Let $ABCD$ be the isosceles trapezium with bases $AB$ and $CD$, such that $AC = BC$. The point $M$ is the midpoint of side $AD$. Prove that $\sphericalangle ACM = \sphericalangle CBD$.

On a cartesian plane a parabola $y = x^2$ is drawn. For a given $k > 0$ we consider all trapezoids inscribed into this parabola with bases parallel to the x-axis, and the product of the lengths of their bases is exactly $k$. Prove that the diagonals of all such trapezoids share a common point.

The horizontal and vertical distances between adjacent points equal $1$ unit. The area of triangle $ABC$ is [asy] for (int a = 0; a < 5; ++a) { for (int b = 0; b < 4; ++b) { dot((a,b)); } } draw((0,0)--(3,2)--(4,3)--cycle); label("$A$",(0,0),SW); label("$B$",(3,2),SE); label("$C$",(4,3),NE); [/asy] $\text{(A)}\ 1/4 \qquad \text{(B)}\ 1/2 \qquad \text{(C)}\ 3/4 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 5/4$

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.

The trapezoid $ABCD$ of bases $AB$ and $CD$, has $\angle A = 90^o, AB = 6, CD = 3$ and $AD = 4$. Let $E, G, H$ be the circumcenters of triangles $ABC, ACD, ABD$, respectively. Find the area of the triangle $EGH$.

There are two triangles $ABC$ and $BKL$ on the plane so that the segment $AK$ is divided into three equal parts by the point of intersection of the medians $\vartriangle ABC$ and the point of intersection of the bisectors $ \vartriangle BKL $ ($AK $ - median $ \vartriangle ABC$, $KA$ - bisector $\vartriangle BKL $) and quadrilateral $KALC $ is trapezoid. Find the angles of the triangle $BKL$. (Bogdan Rublev)

Let $n$ be a positive integer. An equilateral triangle with side $n$ will be denoted by $T_n$ and is divided in $n^2$ unit equilateral triangles with sides parallel to the initial, forming a grid. We will call "trapezoid" the trapezoid which is formed by three equilateral triangles (one base is equal to one and the other is equal to two). Let also $m$ be a positive integer with $m<n$ and suppose that $T_n$ and $T_m$ can be tiled with "trapezoids". Prove that, if from $T_n$ we remove a $T_m$ with the same orientation, then the rest can be tiled with "trapezoids".

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?

In the Cartesian plane, let $S_{i,j} = \{(x,y)\mid i \le x \le j\}$. For $i=0,1,\ldots,2012$, color $S_{i,i+1}$ pink if $i$ is even and gray if $i$ is odd. For a convex polygon $P$ in the plane, let $d(P)$ denote its pink density, i.e. the fraction of its total area that is pink. Call a polygon $P$ [i]pinxtreme[/i] if it lies completely in the region $S_{0,2013}$ and has at least one vertex on each of the lines $x=0$ and $x=2013$. Given that the minimum value of $d(P)$ over all non-degenerate convex pinxtreme polygons $P$ in the plane can be expressed in the form $\frac{(1+\sqrt{p})^2}{q^2}$ for positive integers $p,q$, find $p+q$. [i]Victor Wang.[/i]

The area of the shaded region $\text{BEDC}$ in parallelogram $\text{ABCD}$ is [asy] unitsize(10); pair A,B,C,D,E; A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0); draw(A--B--C--D--cycle); fill(B--E--D--C--cycle,gray); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S); label("$10$",(9,8),N); label("$6$",(7,0),S); label("$8$",(4,4),W); draw((3,0)--(3,1)--(4,1)); [/asy] $\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$

Let $ABCD$ be a trapezium $AB // CD,$ $M$ and $N$ are fixed points on $AB,$ $P$ is a variable point on $CD$. $E = DN \cap AP$, $F = DN \cap MC$, $G = MC \cap PB$, $DP = \lambda \cdot CD$. Find the value of $\lambda$ for which the area of quadrilateral $PEFG$ is maximum.

In trapezoid $ABCD$ the sides $BC$ and $AD$ are parallel, $AC = BC + AD$, and the angle between the diagonals is equal to $ 60^o$. Prove that $AB = CD$. (Stanislav Smirnov, St Petersburg)

In a trapezoid, the sum of the lengths of the side and the diagonal is equal to the sum of the lengths of the other side and the other diagonal. Prove that the trapezoid is isosceles.

In the trapezoid $ABCD$ with bases $AB$ and $CD$, let $M$ be the midpoint of side $DA$. If $BC=a$, $MC=b$ and $\angle MCB=150^\circ$, what is the area of trapezoid $ABCD$ as a function of $a$ and $b$?

In the diagram, $ABDF$ is a trapezoid with $AF$ parallel to $BD$ and $AB$ perpendicular to $BD.$ The circle with center $B$ and radius $AB$ meets $BD$ at $C$ and is tangent to $DF$ at $E.$ Suppose that $x$ is equal to the area of the region inside quadrilateral $ABEF$ but outside the circle, that y is equal to the area of the region inside $\triangle EBD$ but outside the circle, and that $\alpha = \angle EBC.$ Prove that there is exactly one measure $\alpha,$ with $0^\circ \leq \alpha \leq 90^\circ,$ for which $x = y$ and that this value of $\frac 12 < \sin \alpha < \frac{1}{\sqrt 2}.$ [asy] import graph; size(150); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttff = rgb(0,0.2,1); pen fftttt = rgb(1,0.2,0.2); draw(circle((6.04,2.8),1.78),qqttff); draw((6.02,4.58)--(6.04,2.8),fftttt); draw((6.02,4.58)--(6.98,4.56),fftttt); draw((6.04,2.8)--(8.13,2.88),fftttt); draw((6.98,4.56)--(8.13,2.88),fftttt); dot((6.04,2.8),ds); label("$B$", (5.74,2.46), NE*lsf); dot((6.02,4.58),ds); label("$A$", (5.88,4.7), NE*lsf); dot((6.98,4.56),ds); label("$F$", (7.06,4.6), NE*lsf); dot((7.39,3.96),ds); label("$E$", (7.6,3.88), NE*lsf); dot((8.13,2.88),ds); label("$D$", (8.34,2.56), NE*lsf); dot((7.82,2.86),ds); label("$C$", (7.5,2.46), NE*lsf); clip((-4.3,-10.94)--(-4.3,6.3)--(16.18,6.3)--(16.18,-10.94)--cycle); [/asy]

