Found problems: 6
Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the side $BC$ at $A^{\prime}$, and the line $AA^{\prime}$ meets the incircle again at a point $P$. Let the lines $CP$ and $BP$ meet the incircle of triangle $ABC$ again at $N$ and $M$, respectively. Prove that the lines $AA^{\prime}$, $BN$ and $CM$ are concurrent.
Let incircle $(I)$ of triangle $ABC$ touch the sides $BC,CA,AB$ at $D,E,F$ respectively. Let $(O)$ be the circumcircle of $ABC$. Ray $EF$ meets $(O)$ at $M$. Tangents at $M$ and $A$ of $(O)$ meet at $S$. Tangents at $B$ and $C$ of $(O)$ meet at $T$. Line $TI$ meets $OA$ at $J$. Prove that $\angle ASJ=\angle IST$.
Let $ABC$ be an obtuse triangle, and let $H$ denote its orthocenter. Let $\omega_A$ denote the circle with center $A$ and radius $AH$. Let $\omega_B$ and $\omega_C$ be defined in a similar way. For all points $X$ in the plane of triangle $ABC$ let circle $\Omega(X)$ be defined in the following way (if possible): take the polars of point $X$ with respect to circles $\omega_A$, $\omega_B$ and $\omega_C$, and let $\Omega(X)$ be the circumcircle of the triangle defined by these three lines.
With a possible exception of finitely many points find the locus of points $X$ for which point $X$ lies on circle $\Omega(X)$.
[i]Proposed by Vilmos Molnár-Szabó, Budapest[/i]
Let $I$ be the incenter of a non-isosceles triangle $ABC$. The line $AI$ intersects the circumcircle of the triangle $ABC$ at $A$ and $D$. Let $M$ be the middle point of the arc $BAC$. The line through the point $I$ perpendicular to $AD$ intersects $BC$ at $F$. The line $MI$ intersects the circle $BIC$ at $N$.
Prove that the line $FN$ is tangent to the circle $BIC$.
Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the side $BC$ at $A^{\prime}$, and the line $AA^{\prime}$ meets the incircle again at a point $P$. Let the lines $CP$ and $BP$ meet the incircle of triangle $ABC$ again at $N$ and $M$, respectively. Prove that the lines $AA^{\prime}$, $BN$ and $CM$ are concurrent.
Let $H$ be the orthocenter of a triangle $ABC$, and $M$, $N$ be the midpoints of segments $BC$, $AH$ respectively. The perpendicular from $N$ to $MH$ meets $BC$ at point $A^{\prime}$. Points $B^{\prime}$ and $C^{\prime}$ are defined similarly. Prove that $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ are collinear.
Proposed by: F.Ivlev