Let $ M$ and $ N$ be points on the side $ BC$ of triangle $ ABC$, with the point $ M$ lying on the segment $ BN$, such that $ BM \equal{} CN$. Let $ P$ and $ Q$ be points on the segments $ AN$ and $ AM$, respectively, such that $ \measuredangle PMC \equal{}\measuredangle MAB$ and $ \measuredangle QNB \equal{}\measuredangle NAC$. Prove that $ \measuredangle QBC \equal{}\measuredangle PCB$.

A regular octagon $ ABCDEFGH$ has sides of length two. Find the area of $ \triangle{ADG}$. $ \textbf{(A)}\ 4 \plus{} 2 \sqrt{2} \qquad \textbf{(B)}\ 6 \plus{} \sqrt{2} \qquad \textbf{(C)}\ 4 \plus{} 3 \sqrt{2} \qquad \textbf{(D)}\ 3 \plus{} 4 \sqrt{2} \qquad \textbf{(E)}\ 8 \plus{} \sqrt{2}$

Convex quadrilateral $ ABCD$ has $ AB\equal{}9$ and $ CD\equal{}12$. Diagonals $ AC$ and $ BD$ intersect at $ E$, $ AC\equal{}14$, and $ \triangle AED$ and $ \triangle BEC$ have equal areas. What is $ AE$? $ \textbf{(A)}\ \frac{9}{2}\qquad \textbf{(B)}\ \frac{50}{11}\qquad \textbf{(C)}\ \frac{21}{4}\qquad \textbf{(D)}\ \frac{17}{3}\qquad \textbf{(E)}\ 6$

Diagonals of an isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ are perpendicular. Let $DE$ be the perpendicular from $D$ to $AB$, and let $CF$ be the perpendicular from $C$ to $DE$. Prove that angle $DBF$ is equal to half of angle $FCD$.

On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let $ R$ be the region formed by the union of the square and all the triangles, and $ S$ be the smallest convex polygon that contains $ R$. What is the area of the region that is inside $ S$ but outside $ R$? $ \textbf{(A)} \; \frac{1}{4} \qquad \textbf{(B)} \; \frac{\sqrt{2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt{3} \qquad \textbf{(E)} \; 2 \sqrt{3}$

Let $ABC$ be a fixed triangle. Three similar (by point order) isosceles trapezoids are built on its sides: $ABXY, BCZW, CAUV$, such that the sides of the triangle are bases of the respective trapezoids. The circumcircles of triangles $XZU, YWV$ meet at two points $P, Q$. Prove that the line $PQ$ passes through a fixed point independent of the choice of trapezoids.

Let $ABCD$ be a trapezium, $AD\parallel BC$, and let $E,F$ be points on the sides$AB$ and $CD$, respectively. The circumcircle of $AEF$ meets $AD$ again at $A_1$, and the circumcircle of $CEF$ meets $BC$ again at $C_1$. Prove that $A_1C_1,BD,EF$ are concurrent.

Prove that the sum of angles at the longer base of a trapezoid is less than the sum of angles at the shorter base.

Let $ABCD$ be a trapezoid ($AB<CD,AB\parallel CD$) and $P\equiv AD\cap BC$. Suppose that $Q$ be a point inside $ABCD$ such that $\angle QAB=\angle QDC=90-\angle BQC$. Prove that $\angle PQA=2\angle QCD$.

The parallel sides of a trapezoid are $ 3$ and $ 9$. The non-parallel sides are $ 4$ and $ 6$. A line parallel to the bases divides the trapezoid into two trapezoids of equal perimeters. The ratio in which each of the non-parallel sides is divided is: [asy]defaultpen(linewidth(.8pt)); unitsize(2cm); pair A = origin; pair B = (2.25,0); pair C = (2,1); pair D = (1,1); pair E = waypoint(A--D,0.25); pair F = waypoint(B--C,0.25); draw(A--B--C--D--cycle); draw(E--F); label("6",midpoint(A--D),NW); label("3",midpoint(C--D),N); label("4",midpoint(C--B),NE); label("9",midpoint(A--B),S);[/asy]$ \textbf{(A)}\ 4: 3\qquad \textbf{(B)}\ 3: 2\qquad \textbf{(C)}\ 4: 1\qquad \textbf{(D)}\ 3: 1\qquad \textbf{(E)}\ 6: 1$

Trapezoid $ABCD$ with bases $AD$ and $BC$ is inscribed in a circle, $M$ is the intersection of of its diagonals. A straight line passing through $M$ perpendicular to the bases intersects $BC$ at point$ K$, and the circle at point $L$, where $L$ is the one of the two intersection points for which $M$ lies on the segment $KL$. It is known that $MK = a$, $LM = b$. Find the radius of the circle tangent to the segments $AM$, $BM$ and the circle circumscribed around $ABCD$.

