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 $AM$.

Let $ABCDEF$ be a convex hexagon. In the hexagon there is a point $K$, such that $ABCK,DEFK$ are both parallelograms. Prove that the three lines connecting $A,B,C$ to the midpoints of segments $CE,DF,EA$ meet at one point.

Inside the parallelogram $ABCD$ are the circles $\gamma_1$ and $\gamma_2$, which are externally tangent at the point $K$. The circle $\gamma_1$ touches the sides $AD$ and $AB$ of the parallelogram, and the circle $\gamma_2$ touches the sides $CD$ and $CB$. Prove that the point $K$ lies on the diagonal $AC$ of the paralelogram.

The point P lies on the side $CD$ of the parallelogram $ABCD$ with $\angle DBA = \angle CBP$. Point $O$ is the center of the circle passing through the points $D$ and $P$ and tangent to the straight line $AD$ at point $D$. Prove that $AO = OC$.

Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. [i]Proposed by Hossein Karke Abadi, Iran[/i]

$\textbf{Bake Sale}$ Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown. $\circ$ Art's cookies are trapezoids: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label("5 in", (2.5,0), S); label("3 in", (5,1.5), E); label("3 in", (3.5,3), N);[/asy] $\circ$ Roger's cookies are rectangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label("4 in", (2,0), S); label("2 in", (4,1), E);[/asy] $\circ$ Paul's cookies are parallelograms: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label("3 in", (1.5,0), S); label("2 in", (2.5,1), W);[/asy] $\circ$ Trisha's cookies are triangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label("3 in", (1.5,0), S); label("4 in", (3,2), E);[/asy] Each friend uses the same amount of dough, and Art makes exactly 12 cookies. Who gets the fewest cookies from one batch of cookie dough? $ \textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.}$

A rhombus is given with one diagonal twice the length of the other diagonal. Express the side of the rhombus is terms of $K$, where $K$ is the area of the rhombus in square inches. ${{ \textbf{(A)}\ \sqrt{K} \qquad\textbf{(B)}\ \frac{1}{2}\sqrt{2K} \qquad\textbf{(C)}\ \frac{1}{3}\sqrt{3K} \qquad\textbf{(D)}\ \frac{1}{4}\sqrt{4K} }\qquad\textbf{(E)}\ \text{None of these are correct} } $

Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.

Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$. Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,AB$ at $M,N,L$. Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.

Let $ABCD$ be a parallelogram in which angle at $B$ is obtuse and $AD>AB$. Points $K$ and $L$ on $AC$ such that $\angle ADL=\angle KBA$(the points $A, K, C, L$ are all different, with $K$ between $A$ and $L$). The line $BK$ intersects the circumcircle $\omega$ of $ABC$ at points $B$ and $E$, and the line $EL$ intersects $\omega$ at points $E$ and $F$. Prove that $BF||AC$.

$ABCD$ is a parallelogram, line $DF$ is drawn bisecting $BC$ at $E$ and meeting $AB$ (extended) at $F$ from vertex $C$. Line $CH$ is drawn bisecting side $AD$ at $G$ and meeting $AB$ (extended) at $H$. Lines $DF$ and $CH$ intersect at $I$. If the area of parallelogram $ABCD$ is $x$, find the area of triangle $HFI$ in terms of $x$.

Parallelogram $JHMT$ satisfies $JH=11$ and $HM=6,$ and point $P$ lies on $\overline{MT}$ such that $JP$ is an altitude of $JHMT.$ The circumcircles of $\triangle{HMP}$ and $\triangle{JMT}$ intersect at the point $Q\neq M.$ Let $A$ be the point lying on $\overline{JH}$ and the circumcircle of $\triangle{JMT}.$ If $MQ=10,$ then the perimeter of $\triangle{JAM}$ can be expressed in the form $\sqrt{a}+\tfrac{b}{c},$ where $a, \ b,$ and $c$ are positive integers, $a$ is not divisible by the square of any prime, and $b$ and $c$ are relatively prime. Find $a+b+c.$

On the sides $AB,AD$ of the rhombus $ABCD$ are the points $E,F$ such that $AE=DF$. The lines $BC,DE$ intersect at $P$ and $CD,BF$ intersect at $Q$. Prove that: (a) $\frac{PE}{PD} + \frac{QF}{QB} = 1$; (b) $P,A,Q$ are collinear. [i]Virginia Tica, Vasile Tica[/i]

