Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $CA \neq CB$. Let $CH$ be an altitude and $CL$ be an interior angle bisector. Show that for $X \neq C$ on the line $CL$, we have $\angle XAC \neq \angle XBC$. Also show that for $Y \neq C$ on the line $CH$ we have $\angle YAC \neq \angle YBC$. [i]Bulgaria[/i]

Let $AXYZB$ be a regular pentagon with area $5$ inscribed in a circle with center $O$. Let $Y'$ denote the reflection of $Y$ over $\overline{AB}$ and suppose $C$ is the center of a circle passing through $A$, $Y'$ and $B$. Compute the area of triangle $ABC$. [i]Proposed by Evan Chen[/i]

Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$ [i]Proposed by Nazar Serdyuk, Ukraine[/i]

Points $M$ and $N$ lie on the side $BC$ of the regular triangle $ABC$ ($M$ is between $B$ and $N$), and $\angle MAN=30^\circ.$ The circumcircles of triangles $AMC$ and $ANB$ meet at a point $K.$ Prove that the line $AK$ passes through the circumcenter of triangle $AMN.$

Given a triangle $ABC$. Let $M$ and $N$ be the points where the angle bisectors of the angles $ABC$ and $BCA$ intersect the sides $CA$ and $AB$, respectively. Let $D$ be the point where the ray $MN$ intersects the circumcircle of triangle $ABC$. Prove that $\frac{1}{BD}=\frac{1}{AD}+\frac{1}{CD}$.

Assume $ABC$ is an isosceles triangle that $AB=AC$ Suppose $P$ is a point on extension of side $BC$. $X$ and $Y$ are points on $AB$ and $AC$ that: \[PX || AC \ , \ PY ||AB \] Also $T$ is midpoint of arc $BC$. Prove that $PT \perp XY$

In an acute $ABC$ triangle which has a circumcircle center called $O$, there is a line that perpendiculars to $AO$ line cuts $[AB]$ and $[AC]$ respectively on $D$ and $E$ points. There is a point called $K$ that is different from $AO$ and $BC$'s junction point on $[BC]$. $AK$ line cuts the circumcircle of $ADE$ on $L$ that is different from $A$. $M$ is the symmetry point of $A$ according to $DE$ line. Prove that $K$,$L$,$M$,$O$ are circular.

A circle $ C$ with center $ O.$ and a line $ L$ which does not touch circle $ C.$ $ OQ$ is perpendicular to $ L,$ $ Q$ is on $ L.$ $ P$ is on $ L,$ draw two tangents $ L_1, L_2$ to circle $ C.$ $ QA, QB$ are perpendicular to $ L_1, L_2$ respectively. ($ A$ on $ L_1,$ $ B$ on $ L_2$). Prove that, line $ AB$ intersect $ QO$ at a fixed point. [i]Original formulation:[/i] A line $ l$ does not meet a circle $ \omega$ with center $ O.$ $ E$ is the point on $ l$ such that $ OE$ is perpendicular to $ l.$ $ M$ is any point on $ l$ other than $ E.$ The tangents from $ M$ to $ \omega$ touch it at $ A$ and $ B.$ $ C$ is the point on $ MA$ such that $ EC$ is perpendicular to $ MA.$ $ D$ is the point on $ MB$ such that $ ED$ is perpendicular to $ MB.$ The line $ CD$ cuts $ OE$ at $ F.$ Prove that the location of $ F$ is independent of that of $ M.$

A convex $2004$-sided polygon $P$ is given such that no four vertices are cyclic. We call a triangle whose vertices are vertices of $P$ thick if all other $2001$ vertices of $P$ lie inside the circumcircle of the triangle, and thin if they all lie outside its circumcircle. Prove that the number of thick triangles is equal to the number of thin triangles.

Let $ABC$ be an acute triangle. Let $DAC,EAB$, and $FBC$ be isosceles triangles exterior to $ABC$, with $DA=DC, EA=EB$, and $FB=FC$, such that \[ \angle ADC = 2\angle BAC, \quad \angle BEA= 2 \angle ABC, \quad \angle CFB = 2 \angle ACB. \] Let $D'$ be the intersection of lines $DB$ and $EF$, let $E'$ be the intersection of $EC$ and $DF$, and let $F'$ be the intersection of $FA$ and $DE$. Find, with proof, the value of the sum \[ \frac{DB}{DD'}+\frac{EC}{EE'}+\frac{FA}{FF'}. \]

Let $ABC$ be a triangle. $P$, $Q$ and $R$ are the points of contact of the incircle with sides $AB$, $BC$ and $CA$, respectively. Let $L$, $M$ and $N$ be the feet of the altitudes of the triangle $PQR$ from $R$, $P$ and $Q$, respectively. a) Show that the lines $AN$, $BL$ and $CM$ meet at a point. b) Prove that this points belongs to the line joining the orthocenter and the circumcenter of triangle $PQR$. [i]Aarón Ramírez, El Salvador[/i]

Let scalene triangle $ABC$ have orthocentre $H$ and circumcircle $\Gamma$. $AH$ meets $\Gamma$ at $D$ distinct from $A$. $BH$ and $CH$ meet $CA$ and $AB$ at $E$ and $F$ respectively, and $EF$ meets $BC$ at $P$. The tangents to $\Gamma$ at $B$ and $C$ meet at $T$. Show that $AP$ and $DT$ are concurrent on the circumcircle of $AFE$.

