In parallelogram $ABCD$ of the accompanying diagram, line $DP$ is drawn bisecting $BC$ at $N$ and meeting $AB$ (extended) at $P$. From vertex $C$, line $CQ$ is drawn bisecting side $AD$ at $M$ and meeting $AB$ (extended) at $Q$. Lines $DP$ and $CQ$ meet at $O$. If the area of parallelogram $ABCD$ is $k$, then the area of the triangle $QPO$ is equal to [asy] size((400)); draw((0,0)--(5,0)--(6,3)--(1,3)--cycle); draw((6,3)--(-5,0)--(10,0)--(1,3)); label("A", (0,0), S); label("B", (5,0), S); label("C", (6,3), NE); label("D", (1,3), NW); label("P", (10,0), E); label("Q", (-5,0), W); label("M", (.5,1.5), NW); label("N", (5.65, 1.5), NE); label("O", (3.4,1.75)); [/asy] $ \textbf{(A)}\ k \qquad\textbf{(B)}\ \frac{6k}{5} \qquad\textbf{(C)}\ \frac{9k}{8} \qquad\textbf{(D)}\ \frac{5k}{4} \qquad\textbf{(E)}\ 2k $

Let $A,B,C,D,E,F$ be points in space such that the quadrilaterals $ABDE,BCEF, CDFA$ are parallelograms. Prove that the six midpoints of the sides $AB,BC,CD,DE,EF,FA$ are coplanar

Let $ABCD$ be a parallelogram. Let $X \notin AC $ lie inside $ABCD$ so that $\angle AXB = \angle CXD = 90^ {\circ}$ and $\Omega$ denote the circumcircle of $AXC$. Consider a diameter $EF$ of $\Omega$ and assume neither $E, \ X, \ B$ nor $F, \ X, \ D$ are collinear. Let $K \neq X$ be an intersection point of circumcircles of $BXE$ and $DXF$ and $L \neq X$ be such point on $\Omega$ so that $\angle KXL = 90^{\circ}$. Prove that $AB = KL$.

A circle is inscribed in the trapezoid [i]ABCD[/i]. Let [i]K, L, M, N[/i] be the points of tangency of this circle with the diagonals [i]AC[/i] and [i]BD[/i], respectively ([i]K[/i] is between [i]A[/i] and [i]L[/i], and [i]M[/i] is between [i]B[/i] and [i]N[/i]). Given that $AK\cdot LC=16$ and $BM\cdot ND=\frac94$, find the radius of the circle. [color=red][Moderator edit: A solution of this problem can be found on http://www.ajorza.org/math/mathfiles/scans/belarus.pdf , page 20 (the statement of the problem is on page 6). The author of the problem is I. Voronovich.][/color]

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 $ABC$ be a triangle in which no angle is $90^{\circ}$. For any point $P$ in the plane of the triangle, let $A_1, B_1, C_1$ denote the reflections of $P$ in the sides $BC,CA,AB$ respectively. Prove that (i) If $P$ is the incenter or an excentre of $ABC$, then $P$ is the circumenter of $A_1B_1C_1$; (ii) If $P$ is the circumcentre of $ABC$, then $P$ is the orthocentre of $A_1B_1C_1$; (iii) If $P$ is the orthocentre of $ABC$, then $P$ is either the incentre or an excentre of $A_1B_1C_1$.

Let $\Gamma_1$ be a circle and $P$ a point outside of $\Gamma_1$. The tangents from $P$ to $\Gamma_1$ touch the circle at $A$ and $B$. Let $M$ be the midpoint of $PA$ and $\Gamma_2$ the circle through $P$, $A$ and $B$. Line $BM$ cuts $\Gamma_2$ at $C$, line $CA$ cuts $\Gamma_1$ at $D$, segment $DB$ cuts $\Gamma_2$ at $E$ and line $PE$ cuts $\Gamma_1$ at $F$, with $E$ in segment $PF$. Prove lines $AF$, $BP$, and $CE$ are concurrent.