Let $A'$ be a point on one of the sides of the trapezoid $ABCD$ such that line $AA'$ divides the area of the trapezoid in half. Points $B'$, $C'$, $D'$ are defined similarly. Prove that the intersection points of the diagonals of quadrilaterals $ABCD$ and $A'B'C'D'$ are symmetrical wrt the midpoint of midline of trapezoid $ABCD$.

Given $n>4$, prove that every cyclic quadrilateral can be dissected into $n$ cyclic quadrilaterals.

Given an acute triangle $ PBC$ with $ PB\neq PC.$ Points $ A,D$ lie on $ PB,PC,$ respectively. $ AC$ intersects $ BD$ at point $ O.$ Let $ E,F$ be the feet of perpendiculars from $ O$ to $ AB,CD,$ respectively. Denote by $ M,N$ the midpoints of $ BC,AD.$ $ (1)$: If four points $ A,B,C,D$ lie on one circle, then $ EM\cdot FN \equal{} EN\cdot FM.$ $ (2)$: Determine whether the converse of $ (1)$ is true or not, justify your answer.

In trapezoid $ABCD$, $AD$ is perpendicular to $DC$, $AD=AB=3$, and $DC=6$. In addition, E is on $DC$, and $BE$ is parallel to $AD$. Find the area of $\Delta BEC$. [asy] defaultpen(linewidth(0.7)); pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0); draw(E--B--C--D--A--B); draw(rightanglemark(A, D, C)); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, NW); label("$3$", A--D, W); label("$3$", A--B, N); label("$6$", E, S);[/asy] $\textbf{(A)} \: 3\qquad \textbf{(B)} \: 4.5\qquad \textbf{(C)} \: 6\qquad \textbf{(D)} \: 9\qquad \textbf{(E)} \: 18\qquad $

Two circles $ C_1,C_2$ with different radii are given in the plane, they touch each other externally at $ T$. Consider any points $ A\in C_1$ and $ B\in C_2$, both different from $ T$, such that $ \angle ATB \equal{} 90^{\circ}$. (a) Show that all such lines $ AB$ are concurrent. (b) Find the locus of midpoints of all such segments $ AB$.

The diagonals of trapezoid $ABCD$ ($BC\parallel AD$) meet at point $O$. Points $M$ and $N$ lie on the segments $BC$ and $AD$ respectively. The tangent to the circle $AMC$ at $C$ meets the ray $NB$ at point $P$; the tangent to the circle $BND$ at $D$ meets the ray $MA$ at point $R$. Prove that $\angle BOP =\angle AOR$.

Let $ ABC$ be a triangle and $ K$ and $ L$ be two points on $ (AB)$, $ (AC)$ such that $ BK \equal{} CL$ and let $ P \equal{} CK\cap BL$. Let the parallel through $ P$ to the interior angle bisector of $ \angle BAC$ intersect $ AC$ in $ M$. Prove that $ CM \equal{} AB$.

Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.

Let $ABCD$ be a trapezoid (quadrilateral with one pair of parallel sides) such that $AB < CD$. Suppose that $AC$ and $BD$ meet at $E$ and $AD$ and $BC$ meet at $F$. Construct the parallelograms $AEDK$ and $BECL$. Prove that $EF$ passes through the midpoint of the segment $KL$.

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]

Let $ABCD$ be a trapezoid (quadrilateral with one pair of parallel sides) such that $AB < CD$. Suppose that $AC$ and $BD$ meet at $E$ and $AD$ and $BC$ meet at $F$. Construct the parallelograms $AEDK$ and $BECL$. Prove that $EF$ passes through the midpoint of the segment $KL$.

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 an equilateral triangle with the side of $20$ units. Amir divides this triangle into $400$ smaller equilateral triangles with the sides of $1$ unit. Reza then picks $4$ of the vertices of these smaller triangles. The vertices lie inside the triangle $ABC$ and form a parallelogram with sides parallel to the sides of the triangle $ABC.$ There are exactly $46$ smaller triangles that have at least one point in common with the sides of this parallelogram. Find all possible values for the area of this parallelogram. [asy] unitsize(150); defaultpen(linewidth(0.7)); int n = 20; /* # of vertical lines, including BC */ pair A = (0,0), B = dir(-30), C = dir(30); draw(A--B--C--cycle,linewidth(1)); dot(A,UnFill(0)); dot(B,UnFill(0)); dot(C,UnFill(0)); label("$A$",A,W); label("$C$",C,NE); label("$B$",B,SE); for(int i = 1; i < n; ++i) { draw((i*A+(n-i)*B)/n--(i*A+(n-i)*C)/n); draw((i*B+(n-i)*A)/n--(i*B+(n-i)*C)/n); draw((i*C+(n-i)*A)/n--(i*C+(n-i)*B)/n); }[/asy] [Thanks azjps for drawing the diagram.] [hide="Note"][i]Note:[/i] Vid changed to Amir, and Eva change to Reza![/hide]

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".