In a triangle $ABC$, the excircle $\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, while the excircle $\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$ and let $L$ be the projection of $C$ onto $PQ$. Show that the quadrilateral $MKLP$ is cyclic. ([i]Bulgaria[/i])

Two circles $ (O_1)$ and $ (O_2)$ touch externally at the point $C$ and internally at the points $A$ and $B$ respectively with another circle $(O)$. Suppose that the common tangent of $ (O_1)$ and $ (O_2)$ at $C$ meets $(O)$ at $P$ such that $PA=PB$. Prove that $PO$ is perpendicular to $AB$.

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/i]

Let $O$ be the circumcenter of triangle $ABC,$ and $D$ be an arbitrary point on the circumcircle of $ABC.$ Let points $X, Y$ and $Z$ be the orthogonal projections of point $D$ onto lines $OA, OB$ and $OC,$ respectively. Prove that the incenter of triangle $XYZ$ is on the Simson-Wallace line of triangle $ABC$ corresponding to point $D.$

$ABC$ is an acute-angled triangle with circumcenter $O$. The circumcircle of $ABO$ intersects$ AC$ and $BC$ at $M$ and $N$. Show that the circumradii of $ABO$ and $MNC$ are the same.

In $\triangle ABC$, $AB = 40$, $BC = 60$, and $CA = 50$. The angle bisector of $\angle A$ intersects the circumcircle of $\triangle ABC$ at $A$ and $P$. Find $BP$. [i]Proposed by Eugene Chen[/i]

Let $I$ be the incenter of triangle $ABC$, $O_1$ a circle through $B$ tangent to $CI$, and $O_2$ a circle through $C$ tangent to $BI$. Prove that $O_1$,$O_2$ and the circumcircle of $ABC$ have a common point.

Consider a circle $C$ of center $O$ and its inner point $Q$, different from $O$. Where we must place a point $P$ on the circle $C$ so that the angle $\angle OPQ$ is the largest possible?

Points $B_1,C_1$ are on $AC$ and $AB$ and $B_1C_1 \parallel BC$. Circumcircle of $ABB_1$ intersect $CC_1$ at $L$. Circumcircle $CLB_1$ is tangent to $AL$. Prove $AL \leq \frac{AC+AC_1}{2}$

Let $\triangle ABC$ be an acute triangle, and let $D$ be the projection of $A$ on $BC$. Let $M,N$ be the midpoints of $AB$ and $AC$ respectively. Let $\Gamma_1$ and $\Gamma_2$ be the circumcircles of $\triangle BDM$ and $\triangle CDN$ respectively, and let $K$ be the other intersection point of $\Gamma_1$ and $\Gamma_2$. Let $P$ be an arbitrary point on $BC$ and $E,F$ are on $AC$ and $AB$ respectively such that $PEAF$ is a parallelogram. Prove that if $MN$ is a common tangent line of $\Gamma_1$ and $\Gamma_2$, then $K,E,A,F$ are concyclic.

Let $ABC$ be a triangle with incenter $I$. The line $AI$ intersects $BC$ and the circumcircle of $ABC$ at the points $T$ and $S$, respectively. Let $K$ and $L$ be the incenters of $SBT$ and $SCT$, respectively, $M$ be the midpoint of $BC$ and $P$ be the reflection of $I$ with respect to $KL$. a) Prove that $M$, $T$, $K$ and $L$ are concyclic. b) Determine the measure of $\angle BPC$.

Point $ I_1$ is the reflection of incentre $ I$ of triangle $ ABC$ across the side $ BC$. The circumcircle of $ BCI_1$ intersects the line $ II_1$ again at point $ P$. It is known that $ P$ lies outside the incircle of the triangle $ ABC$. Two tangents drawn from $ P$ to the latter circle touch it at points $ X$ and $ Y$. Prove that the line $ XY$ contains a medial line of the triangle $ ABC$. [i]Author: L. Emelyanov[/i]

Let $ABC$ be an acute-angled triangle with $AB < AC$ and $\Gamma$ the circumference that passes through $A,\ B$ and $C$. Let $D$ be the point diametrically opposite $A$ on $\Gamma$ and $\ell$ the tangent through $D$ to $\Gamma$. Let $P, Q$ and $R$ be the intersection points of $B C$ with $\ell$, of $A P$ with $\Gamma$ such that $Q \neq A$ and of $Q D$ with the $A$-altitude of the triangle $ABC$, respectively. Define $S$ to be the intersection of $AB$ with $\ell$ and $T$ to be the intersection of $A C$ with $\ell$. Show that $S$ and $T$ lie on the circumference that passes through $A, Q$ and $R$.