Let $ABC$ be a triangle, $H_a, H_b$ and $H_c$ the feet of its altitudes from $A, B$ and $C$, respectively, $T_a, T_b, T_c$ its touchpoints of the incircle with the sides $BC, CA$ and $AB$, respectively. The circumcircles of triangles $AH_bH_c$ and $AT_bT_c$ intersect again at $A'$. The circumcircles of triangles $BH_cH_a$ and $BT_cT_a$ intersect again at $B'$. The circumcircles of triangles $CH_aH_b$ and $CT_aT_b$ intersect again at $C'$. Prove that the points $A',B',C'$ are collinear. Malik Talbi

Circles ${{w} _ {1}}$ and ${{w} _ {2}}$ with centers ${{O} _ {1}}$ and ${{O} _ {2}}$ intersect at points $A$ and $B$, respectively. The line ${{O} _ {1}} {{O} _ {2}}$ intersects ${{w} _ {1}}$ at the point $Q$, which does not lie inside the circle ${{w} _ {2}}$, and ${{w} _ {2}}$ at the point $X$ lying inside the circle ${{w} _ {1} }$. Around the triangle ${{O} _ {1}} AX$ circumscribe a circle ${{w} _ {3}}$ intersecting the circle ${{w} _ {1}}$ for the second time in point $T$. The line $QT$ intersects the circle ${{w} _ {3}}$ at the point $K$, and the line $QB$ intersects ${{w} _ {2}}$ the second time at the point $H$. Prove that a) points $T, \, \, X, \, \, B$ lie on one line; b) points $K, \, \, X, \, \, H$ lie on one line. (Vadym Mitrofanov)

In triangle $ABC$ let $I_a$ be the $A$-excenter. Let $\omega$ be an arbitrary circle that passes through $A,I_a$ and intersects the extensions of sides $AB,AC$ (extended from $B,C$) at $X,Y$ respectively. Let $S,T$ be points on segments $I_aB,I_aC$ respectively such that $\angle AXI_a=\angle BTI_a$ and $\angle AYI_a=\angle CSI_a$.Lines $BT,CS$ intersect at $K$. Lines $KI_a,TS$ intersect at $Z$. Prove that $X,Y,Z$ are collinear. [i]Proposed by Hooman Fattahi[/i]

In triangle $ABC$, the points $M$ and $N$ are the midpoints of the sides $AB$ and $AC$, respectively. The bisectors of interior angles $\angle ABC$ and $\angle BCA$ intersect the line $MN$ at points $P$ and $Q$, respectively. Let $p$ be the tangent to the circumscribed circle of the triangle $AMP$ passing through point $P$, and $q$ be the tangent to the circumscribed circle of the triangle $ANQ$ passing through point $Q$. Prove that the lines $p$ and $q$ intersect on line $BC$.

Let $A$ be a point on the circle $\omega$ centred at $B$ and $\Gamma$ a circle centred at $A$. For $i=1,2,3$, a chord $P_iQ_i$ of $\omega$ is tangent to $\Gamma$ at $S_i$ and another chord $P_iR_i$ of $\omega$ is perpendicular to $AB$ at $M_i$. Let $Q_iT_i$ be the other tangent from $Q_i$ to $\Gamma$ at $T_i$ and $N_i$ be the intersection of $AQ_i$ with $M_iT_i$. Prove that $N_1,N_2,N_3$ are collinear.

The circle $ \omega $ inscribed in triangle $ABC$ touches its sides $AB, BC, CA$ at points $K, L, M$ respectively. In the arc $KL$ of the circle $ \omega $ that does not contain the point $M$, we select point $S$. Denote by $P, Q, R, T$ the intersection points of straight $AS$ and $KM, ML$ and $SC, LP$ and $KQ, AQ$ and $PC$ respectively. It turned out that the points $R, S$ and $M$ are collinear. Prove that the point $T$ also lies on the line $SM$.

Let $ABC$ be an acute-angled triangle. $M ,N$ are any two points on the sides $AB , AC$ respectively. The circles with the diameters $BN$ and $CM$ intersect at points $P$ and $Q$. Show that the points $P, Q$ and the orthocenter of the triangle $ABC$ lie on a straight line.

Let $ABCDA' B' C' D '$ be a cube , $M$ the foot of the perpendicular from $A$ on the plane $(A'CD)$, $N$ the foot of the perpendicular from $B$ on the diagonal $A'C$ and $P$ is symmetric of the point $D$ with respect to $C$. Show that the points $M, N, P$ are collinear.

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)

On the diagonal $AC$ of the rhombus $ABCD$, a point $E$ is taken, which is different from points $A$ and $C$, and on the lines $AB$ and $BC$ are points $N$ and $M$, respectively, with $AE = NE$ and $CE = ME$. Let $K$ be the intersection point of lines $AM$ and $CN$. Prove that points $K, E$ and $D$ are collinear.

Let $ABC$ be an acute triangle. Suppose that circle $\Gamma_1$ has it's center on the side $AC$ and is tangent to the sides $AB$ and $BC$, and circle $\Gamma_2$ has it's center on the side $AB$ and is tangent to the sides $AC$ and $BC$. The circles $\Gamma_1$ and $ \Gamma_2$ intersect at two points $P$ and $Q$. Show that if $A, P, Q$ are collinear, then $AB = AC$.

Let $ABCD$ be a convex quadrilateral and let $P$ and $Q$ be variable points inside this quadrilateral so that $\angle APB=\angle CPD=\angle AQB=\angle CQD$. Prove that the lines $PQ$ obtained in this way all pass through a fixed point , or they are all parallel.

Let $\vartriangle ABC$ be a right-angled triangle with $\angle A = 90^o$ and circumcircle $\Gamma$. The inscribed circle is tangent to $BC$ in point $D$. Let $E$ be the midpoint of the arc $AB$ of $\Gamma$ not containing $C$ and let $F$ be the midpoint of the arc $AC$ of $\Gamma$ not containing $B$. (a) Prove that $\vartriangle ABC \sim \vartriangle DEF$. (b) Prove that $EF$ goes through the points of tangency of the incircle to $AB$ and $AC$.

Segments connect vertices $A, B, C$ of $\vartriangle ABC$ with respective points $A_1, B_1, C_1$ on the opposite sides of the triangle. Prove that the midpoints of segments $AA_1, BB_1, CC_1$ do not belong to one straight line.

Consider the set of all integer points in $Z^3$. Sasha and Masha play such a game. At first, Masha marks an arbitrary point. After that, Sasha marks all the points on some a plane perpendicular to one of the coordinate axes and at no point, which Masha noted. Next, they continue to take turns (Masha can't to select previously marked points, Sasha cannot choose the planes on which there are points said Masha). Masha wants to mark $n$ consecutive points on some line that parallel to one of the coordinate axes, and Sasha seeks to interfere with it. Find all $n$, in which Masha can achieve the desired result.