$ABCD$ is a convex quadrilateral; half-lines $DA$ and $CB$ meet at point $Q$; half-lines $BA$ and $CD$ meet at point $P$. It is known that $\angle AQB=\angle APD$. The bisector of angle $\angle AQB$ meets the sides $AB$ and $CD$ of the quadrilateral at points $X$ and $Y$, respectively; the bisector of angle $\angle APD$ meets the sides $AD$ and $BC$ at points $Z$ and $T$, respectively. The circumcircles of triangles $ZQT$ and $XPY$ meet at point $K$ inside the quadrilateral. Prove that $K$ lies on the diagonal $AC$. [i]Proposed by S. Berlov[/i]

Let $ABCD$ be a parallelogram. Half-circles $\omega_1,\omega_2,\omega_3$ and $\omega_4$ with diameters $AB,BC,CD$ and $DA$, respectively, are erected on the exterior of $ABCD$. Line $l_1$ is parallel to $BC$ and cuts $\omega_1$ at $X$, segment $AB$ at $P$, segment $CD$ at $R$ and $\omega_3$ at $Z$. Line $l_2$ is parallel to $AB$ and cuts $\omega_2$ at $Y$, segment $BC$ at $Q$, segment $DA$ at $S$ and $\omega_4$ at $W$. If $XP\cdot RZ=YQ\cdot SW$, prove that $PQRS$ is cyclic. [i]Proposed by José Alejandro Reyes González[/i]

Let $ AD$ be height of triangle $ \triangle ABC$ and $ R$ circumradius. Denote by $ E$ and $ F$ feet of perpendiculars from point $ D$ to sides $ AB$ and $ AC$. If $ AD\equal{}R\sqrt{2}$, prove that circumcenter of triangle $ \triangle ABC$ lies on line $ EF$.

Let $ ABCD$ be a quadrilateral inscribed in a circle of center $ O$. Let M and N be the midpoints of diagonals $ AC$ and $ BD$, respectively and let $ P$ be the intersection point of the diagonals $ AC$ and $ BD$ of the given quadrilateral .It is known that the points $ O,M,Np$ are distinct. Prove that the points $ O,N,A,C$ are concyclic if and only if the points $ O,M,B,D$ are concyclic. [i]Proposer[/i]: [b]Dorian Croitoru[/b]

$AB < AC$ on $\triangle ABC$. The midpoint of arc $BC$ which doesn't include $A$ is $T$ and which includes $A$ is $S$. On segment $AB,AC$, $D,E$ exist so that $DE$ and $BC$ are parallel. The outer angle bisector of $\angle ABE$ and $\angle ACD$ meets $AS$ at $P$ and $Q$. Prove that the circumcircle of $\triangle PBE$ and $\triangle QCD$ meets on $AT$.

$ABC$ is an acute triangle. $P$(different from $B,C$) is a point on side $BC$. $H$ is an orthocenter, and $D$ is a foot of perpendicular from $H$ to $AP$. The circumcircle of the triangle $ABD$ and $ACD$ is $O _1$ and $O_2$, respectively. A line $l$ parallel to $BC$ passes $D$ and meet $O_1$ and $O_2$ again at $X$ and $Y$, respectively. $l$ meets $AB$ at $E$, and $AC$ at $F$. Two lines $XB$ and $YC$ intersect at $Z$. Prove that $ZE=ZF$ is a necessary and sufficient condition for $BP=CP$.

Let $ABC$ be a triangle, and $B_1,C_1$ be its excenters opposite $B,C$. $B_2,C_2$ are reflections of $B_1,C_1$ across midpoints of $AC,AB$. Let $D$ be the extouch at $BC$. Show that $AD$ is perpendicular to $B_2C_2$

Let $O, I$ be the circumcenter and incenter of triangle $ABC$. The incircle of $\triangle ABC$ touches $BC, CA, AB$ at points $D, E, F$ repsectively. $FD$ meets $CA$ at $P$, $ED$ meets $AB$ at $Q$. $M$ and $N$ are midpoints of $PE$ and $QF$ respectively. Show that $OI \perp MN$.

For a given triangle ABC, let X be a variable point on the line BC such that the point C lies between the points B and X. Prove that the radical axis of the incircles of the triangles ABX and ACX passes through a point independent of X. This is a slight extension of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41033]IMO Shortlist 2004 geometry problem 7[/url] and can be found, together with the proposed solution, among the files uploaded at http://www.mathlinks.ro/Forum/viewtopic.php?t=15622 . Note that the problem was proposed by Russia. I could not find the names of the authors, but I have two particular persons under suspicion. Maybe somebody could shade some light on this... Darij

Let $ABC$ be an acute triangle and $\Gamma$ its circumcircle. Let $l$ be the line tangent to $\Gamma$ at $A$. Let $D$ and $E$ be the intersections of the circumference with center $B$ and radius $AB$ with lines $l$ and $AC$, respectively. Prove the orthocenter of $ABC$ lies on line $DE$.

