Inside the parallelogram $ABCD$ is a point $X$. Make a line that passes through point $X$ and divides the parallelogram into two parts whose areas differ from each other the most.

In a paralleogram $ABCD$ , a point $P$ on the segment $AB$ is taken such that $\frac{AP}{AB}=\frac{61}{2022}$\\ and a point $Q$ on the segment $AD$ is taken such that $\frac{AQ}{AD}=\frac{61}{2065}$.If $PQ$ intersects $AC$ at $T$, find $\frac{AC}{AT}$ to the nearest integer

A triangle $ T$ is contained inside a point-symmetrical polygon $ M.$ The triangle $ T'$ is the mirror image of the triangle $ T$ with the reflection at one point $ P$, which inside the triangle $ T$ lies. Prove that at least one of the vertices of the triangle $ T'$ lies in inside or on the boundary of the polygon $ M.$

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 $ABCD$ be a square. Let $E, Z$ be points on the sides $AB, CD$ of the square respectively, such that $DE\parallel BZ$. Assume that the triangles $\triangle EAD, \triangle ZCB$ and the parallelogram $BEDZ$ have the same area. If the distance between the parallel lines $DE$ and $BZ$ is equal to $1$, determine the area of the square.

Let $ABCM$ be a quadrilateral and $D$ be an interior point such that $ABCD$ is a parallelogram. It is known that $\angle AMB =\angle CMD$. Prove that $\angle MAD =\angle MCD$.

In $ \triangle{ABC}$, we have $ AB\equal{}1$ and $ AC\equal{}2$. Side $ BC$ and the median from $ A$ to $ BC$ have the same length. What is $ BC$? $ \textbf{(A)}\ \frac{1\plus{}\sqrt2}{2} \qquad \textbf{(B)}\ \frac{1\plus{}\sqrt3}{2} \qquad \textbf{(C)}\ \sqrt2 \qquad \textbf{(D)}\ \frac{3}{2} \qquad \textbf{(E)}\ \sqrt3$

Two circles are internally tangent at $N$. The chords $BA$ and $BC$ of the larger circle are tangent to the smaller circle at $K$ and $M$ respectively. $Q$ and $P$ are midpoint of arcs $AB$ and $BC$ respectively. Circumcircles of triangles $BQK$ and $BPM$ are intersect at $L$. Show that $BPLQ$ is a parallelogram.

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.)

Let $\omega$ be a semicircle with diamater $AB$. Let $M$ be the midpoint of $AB$. Let $X,Y$ be points on the same semiplane with $\omega$ with respect to the line $AB$ such that $AMXY$ is a parallelogram. Let $XM\cap \omega = C$ and $YM \cap \omega = D$. Let $I$ be the incenter of $\triangle XYM$. Let $AC \cap BD= E$ and $ME$ intersects $XY$ at $T$. Let the intersection point of $TI$ and $AB$ be $Q$ and let the perpendicular projection of $T$ onto $AB$ be $P$. Prove that $M$ is midpoint of $PQ$

Let $ABCD$ be a parallelogram. Points $P$ and $Q$ lie inside $ABCD$ such that $\bigtriangleup ABP$ and $\bigtriangleup{BCQ}$ are equilateral. Prove that the intersection of the line through $P$ perpendicular to $PD$ and the line through $Q$ perpendicular to $DQ$ lies on the altitude from $B$ in $\bigtriangleup{ABC}$.

Let a triangle $ABC$ , $O$ it's circumcenter , $H$ ortocenter and $M$ the midpoint of $AH$. The perpendicular at $M$ to line $OM$ meets $AB$ and $AC$ at points $P$, respective $Q$. Prove that $MP=MQ$. Babis

In parallelogram $ABCD$, $\overline{DE}$ is the altitude to the base $\overline{AB}$ and $\overline{DF}$ is the altitude to the base $\overline{BC}$. ['''Note:''' ''Both pictures represent the same parallelogram.''] If $DC=12$, $EB=4$, and $DE=6$, then $DF=$ [asy] unitsize(12); pair A,B,C,D,P,Q,W,X,Y,Z; A = (0,0); B = (12,0); C = (20,6); D = (8,6); W = (18,0); X = (30,0); Y = (38,6); Z = (26,6); draw(A--B--C--D--cycle); draw(W--X--Y--Z--cycle); P = (8,0); Q = (758/25,6/25); dot(A); dot(B); dot(C); dot(D); dot(W); dot(X); dot(Y); dot(Z); dot(P); dot(Q); draw(A--B--C--D--cycle); draw(W--X--Y--Z--cycle); draw(D--P); draw(Z--Q); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",P,S); label("$A$",W,SW); label("$B$",X,S); label("$C$",Y,NE); label("$D$",Z,NW); label("$F$",Q,E); [/asy] $\text{(A)}\ 6.4 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 7.2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$