Let $L$ be a subset in the coordinate plane defined by $L = \{(41x + 2y, 59x + 15y) | x, y \in \mathbb Z \}$, where $\mathbb Z$ is set of integers. Prove that for any parallelogram with center in the origin of coordinate and area $1990$, there exist at least two points of $L$ located in it.

Consider an acute triangle $ABC$ of area $S$. Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $DN \perp BC$ ($N \in BC$). Denote by $H_1$ and $H_2$ the orthocentres of the triangles $MNC$, respectively $MND$. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$.

The quadrilateral $ABCD$ is circumscribed about the circle, touches the sides $AB, BC, CD, DA$ in the points $K, L, M, N,$ respectively. Let $P, Q, R, S$ midpoints of the sides $KL, LM, MN, NK$. Prove that $PR = QS$ if and only if $ABCD$ is inscribed.

Let $ABCD$ be a convex quadrilateral with $AB = BC = CD$ and $P$ its intersection of diagonals. Denote by $O_1$, $O_2$ the circumcenters of triangles $ABP$, $CDP$, respectively. Prove that $O_1BCO_2$ is a parallelogram. [i] (Patrik Bak)[/i]

Let $P$ be a point inside a triangle $ABC$ such that $\angle PAC= \angle PCB$. Let the projections of $P$ onto $BC$, $CA$, and $AB$ be $X,Y,Z$ respectively. Let $O$ be the circumcenter of $\triangle XYZ$, $H$ be the foot of the altitude from $B$ to $AC$, $N$ be the midpoint of $AC$, and $T$ be the point such that $TYPO$ is a parallelogram. Show that $\triangle THN$ is similar to $\triangle PBC$. [i]Proposed by Sammy Luo[/i]

Let $ ABCD $ be an inscriptible quadrilateral and $ M $ be a point on its circumcircle, distinct from its vertices. Let $ H_1,H_2,H_3,H_4 $ be the orthocenters of $ MAB,MBC, MCD, $ respectively, $ MDA, $ and $ E,F, $ the midpoints of the segments $ AB, $ respectivley, $ CD. $ Prove that: [b]a)[/b] $ H_1H_2H_3H_4 $ is a parallelogram. [b]b)[/b] $ H_1H_3=2\cdot EF. $

Three $ \Delta $'s and a $ \diamondsuit $ will balance nine $ \bullet $'s. One $ \Delta $ will balance a $ \diamondsuit $ and a $ \bullet $. [asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,2)--(-12,0)--(12,0)--(12,2)); draw(ellipse((-12,5),8,3)); draw(ellipse((12,5),8,3)); label("$\Delta \hspace{2 mm}\Delta \hspace{2 mm}\Delta \hspace{2 mm}\diamondsuit $",(-12,6.5),S); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm} \bullet $",(12,5.2),N); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet $",(12,5.2),S); fill((44,0)--(40,-2)--(48,-2)--cycle,black); draw((34,2)--(34,0)--(54,0)--(54,2)); draw(ellipse((34,5),6,3)); draw(ellipse((54,5),6,3)); label("$\Delta $",(34,6.5),S); label("$\bullet \hspace{2 mm}\diamondsuit $",(54,6.5),S);[/asy] How many $ \bullet $'s will balance the two $ \diamondsuit $'s in this balance? [asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,4)--(-12,2)--(12,-2)--(12,0)); draw(ellipse((-12,7),6.5,3)); draw(ellipse((12,3),6.5,3)); label("$?$",(-12,8.5),S); label("$\diamondsuit \hspace{2 mm}\diamondsuit $",(12,4.5),S);[/asy] $ \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5 $

Let $ ABCD$ be a trapezoid with $ AB$ and $ CD$ parallel, $ \angle D \equal{} 2 \angle B, AD \equal{} 5,$ and $ CD \equal{} 2.$ Then $ AB$ equals A. 7 B. 8 C. 13/2 D. 27/4 E. $ 5 \plus{} \frac{3 \sqrt{2}}{2}$