In a trapezoid $ABCD$ with a base $AD$, point $L$ is the orthogonal projection of $C$ on $AB$, and $K$ is the point on $BC$ such that $AK$ is perpendicular to $AD$. Let $O$ be the circumcenter of triangle $ACD$. Suppose that the lines $AK , CL$ and $DO$ have a common point. Prove that $ABCD$ is a parallelogram.

Quadrilateral $ABCD$ has right angles at $B$ and $C$, $\triangle ABC \sim \triangle BCD$, and $AB > BC$. There is a point $E$ in the interior of $ABCD$ such that $\triangle ABC \sim \triangle CEB$ and the area of $\triangle AED$ is $17$ times the area of $\triangle CEB$. What is $\tfrac{AB}{BC}$? $\textbf{(A) \ } 1+\sqrt{2} \qquad \textbf{(B) \ } 2+\sqrt{2}\qquad \textbf{(C) \ } \sqrt{17}\qquad \textbf{(D) \ } 2+\sqrt{5} \qquad \textbf{(E) \ } 1+2\sqrt{3}$

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} $

Through the point of intersection of the diagonals of the trapezoid, a straight line is drawn parallel to its bases. Determine the length of the segment of this straight line, enclosed between the lateral sides of the trapezoid, if the lengths of the bases of the trapezoid are equal to $a$ and $b$.

Let $ABCD$ be an isosceles trapezoid, $AD=BC$, $AB \parallel CD$. The diagonals of the trapezoid intersect at the point $O$, and the point $M$ is the midpoint of the side $AD$. The circle circumscribed around the triangle $BCM$ intersects the side $AD$ at the point $K$. Prove that $OK \parallel AB$.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a three times differentiable function. Prove that there exists $w \in [-1,1]$ such that \[ \frac{f'''(w)}{6} = \frac{f(1)}{2}-\frac{f(-1)}{2}-f'(0). \]

The diagram below shows a regular hexagon with an inscribed square where two sides of the square are parallel to two sides of the hexagon. There are positive integers $m$, $n$, and $p$ such that the ratio of the area of the hexagon to the area of the square can be written as $\tfrac{m+\sqrt{n}}{p}$ where $m$ and $p$ are relatively prime. Find $m + n + p$. [asy] import graph; size(4cm); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((0,1)--(1,1)--(1.5,1.87)--(1,2.73)--(0,2.73)--(-0.5,1.87)--cycle); filldraw((1.13,2.5)--(-0.13,2.5)--(-0.13,1.23)--(1.13,1.23)--cycle,grey); draw((0,1)--(1,1)); draw((1,1)--(1.5,1.87)); draw((1.5,1.87)--(1,2.73)); draw((1,2.73)--(0,2.73)); draw((0,2.73)--(-0.5,1.87)); draw((-0.5,1.87)--(0,1)); draw((1.13,2.5)--(-0.13,2.5)); draw((-0.13,2.5)--(-0.13,1.23)); draw((-0.13,1.23)--(1.13,1.23)); draw((1.13,1.23)--(1.13,2.5)); [/asy]

Given a circle $\Gamma$ and the positive numbers $h$ and $m$, construct with straight edge and compass a trapezoid inscribed in $\Gamma$, such that it has altitude $h$ and the sum of its parallel sides is $m$.

Triangle $ABC$ is given. Let $M$ be the midpoint of the segment $AB$ and $T$ be the midpoint of the arc $BC$ not containing $A$ of the circumcircle of $ABC.$ The point $K$ inside the triangle $ABC$ is such that $MATK$ is an isosceles trapezoid with $AT\parallel MK.$ Show that $AK = KC.$

Let $ABC$ be an acute triangle and $H$ its orthocenter. Let $D$ be the intersection of the altitude from $A$ to $BC$. Let $M$ and $N$ be the midpoints of $BH$ and $CH$, respectively. Let the lines $DM$ and $DN$ intersect $AB$ and $AC$ at points $X$ and $Y$ respectively. If $P$ is the intersection of $XY$ with $BH$ and $Q$ the intersection of $XY$ with $CH$, show that $H, P, D, Q$ lie on a circumference.

