ$. Also, let $O$ be the circuncenter of $HMN$ and $D$ be a point on the semiplane determined by $BC$ that contains $A$ in such a way that $DBC$ is equilateral. Prove that $H$, $O$ and $D$ are collinear.

In triangle $ABC$, $M,N$ and $P$ are midpoints of sides $BC,CA$ and $AB$. Point $K$ lies on segment $NP$ so that $AK$ bisects $\angle BKC$. Lines $MN,BK$ intersects at $E$ and lines $MP,CK$ intersects at $F$. Suppose that $H$ be the foot of perpendicular line from $A$ to $BC$ and $L$ the second intersection of circumcircle of triangles $AKH, HEF$. Prove that $MK,EF$ and $HL$ are concurrent. [i]Proposed by Alireza Dadgarnia[/i]

Let $ABCD$ be a rectangle. Determine the set of all points $P$ from the region between the parallel lines $AB$ and $CD$ such that $\angle APB=\angle CPD$.

Two triangles $ABC, A’B’C’$ have the same orthocenter $H$ and the same circumcircle with center $O$. Letting $PQR$ be the triangle formed by $AA’, BB’, CC’$, prove that the circumcenter of $PQR$ lies on $OH$.

Let $ABC$ be a triangle inscribed in circle $C$. Circles $C_1, C_2, C_3$ are tangent internally with circle $C$ in $A_1, B_1, C_1$ and tangent to sides $[BC], [CA], [AB]$ in points $A_2, B_2, C_2$ respectively, so that $A, A_1$ are on one side of $BC$ and so on. Lines $A_1A_2, B_1B_2$ and $C_1C_2$ intersect the circle $C$ for second time at points $A’,B’$ and $C’$, respectively. If $ M = BB’ \cap CC’$, prove that $m (\angle MAA’) = 90^\circ$ .

Let $\Gamma$ be a circle and let $d$ be a line such that $\Gamma$ and $d$ have no common points. Further, let $AB$ be a diameter of the circle $\Gamma$; assume that this diameter $AB$ is perpendicular to the line $d$, and the point $B$ is nearer to the line $d$ than the point $A$. Let $C$ be an arbitrary point on the circle $\Gamma$, different from the points $A$ and $B$. Let $D$ be the point of intersection of the lines $AC$ and $d$. One of the two tangents from the point $D$ to the circle $\Gamma$ touches this circle $\Gamma$ at a point $E$; hereby, we assume that the points $B$ and $E$ lie in the same halfplane with respect to the line $AC$. Denote by $F$ the point of intersection of the lines $BE$ and $d$. Let the line $AF$ intersect the circle $\Gamma$ at a point $G$, different from $A$. Prove that the reflection of the point $G$ in the line $AB$ lies on the line $CF$.

Let $ABCD$ be circumscribed quadrilateral such that the midpoints of $AB$,$BC$,$CD$ and $DA$ are $K$, $L$, $M$, $N$ respectively. Let the reflections of the point $M$ wrt the lines $AD$ and $BC$ be $P$ and $Q$ respectively. Let the circumcenter of the triangle $KPQ$ be $R$. Prove that $RN=RL$

Let $ABC$ be a triangle, and let $X$, $Y$, and $Z$ be collinear points such that $AY=AZ$, $BZ=BX$, and $CX=CY$. Points $X'$, $Y'$, and $Z'$ are the reflections of $X$, $Y$, and $Z$ over $BC$, $CA$, and $AB$, respectively. Prove that if $X'Y'Z'$ is a nondegenerate triangle, then its circumcenter lies on the circumcircle of $ABC$. [i]Michael Ren[/i]

Let $\Delta ABC$ be a triangle with an inscribed circle centered at $I$. The line perpendicular to $AI$ at $I$ intersects $\odot (ABC)$ at $P,Q$ such that, $P$ lies closer to $B$ than $C$. Let $\odot (BIP) \cap \odot (CIQ) =S$. Prove that, $SI$ is the angle bisector of $\angle PSQ$

Let $ABC$ be a triangle with $AB = 42$, $AC = 39$, $BC = 45$. Let $E$, $F$ be on the sides $\overline{AC}$ and $\overline{AB}$ such that $AF = 21, AE = 13$. Let $\overline{CF}$ and $\overline{BE}$ intersect at $P$, and let ray $AP$ meet $\overline{BC}$ at $D$. Let $O$ denote the circumcenter of $\triangle DEF$, and $R$ its circumradius. Compute $CO^2-R^2$. [i]Proposed by Yang Liu[/i]

Inside the inscribed quadrilateral $ ABCD $, a point $ P $ is marked such that $ \angle PBC = \angle PDA $, $ \angle PCB = \angle PAD $. Prove that there exists a circle that touches the straight lines $ AB $ and $ CD $, as well as the circles circumscribed by the triangles $ ABP $ and $ CDP $.

Let $ AD$ is centroid of $ ABC$ triangle. Let $ (d)$ is the perpendicular line with $ AD$. Let $ M$ is a point on $ (d)$. Let $ E, F$ are midpoints of $ MB, MC$ respectively. The line through point $ E$ and perpendicular with $ (d)$ meet $ AB$ at $ P$. The line through point $ F$ and perpendicular with $ (d)$ meet $ AC$ at $ Q$. Let $ (d')$ is a line through point $ M$ and perpendicular with $ PQ$. Prove $ (d')$ always pass a fixed point.

In $\Delta ABC$, $AD \perp BC$ at $D$. $E,F$ lie on line $AB$, such that $BD=BE=BF$. Let $I,J$ be the incenter and $A$-excenter. Prove that there exist two points $P,Q$ on the circumcircle of $\Delta ABC$ , such that $PB=QC$, and $\Delta PEI \sim \Delta QFJ$ .

Let $ABC$ be an acute-angled triangle such that $AB<AC$. Let $D$ be the point of intersection of the perpendicular bisector of the side $BC$ with the side $AC$. Let $P$ be a point on the shorter arc $AC$ of the circumcircle of the triangle $ABC$ such that $DP \parallel BC$. Finally, let $M$ be the midpoint of the side $AB$. Prove that $\angle APD=\angle MPB$. [i]Proposed by Dominik Burek, Poland[/i]

A circle which is tangent to the sides $ [AB]$ and $ [BC]$ of $ \triangle ABC$ is also tangent to its circumcircle at the point $ T$. If $ I$ is the incenter of $ \triangle ABC$ , show that $ \widehat{ATI}\equal{}\widehat{CTI}$

Let $A,B,C$ be points on the sides $B_1C_1, C_1A_1,A_1B_1$ of a triangle $A_1B_1C_1$ such that $A_1A,B_1B,C_1C$ are the bisectors of angles of the triangle. We have that $AC = BC$ and $A_1C_1 \neq B_1C_1.$ $(a)$ Prove that $C_1$ lies on the circumcircle of the triangle $ABC$. $(b)$ Suppose that $\angle BAC_1 =\frac{\pi}{6};$ find the form of triangle $ABC$.