Suppose $O$ is circumcenter of triangle $ABC$. Suppose $\frac{S(OAB)+S(OAC)}2=S(OBC)$. Prove that the distance of $O$ (circumcenter) from the radical axis of the circumcircle and the 9-point circle is \[\frac {a^2}{\sqrt{9R^2-(a^2+b^2+c^2)}}\]

In $\triangle ABC$, $H$ is the orthocenter, and $AD,BE$ are arbitrary cevians. Let $\omega_1, \omega_2$ denote the circles with diameters $AD$ and $BE$, respectively. $HD,HE$ meet $\omega_1,\omega_2$ again at $F,G$. $DE$ meets $\omega_1,\omega_2$ again at $P_1,P_2$ respectively. $FG$ meets $\omega_1,\omega_2$ again $Q_1,Q_2$ respectively. $P_1H,Q_1H$ meet $\omega_1$ at $R_1,S_1$ respectively. $P_2H,Q_2H$ meet $\omega_2$ at $R_2,S_2$ respectively. Let $P_1Q_1\cap P_2Q_2 = X$, and $R_1S_1\cap R_2S_2=Y$. Prove that $X,Y,H$ are collinear. [i]Ray Li.[/i]

Let $M$ be the midpoint of the side $AC$ of $ \triangle ABC$. Let $P\in AM$ and $Q\in CM$ be such that $PQ=\frac{AC}{2}$. Let $(ABQ)$ intersect with $BC$ at $X\not= B$ and $(BCP)$ intersect with $BA$ at $Y\not= B$. Prove that the quadrilateral $BXMY$ is cyclic. [i]F. Ivlev, F. Nilov[/i]

If circular arcs $ AC$ and $ BC$ have centers at $ B$ and $ A$, respectively, then there exists a circle tangent to both $ \stackrel{\frown}{AC}$ and $ \stackrel{\frown}{BC}$, and to $ \overline{AB}$. If the length of $ \stackrel{\frown}{BC}$ is $ 12$, then the circumference of the circle is [asy]unitsize(4cm); defaultpen(fontsize(8pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,.375); pair A=(-.5,0); pair B=(.5,0); pair C=shift(-.5,0)*dir(60); draw(Arc(A,1,0,60)); draw(Arc(B,1,120,180)); draw(A--B); draw(Circle(O,.375)); dot(A); dot(B); dot(C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N);[/asy]$ \textbf{(A)}\ 24 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 28$

The incircle $\odot (I)$ of $\triangle ABC$ touch $AC$ and $AB$ at $E$ and $F$ respectively. Let $H$ be the foot of the altitude from $A$, if $R \equiv IC \cap AH, \ \ Q \equiv BI \cap AH$ prove that the midpoint of $AH$ lies on the radical axis between $\odot (REC)$ and $\odot (QFB)$ I hope that this is not repost :)

Let $ ABC$ be a triangle. Point $ D$ lies on its sideline $ BC$ such that $ \angle CAD \equal{} \angle CBA.$ Circle $ (O)$ passing through $ B,D$ intersects $ AB,AD$ at $ E,F$, respectively. $ BF$ meets $ DE$ at $ G$.Denote by$ M$ the midpoint of $ AG.$ Show that $ CM\perp AO.$

Let $\triangle ABC$ be a triangle with circumcircle $\Gamma$, and let $I$ be the center of the incircle of $\triangle ABC$. The lines $AI$, $BI$ and $CI$ intersect $\Gamma$ in $D \ne A$, $E \ne B$ and $F \ne C$. The tangent lines to $\Gamma$ in $F$, $D$ and $E$ intersect the lines $AI$, $BI$ and $CI$ in $R$, $S$ and $T$, respectively. Prove that \[\vert AR\vert \cdot \vert BS\vert \cdot \vert CT\vert = \vert ID\vert \cdot \vert IE\vert \cdot \vert IF\vert.\]

