Let $B$ and $C$ be two fixed points on a circle centered at $O$ that are not diametrically opposed. Let $A$ be a variable point on the circle distinct from $B$ and $C$ and not belonging to the perpendicular bisector of $BC$. Let $H$ be the orthocenter of $\triangle ABC$, and $M$ and $N$ be the midpoints of the segments $BC$ and $AH$, respectively. The line $AM$ intersects the circle again at $D$, and finally, $NM$ and $OD$ intersect at $P$. Determine the locus of points $P$ as $A$ moves around the circle.

Let $ABC$ be an acute triangle and let $\omega$ be its circumcircle. Let the tangents to $\omega$ through $B,C$ meet each other at point $P$. Prove that the perpendicular bisector of $AB$ and the parallel to $AB$ through $P$ meet at line $AC$.

Let $ABC$ and $PQR$ be two triangles such that [list] [b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$. [b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$ [/list] Prove that $AB+AC=PQ+PR$.

Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX \equal{} \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX \equal{} \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB \equal{} 2\cdot\measuredangle ADX$.

Quadrangle $ABCD$ is inscribed in a circle with diameter $AC$. Points $K$ and $M$ are projections of vertices $A$ and $C$, respectively, onto line $BD$. A line parallel to $BC$ is drawn through point $K$ and intersecting $AC$ at point $P$. Prove that angle $KPM$ is a right angle.

Let $B$ and $C$ be two fixed points on a circle centered at $O$ that are not diametrically opposed. Let $A$ be a variable point on the circle distinct from $B$ and $C$ and not belonging to the perpendicular bisector of $BC$. Let $H$ be the orthocenter of $\triangle ABC$, and $M$ and $N$ be the midpoints of the segments $BC$ and $AH$, respectively. The line $AM$ intersects the circle again at $D$, and finally, $NM$ and $OD$ intersect at $P$. Determine the locus of points $P$ as $A$ moves around the circle.

Let $ABC$ be a isosceles triangle, with $AB=AC$. A line $r$ that pass through the incenter $I$ of $ABC$ touches the sides $AB$ and $AC$ at the points $D$ and $E$, respectively. Let $F$ and $G$ be points on $BC$ such that $BF=CE$ and $CG=BD$. Show that the angle $\angle FIG$ is constant when we vary the line $r$.

[b]i)[/b] Let $0<a<b$.Prove that amongst all triangles having base $a$ and perimeter $a+b$ the triangle having two sides(other than the base) equal to $\frac {b}{2}$ has the maximum area. [b]ii)[/b]Using $i)$ or otherwise, prove that amongst all quadrilateral having give perimeter the square has the maximum area.

Let $ABCD$ be a quadrilateral inscribed in a circle $k$. $AC$ and $BD$ meet at $E$. The rays $\overrightarrow{CB}, \overrightarrow{DA}$ meet at $F$. Prove that the line through the incenters of $\triangle ABE\,,\, \triangle ABF$ and the line through the incenters of $\triangle CDE\,,\, \triangle CDF$ meet at a point lying on the circle $k$. [i]Proposed by N. Beluhov[/i]

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? $ \textbf{(A)}\ \frac{3\sqrt2}{\pi} \qquad \textbf{(B)}\ \frac{3\sqrt3}{\pi} \qquad \textbf{(C)}\ \sqrt3 \qquad \textbf{(D)}\ \frac{6}{\pi} \qquad \textbf{(E)}\ \sqrt3\pi$

Let $ABC$ be an obtuse triangle with $AB = AC$, and let $\Gamma$ be the circle that is tangent to $AB$ at $B$ and to $AC$ at $C$. Let $D$ be the point on $\Gamma$ furthest from $A$ such that $AD$ is perpendicular to $BC$. Point $E$ is the intersection of $AB$ and $DC$, and point $F$ lies on line $AB$ such that $BC = BF$ and $B$ lies on segment $AF$. Finally, let $P$ be the intersection of lines $AC$ and $DB$. Show that $PE = PF$.

According to the estimated number of participants who gave a correct solution, this was the hardest (!) problem from today's paper. So here is this great German killer - be warned! Given a circle k and three pairwisely distinct points A, B, C on this circle. Let h and g be the perpendiculars to the line BC at the points B and C. The perpendicular bisector of the segment AB meets the line h at a point F; the perpendicular bisector of the segment AC meets the line g at a point G. Prove that the product $BF\cdot CG$ is independent from the position of the point A, as long as the points B and C stay fixed. The actual problem behind the problem: Why on hell should the points B and C stay fixed? Darij

In a right triangle rectangle $ABC$ such that $AB = AC$, $M$ is the midpoint of $BC$. Let $P$ be a point on the perpendicular bisector of $AC$, lying in the semi-plane determined by $BC$ that does not contain $A$. Lines $CP$ and $AM$ intersect at $Q$. Calculate the angles that form the lines $AP$ and $BQ$.

The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$