Triangle $ABC$ is inscribed in circle $\Omega$. The interior angle bisector of angle $A$ intersects side $BC$ and $\Omega$ at $D$ and $L$ (other than $A$), respectively. Let $M$ be the midpoint of side $BC$. The circumcircle of triangle $ADM$ intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. Let $N$ be the midpoint of segment $PQ$, and let $H$ be the foot of the perpendicular from $L$ to line $ND$. Prove that line $ML$ is tangent to the circumcircle of triangle $HMN$.

Find the least positive integer $n$ ($n\geq 3$), such that among any $n$ points (no three are collinear) in the plane, there exist three points which are the vertices of a non-isoscele triangle.

Let $O$ denote the circumcentre of an acute-angled triangle $ABC$. Let point $P$ on side $AB$ be such that $\angle BOP = \angle ABC$, and let point $Q$ on side $AC$ be such that $\angle COQ = \angle ACB$. Prove that the reflection of $BC$ in the line $PQ$ is tangent to the circumcircle of triangle $APQ$.

Hexagon $ABCDEF$ is divided into four rhombuses, $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent, and each has area $\sqrt{2006}$. Let $K$ be the area of rhombus $\mathcal{T}$. Given that $K$ is a positive integer, find the number of possible values for $K$. [asy] size(150);defaultpen(linewidth(0.7)+fontsize(10)); draw(rotate(45)*polygon(4)); pair F=(1+sqrt(2))*dir(180), C=(1+sqrt(2))*dir(0), A=F+sqrt(2)*dir(45), E=F+sqrt(2)*dir(-45), B=C+sqrt(2)*dir(180-45), D=C+sqrt(2)*dir(45-180); draw(F--(-1,0)^^C--(1,0)^^A--B--C--D--E--F--cycle); pair point=origin; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$\mathcal{P}$", intersectionpoint( A--(-1,0), F--(0,1) )); label("$\mathcal{S}$", intersectionpoint( E--(-1,0), F--(0,-1) )); label("$\mathcal{R}$", intersectionpoint( D--(1,0), C--(0,-1) )); label("$\mathcal{Q}$", intersectionpoint( B--(1,0), C--(0,1) )); label("$\mathcal{T}$", point); dot(A^^B^^C^^D^^E^^F);[/asy]

A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is: $ \textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\\ \textbf{(B)}\ \frac{1}{2} \text{ the area of the original square}\\ \textbf{(C)}\ \frac{1}{2} \text{ the area of the circular piece}\\ \textbf{(D)}\ \frac{1}{4} \text{ the area of the circular piece}\\ \textbf{(E)}\ \text{none of these}$

Point $T$ is chosen in the plane of a rhombus $ABCD$ so that $\angle ATC + \angle BTD = 180^\circ$, and circumcircles of triangles $ATC$ and $BTD$ are tangent to each other. Show that $T$ is equidistant from diagonals of $ABCD$. [i]Proposed by Fedir Yudin[/i]

On the sides $AB,AD$ of the rhombus $ABCD$ are the points $E,F$ such that $AE=DF$. The lines $BC,DE$ intersect at $P$ and $CD,BF$ intersect at $Q$. Prove that: (a) $\frac{PE}{PD} + \frac{QF}{QB} = 1$; (b) $P,A,Q$ are collinear. [i]Virginia Tica, Vasile Tica[/i]

Let $ \mathcal{F}$ be a family of hexagons $ H$ satisfying the following properties: i) $ H$ has parallel opposite sides. ii) Any 3 vertices of $ H$ can be covered with a strip of width 1. Determine the least $ \ell\in\mathbb{R}$ such that every hexagon belonging to $ \mathcal{F}$ can be covered with a strip of width $ \ell$. Note: A strip is the area bounded by two parallel lines separated by a distance $ \ell$. The lines belong to the strip, too.

Rhombuses $ABCD$ and $A_1B_1C_1D_1$ are equal. Side $BC$ intersects sides $B_1C_1$ and $C_1D_1$ at points $M$ and $N$ respectively. Side $AD$ intersects sides $A_1B_1$ and $A_1D_1$ at points $Q$ and $P$ respectively. Let $O$ be the intersection point of lines $MP$ and $QN$. Find $\angle A_1B_1C_1$ , if $\angle QOP = \frac12 \angle B_1C_1D_1$.

In a triangle $ABC$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed into the quadrilateral $AEDB$ (all vertices of the rhombus lie on different sides of $AEDB$). Let $\varphi$ be the non-obtuse angle of the rhombus. Prove that $\varphi \le \max \{ \angle BAC, \angle ABC \}$.

Rhombus $CKLN$ is inscribed into triangle $ABC$ in such way that point $L$ lies on side $AB$, point $N$ lies on side $AC$, point $K$ lies on side $BC$. $O_1, O_2$ and $O$ are the circumcenters of triangles $ACL, BCL$ and $ABC$ respectively. Let $P$ be the common point of circles $ANL$ and $BKL$, distinct from $L$. Prove that points $O_1, O_2, O$ and $P$ are concyclic. (D.Prokopenko)

Let $n>3$ be a positive integer. Equilateral triangle ABC is divided into $n^2$ smaller congruent equilateral triangles (with sides parallel to its sides). Let $m$ be the number of rhombuses that contain two small equilateral triangles and $d$ the number of rhombuses that contain eight small equilateral triangles. Find the difference $m-d$ in terms of $n$.

