We consider $\Gamma_1$ and $\Gamma_2$ two circles that intersect at points $P$ and $Q$ . Let $A , B$ and $C$ be points on the circle $\Gamma_1$ and $D , E$ and $F$ points on the circle $\Gamma_2$ so that the lines $A E$ and $B D$ intersect at $P$ and the lines $A F$ and $C D$ intersect at $Q$. Denote $M$ and $N$ the intersections of lines $A B$ and $D E$ and of lines $A C$ and $D F$ , respectively. Show that $A M D N$ is a parallelogram.

It is known that $\triangle ABC$ is an acute triangle. Let $C'$ be the foott of the perpendicular from $C$ to $AB$, and $D$, $E$ two distinct points on $CC'$. The feet of the perpendiculars from $D$ to $AC$ and $BC$ are $F$ and $G$, respectively. Show that if $DGEF$ is a parallelogram then $ABC$ is isosceles.

In a parallelogram $ABCD$ , $BD$ is the largest diagonal. By matching $B$ with $D$ by a bend, a regular pentagon is formed. Calculate the measures of the angles formed by the diagonal $BD$ with each of the sides of the parallelogram.

Let $ ABCD $ be a parallelogram with $ BC = 17 $. Let $ M $ be the midpoint of $ \overline{BC} $ and let $ N $ be the point such that $ DANM $ is a parallelogram. What is the length of segment $ \overline{NC} $?

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]

The bisectors of trapezoid's angles form a quadrilateral with perpendicular diagonals. Prove that this trapezoid is isosceles.

On the sides (excluding its endpoints) $ AB,BC,CD,DA $ of a parallelogram consider the points $ M,N,P,Q, $ respectively, such that $ \overrightarrow{AP} +\overrightarrow{AN} +\overrightarrow{CQ} +\overrightarrow{CM} = 0. $ Show that $ QN, PM,AC $ are concurrent. [i]Adrian Ivan[/i]

Equilateral $\triangle ABC$ is inscribed in a circle of radius 2. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13$, and extend $\overline{AC}$ through $C$ to point $E$ so that $AE=11$. Through $D$, draw a line $l_1$ parallel to $\overline{AE}$, and through $E$, draw a line ${l}_2$ parallel to $\overline{AD}$. Let $F$ be the intersection of ${l}_1$ and ${l}_2$. Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A$. Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r}$, where $p$, $q$, and $r$ are positive integers, $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime, find $p+q+r$.

A quadrilateral $ABCD$ is inscribed into a circle with center $O.$ Points $P$ and $Q$ are opposite to $C$ and $D$ respectively. Two tangents drawn to that circle at these points meet the line $AB$ in points $E$ and $F.$ ($A$ is between $E$ and $B$, $B$ is between $A$ and $F$). The line $EO$ meets $AC$ and $BC$ in points $X$ and $Y$ respectively, and the line $FO$ meets $AD$ and $BD$ in points $U$ and $V$ respectively. Prove that $XV=YU.$

In a triangle $ABC$, with $\widehat{A} > 90^\circ$, let $O$ and $H$ denote its circumcenter and orthocenter, respectively. Let $K$ be the reflection of $H$ with respect to $A$. Prove that $K, O$ and $C$ are collinear if and only if $\widehat{A} - \widehat{B} = 90^\circ$.

$O$ is the circumcenter of acute $\Delta ABC$, $H$ is the Orthocenter. $AD \bot BC$, $EF$ is the perpendicular bisector of $AO$,$D,E$ on the $BC$. Prove that the circumcircle of $\Delta ADE$ through the midpoint of $OH$.

Triangle $ABC$ is given. Let $M$ be the midpoint of the segment $AB$ and $T$ be the midpoint of the arc $BC$ not containing $A$ of the circumcircle of $ABC.$ The point $K$ inside the triangle $ABC$ is such that $MATK$ is an isosceles trapezoid with $AT\parallel MK.$ Show that $AK = KC.$

Let $a,b,c,$ and $d$ be the lengths of sides $MN,NP,PQ,$ and $QM$, respectively, of quadrilateral $MNPQ$. If $A$ is the area of $MNPQ$, then $\textbf{(A) }A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is convex}$ $\textbf{(B) }A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a rectangle}$ $\textbf{(C) }A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a rectangle}$ $\textbf{(D) }A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a parallelogram}$ $\textbf{(E) }A\ge\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a parallelogram}$

Points $E, F, G$ lie, and on the sides $BC, CA, AB$, respectively of a triangle $ABC$, with $2AG=GB, 2BE=EC$ and $2CF=FA$. Points $P$ and $Q$ lie on segments $EG$ and $FG$, respectively such that $2EP = PG$ and $2GQ=QF$. Prove that the quadrilateral $AGPQ$ is a parallelogram.

