Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. [i]Proposed by Charles Leytem, Luxembourg[/i]

Let $A'$ and $B'$ be feet of altitudes from $A$ and $B$, respectively, in acute-angled triangle $ABC$ ($AC\not = BC$). Circle $k$ contains points $A'$ and $B'$ and touches segment $AB$ in $D$. If triangles $ADA'$ and $BDB'$ have the same area, prove that \[\angle A'DB'= \angle ACB.\]

Let $ABCD$ be a cyclie quadrilateral, $\omega$ be it's circumcircle and $M$ be the midpoint of the arc $AB$ of $\omega$ which does not contain the vertices $C$ and $D$. The line that passes through $M$ and the intersection point of segments $AC$ and $BD$, intersects again $\omega$ in $N$. Let $P$ and $Q$ be points in the $CD$ segment such that $\angle AQD = \angle DAP$ and $\angle BPC = \angle CBQ$. Prove that the circumcircle of $NPQ$ and $\omega$ are tangent to each other.

In triangle $\triangle ABC$ , $I$ is the incenter and $M$ is the midpoint of arc $(BC)$ in the circumcircle of $(ABC)$not containing $A$. Let $X$ be an arbitrary point on the external angle bisector of $A$. Let $BX \cap (BIC) = T$. $Y$ lies on $(AXC)$ , different from $A$ , st $MA=MY$ . Prove that $TC || AY$ (Assume that $X$ is not on $(ABC)$ or $BC$)

Let $\Gamma$ be a circle and let $d$ be a line such that $\Gamma$ and $d$ have no common points. Further, let $AB$ be a diameter of the circle $\Gamma$; assume that this diameter $AB$ is perpendicular to the line $d$, and the point $B$ is nearer to the line $d$ than the point $A$. Let $C$ be an arbitrary point on the circle $\Gamma$, different from the points $A$ and $B$. Let $D$ be the point of intersection of the lines $AC$ and $d$. One of the two tangents from the point $D$ to the circle $\Gamma$ touches this circle $\Gamma$ at a point $E$; hereby, we assume that the points $B$ and $E$ lie in the same halfplane with respect to the line $AC$. Denote by $F$ the point of intersection of the lines $BE$ and $d$. Let the line $AF$ intersect the circle $\Gamma$ at a point $G$, different from $A$. Prove that the reflection of the point $G$ in the line $AB$ lies on the line $CF$.

In the triangle $ABC$ let $D,E$ and $F$ be the mid-points of the three sides, $X,Y$ and $Z$ the feet of the three altitudes, $H$ the orthocenter, and $P,Q$ and $R$ the mid-points of the line segment joining $H$ to the three vertices. Show that the nine points $D,E,F,P,Q,R,X,Y,Z$ lie on a circle.

Let $A'\in(BC),$ $B'\in(CA),C'\in(AB)$ be the points of tangency of the excribed circles of triangle $\triangle ABC$ with the sides of $\triangle ABC.$ Let $R'$ be the circumradius of triangle $\triangle A'B'C'.$ Show that \[ R'=\frac{1}{2r}\sqrt{2R\left(2R-h_{a}\right)\left(2R-h_{b}\right)\left(2R-h_{c}\right)}\] where as usual, $R$ is the circumradius of $\triangle ABC,$ r is the inradius of $\triangle ABC,$ and $h_{a},h_{b},h_{c}$ are the lengths of altitudes of $\triangle ABC.$

Let $ABC$ be a triangle with $\angle BAC= 45^o$ . Altitudes $AD, BE, CF$ meet at $H$. $EF$ cuts $BC$ at $P$. $I$ is the midpoint of $BC$, $IF$ cuts $PH$ in $Q$. a) Prove that $\angle IQH = \angle AIE$. b) Let $(K)$ be the circumcircle of triangle $ABC$, $(J)$ be the circumcircle of triangle $KPD$. $CK$ cuts circle $(J)$ at $G$, $IG$ cuts $(J)$ at $M$, $JC$ cuts circle of diameter $BC$ at $N$. Prove that $G, N, M, C$ lie on the same circle.

It is given isosceles triangle $ABC$ with $AB = AC$. $AD$ is diameter of circumcircle of triangle $ABC$. On the side $BC$ is chosen point $E$. On the sides $AC, AB$ there are points $F, G$ respectively such that $AFEG$ is parallelogram. Prove that $DE$ is perpendicular to $FG$.

Let $O$ be the circumcenter of an acute triangle $ABC$. Let $D$ be the intersection of the bisector of the angle $A$ and $BC$. Suppose that $\angle ODC = 2 \angle DAO$. The circumcircle of $ABD$ meets the line segment $OA$ and the line $OD$ at $E (\neq A,O)$, and $F(\neq D)$, respectively. Let $X$ be the intersection of the line $DE$ and the line segment $AC$. Let $Y$ be the intersection of the bisector of the angle $BAF$ and the segment $BE$. Prove that $\frac{\overline{AY}}{\overline{BY}}= \frac{\overline{EX}}{\overline{EO}}$.

Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.

Construct a quadrilateral which is inscribed and circumscribed, given the radii of the respective circles and the angle between the diagonals of quadrilateral.

Let $ABC$ be a triangle with $AC> AB$. Let $P$ be the intersection of $BC$ and the tangent through $A$ around the triangle $ABC$. Let $Q$ be the point on the straight line $AC$, so that $AQ = AB$ and $A$ is between $C$ and $Q$. Let $X$ and $Y$ be the center of $BQ$ and $AP$. Let $R$ be the point on $AP$ so that $AR = BP$ and $R$ is between $A$ and $P$. Show that $BR = 2XY$.