On the sides $ AB, BC, CD$ and $ DA$ of the rhombus $ ABCD$, respectively, are chosen points $ E, F, G$ and $ H$ so, that $ EF$ and $ GH$ touch the incircle of the rhombus. Prove that the lines $ EH$ and $ FG$ are parallel.

Let $ABCD$ be a quadrilateral $AB = BC = CD = DA$. Let $MN$ and $PQ$ be two segments perpendicular to the diagonal $BD$ and such that the distance between them is $d > \frac{BD}{2}$, with $M \in AD$, $N \in DC$, $P \in AB$, and $Q \in BC$. Show that the perimeter of hexagon $AMNCQP$ does not depend on the position of $MN$ and $PQ$ so long as the distance between them remains constant.

Let $ABCD$ be a parallelogram. The circle with diameter $AC$ intersects the line $BD$ at points $P$ and $Q$. The perpendicular to the line $AC$ passing through the point $C$ intersects the lines $AB$ and $AD$ at points $X$ and $Y$, respectively. Prove that the points $P,Q,X$ and $Y$ lie on the same circle.

A holey triangle is an upward equilateral triangle of side length $n$ with $n$ upward unit triangular holes cut out. A diamond is a $60^\circ-120^\circ$ unit rhombus. Prove that a holey triangle $T$ can be tiled with diamonds if and only if the following condition holds: Every upward equilateral triangle of side length $k$ in $T$ contains at most $k$ holes, for $1\leq k\leq n$. [i]Proposed by Federico Ardila, Colombia [/i]

A point $M$ lies on the side $BC$ of square $ABCD$. Let $X$, $Y$ , and $Z$ be the incenters of triangles $ABM$, $CMD$, and $AMD$ respectively. Let $H_x$, $H_y$, and $H_z$ be the orthocenters of triangles $AXB$, $CY D$, and $AZD$. Prove that $H_x$, $H_y$, and $H_z$ are collinear.

Find the least positive integer $n$ ($n\geq 3$), such that among any $n$ points (no three are collinear) in the plane, there exist three points which are the vertices of a non-isoscele triangle.

Let $ABCD$ be a rhombus where $\angle DAB = 60^\circ$, and $P$ be the intersection between $AC$ and $BD$. Let $Q,R,S$ be three points on the boundary of $ABCD$ such that $PQRS$ is a rhombus. Prove that exactly one of $Q,R,S$ lies on one of $A,B,C,D$.

The points $E, F, G$ and $H{}$ are located on the sides $DA, AB, BC$ and $CD$ of the rhombus $ABCD$ respectively, so that the segments $EF$ and $GH$ touch the circle inscribed in the rhombus. Prove that $FG\parallel HE$. [i]Proposed by V. Eisenstadt[/i]

Prove or disprove: given any quadrilateral inscribed in a convex polygon, we can find a rhombus inscribed in the polygon with side not less than the shortest side of the quadrilateral.

Let $ABCD$ be a rhombus. Eight additional points $X_1$, $X_2$, $Y_1$, $Y_2$, $Z_1$, $Z_2$, $W_1$, $W_2$ were chosen so that the quadrilaterals $AX_1BX_2$, $BY_1CY_2$, $CZ_1DZ_2$, $DW_1AW_2$ are squares. Prove that the eight new points lie on two straight lines.

Let $ABCD$ be a convex quadrilateral with $\angle ABC = 120^{\circ}$ and $\angle BCD = 90^{\circ}$, and let $M$ and $N$ denote the midpoints of $\overline{BC}$ and $\overline{CD}$. Suppose there exists a point $P$ on the circumcircle of $\triangle CMN$ such that ray $MP$ bisects $\overline{AD}$ and ray $NP$ bisects $\overline{AB}$. If $AB + BC = 444$, $CD = 256$ and $BC = \frac mn$ for some relatively prime positive integers $m$ and $n$, compute $100m+n$. [i]Proposed by Michael Ren[/i]

A figure is an equiangular parallelogram if and only if it is a $ \textbf{(A)}\ \text{rectangle}\qquad\textbf{(B)}\ \text{regular polygon}\qquad\textbf{(C)}\ \text{rhombus}\qquad\textbf{(D)}\ \text{square}\qquad\textbf{(E)}\ \text{trapezoid} $

It is allowed to perform the following transformations in the plane with any integers $a$: (1) Transform every point $(x, y)$ to the corresponding point $(x + ay, y)$, (2) Transform every point $(x, y)$ to the corresponding point $(x, y + ax)$. Does there exist a non-square rhombus whose all vertices have integer coordinates and which can be transformed to: a) Vertices of a square, b) Vertices of a rectangle with unequal side lengths?