In $3$ dimensional space we are given a parallelogram $ABCD$ and plane $M$. The distances from vertices $A, B$ and $C$ to plane $M$ are $a, b$ and $c$ respectively. Find the distance $d$ from vertex $D$ to the plane $M$ .

Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.

Prove that the parallelogram $ABCD$ with relation $\angle ABD + \angle DAC = 90^o$, is either a rectangle or a rhombus.

(A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$.

Let $H$ be the orthocenter of a triangle $ABC$, and let $D$ be the midpoint of the segment $AH$. The altitude $BH$ of triangle $ABC$ intersects the perpendicular to the line $AB$ through the point $A$ at the point $M$. The altitude $CH$ of triangle $ABC$ intersects the perpendicular to the line $CA$ through the point $A$ at the point $N$. The perpendicular bisector of the segment $AB$ intersects the perpendicular to the line $BC$ through the point $B$ at the point $U$. The perpendicular bisector of the segment $CA$ intersects the perpendicular to the line $BC$ through the point $C$ at the point $V$. Finally, let $E$ be the midpoint of the side $BC$ of triangle $ABC$. Prove that the points $D$, $M$, $N$, $U$, $V$ all lie on one and the same perpendicular to the line $AE$. [i]Extensions.[/i] In other words, we have to show that the points $M$, $N$, $U$, $V$ lie on the perpendicular to the line $AE$ through the point $D$. Additionally, one can find two more points on this perpendicular: [b](a)[/b] The nine-point circle of triangle $ABC$ is known to pass through the midpoint $E$ of its side $BC$. Let $D^{\prime}$ be the point where this nine-point circle intersects the line $AE$ apart from $E$. Then, the point $D^{\prime}$ lies on the perpendicular to the line $AE$ through the point $D$. [b](b)[/b] Let the tangent to the circumcircle of triangle $ABC$ at the point $A$ intersect the line $BC$ at a point $X$. Then, the point $X$ lies on the perpendicular to the line $AE$ through the point $D$. [i]Comment.[/i] The actual problem was created by Victor Thébault around 1950 (cf. Hyacinthos messages #1102 and #1551). The extension [b](a)[/b] initially was a (pretty trivial) lemma in Thébault's solution of the problem. Extension [b](b)[/b] is rather new; in the form "prove that $X\in UV$", it was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3659]proposed by Valentin Vornicu for the Balkan MO 2003[/url], however it circulated in the Hyacinthos newsgroup before (Hyacinthos messages #7240 and #7242), where different solutions of the problem were discussed as well. Hereby, "Hyacinthos" always refers to the triangle geometry newsgroup "Hyacinthos", which can be found at http://groups.yahoo.com/group/Hyacinthos . I proposed the problem for the QEDMO math fight wishing to draw some attention to it. It has a rather short and elementary solution, by the way (without using radical axes or inversion like the standard solutions). Darij

Let $AC$ and $BD$ be two chords of a circle with center $O$ such that they intersect at right angles inside the circle at the point $M$. Suppose $K$ and $L$ are midpoints of the chords $AB$ and $CD$ respectively. Prove that $OKML$ is a parallelogram.

Let $N$ be a point on the longest side $AC$ of a triangle $ABC$. The perpendicular bisectors of $AN$ and $NC$ intersect $AB$ and $BC$ respectively in $K$ and $M$. Prove that the circumcenter $O$ of $\triangle ABC$ lies on the circumcircle of triangle $KBM$.

Let $n\ge 1$ be a positive integer. A square of side length $n$ is divided by lines parallel to each side into $n^2$ squares of side length $1$. Find the number of parallelograms which have vertices among the vertices of the $n^2$ squares of side length $1$, with both sides smaller or equal to $2$, and which have tha area equal to $2$. (Greece)

We divided a regular octagon into parallelograms. Prove that there are at least $2$ rectangles between the parallelograms.

Let $\omega$ be the circumcircle of $\triangle{ABC}$. $P$ is an interior point of $\triangle{ABC}$. $A_{1}, B_{1}, C_{1}$ are the intersections of $AP, BP, CP$ respectively and $A_{2}, B_{2}, C_{2}$ are the symmetrical points of $A_{1}, B_{1}, C_{1}$ with respect to the midpoints of side $BC, CA, AB$. Show that the circumcircle of $\triangle{A_{2}B_{2}C_{2}}$ passes through the orthocentre of $\triangle{ABC}$.

In a village, there are $n$ houses with $n>2$ and all of them are not collinear. We want to generate a water resource in the village. For doing this, point $A$ is [i]better[/i] than point $B$ if the sum of the distances from point $A$ to the houses is less than the sum of the distances from point $B$ to the houses. We call a point [i]ideal[/i] if there doesn’t exist any [i]better[/i] point than it. Prove that there exist at most $1$ [i]ideal[/i] point to generate the resource.