Let $P$ and $Q$ be isogonal conjugates inside triangle $ABC$. Let $\omega$ be the circumcircle of $ABC$. Let $A_1$ be a point on arc $BC$ of $\omega$ satisfying $\angle BA_1P = \angle CA_1Q$. Points $B_1$ and $C_1$ are defined similarly. Prove that $AA_1$, $BB_1$, $CC_1$ are concurrent.

Consider an acute-angled triangle $ABC$, with $AC>AB$, and let $\Gamma$ be its circumcircle. Let $E$ and $F$ be the midpoints of the sides $AC$ and $AB$, respectively. The circumcircle of the triangle $CEF$ and $\Gamma$ meet at $X$ and $C$, with $X\neq C$. The line $BX$ and the tangent to $\Gamma$ through $A$ meet at $Y$. Let $P$ be the point on segment $AB$ so that $YP = YA$, with $P\neq A$, and let $Q$ be the point where $AB$ and the parallel to $BC$ through $Y$ meet each other. Show that $F$ is the midpoint of $PQ$.

The circle $\omega$ inscribed in an isosceles triangle $ABC$ ($AC = BC$) touches the side $BC$ at point $D$ .On the extensions of the segment $AB$ beyond points $A$ and $B$, respectively mark the points $K$ and $L$ so that $AK = BL$, The lines $KD$ and $LD$ intersect the circle $\omega$ for second time at points $G$ and $H$, respectively. Prove that point $A$ belongs to the line $GH$.

Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$. [color=red][Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .][/color]

Let $ABC$ be an acute triangle, with $AB \ne AC$. Let $D$ be the midpoint of the line segment $BC$, and let $E$ and $F$ be the projections of $D$ onto the sides $AB$ and $AC$, respectively. If $M$ is the midpoint of the line segment $EF$, and $O$ is the circumcenter of triangle $ABC$, prove that the lines $DM$ and $AO$ are parallel. [hide=PS] As source was given [url=https://artofproblemsolving.com/community/c629086_caucasus_mathematical_olympiad]Caucasus MO[/url], but I was unable to find this problem in the contest collections [/hide]

Medians $AM_A,BM_B,CM_C$ of triangle $ABC$ intersect at $M$.Let $\Omega_A$ be circumcircle of triangle passes through midpoint of $AM$ and tangent to $BC$ at $M_A$.Define $\Omega_B$ and $\Omega_C$ analogusly.Prove that $\Omega_A,\Omega_B$ and $\Omega_C$ intersect at one point.(A.Yakubov) [hide=P.S]sorry for my mistake in translation :blush: :whistling: .thank you jred for your help :coolspeak: [/hide]

Let $\triangle ABC$ be scalene, with $BC$ as the largest side. Let $D$ be the foot of the perpendicular from $A$ on side $BC$. Let points $K,L$ be chosen on the lines $AB$ and $AC$ respectively, such that $D$ is the midpoint of segment $KL$. Prove that the points $B,K,C,L$ are concyclic if and only if $\angle BAC=90^{\circ}$.

Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$ . Let the perpendicular bisector of $EF$ intersect the line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect the lines $AC$ and $AB$ at $S$ and $T$ , respectively. We call the pair $(E, F )$ $\textit{interesting}$, if the quadrilateral $KSAT$ is cyclic. Suppose that the pairs $(E_1 , F_1 )$ and $(E_2 , F_2 )$ are interesting. Prove that $\displaystyle\frac{E_1 E_2}{AB}=\frac{F_1 F_2}{AC}$ [i]Proposed by Ali Zamani, Iran[/i]

Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.

Let $D,E,F$ denote the tangent points of the incircle of $ABC$ with sides $BC,AC,AB$ respectively. Let $M$ be the midpoint of the segment $EF$. Let $L$ be the intersection point of the circle passing through $D,M,F$ and the segment $AB$, $K$ be the intersection point of the circle passing through $D,M,E$ and the segment $AC$. Prove that the circle passing through $A,K,L$ touches the line $BC$

An isosceles triangle $ABC$ with $AC = BC$ is given. Let $M$ be the midpoint of the side $AB$ and let $P$ be a point inside the triangle such that $\angle PAB = \angle PBC$. Prove that $\angle APM + \angle BPC = 180 \textdegree $

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]

A triangle $ABC$, its circumcircle, and its incenter $I$ are drawn on the plane. Construct the circumcenter of $ABC$ using only a ruler.

Consider an acute-angled triangle $ABC$ with $AB<AC$ and let $\omega$ be its circumscribed circle. Let $t_B$ and $t_C$ be the tangents to the circle $\omega$ at points $B$ and $C$, respectively, and let $L$ be their intersection. The straight line passing through the point $B$ and parallel to $AC$ intersects $t_C$ in point $D$. The straight line passing through the point $C$ and parallel to $AB$ intersects $t_B$ in point $E$. The circumcircle of the triangle $BDC$ intersects $AC$ in $T$, where $T$ is located between $A$ and $C$. The circumcircle of the triangle $BEC$ intersects the line $AB$ (or its extension) in $S$, where $B$ is located between $S$ and $A$. Prove that $ST$, $AL$, and $BC$ are concurrent. $\text{Vangelis Psychas and Silouanos Brazitikos}$