$ABCD$ is a parallelogram. The excircle of $ABC$ opposite $A$ has center $E$ and touches the line $AB$ at $X$. The excircle of $ADC$ opposite $A$ has center $F$ and touches the line $AD$ at $Y$. The line $FC$ meets the line$ AB$ at $W$, and the line $EC$ meets the line $AD$ at $Z$. Show that $WX = YZ$.

Let $ABCD$ be a parallelogram with the angle at $A$ obtuse. Let $P$ be a point on segment $BD$. The circle with center $P$ passing through $A$ cuts line $AD$ at $A$ and $Y$ and cuts line $AB$ at $A$ and $X$. Line $AP$ intersects $BC$ at $Q$ and $CD$ at $R$. Prove $\angle XPY = \angle XQY + \angle XRY$.

On the sides $AB$ and $BC$ arbitrarily mark points $M$ and $N$, respectively. Let $P$ be the point of intersection of segments $AN$ and $BM$. In addition, we note the points $Q$ and $R$ such that quadrilaterals $MCNQ$ and $ACBR$ are parallelograms. Prove that the points $P,Q$ and $R$ lie on one line.

Let $ H$ be the orthocenter, $ O$ the circumcenter, and $ R$ the circumradius of an acute-angled triangle $ ABC$. Consider the circles $ k_a,k_b,k_c,k_h,k$, all with radius $ R$, centered at $ A,B,C,H,M,$ respectively. Circles $ k_a$ and $ k_b$ meet at $ M$ and $ F$; $ k_a$ and $ k_c$ meet at $ M$ and $ E$; and $ k_b$ and $ k_c$ meet at $ M$ and $ D$. $ (a)$ Prove that the points $ D,E,F$ lie on the circle $ k_h$. $ (b)$ Prove that the set of the points inside $ k_h$ that are inside exactly one of the circles $ k_a,k_b,k_c$ has the area twice the area of $ \triangle ABC$.

Let $ABCD$ be a parallelogram and let $K$ and $L$ be points on the segments $BC$ and $CD$, respectively, such that $BK\cdot AD=DL\cdot AB$. Let the lines $DK$ and $BL$ intersect at $P$. Show that $\measuredangle DAP=\measuredangle BAC$.

Legs $L_1, L_2, L_3, L_4$ of a square table each have length $n$, where $n$ is a positive integer. For how many ordered 4-tuples $(k_1, k_2, k_3, k_4)$ of nonnegative integers can we cut a piece of length $k_i$ from the end of leg $L_i \; (i=1,2,3,4)$ and still have a stable table? (The table is [i]stable[/i] if it can be placed so that all four of the leg ends touch the floor. Note that a cut leg of length 0 is permitted.)

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 $\widehat{A} - \widehat{B} = 90^\circ$.

The triangle $ ABC$ is given. On the extension of the side $ AB$ we construct the point $ R$ with $ BR \equal{} BC$, where $ AR > BR$ and on the extension of the side $ AC$ we construct the point $ S$ with $ CS \equal{} CB$, where $ AS > CS$. Let $ A_1$ be the point of intersection of the diagonals of the quadrilateral $ BRSC$. Analogous we construct the point $ T$ on the extension of the side $ BC$, where $ CT \equal{} CA$ and $ BT > CT$ and on the extension of the side $ BA$ we construct the point $ U$ with $ AU \equal{} AC$, where $ BU > AU$. Let $ B_1$ be the point of intersection of the diagonals of the quadrilateral $ CTUA$. Likewise we construct the point $ V$ on the extension of the side $ CA$, where $ AV \equal{} AB$ and $ CV > AV$ and on the extension of the side $ CB$ we construct the point $ W$ with $ BW \equal{} BA$ and $ CW > BW$. Let $ C_1$ be the point of intersection of the diagonals of the quadrilateral $ AVWB$. Show that the area of the hexagon $ AC_1BA_1CB_1$ is equal to the sum of the areas of the triangles $ ABC$ and $ A_1B_1C_1$.

In the plane there are six different points $A, B, C, D, E, F$ such that $ABCD$ and $CDEF$ are parallelograms. What is the maximum number of those points that can be located on one circle?

