Let $ABC$ be an acute angled scalene triangle with circumcentre $O$ and orthocentre $H.$ If $M$ is the midpoint of $BC,$ then show that $AO$ and $HM$ intersect on the circumcircle of $ABC.$

Consider $2011^2$ points arranged in the form of a $2011 \times 2011$ grid. What is the maximum number of points that can be chosen among them so that no four of them form the vertices of either an isosceles trapezium or a rectangle whose parallel sides are parallel to the grid lines?

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form: $ \textbf{(A)}\ \text{a square} \qquad\textbf{(B)}\ \text{a rectangle} \qquad\textbf{(C)}\ \text{a parallelogram with unequal sides} \\ \textbf{(D)}\ \text{a rhombus} \qquad\textbf{(E)}\ \text{a quadrilateral with no special properties}$

Consider a triangle $XYZ$ and a point $O$ in its interior. Three lines through $O$ are drawn, parallel to the respective sides of the triangle. The intersections with the sides of the triangle determine six line segments from $O$ to the sides of the triangle. The lengths of these segments are integer numbers $a, b, c, d, e$ and $f$ (see figure). Prove that the product $a \cdot b \cdot c\cdot d \cdot e \cdot f$ is a perfect square. [asy] unitsize(1 cm); pair A, B, C, D, E, F, O, X, Y, Z; X = (1,4); Y = (0,0); Z = (5,1.5); O = (1.8,2.2); A = extension(O, O + Z - X, X, Y); B = extension(O, O + Y - Z, X, Y); C = extension(O, O + X - Y, Y, Z); D = extension(O, O + Z - X, Y, Z); E = extension(O, O + Y - Z, Z, X); F = extension(O, O + X - Y, Z, X); draw(X--Y--Z--cycle); draw(A--D); draw(B--E); draw(C--F); dot("$A$", A, NW); dot("$B$", B, NW); dot("$C$", C, SE); dot("$D$", D, SE); dot("$E$", E, NE); dot("$F$", F, NE); dot("$O$", O, S); dot("$X$", X, N); dot("$Y$", Y, SW); dot("$Z$", Z, dir(0)); label("$a$", (A + O)/2, SW); label("$b$", (B + O)/2, SE); label("$c$", (C + O)/2, SE); label("$d$", (D + O)/2, SW); label("$e$", (E + O)/2, SE); label("$f$", (F + O)/2, NW); [/asy]

The acute triangle $ABC$ with $AB\neq AC$ has circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. The midpoint of $BC$ is $M$, and the extension of the median $AM$ intersects $\Gamma$ at $N$. The circle of diameter $AM$ intersects $\Gamma$ again at $A$ and $P$. Show that the lines $AP$, $BC$, and $OH$ are concurrent if and only if $AH = HN$.

Let $ ABC$ be an acute triangle. $ H$ is the orthocenter and $ M$ is the middle of the side $ BC$. A line passing through $ H$ and perpendicular to $ HM$ intersect the segment $ AB$ and $ AC$ in $ P$ and $ Q$. Prove that $ MP \equal{} MQ$

Point $E$ is marked on side $BC$ of parallelogram $ABCD$, and on the side $AD$ - point $F$ so that the circumscribed circle of $ABE$ is tangent to line segment $CF$. Prove that the circumcircle of triangle $CDF$ is tangent to line $AE$.

Two parallelograms $ABCD$ and $A'B'C'D'$ are given in space. Points $A'',B'',C'',D''$ divide the segments $AA',BB',CC',DD'$ in the same ratio. What can be said about the quadrilateral $A''B''C''D''$?

In an acute triangle $\triangle{ABC}$, let $AE$ and $BF$ be highs of it, and $H$ its orthocenter. The symmetric line of $AE$ with respect to the angle bisector of $\sphericalangle{A}$ and the symmetric line of $BF$ with respect to the angle bisector of $\sphericalangle{B}$ intersect each other on the point $O$. The lines $AE$ and $AO$ intersect again the circuncircle to $\triangle{ABC}$ on the points $M$ and $N$ respectively. Let $P$ be the intersection of $BC$ with $HN$; $R$ the intersection of $BC$ with $OM$; and $S$ the intersection of $HR$ with $OP$. Show that $AHSO$ is a paralelogram.

Consider a parallelogram $ABCD$. a) Prove that if the incenter of the triangle $ABC$ is located on the diagonal $BD$, then the parallelogram $ABCD$ is a rhombus. b) Is the parallelogram $ABCD$ a rhombus whenever the circumcenter of the triangle $ABC$ is located on the diagonal $BD$?

Let $ABCD$ be a parallelogram and $P$ be an arbitrary point of the circumcircle of $\Delta ABD$, different from the vertices. Line $PA$ intersects the line $CD$ at point $Q$. Let $O$ be the center of the circumcircle $\Delta PCQ$. Prove that $\angle ADO = 90^o$.

Inside the parallelogram $ABCD$, point $M$ is chosen, and inside the triangle $AMD$, point $N$ is chosen in such a way that $$\angle MNA + \angle MCB =\angle MND + \angle MBC = 180^o.$$ Prove that lines $MN$ and $AB$ are parallel.

In a given plane $\varrho$ consider a convex quadrilateral $ABCD$ and denote $E=AC\cap BD.$ Moreover, consider a point $V\notin\varrho$. On rays $VA,VB,VC,VD$ find points $A',B',C',D'$ respectively such that $E,A',B',C',D'$ are coplanar and $A'B'C'D'$ is a parallelogram. Discuss conditions of solvability.

Let be a point $ P $ in the interior of a parallelogram $ ABCD $ such that $ \angle PAD=\angle PCD. $ Prove that the bisectors of $ \angle BAD $ and $ \angle BPD $ are parallel.

Given $\vartriangle ABC$ and a point $P$ inside that triangle. The parallelograms $CPBL$, $APCM$ and $BPAN$ are constructed. Prove that $AL$, $BM$ and $CN$ pass through one point $S$, and that $S$ is the midpoint of $AL$, $BM$ and $CN$.