We construct similar isosceles triangles on the sides of the triangle $ ABC $: triangle $ APB $ outside the triangle $ ABC $ ($ AP = PB $), triangle $ CQA $ outside the triangle $ ABC $ ($ CQ = QA $), triangle $ CRB $ inside the triangle $ ABC $ ($ CR = RB $). Prove that $ APRQ $ is a parallelogram or that the points $ A, P, R, Q $ lie on a straight line.

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.

Given $ \triangle ABC$ with point $ D$ inside. Let $ A_0\equal{}AD\cap BC$, $ B_0\equal{}BD\cap AC$, $ C_0 \equal{}CD\cap AB$ and $ A_1$, $ B_1$, $ C_1$, $ A_2$, $ B_2$, $ C_2$ are midpoints of $ BC$, $ AC$, $ AB$, $ AD$, $ BD$, $ CD$ respectively. Two lines parallel to $ A_1A_2$ and $ C_1C_2$ and passes through point $ B_0$ intersects $ B_1B_2$ in points $ A_3$ and $ C_3$respectively. Prove that $ \frac{A_3B_1}{A_3B_2}\equal{}\frac{C_3B_1}{C_3B_2}$.

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]

$ABCD$ is a parallelogram. $K$ is the circumcircle of $ABD$. The lines $BC$ and $CD$ meet $K$ again at $E$ and $F$. Show that the circumcenter of $CEF$ lies on $K$.

Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.

Two circles $ \omega_{1}$ and $ \omega_{2}$ intersect in points $ A$ and $ B$. Let $ PQ$ and $ RS$ be segments of common tangents to these circles (points $ P$ and $ R$ lie on $ \omega_{1}$, points $ Q$ and $ S$ lie on $ \omega_{2}$). It appears that $ RB\parallel PQ$. Ray $ RB$ intersects $ \omega_{2}$ in a point $ W\ne B$. Find $ RB/BW$. [i]S. Berlov [/i]

Consider a triangle $ABC$. For every point M $\in BC$ ,we define $N \in CA$ and $P \in AB$ such that $APMN$ is a parallelogram. Let $O$ be the intersection of $BN$ and $CP$. Find $M \in BC$ such that $\angle PMO=\angle OMN$

Let $D$ be an interior point of triangle $ABC$. Lines $BD$ and $CD$ intersect sides $AC$ and $AB$ at points $E$ and $F$, respectively. Points $X$ and $Y$ are on the plane such that $BFEX$ and $CEFY$ are parallelograms. Suppose lines $EY$ and $FX$ intersect at a point $T$ inside triangle $ABC$. Prove that points $B$, $C$, $E$, and $F$ are concyclic if and only if $\angle BAD = \angle CAT$.

Let $ABC$ be triangle with $H$ is the orthocenter and $I$ is incenter. Denote $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ be the points on the rays $AB, AC, BC, CA, CB$, respectively such that $$AA_{1} = AA_{2} = BC, BB_{1} = BB_{2} = CA, CC_{1} = CC_{2} = AB.$$ Suppose that $B_{1}B_{2}$ cuts $C_{1}C_{2}$ at $A'$, $C_{1}C_{2}$ cuts $A_{1}A_{2}$ at $B'$ and $A_{1}A_{2}$ cuts $B_{1}B_{2}$ at $C'$. a) Prove that area of triangle $A'B'C'$ is smaller than or equal to the area of triangle $ABC$. b) Let $J$ be circumcenter of triangle $A'B'C'$. $AJ$ cuts $BC$ at $R$, $BJ$ cuts $CA$ at $S$ and $CJ$ cuts $AB$ at $T$. Suppose that $(AST), (BTR), (CRS)$ intersect at $K$. Prove that if triangle $ABC$ is not isosceles then $HIJK$ is a parallelogram.

A circle can be inscribed in the convex quadrilateral $ ABCD$. The incircle touches the sides $ AB, BC, CD, DA$ at $ A_1, B_1, C_1, D_1$ respectively. The points $ E, F, G, H$ are the midpoints of $ A_1B_1, B_1C_1, C_1D_1, D_1A_1$ respectively. Prove that the quadrilateral $ EFGH$ is a rectangle if and only if $ A, B, C, D$ are concyclic.

