Given an oriented line $\Delta$ and a fixed point $A$ on it, consider all trapezoids $ABCD$ one of whose bases $AB$ lies on $\Delta$, in the positive direction. Let $E,F$ be the midpoints of $AB$ and $CD$ respectively. Find the loci of vertices $B,C,D$ of trapezoids that satisfy the following: [i](i) [/i] $|AB| \leq a$ ($a$ fixed); [i](ii) [/i] $|EF| = l$ ($l$ fixed); [i](iii)[/i] the sum of squares of the nonparallel sides of the trapezoid is constant. [hide="Remark"] [b]Remark.[/b] The constants are chosen so that such trapezoids exist.[/hide]

Let $I$ be the incentre of triangle $ABC$. A circle containing the points $B$ and $C$ meets the segments $BI$ and $CI$ at points $P$ and $Q$ respectively. It is known that $BP\cdot CQ=PI\cdot QI$. Prove that the circumcircle of the triangle $PQI$ is tangent to the circumcircle of $ABC$. [i]Proposed by S. Berlov[/i]

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.

In parallelogram $ABCD$, point $K$ is marked on diagonal $AC$. Circle $s_1$ passes through point $K$ and touches lines $AB$ and $AD$ ($s_1$ intersects the diagonal $AC$ for the second time on the segment $AK$). Circle $s_2$ passes through point $K$ and touches lines $CB$ and $CD$ ($s_2$ intersects for the second time diagonal $AC$ on segment $KC$). Prove that for all positions of the point $K$ on the diagonal $AC$, the straight lines connecting the centers of circles $ s_1$ and $s_2$, will be parallel to each other.

The vertex $ A$ of the acute triangle $ ABC$ is equidistant from the circumcenter $ O$ and the orthocenter $ H.$ Determine all possible values for the measure of angle $ A.$

Prove that, for all convex pentagons $P_1 P_2 P_3 P_4 P_5$ with area 1, there are indices $i$ and $j$ (assume $P_7 = P_2$ and $P_6 = P_1$) such that: \[ \text{Area of} \ \triangle P_i P_{i+1} P_{i+2} \le \frac{5 - \sqrt 5}{10} \le \text{Area of} \ \triangle P_j P_{j+1} P_{j+2}\]

Consider $\triangle XPQ$ and $\triangle YPQ$ such that $X$ and $Y$ are on the opposite sides of $PQ$ and the circumradius of $\triangle XPQ$ and the circumradius of $\triangle YPQ$ are the same. $I$ and $J$ are the incenters of $\triangle XPQ$ and $\triangle YPQ$ respectively. Let $M$ be the midpoint of $PQ$. Suppose $I, M, J$ are collinear. Prove that $XPYQ$ is a parallelogram.

The quadrangles $AMBE, AHBT, BKXM$, and $CKXP$ are parallelograms. Prove that the quadrangle $ABTE$ is also parallelogram. (the vertices are mentioned counterclockwise)

We have an acute-angled triangle $ABC$, and $AA',BB'$ are its altitudes. A point $D$ is chosen on the arc $ACB$ of the circumcircle of $ABC$. If $P=AA'\cap BD,Q=BB'\cap AD$, show that the midpoint of $PQ$ lies on $A'B'$.

A regular hexagon with side length $1$ is given. Using a ruler construct points in such a way that among the given and constructed points there are two such points that the distance between them is $\sqrt7$. Notes: ''Using a ruler construct points $\ldots$'' means: Newly constructed points arise only as the intersection of straight lines connecting two points that are given or already constructed. In particular, no length can be measured by the ruler.

Let $ABCD$ be the parallelogram, such that angle at vertex $A$ is acute. Perpendicular bisector of the segment $AB$ intersects the segment $CD$ in the point $X$. Let $E$ be the intersection point of the diagonals of the parallelogram $ABCD$. Prove that $XE = \frac{1}{2}AD$.

Let \( AL \) be the bisector of triangle \( ABC \), \( O \) the center of its circumcircle, and \( D \) and \( E \) the midpoints of \( BL \) and \( CL \), respectively. Points \( P \) and \( Q \) are chosen on segments \( AD \) and \( AE \) such that \( APLQ \) is a parallelogram. Prove that \( PQ \perp AO \). [i]Proposed by Mykhailo Plotnikov[/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 $PB\perp MK$.

Outside the parallelogram $ABCD$ on its sides $AB$ and $BC$ are constructed equilateral triangles $ABK$, and $BCM$. Prove that the triangle $KMD$ is equilateral.

