A circle is tangential to sides $AB$ and $AD$ of convex quadrilateral $ABCD$ at $G$ and $H$ respectively, and cuts diagonal $AC$ at $E$ and $F$. What are the necessary and sufficient conditions such that there exists another circle which passes through $E$ and $F$, and is tangential to $DA$ and $DC$ extended?

$ AB$ is a fixed diameter of a circle whose center is $ O$. From $ C$, any point on the circle, a chord $ CD$ is drawn perpendicular to $ AB$. Then, as $ C$ moves over a semicircle, the bisector of angle $ OCD$ cuts the circle in a point that always: $ \textbf{(A)}\ \text{bisects the arc } AB \qquad\textbf{(B)}\ \text{trisects the arc } AB \qquad\textbf{(C)}\ \text{varies}$ $ \textbf{(D)}\ \text{is as far from }AB \text{ as from } D \qquad\textbf{(E)}\ \text{is equidistant from }B \text{ and } C$

Let $ABCD$ be a convex quadrilateral with $\angle CBD = 2 \angle ADB$, $\angle ABD = 2 \angle CDB$ and $AB = CB$. Prove that $AD = CD$.

In triangle $ABC$, $\omega$ is its circumcircle and $O$ is the center of this circle. Points $M$ and $N$ lie on sides $AB$ and $AC$ respectively. $\omega$ and the circumcircle of triangle $AMN$ intersect each other for the second time in $Q$. Let $P$ be the intersection point of $MN$ and $BC$. Prove that $PQ$ is tangent to $\omega$ iff $OM=ON$. [i]proposed by Mr.Etesami[/i]

A triangle $ABC$ is given with barycentre $G$ and circumcentre $O$. The perpendicular bisectors of $GA, GB$ meet at $C_1$,of $GB,GC$ meet at $A _1$, and $GC,GA$ meet at $B_1$. Prove that $O$ is the barycenter of the triangle $A_1B_1C_1$.

Let $O$ be the center of the circumcircle $\omega$ of an acute-angle triangle $ABC$. A circle $\omega_1$ with center $K$ passes through $A$, $O$, $C$ and intersects $AB$ at $M$ and $BC$ at $N$. Point $L$ is symmetric to $K$ with respect to line $NM$. Prove that $BL \perp AC$.

Let $\gamma$ be the circumcircle of the acute triangle $ABC$. Let $P$ be the midpoint of the minor arc $BC$. The parallel to $AB$ through $P$ cuts $BC, AC$ and $\gamma$ at points $R,S$ and $T$, respectively. Let $K \equiv AP \cap BT$ and $L \equiv BS \cap AR$. Show that $KL$ passes through the midpoint of $AB$ if and only if $CS = PR$.

Let $ABC$ be a triangle and $I$ its incenter. The line $AI$ intersects the side $BC$ at $D$ and the perpendicular bisector of $BC$ at $E$. Let $J$ be the incenter of triangle $CDE$. Prove that triangle $CIJ$ is isosceles.

Two circles $C_1$ and $C_2$ with the respective radii $r_1$ and $r_2$ intersect in $A$ and $B.$ A variable line $r$ through $B$ meets $C_1$ and $C_2$ again at $P_r$ and $Q_r$ respectively. Prove that there exists a point $M,$ depending only on $C_1$ and $C_2,$ such that the perpendicular bisector of each segment $P_rQ_r$ passes through $M.$

Let $\triangle ABC$ which $\angle ABC$ are right angle, Let $D$ be point on $AB$ ( $D \neq A , B$ ), Let $E$ be point on line $AB$ which $B$ is the midpoint of $DE$, Let $I$ be incenter of $\triangle ACE$ and $J$ be $A$-excenter of $\triangle ACD$. Prove that perpendicular bisector of $BC$ bisects $IJ$

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 $BQ$.

Let $ABCC_1B_1A_1$ be a convex hexagon such that $AB=BC$, and suppose that the line segments $AA_1, BB_1$, and $CC_1$ have the same perpendicular bisector. Let the diagonals $AC_1$ and $A_1C$ meet at $D$, and denote by $\omega$ the circle $ABC$. Let $\omega$ intersect the circle $A_1BC_1$ again at $E \neq B$. Prove that the lines $BB_1$ and $DE$ intersect on $\omega$.

In the triangle $ABC$ let $AD$ be the interior angle bisector of $\angle ACB$, where $D\in AB$. The circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $BCD$. Prove that $AC^2 = AD\cdot AB$.