[asy] //size(100);//local size(200); real r1=2; pair O=(0,0), D=(.5,.5*sqrt(3)), C=(D.x+.5*3,D.y), B, B_prime=endpoint(arc(D, 3, 0,-2)); B=B_prime; path c1=circle(O, r1); pair C=midpoint(D--B_prime); path arc2=arc(B_prime, 6/2, 158.25,250); draw(c1); draw(O--D); draw(D--C); draw(C--B_prime); pair A=beginpoint(arc2); draw(B_prime--A); //dot(O^^D^^C^^A); //dot(B_prime); label("\scriptsize{$O$}",O,.6dir(D--O)); label("\scriptsize{$C$}",C,.5dir(-55)); label("\scriptsize{$D$}", D,.2NW); //label("\scriptsize{$B$}",B,S); label("\scriptsize{$B$}", B_prime, .5*dir(D--B_prime)); label("\scriptsize{$A$}",A,.5dir(NE)); label("\tiny{2}", O--D, .45*LeftSide); label("\tiny{3}", D--C, .45*LeftSide); label("\tiny{6}", B_prime--A, .45*RightSide); label("\tiny{3}", waypoint(C--B_prime,.1), .45*N); //Credit to Klaus-Anton for the diagram[/asy] In the adjoining figure, $AB$ is tangent at $A$ to the circle with center $O$; point $D$ is interior to the circle; and $DB$ intersects the circle at $C$. If $BC=DC=3$, $OD=2$, and $AB=6$, then the radius of the circle is $\textbf{(A) }3+\sqrt{3}\qquad\textbf{(B) }15/\pi\qquad\textbf{(C) }9/2\qquad\textbf{(D) }2\sqrt{6}\qquad \textbf{(E) }\sqrt{22}$

In triangle $ABC$, let $\omega$ be the excircle opposite to $A$. Let $D, E$ and $F$ be the points where $\omega$ is tangent to $BC, CA$, and $AB$, respectively. The circle $AEF$ intersects line $BC$ at $P$ and $Q$. Let $M$ be the midpoint of $AD$. Prove that the circle $MPQ$ is tangent to $\omega$.

Circles $S_1$ and $S_2$ meet at points $A$ and $B$. A line through $A$ is parallel to the line through the centers of $S_1$ and $S_2$ and meets $S_1$ and $S_2$ again $C$ and $D$ respectively. Circle $S_3$ having $CD$ as its diameter meets $S_1$ and $S_2$ again at $P$ and $Q$ respectively. Prove that lines $CP$, $DQ$, and $AB$ are concurent.

Let $\Delta ABC$ be an acute triangle and $\omega$ be its circumcircle. Perpendicular from $A$ to $BC$ intersects $BC$ at $D$ and $\omega$ at $K$. Circle through $A$, $D$ and tangent to $BC$ at $D$ intersect $\omega$ at $E$. $AE$ intersects $BC$ at $T$. $TK$ intersects $\omega$ at $S$. Assume, $SD$ intersects $\omega$ at $X$. Prove that $X$ is the reflection of $A$ with respect to the perpendicular bisector of $BC$.

Let $ABC$ be a triangle and $X,Y,Z$ be the tangency points of its inscribed circle with the sides $BC, CA, AB$, respectively. Suppose that $C_1, C_2, C_3$ are circle with chords $YZ, ZX, XY$, respectively, such that $C_1$ and $C_2$ intersect on the line $CZ$ and that $C_1$ and $C_3$ intersect on the line $BY$. Suppose that $C_1$ intersects the chords $XY$ and $ZX$ at $J$ and $M$, respectively; that $C_2$ intersects the chords $YZ$ and $XY$ at $L$ and $I$, respectively; and that $C_3$ intersects the chords $YZ$ and $ZX$ at $K$ and $N$, respectively. Show that $I, J, K, L, M, N$ lie on the same circle.

Let $PQRS$ be a cyclic quadrilateral such that the segments $PQ$ and $RS$ are not parallel. Consider the set of circles through $P$ and $Q$, and the set of circles through $R$ and $S$. Determine the set $A$ of points of tangency of circles in these two sets.

Let $I$ be the incentre of triangle $ABC$. A circle containing the points $B$ and $C$ meets the segments $BI$ and $CI$ at points $P$ and $Q$ respectively. It is known that $BP\cdot CQ=PI\cdot QI$. Prove that the circumcircle of the triangle $PQI$ is tangent to the circumcircle of $ABC$. [i]Proposed by S. Berlov[/i]

Let $ P$ be the the isogonal conjugate of $ Q$ with respect to triangle $ ABC$, and $ P,Q$ are in the interior of triangle $ ABC$. Denote by $ O_{1},O_{2},O_{3}$ the circumcenters of triangle $ PBC,PCA,PAB$, $ O'_{1},O'_{2},O'_{3}$ the circumcenters of triangle $ QBC,QCA,QAB$, $ O$ the circumcenter of triangle $ O_{1}O_{2}O_{3}$, $ O'$ the circumcenter of triangle $ O'_{1}O'_{2}O'_{3}$. Prove that $ OO'$ is parallel to $ PQ$.