Let $O$ be the circumcenter of the triangle $ABC$. The segment $XY$ is the diameter of the circumcircle perpendicular to $BC$ and it meets $BC$ at $M$. The point $X$ is closer to $M$ than $Y$ and $Z$ is the point on $MY$ such that $MZ = MX$. The point $W$ is the midpoint of $AZ$. a) Show that $W$ lies on the circle through the midpoints of the sides of $ABC$; b) Show that $MW$ is perpendicular to $AY$.

Let $ABCD$ a trapezium with $AB$ parallel to $CD, \Omega$ the circumcircle of $ABCD$ and $A_1,B_1$ points on segments $AC$ and $BC$ respectively, such that $DA_1B_1C$ is a cyclic cuadrilateral. Let $A_2$ and $B_2$ the symmetric points of $A_1$ and $B_1$ with respect of the midpoint of $AC$ and $BC$, respectively. Prove that points $A, B, A_2, B_2$ are concyclic.

Let $n=3k$ be a positive integer (with $k\geq 2$). An equilateral triangle 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). We colour the points of the grid with three colours (red, blue and green) such that each two neighboring points have different colour. Finally, the colour of a "trapezoid" will be the colour of the midpoint of its big base. Find the number of all "trapezoids" in the grid (not necessarily disjoint) and determine the number of red, blue and green "trapezoids".

In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of angle $D$ is twice the measure of angle $B$, and the measures of segments $AB$ and $CD$ are $a$ and $b$ respectively. Then the measure of $AB$ is equal to $\textbf{(A) }\dfrac{1}{2}a+2b\qquad\textbf{(B) }\dfrac{3}{2}b+\dfrac{3}{4}a\qquad\textbf{(C) }2a-b\qquad\textbf{(D) }4b-\dfrac{1}{2}a\qquad \textbf{(E) }a+b$ [asy] size(175); defaultpen(linewidth(0.8)); real r=50, a=4,b=2.5,c=6.25; pair A=origin,B=c*dir(r),D=(a,0),C=shift(b*dir(r))*D; draw(A--B--C--D--cycle); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S); label("$a$",D/2,N); label("$b$",(C+D)/2,NW); //Credit to djmathman for the diagram[/asy]

A non-isosceles trapezoid $ABCD$ ($AB // CD$) is given. An arbitrary circle passing through points $A$ and $B$ intersects the sides of the trapezoid at points $P$ and $Q$, and the intersect the diagonals at points $M$ and $N$. Prove that the lines $PQ, MN$ and $CD$ are concurrent.

Let $ ABCD$ be a (not self-intersecting) quadrilateral satisfying $ \measuredangle DAB \equal{} \measuredangle BCD\neq 90^{\circ}$. Let $ X$ and $ Y$ be the orthogonal projections of the point $ D$ on the lines $ AB$ and $ BC$, and let $ Z$ and $ W$ be the orthogonal projections of the point $ B$ on the lines $ CD$ and $ DA$. Establish the following facts: [b]a)[/b] The quadrilateral $ XYZW$ is an isosceles trapezoid such that $ XY\parallel ZW$. [b]b)[/b] Let $ M$ be the midpoint of the segment $ AC$. Then, the lines $ XZ$ and $ YW$ pass through the point $ M$. [b]c)[/b] Let $ N$ be the midpoint of the segment $ BD$, and let $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$, $ W^{\prime}$ be the midpoints of the segments $ AB$, $ BC$, $ CD$, $ DA$. Then, the point $ M$ lies on the circumcircles of the triangles $ W^{\prime}X^{\prime}N$ and $ Y^{\prime}Z^{\prime}N$. [hide="Notice"][i]Notice.[/i] This problem has been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=172417 .[/hide]

Let $ABC$ be a nonisosceles triangle. Point $O$ is its circumcenter, and point $K$ is the center of the circumcircle $w$ of triangle $BCO$. The altitude of $ABC$ from $A$ meets $w$ at a point $P$. The line $PK$ intersects the circumcircle of $ABC$ at points $E$ and $F$. Prove that one of the segments $EP$ and $FP$ is equal to the segment $PA$.

A trapezoid has side lengths as indicated in the figure (the sides with length $11$ and $36$ are parallel). Calculate the area of the trapezoid.[img]https://1.bp.blogspot.com/-5PKrqDG37X4/XzcJtCyUv8I/AAAAAAAAMY0/tB0FObJUJdcTlAJc4n6YNEaVIDfQ91-eQCLcBGAsYHQ/s0/1995%2BMohr%2Bp1.png[/img]

Nine lines are given such that every one of them intersects given square $ABCD$ on two trapezoids, which area ratio is $2 : 3$. Prove that at least $3$ of those $9$ lines pass through the same point

In quadrilateral $ABCD$, $AC = BD$ and $\measuredangle B = 60^\circ$. Denote by $M$ and $N$ the midpoints of $\overline{AB}$ and $\overline{CD}$, respectively. If $MN = 12$ and the area of quadrilateral $ABCD$ is 420, then compute $AC$. [i]Proposed by Aaron Lin[/i]

Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. [i]Proposed by Charles Leytem, Luxembourg[/i]

Point $X$ is chosen inside the non trapezoid quadrilateral $ABCD$ such that $\angle AXD +\angle BXC=180$. Suppose the angle bisector of $\angle ABX$ meets the $D$-altitude of triangle $ADX$ in $K$, and the angle bisector of $\angle DCX$ meets the $A$-altitude of triangle $ADX$ in $L$.We know $BK \perp CX$ and $CL \perp BX$. If the circumcenter of $ADX$ is on the line $KL$ prove that $KL \perp AD$. Proposed by [i]Alireza Dadgarnia[/i]

In triangle $ABC$, $\angle B=2\angle C$ and $\angle A>90^\circ$. Let $D$ be the point on the line $AB$ such that $CD$ is perpendicular to $AC$, and let $M$ be the midpoint of $BC$. Prove that $\angle AMB=\angle DMC$.

Label the vertices of a trapezoid $T$ inscribed in the unit circle as $A,B,C,D$ counterclockwise with $AB\parallel CD$. Let $s_1,s_2,$ and $d$ denote the lengths of $AB$, $CD$, and $OE$, where $E$ is the intersection of the diagonals of $T$, and $O$ is the center of the circle. Determine the least upper bound of $\frac{s_1-s_2}d$ over all $T$ for which $d\ne0$, and describe all cases, if any, in which equality is attained.

Let side $BC$ of $\bigtriangleup ABC$ be the diameter of a semicircle which cuts $AB$ and $AC$ at $D$ and $E$ respectively. $F$ and $G$ are the feet of the perpendiculars from $D$ and $E$ to $BC$ respectively. $DG$ and $EF$ intersect at $M$. Prove that $AM \perp BC$.

Two circles $O_1$ and $O_2$ intersect each other at $M$ and $N$. The common tangent to two circles nearer to $M$ touch $O_1$ and $O_2$ at $A$ and $B$ respectively. Let $C$ and $D$ be the reflection of $A$ and $B$ respectively with respect to $M$. The circumcircle of the triangle $DCM$ intersect circles $O_1$ and $O_2$ respectively at points $E$ and $F$ (both distinct from $M$). Show that the circumcircles of triangles $MEF$ and $NEF$ have same radius length.