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 $% P^{\prime }.$

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 $P.$

Point $P$ lies inside of parallelogram $ABCD$. Perpendicular lines to $PA,PB,PC$ and $PD$ through $A,B,C$ and $D$ construct convex quadrilateral $XYZT$. Prove that $S_{XYZT}\geq 2S_{ABCD}$. [i]Proposed by Siamak Ahmadpour[/i]

$ABCD$ is a parallelogram. A point $M$ is found on the side $AB$ or its extension such that $\angle MAD = \angle AMO$ where $O$ is the intersection point of the diagonals of the parallelogram. Prove that $MD = MG$. (M Smurov)

Given a non-isoceles triangle $ABC$ inscribes a circle $(O,R)$ (center $O$, radius $R$). Consider a varying line $l$ such that $l\perp OA$ and $l$ always intersects the rays $AB,AC$ and these intersectional points are called $M,N$. Suppose that the lines $BN$ and $CM$ intersect, and if the intersectional point is called $K$ then the lines $AK$ and $BC$ intersect. $1$, Assume that $P$ is the intersectional point of $AK$ and $BC$. Show that the circumcircle of the triangle $MNP$ is always through a fixed point. $2$, Assume that $H$ is the orthocentre of the triangle $AMN$. Denote $BC=a$, and $d$ is the distance between $A$ and the line $HK$. Prove that $d\leq\sqrt{4R^2-a^2}$ and the equality occurs iff the line $l$ is through the intersectional point of two lines $AO$ and $BC$.

Consider an $ABCD$ parallelogram of area $1$. Let $E$ be the center of gravity of the triangle $ABC, F$ the center of gravity of the triangle $BCD, G$ the center of gravity of the triangle $CDA$ and $H$ the center of gravity of the triangle $DAB$. Calculate the area of quadrilateral $EFGH$.

Let $ABC$ and $XYZ$ be two triangles. Define \[A_1=BC\cap ZX, A_2=BC\cap XY,\]\[B_1=CA\cap XY, B_2=CA\cap YZ,\]\[C_1=AB\cap YZ, C_2=AB\cap ZX.\] Hereby, the abbreviation $g\cap h$ means the point of intersection of two lines $g$ and $h$. Prove that $\frac{C_1C_2}{AB}=\frac{A_1A_2}{BC}=\frac{B_1B_2}{CA}$ holds if and only if $\frac{A_1C_2}{XZ}=\frac{C_1B_2}{ZY}=\frac{B_1A_2}{YX}$.

Consider an $n\times{n}$ grid formed by $n^2$ unit squares. We define the centre of a unit square as the intersection of its diagonals. Find the smallest integer $m$ such that, choosing any $m$ unit squares in the grid, we always get four unit squares among them whose centres are vertices of a parallelogram.

Let $I$ be the center of the incircle of non-isosceles triangle $ABC,A_{1}=AI\cap BC$ and $B_{1}=BI\cap AC.$ Let $l_{a}$ be the line through $A_{1}$ which is parallel to $AC$ and $l_{b}$ be the line through $B_{1}$ parallel to $BC.$ Let $l_{a}\cap CI=A_{2}$ and $l_{b}\cap CI=B_{2}.$ Also $N=AA_{2}\cap BB_{2}$ and $M$ is the midpoint of $AB.$ If $CN\parallel IM$ find $\frac{CN}{IM}$.

Let $H$ be the orthocenter of an acute trangle $ABC$ with circumcircle $\Gamma$. Let $P$ be a point on the arc $BC$ (not containing $A$) of $\Gamma$, and let $M$ be a point on the arc $CA$ (not containing $B$) of $\Gamma$ such that $H$ lies on the segment $PM$. Let $K$ be another point on $\Gamma$ such that $KM$ is parallel to the Simson line of $P$ with respect to triangle $ABC$. Let $Q$ be another point on $\Gamma$ such that $PQ \parallel BC$. Segments $BC$ and $KQ$ intersect at a point $J$. Prove that $\triangle KJM$ is an isosceles triangle.

Points $ M,N$ are taken on sides $ BC,CD$ respectively of parallelogram $ ABCD$. Let $ E\equal{}BD\cap AM, F\equal{}BD\cap AN$. Diagonal $ BD$ cuts triangle $ AMN$ into two parts. Prove that these two parts have equal area if and only if the point $ K$ given by $ EK\parallel{}AD, FK\parallel{}AB$ lies on segment $ MN$.