Let $ABCD$ be an inscribed quadrilateral in a circumference with center $O$ such that it lies inside $ABCD$ and $\angle{BAC} = \angle{ODA}$. Let $E$ be the intersection of $AC$ with $BD$. Lines $r$ and $s$ are drawn through $E$ such that $r$ is perpendicular to $BC$, and $s$ is perpendicular to $AD$. Let $P$ be the intersection of $r$ with $AD$, and $M$ the intersection of $s$ with $BC$. Let $N$ be the midpoint of $EO$. Prove that $M$, $N$, and $P$ lie on a line.

An equilateral triangle with side length 9 is divided into 81 congruent triangles with segments which are parallel to the sides of the triangle. Prove that it cannot be divided into more than 18 parallelograms with sides 1 and 2. [I]Proposed by O.Vanyushina[/i]

Triangle $ABC$ has sides $AB = 25$, $BC = 30$, and $CA=20$. Let $P,Q$ be the points on segments $AB,AC$, respectively, such that $AP=5$ and $AQ=4$. Suppose lines $BQ$ and $CP$ intersect at $R$ and the circumcircles of $\triangle{BPR}$ and $\triangle{CQR}$ intersect at a second point $S\ne R$. If the length of segment $SA$ can be expressed in the form $\frac{m}{\sqrt{n}}$ for positive integers $m,n$, where $n$ is not divisible by the square of any prime, find $m+n$. [i]Victor Wang[/i]

Let $A,C, P$ be three distinct points in the plane. Construct all parallelograms $ABCD$ such that point $P$ lies on the bisector of angle $DAB$ and $\angle APD = 90^\circ$.

$ABCD$ is a quadrilateral such that $AB$ is not parallel to $CD$, and $BC$ is not parallel to $AD$. Variable points $P, Q, R, S$ are taken on $AB, BC, CD, DA$ respectively so that $PQRS$ is a parallelogram. Find the locus of its center.

In rectangle $ABCD$, the points $M,N,P, Q$ lie on $AB$, $BC$, $CD$, $DA$ respectively such that the area of triangles $AQM$, $BMN$, $CNP$, $DPQ$ are equal. Prove that the quadrilateral $MNPQ$ is parallelogram. by Mahdi Etesami Fard

Let $A$ be a set consisting of 6 points in the plane. denoted $n(A)$ as the number of the unit circles which meet at least three points of $A$. Find the maximum of $n(A)$

Let $ABCD$ be a parallelogram. On side $AB$, point $M$ is taken so that $AD = DM$. On side $AD$ point $N$ is taken so that $AB = BN$. Prove that $CM = CN$.

In a triangle $ABC$ with $AB<AC<BC$, the perpendicular bisectors of $AC$ and $BC$ intersect $BC$ and $AC$ at $K$ and $L$, respectively. Let $O$, $O_1$, and $O_2$ be the circumcentres of triangles $ABC$, $CKL$, and $OAB$, respectively. Prove that $OCO_1O_2$ is a parallelogram.

Consider a latticelike packing of translates of a convex region $K$. Let $t$ be the area of the fundamental parallelogram of the lattice defining the packing, and let $t_{\min} (K)$ denote the minimal value of $t$ taken for all latticelike packings. Is there a natural number $N$ such that for any $n>N$ and for any $K$ different from a parallelogram, $nt_{\min} (K)$ is smaller that the area of any convex domain in which $n$ translates to $K$ can be placed without overlapping? (By a [i]latticelike packing[/i] of $K$ we mean a set of nonoverlapping translates of $K$ obtained from $K$ by translations with all vectors of a lattice.) [G. and L. Fejes-Toth]

Let $P$ be any point inside the parallelogram $ABCD$ and let $R$ be the radius of the circle through $A$, $B$, and $C$. Show that the distance from $P$ to the nearest vertex is not greater than $R$.

In the parallelogram $CMNP$ extend the bisectors of angles $MCN$ and $PCN$ and intersect with extensions of sides PN and $MN$ at points $A$ and $B$, respectively. Prove that the bisector of the original angle $C$ of the the parallelogram is perpendicular to $AB$. [img]https://cdn.artofproblemsolving.com/attachments/f/3/fde8ef133758e06b1faf8bdd815056173f9233.png[/img]