Let $ ABP, BCQ, CAR$ be three non-overlapping triangles erected outside of acute triangle $ ABC$. Let $ M$ be the midpoint of segment $ AP$. Given that $ \angle PAB \equal{} \angle CQB \equal{} 45^\circ$, $ \angle ABP \equal{} \angle QBC \equal{} 75^\circ$, $ \angle RAC \equal{} 105^\circ$, and $ RQ^2 \equal{} 6CM^2$, compute $ AC^2/AR^2$. [i]Zuming Feng.[/i]

Parallelogram $ ABCD$ has area $ 1,\!000,\!000$. Vertex $ A$ is at $ (0,0)$ and all other vertices are in the first quadrant. Vertices $ B$ and $ D$ are lattice points on the lines $ y\equal{}x$ and $ y\equal{}kx$ for some integer $ k>1$, respectively. How many such parallelograms are there? $ \textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048$

Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$. [i]Calvin Deng.[/i]

Let $ABCD$ be a parallelogram. $E$ and $F$ are on $BC, CD$ respectively such that the triangles $ABE$ and $BCF$ have the same area. Let $BD$ intersect $AE, AF$ at $M, N$ respectively. Prove there exists a triangle whose side lengths are $BM, MN, ND$.

Prove that in any parallelepiped the sum of the lengths of the edges is less than or equal to twice the sum of the lengths of the four diagonals.

Let $ABC$ be an acute-angled triangle, $H$ the intersection point of its altitudes , and $AA'$ the diameter of the circumcircle of triangle $ABC$. Prove that the quadrilateral $HB A'C$ is a parallelogram.

Let $ABCD$ be a quadrilateral such that $\angle ABC = \angle ADC = 90º$ and $\angle BCD$ > $90º$. Let $P$ be a point inside of the $ABCD$ such that $BCDP$ is parallelogram, the line $AP$ intersects $BC$ in $M$. If $BM = 2, MC = 5, CD = 3$. Find the length of $AM$.

Given an acute angles triangle $ABC$, and $H$ is its orthocentre. The external bisector of the angle $\angle BHC$ meets the sides $AB$ and $AC$ at the points $D$ and $E$ respectively. The internal bisector of the angle $\angle BAC$ meets the circumcircle of the triangle $ADE$ again at the point $K$. Prove that $HK$ is through the midpoint of the side $BC$.

Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]

$BC$ is a diameter of a circle and the points $X,Y$ are on the circle such that $XY\perp BC$. The points $P,M$ are on $XY,CY$ (or their stretches), respectively, such that $CY||PB$ and $CX||PM$. Let $K$ be the meet point of the lines $XC,BP$. Prove that $PB\perp MK$.

For an interior point $D$ of a triangle $ABC,$ let $\Gamma_D$ denote the circle passing through the points $A, \: E, \: D, \: F$ if these points are concyclic where $BD \cap AC=\{E\}$ and $CD \cap AB=\{F\}.$ Show that all circles $\Gamma_D$ pass through a second common point different from $A$ as $D$ varies.

Let $ABCD$ be a parallelogram. $P \in (CD), Q \in (AB)$, $M= AP \cap DQ$, $N=BP \cap CQ$, $ K=MN \cap AD$, $L= MN \cap BC$. Prove that $BL=DK$.

Let $w$-circumcircle of triangle $ABC$ with an obtuse angle $C$ and $C '$symmetric point of point $C$ with respect to $AB$. $M$ midpoint of $AB$. $C'M$ intersects $w$ at $N$ ($C '$ between $M$ and $N$). Let $BC'$ second crossing point $w$ in $F$, and $AC'$ again crosses the $w$ at point $E$. $K$-midpoint $EF$. Prove that the lines $AB, CN$ and$ KC'$are concurrent.

Prove that if a plane section of a tetrahedron is a parallelogram, then half of its perimeter is contained between the length of the smallest and the length of the largest edge of the tetrahedron.

$ABC$ is an isosceles triangle with $AB=AC$ and the angle in $A$ is less than $60^{\circ}$. Let $D$ be a point on $AC$ such that $\angle{DBC}=\angle{BAC}$. $E$ is the intersection between the perpendicular bisector of $BD$ and the line parallel to $BC$ passing through $A$. $F$ is a point on the line $AC$ such that $FA=2AC$ ($A$ is between $F$ and $C$). Show that $EB$ and $AC$ are parallel and that the perpendicular from $F$ to $AB$, the perpendicular from $E$ to $AC$ and $BD$ are concurrent.

In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.