Given a triangle $ABC$, a point $P$ is chosen on side $BC$. Points $M$ and $N$ lie on sides $AB$ and $AC$, respectively, such that $MP \parallel AC$ and $NP \parallel AB$. Point $P$ is reflected across $MN$ to point $Q$. Show that triangle $QMB$ is similar to triangle $CNQ$. [i]Brian Hamrick.[/i]

Let $ABCD$ be a trapezoid with the measure of base $AB$ twice that of base $DC$, and let $E$ be the point of intersection of the diagonals. If the measure of diagonal $AC$ is $11$, then that of segment $EC$ is equal to $\textbf{(A) }3\textstyle\frac{2}{3}\qquad\textbf{(B) }3\frac{3}{4}\qquad\textbf{(C) }4\qquad\textbf{(D) }3\frac{1}{2}\qquad \textbf{(E) }3$

Find the necessary and sufficient condition that a trapezoid can be formed out of a given four-bar linkage.

Let $ABC$ be an acute,non-isosceles triangle with $AB<AC<BC$.Let $D,E,Z$ be the midpoints of $BC,AC,AB$ respectively and segments $BK,CL$ are altitudes.In the extension of $DZ$ we take a point $M$ such that the parallel from $M$ to $KL$ crosses the extensions of $CA,BA,DE$ at $S,T,N$ respectively (we extend $CA$ to $A$-side and $BA$ to $A$-side and $DE$ to $E$-side).If the circumcirle $(c_{1})$ of $\triangle{MBD}$ crosses the line $DN$ at $R$ and the circumcirle $(c_{2})$ of $\triangle{NCD}$ crosses the line $DM$ at $P$ prove that $ST\parallel PR$.

Let $ABC$ be a triangle each of whose angles is greater than $30^{\circ}$. Suppose a circle centered with $P$ cuts segments $BC$ in $T,Q; CA$ in $K,L$ and $AB$ in $M,N$ such that they are on a circle in counterclockwise direction in that order.Suppose further $PQK,PLM,PNT$ are equilateral. Prove that: $a)$ The radius of the circle is $\frac{2abc}{a^2+b^2+c^2+4\sqrt{3}S}$ where $S$ is area. $b) a\cdot AP=b\cdot BP=c\cdot PC.$

The trapezoid $ABCD$, of bases $AB$ and $CD$, is inscribed in a circumference $\Gamma$. Let $X$ a variable point of the arc $AB$ of $\Gamma$ that does not contain $C$ or $D$. We denote $Y$ to the point of intersection of $AB$ and $DX$, and let Z be the point of the segment $CX$ such that $\frac{XZ}{XC}=\frac{AY}{AB}$ . Prove that the measure of $\angle AZX$ does not depend on the choice of $X.$

In the $xy$-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at $(0,0), (0,3), (3,3), (3,1), (5,1)$ and $(5,0)$. The slope of the line through the origin that divides the area of this region exactly in half is [asy] size(200); Label l; l.p=fontsize(6); xaxis("$x$",0,6,Ticks(l,1.0,0.5),EndArrow); yaxis("$y$",0,4,Ticks(l,1.0,0.5),EndArrow); draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));[/asy] $ \textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9} $

$ABCD$ is a parallelogram in which angle $DAB$ is acute. Points $A, P, B, D$ lie on one circle in exactly this order. Lines $AP$ and $CD$ intersect in $Q$. Point $O$ is the circumcenter of the triangle $CPQ$. Prove that if $D \neq O$ then the lines $AD$ and $DO$ are perpendicular.

In this diagram the center of the circle is $O$, the radius is $a$ inches, chord $EF$ is parallel to chord $CD, O, G, H, J$ are collinear, and $G$ is the midpoint of $CD$. Let $K$ (sq. in.) represent the area of trapezoid $CDFE$ and let $R$ (sq. in.) represent the area of rectangle $ELMF$. Then, as $CD$ and $EF$ are translated upward so that $OG$ increases toward the value $a$, while $JH$ always equals $HG$, the ratio $K:R$ become arbitrarily close to: [asy] size((270)); draw((0,0)--(10,0)..(5,5)..(0,0)); draw((5,0)--(5,5)); draw((9,3)--(1,3)--(1,1)--(9,1)--cycle); draw((9.9,1)--(.1,1)); label("O", (5,0), S); label("a", (7.5,0), S); label("G", (5,1), SE); label("J", (5,5), N); label("H", (5,3), NE); label("E", (1,3), NW); label("L", (1,1), S); label("C", (.1,1), W); label("F", (9,3), NE); label("M", (9,1), S); label("D", (9.9,1), E); [/asy] $\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{1}{\sqrt{2}}+\frac{1}{2} \qquad\textbf{(E)}\ \frac{1}{\sqrt{2}}+1$

Let $ABCD$ be an isosceles trapezoid with $AB=AD=BC, AB//CD, AB>CD$. Let $E= AC \cap BD$ and $N$ symmetric to $B$ wrt $AC$. Prove that the quadrilateral $ANDE$ is cyclic.