Let $ ABCD$ be a parallelogram with $ \angle BAD \equal{} 60$ and denote by $ E$ the intersection of its diagonals. The circumcircle of triangle $ ACD$ meets the line $ BA$ at $ K \ne A$, the line $ BD$ at $ P \ne D$ and the line $ BC$ at $ L\ne C$. The line $ EP$ intersects the circumcircle of triangle $ CEL$ at points $ E$ and $ M$. Prove that triangles $ KLM$ and $ CAP$ are congruent.

Two medians of a triangle with unequal sides are $ 3$ inches and $ 6$ inches. Its area is $ 3 \sqrt{15}$ square inches. The length of the third median in inches, is: $ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \sqrt{3} \qquad \textbf{(C)}\ 3 \sqrt{6} \qquad \textbf{(D)}\ 6 \sqrt{3} \qquad \textbf{(E)}\ 6 \sqrt{6}$

In triangle $ABC$, point $I$ is the center of the inscribed circle. $AT$ is a segment tangent to the circle circumscribed around the triangle $BIC$ . On the ray $AB$ beyond the point$ B$ and on the ray $AC$ beyond the point $C$, we draw the segments $BD$ and $CE$, respectively, such that $BD = CE = AT$. Let the point $F$ be such that $ABFC$ is a parallelogram. Prove that points $D, E$ and $F$ lie on the same line. (Dmitry Prokopenko)

In the acute-angled triangle $ABC$ all angles are greater than $45^o$. Let $AM$ and $BN$ be the heights of this triangle and let $X$ and $Y$ be the points on $MA$ and $NB$, respecively, such that $|MX| =|MB|$ and $|NY| =|NA|$. Prove that $MN$ and $XY$ are parallel.

Call an $n$-gon to be [i]lattice[/i] if its vertices are lattice points. Prove that inside every lattice convex pentagon there exists a lattice point.

A parallelogram with an angle of $60^o$ has $a$ as the longest side and a shortest side $b$. Let's take the perpendiculars down from the vertices of the obtuse angles to the longest diagonal, then it is divided into three equal parts. Determine the ratio $\frac{a}{b}$.

Assume $ABC$ is an isosceles triangle that $AB=AC$ Suppose $P$ is a point on extension of side $BC$. $X$ and $Y$ are points on $AB$ and $AC$ that: \[PX || AC \ , \ PY ||AB \] Also $T$ is midpoint of arc $BC$. Prove that $PT \perp XY$

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 $ABC$ points $M, N$ are midpoints of $BC, CA$ respectively. Point $P$ is inside $ABC$ such that $\angle BAP = \angle PCA = \angle MAC .$ Prove that $\angle PNA = \angle AMB .$

A quadrilateral $ABCD$ is inscribed into a circle with center $O.$ Points $P$ and $Q$ are opposite to $C$ and $D$ respectively. Two tangents drawn to that circle at these points meet the line $AB$ in points $E$ and $F.$ ($A$ is between $E$ and $B$, $B$ is between $A$ and $F$). The line $EO$ meets $AC$ and $BC$ in points $X$ and $Y$ respectively, and the line $FO$ meets $AD$ and $BD$ in points $U$ and $V$ respectively. Prove that $XV=YU.$

Let $PQ$ be a line segment of constant length $\lambda$ taken on the side $BC$ of a triangle $ABC$ with the order $B,P,Q,C$, and let the lines through $P$ and $Q$ parallel to the lateral sides meet $AC$ at $P_1$ and $Q_1$ and $AB$ at $P_2$ and $Q_2$ respectively. Prove that the sum of the areas of the trapezoids $PQQ_1P_1$ and $PQQ_2P_2$ is independent of the position of $PQ$ on $BC.$

Let $ABCD$ be a convex quadrilateral, $M$ the midpoint of $AB$, $N$ the midpoint of $BC$, $E$ the intersection of the segments $AN$ and $BD$, $F$ the intersection of the segments $DM$ and $AC$. Prove that if $BE = \frac 13 BD$ and $AF = \frac 13 AC$, then $ABCD$ is a parallelogram.

In parallelogram $ABCD,$ let $O$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}.$ Angles $CAB$ and $DBC$ are each twice as large as angle $DBA,$ and angle $ACB$ is $r$ times as large as angle $AOB.$ Find the greatest integer that does not exceed $1000r.$

Let $ABCD$ be a parallelogram. An interior circle of the $ABCD$ is tangent to the lines $AB$ and $AD$ and intersects the diagonal $BD$ at $E$ and $F$. Prove that exists a circle that passes through $E$ and $F$ and is tangent to the lines $CB$ and $CD$.