$ABC$ is an acute triangle with $AB> AC$. $\Gamma_B$ is a circle that passes through $A,B$ and is tangent to $AC$ on $A$. Define similar for $ \Gamma_C$. Let $D$ be the intersection $\Gamma_B$ and $\Gamma_C$ and $M$ be the midpoint of $BC$. $AM$ cuts $\Gamma_C$ at $E$. Let $O$ be the center of the circumscibed circle of the triangle $ABC$. Prove that the circumscibed circle of the triangle $ODE$ is tangent to $\Gamma_B$.

Let $D$ and $E$ be points on side $[AB]$ of a right triangle with $m(\widehat{C})=90^\circ$ such that $|AD|=|AC|$ and $|BE|=|BC|$. Let $F$ be the second intersection point of the circumcircles of triangles $AEC$ and $BDC$. If $|CF|=2$, what is $|ED|$? $ \textbf{(A)}\ \sqrt 2 \qquad\textbf{(B)}\ 1+\sqrt 2 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 2\sqrt 2 \qquad\textbf{(E)}\ \text{None of above} $

Let $ABC$ be an acute angled triangle with orthocenter $H$. centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $AH$ and $AG$ with side $BC$. Rays $MH$ and $DG$ interect $ \omega$ again at $P$ and $Q$ respectively. Prove that $PD$ and $QM$ intersect on $\omega$.

$ABCD$ is a square. Describe the locus of points $P$, different from $A, B, C, D$, on that plane for which \[\widehat{APB}+\widehat{CPD}=180^\circ\]

Let $O_1$ be the center of a semicircle $\omega_1$ with diameter $AB$ and let $O_2$ be the center of a circle $\omega_2$ inscribed in $\omega_1$ and which is tangent to $AB$ at $O_1$. Let $O_3$ be a point on $AB$ that is the center of a semicircle $\omega_3$ which is tangent to both $\omega_1$ and $\omega_2$. Let $P$ be the intersection of the line through $O_3$ perpendicular to $AB$ and the line through $O_2$ parallel to $AB$. Show that $P$ is the center of a circle $\Gamma$ tangent to all of $\omega_1, \omega_2$ and $\omega_3$.

(a) Triangles $A_1B_1C_1$ and $A_2B_2C_2$ are inscribed into triangle $ABC$ so that $C_1A_1 \perp BC$, $A_1B_1 \perp CA$, $B_1C_1 \perp AB$, $B_2A_2 \perp BC$, $C_2B_2 \perp CA$, $A_2C_2 \perp AB$. Prove that these triangles are equal. (b) Points $A_1$, $B_1$, $C_1$, $A_2$, $B_2$, $C_2$ lie inside a triangle $ABC$ so that $A_1$ is on segment $AB_1$, $B_1$ is on segment $BC_1$, $C_1$ is on segment $CA_1$, $A_2$ is on segment $AC_2$, $B_2$ is on segment $BA_2$, $C_2$ is on segment $CB_2$, and the angles $BAA_1$, $CBB_2$, $ACC_1$, $CAA_2$, $ABB_2$, $BCC_2$ are equal. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are equal.

In $\triangle ABC$ points $M,O$ are midpoint of $AB$ and circumcenter. It is true, that $OM=R-r$. Bisector of external $\angle A$ intersect $BC$ at $D$ and bisector of external $\angle C$ intersect $AB$ at $E$. Find possible values of $\angle CED$ [i]D. Shiryaev [/i]

Each cell of a beehive is constructed from a right regular 6-angled prism, open at the bottom and closed on the top by a regular 3-sided pyramidical mantle. The edges of this pyramid are connected to three of the rising edges of the prism and its apex $T$ is on the perpendicular line through the center $O$ of the base of the prism (see figure). Let $s$ denote the side of the base, $h$ the height of the cell and $\theta$ the angle between the line $TO$ and $TV$. (a) Prove that the surface of the cell consists of 6 congruent trapezoids and 3 congruent rhombi. (b) the total surface area of the cell is given by the formula $6sh - \dfrac{9s^2}{2\tan\theta} + \dfrac{s^2 3\sqrt{3}}{2\sin\theta}$ [img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=286[/img]

Let $ABCD$ be a parallelogram whose diagonals intersect in $M$. Suppose that the circumcircle of $ABM$ intersects the segment $AD$ in a point $E \ne A$ and that the circumcircle of $EMD$ intersects the segment $BE$ in a point $F \ne E$. Show that $\angle ACB=\angle DCF$.

Given a real number $\lambda > 1$, let $P$ be a point on the arc $BAC$ of the circumcircle of $\bigtriangleup ABC$. Extend $BP$ and $CP$ to $U$ and $V$ respectively such that $BU = \lambda BA$, $CV = \lambda CA$. Then extend $UV$ to $Q$ such that $UQ = \lambda UV$. Find the locus of point $Q$.

Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$ [i]Proposed by Nazar Serdyuk, Ukraine[/i]

Let $ABC$ be a triangle. $I$ is the incenter of $\Delta ABC$ and $D$ is the meet point of $AI$ and the circumcircle of $\Delta ABC$. Let $E,F$ be on $BD,CD$, respectively such that $IE,IF$ are perpendicular to $BD,CD$, respectively. If $IE+IF=\frac{AD}{2}$, find the value of $\angle BAC$.