Let be given a parallelogram $ ABCD$ and two points $ A_1$, $ C_1$ on its sides $ AB$, $ BC$, respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$. Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$. Prove that $ \angle PDA \equal{} \angle QBA$.

In a convex quadrilateral $ABCD$ with $\angle ABC = \angle ADC = 135^\circ$, points $M$ and $N$ are taken on the rays $AB$ and $AD$ respectively such that $\angle MCD = \angle NCB = 90^\circ$. The circumcircles of triangles $AMN$ and $ABD$ intersect at $A$ and $K$. Prove that $AK \perp KC.$

Two parallelograms $ABCD$ and $A'B'C'D'$ are given in space. Points $A'',B'',C'',D''$ divide the segments $AA',BB',CC',DD'$ in the same ratio. What can be said about the quadrilateral $A''B''C''D''$?

Given a circle $K$, find the locus of vertices $A$ of parallelograms $ABCD$ with diagonals $AC \leq BD$, such that $BD$ is inside $K$.

Let $ABC$ be an acute triangle. Let $D$ and $E$ be the feet of the altitudes of the triangle $ABC$ from $A$ and $B$, respectively. Let $F$ be the reflection of the point $A$ over $BC$. Let $G$ be a point such that the quadrilateral $ABCG$ is a parallelogram. Show that the circumcircles of triangles $BCF$ , $ACG$ and $CDE$ are concurrent on a point different from $C$.

Let be given a tetrahedron whose circumcenter is $O$. Draw diameters $AA_{1},BB_{1},CC_{1},DD_{1}$ of the circumsphere of $ABCD$. Let $A_{0},B_{0},C_{0},D_{0}$ be the centroids of triangle $BCD,CDA,DAB,ABC$. Prove that $A_{0}A_{1},B_{0}B_{1},C_{0}C_{1},D_{0}D_{1}$ are concurrent at a point, say, $F$. Prove that the line through $F$ and a midpoint of a side of $ABCD$ is perpendicular to the opposite side.

Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$

Let $ABCD$ be a parallelogram and $\omega$ be the circumcircle of the triangle $ABD$. Let $E ,F$ be the intersections of $\omega$ with lines $BC ,CD$ respectively . Prove that the circumcenter of the triangle $CEF$ lies on $\omega$.

Convex quadrilateral $ ABCD$ has $ AB\equal{}9$ and $ CD\equal{}12$. Diagonals $ AC$ and $ BD$ intersect at $ E$, $ AC\equal{}14$, and $ \triangle AED$ and $ \triangle BEC$ have equal areas. What is $ AE$? $ \textbf{(A)}\ \frac{9}{2}\qquad \textbf{(B)}\ \frac{50}{11}\qquad \textbf{(C)}\ \frac{21}{4}\qquad \textbf{(D)}\ \frac{17}{3}\qquad \textbf{(E)}\ 6$

Let $\triangle ABC$ be a triangle with $BC = 4, CA= 5, AB= 6$, and let $O$ be the circumcenter of $\triangle ABC$. Let $O_b$ and $O_c$ be the reflections of $O$ about lines $CA$ and $AB$ respectively. Suppose $BO_b$ and $CO_c$ intersect at $T$, and let $M$ be the midpoint of $BC$. Given that $MT^2 = \frac{p}{q}$ for some coprime positive integers $p$ and $q$, find $p+q$. [i]Proposed by Sreejato Bhattacharya[/i]

Let $ABC$ be an isosceles triangle with $AC=BC$. Let $P,Q,R$ be points on the sides $AB, BC, CA$ of the triangle such that $CQPR$ is a parallelogram. Show that the reflection of $P$ over $QR$ lies on the circumcircle of $ABC$.

In a parallelogram $ABCD$ with $\angle A < 90$, the circle with diameter $AC$ intersects the lines $CB$ and $CD$ again at $E$ and $F$ , and the tangent to this circle at $A$ meets the line $BD$ at $P$ . Prove that the points $P$, $E$, $F$ are collinear.

Prove that a triangle inscribed in a parallelogram has at most half the area of the parallelogram.

Let $OFT$ and $NOT$ be two similar triangles (with the same orientation) and let $FANO$ be a parallelogram. Show that \[\vert OF\vert \cdot \vert ON\vert=\vert OA\vert \cdot \vert OT\vert.\]

Let $ABCD$ be a parallelogram and point $M$ be the midpoint of $[AB]$ so that the quadrilateral $MBCD$ is cyclic. If $N$ is the point of intersection of the lines $DM$ and $BC$, and $P \in BC$, then prove that the ray $(DP$ is the angle bisector of $\angle ADM$ if and only if $PC = 4BC$.

Consider a convex quadrilateral, and the incircles of the triangles determined by one of its diagonals. Prove that the tangency points of the incircles with the diagonal are symmetrical with respect to the midpoint of the diagonal if and only if the line of the incenters passes through the crossing point of the diagonals. [i]Dan Schwarz[/i]

Let $n$ be a positive integer and let $G$ be the set of points $(x, y)$ in the plane such that $x$ and $y$ are integers with $1 \leq x, y \leq n$. A subset of $G$ is called [i]parallelogram-free[/i] if it does not contains four non-collinear points, which are the vertices of a parallelogram. What is the largest number of elements a parallelogram-free subset of $G$ can have?

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 $\textbf{(A) }A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is convex}$ $\textbf{(B) }A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a rectangle}$ $\textbf{(C) }A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a rectangle}$ $\textbf{(D) }A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a parallelogram}$ $\textbf{(E) }A\ge\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a parallelogram}$

On the sides of the parallelogram $ABCD$ outside it are constructed equilateral triangles $ABM$, $BCN$, $CDP$, $ADQ$. Prove that $MNPQ$ is a parallelogram.

In a given tedrahedron $ ABCD$ let $ K$ and $ L$ be the centres of edges $ AB$ and $ CD$ respectively. Prove that every plane that contains the line $ KL$ divides the tedrahedron into two parts of equal volume.

Let $ABCD$ be a convex quadrilateral. Assume line $AB$ and $CD$ intersect at $E$, and $B$ lies between $A$ and $E$. Assume line $AD$ and $BC$ intersect at $F$, and $D$ lies between $A$ and $F$. Assume the circumcircles of $\triangle BEC$ and $\triangle CFD$ intersect at $C$ and $P$. Prove that $\angle BAP=\angle CAD$ if and only if $BD\parallel EF$.

Let $ABC$ be a triangle. Let $X$ be the point on side $AB$ such that $\angle{BXC} = 60^{\circ}$. Let $P$ be the point on segment $CX$ such that $BP\bot AC$. Given that $AB = 6, AC = 7,$ and $BP = 4,$ compute $CP$.

$ABCD$ is parallelogram, $M$ is the midpoint of side $CD$ and $H$ is the foot of the perpendicular from $B$ to line $AM$. Prove that $BCH$ is an isosceles triangle. (M Volchkevich)

A convex polygon is partitioned into parallelograms. A vertex of the polygon is called [i]good[/i] if it belongs to exactly one parallelogram. Prove that there are more than two good vertices.

Let $A$ be a point on a circle with center $O$ and $B$ be the midpoint of $[OA]$. Let $C$ and $D$ be points on the circle such that they lie on the same side of the line $OA$ and $\widehat{CBO} = \widehat{DBA}$. Show that the reflection of the midpoint of $[CD]$ over $B$ lies on the circle.