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

Let $ABCD$ by a cyclic quadrilateral with circumcenter $O$. Let $P$ be the point of intersection of the diagonals $AC$ and $BD$, and $K, L, M, N$ the circumcenters of triangles $AOP, BOP$, $COP, DOP$, respectively. Prove that $KL = MN$.

Let $ABC$ to be an acute triangle. Also, let $K$ and $L$ to be the two intersections of the perpendicular from $B$ with respect to side $AC$ with the circle of diameter $AC$, with $K$ closer to $B$ than $L$. Analogously, $X$ and $Y$ are the two intersections of the perpendicular from $C$ with respect to side $AB$ with the circle of diamter $AB$, with $X$ closer to $C$ than $Y$. Prove that the intersection of $XL$ and $KY$ lies on $BC$.

The set of all points $P$ such that the sum of the (undirected) distances from $P$ to two fixed points $A$ and $B$ equals the distance between $A$ and $B$ is $\textbf{(A) }\text{the line segment from }A\text{ to }B\qquad$ $\textbf{(B) }\text{the line passing through }A\text{ and }B\qquad$ $\textbf{(C) }\text{the perpendicular bisector of the line segment from }A\text{ to }B\qquad$ $\textbf{(D) }\text{an elllipse having positive area}\qquad$ $\textbf{(E) }\text{a parabola}$

Let $ABC$ be an acute triangle with $AB<AC$. Point $M{}$ from the side $(BC)$ is the foot of the bisector from the vertex $A{}$. The perpendicular bisector of the segment $[AM]$ intersects the side $(AC)$ in $E{}$, the side $(AB)$ in $D$ and the line $(BC)$ in $F{}$. Prove that $\frac{DB}{CE}=\frac{FB}{FC}=\left(\frac{AB}{AC}\right)^2$.

Perpendicular bisectors of the sides $BC$ and $AC$ of an acute-angled triangle $ABC$ intersect lines $AC$ and $BC$ at points $M$ and $N$. Let point $C$ move along the circumscribed circle of triangle $ABC$, remaining in the same half-plane relative to $AB$ (while points $A$ and $B$ are fixed). Prove that line $MN$ touches a fixed circle.

A $30$-gon $A_1A_2\cdots A_{30}$ is inscribed in a circle of radius $2$. Prove that one can choose a point $B_k$ on the arc $A_kA_{k+1}$ for $1 \leq k \leq 29$ and a point $B_{30}$ on the arc $A_{30}A_1$, such that the numerical value of the area of the $60$-gon $A_1B_1A_2B_2 \dots A_{30}B_{30}$ is equal to the numerical value of the perimeter of the original $30$-gon.

Let $ABC$ be a triangle with height $AH$. $P$ lies on the circle over 3 midpoint of $AB,BC,CA$ ($P \notin BC$). Prove that the line connect 2 center of $(PBH)$ and $(PCH)$ go through a fixed point. (where $(XYZ)$ be a circumscribed circle of triangle $XYZ$)

Consider a cyclic quadrilateral $ABCD$, and let $S$ be the intersection of $AC$ and $BD$. Let $E$ and $F$ the orthogonal projections of $S$ on $AB$ and $CD$ respectively. Prove that the perpendicular bisector of segment $EF$ meets the segments $AD$ and $BC$ at their midpoints.

A triangle $ ABC$ is given in the plane. Let $ M$ be a point inside the triangle and $ A'$, $ B'$, $ C'$ be its projections on the sides $ BC$, $ CA$, $ AB$, respectively. Find the locus of $ M$ for which $ MA \cdot MA' \equal{} MB \cdot MB' \equal{} MC \cdot MC'$.

Given Triangle $ABC$, $\angle B= 2 \angle C$, and $\angle A>90^\circ$. Let $M$ be midpoint of $BC$. Perpendicular of $AC$ at $C$ intersects $AB$ at $D$. Show $\angle AMB = \angle DMC$ [hide]If possible, don't use projective geometry[/hide]

Let $ ABC$ be a triangle, let $ E, F$ be the tangency points of the incircle $ \Gamma(I)$ to the sides $ AC$, respectively $ AB$, and let $ M$ be the midpoint of the side $ BC$. Let $ N \equal{} AM \cap EF$, let $ \gamma(M)$ be the circle of diameter $ BC$, and let $ X, Y$ be the other (than $ B, C$) intersection points of $ BI$, respectively $ CI$, with $ \gamma$. Prove that \[ \frac {NX} {NY} \equal{} \frac {AC} {AB}. \] [i]Cosmin Pohoata[/i]