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 $P$ and $P^{\prime }$ be two parallelograms with equal area, and let their sidelengths be $a,$ $b$ and $a^{\prime },$ $b^{\prime }.$ Assume that $a^{\prime }\leq a\leq b\leq b^{\prime },$ and moreover, it is possible to place the segment $b^{\prime }$ such that it completely lies in the interior of the parallelogram $P.$ Show that the parallelogram $P$ can be partitioned into four polygons such that these four polygons can be composed again to form the parallelogram $% P^{\prime }.$

Let $m$ and $n$ be positive integers. A sequence of points $(A_0,A_1,\ldots,A_n)$ on the Cartesian plane is called [i]interesting[/i] if $A_i$ are all lattice points, the slopes of $OA_0,OA_1,\cdots,OA_n$ are strictly increasing ($O$ is the origin) and the area of triangle $OA_iA_{i+1}$ is equal to $\frac{1}{2}$ for $i=0,1,\ldots,n-1$. Let $(B_0,B_1,\cdots,B_n)$ be a sequence of points. We may insert a point $B$ between $B_i$ and $B_{i+1}$ if $\overrightarrow{OB}=\overrightarrow{OB_i}+\overrightarrow{OB_{i+1}}$, and the resulting sequence $(B_0,B_1,\ldots,B_i,B,B_{i+1},\ldots,B_n)$ is called an [i]extension[/i] of the original sequence. Given two [i]interesting[/i] sequences $(C_0,C_1,\ldots,C_n)$ and $(D_0,D_1,\ldots,D_m)$, prove that if $C_0=D_0$ and $C_n=D_m$, then we may perform finitely many [i]extensions[/i] on each sequence until the resulting two sequences become identical.

Let $P$ be an interior point of the parallelogram $ABCD$. Prove that $\angle APB+ \angle CPD = 180^\circ$ if and only if $\angle PDC = \angle PBC$.

Three vertices of parallelogram $PQRS$ are $P(-3,-2)$, $Q(1,-5)$, $R(9,1)$ with $P$ and $R$ diagonally opposite. The sum of the coordinates of vertex $S$ is: $\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 9$

Let $ABCD$ be a trapezoid (quadrilateral with one pair of parallel sides) such that $AB < CD$. Suppose that $AC$ and $BD$ meet at $E$ and $AD$ and $BC$ meet at $F$. Construct the parallelograms $AEDK$ and $BECL$. Prove that $EF$ passes through the midpoint of the segment $KL$.

Prove that if a lattice parallelogram contains at most three lattice points in addition to its vertices, then those are on one of the diagonals.

Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P\subset R$ we have \[\frac{|R|}{|P|}\leq \sqrt 2\] where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively. [i]Proposed by Witold Szczechla, Poland[/i]

Let $ABC$ be a triangle inscribed in circle $\omega$, and let the medians from $B$ and $C$ intersect $\omega$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $D$ tangent to $AC$ at $C$, and let $O_2$ be the center of the circle through $E$ tangent to $AB$ at $B$. Prove that $O_1$, $O_2$, and the nine-point center of $ABC$ are collinear. [i]Proposed by Michael Kural[/i]

On the sides $AB$ and $CD$ of the parallelogram $ABCD$ mark points $E$ and $F$, respectively. On the diagonals $AC$ and $BD$ chose the points $M$ and $N$ so that $EM\parallel BD$ and $FN\parallel AC$. Prove that the lines $AF, DE$ and $MN$ intersect at one point. (B. Rublev)

Let $ABCD$ be a convex quadrilateral, let $E$ and $F$ be the midpoints of the sides $AD$ and $BC$, respectively. The segment $CE$ meets $DF$ in $O$. Show that if the lines $AO$ and $BO$ divide the side $CD$ in 3 equal parts, then $ABCD$ is a parallelogram.

Let $ABCD$ be a quadrilateral, and $P,Q,R,S$ are the midpoints of $AB, BC, CD, DA$ respectively. Let $O$ be the intersection between $PR$ and $QS$. Prove that $PO = OR$ and $QO = OS$.