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 $n$ 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".

A hexagon inscribed in a circle has three consecutive sides each of length $3$ and three consecutive sides each of length $5$. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length $3$ and the other with three sides each of length $5$, has length equal to $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. $\text{(A)}\ 309 \qquad \text{(B)}\ 349 \qquad \text{(C)}\ 369 \qquad \text{(D)}\ 389\qquad \text{(E)}\ 409$

Let $ABCD$ be a trapezoid such that $AB \parallel CD$, $|AB|<|CD|$, and $\text{Area}(ABC)=30$. Let the line through $B$ parallel to $AD$ meet $[AC]$ at $E$. If $|AE|:|EC|=3:2$, then what is the area of trapezoid $ABCD$? $ \textbf{(A)}\ 45 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 72 \qquad\textbf{(D)}\ 80 \qquad\textbf{(E)}\ 90 $

Let $ABCD$ be an isosceles trapezoid such that its diagonal is $\sqrt 3$ and its base angle is $60^\circ$, where $AD \parallel BC$. Let $P$ be a point on the plane of the trapezoid such that $|PA|=1$ and $|PD|=3$. Which of the following can be the length of $[PC]$? $ \textbf{(A)}\ \sqrt 6 \qquad\textbf{(B)}\ 2\sqrt 2 \qquad\textbf{(C)}\ 2 \sqrt 3 \qquad\textbf{(D)}\ 3\sqrt 3 \qquad\textbf{(E)}\ \sqrt 7 $

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]

All edges of the triangular pyramid $ABCD$ are equal in length. Let $M$ be the midpoint of $DB$, $N$ is the point on $\overline{AB}$, such that $2NA=NB$ and $N\not\in AB$ and $P$ is a point on the altitude through point $D$ in $\triangle BCD$. Find $\angle MPD$ if the intersection of the pyramid with the plane $(NMP)$ is a trapezoid.

Let $ABC$ be a triangle. Prove that there is a line $\ell$ (in the plane of triangle $ABC$) such that the intersection of the interior of triangle $ABC$ and the interior of its reflection $A'B'C'$ in $\ell$ has area more than $\frac23$ the area of triangle $ABC$.

Quadrilateral $ABCD$ is a trapezoid, $AD = 15$, $AB = 50$, $BC = 20$, and the altitude is $12$. What is the area of the trapezoid? [asy] pair A,B,C,D; A=(3,20); B=(35,20); C=(47,0); D=(0,0); draw(A--B--C--D--cycle); dot((0,0)); dot((3,20)); dot((35,20)); dot((47,0)); label("A",A,N); label("B",B,N); label("C",C,S); label("D",D,S); draw((19,20)--(19,0)); dot((19,20)); dot((19,0)); draw((19,3)--(22,3)--(22,0)); label("12",(21,10),E); label("50",(19,22),N); label("15",(1,10),W); label("20",(41,12),E);[/asy] $ \textbf{(A)}600\qquad\textbf{(B)}650\qquad\textbf{(C)}700\qquad\textbf{(D)}750\qquad\textbf{(E)}800 $

Points $E$ and $F$ are chosen on the base side $AD$ and the lateral side $AB$ of an isosceles trapezoid $ABCD$, respectively. Quadrilateral $CDEF$ is an isosceles trapezoid as well. Prove that $AE \cdot ED = AF \cdot FB$.

Let $\Omega$ be the circumcircle of an isosceles trapezoid $ABCD$, in which $AD$ is parallel to $BC$. Let $X$ be the reflection point of $D$ with respect to $BC$. Point $Q$ is on the arc $BC$ of $\Omega$ that does not contain $A$. Let $P$ be the intersection of $DQ$ and $BC$. A point $E$ satisfies that $EQ$ is parallel to $PX$, and $EQ$ bisects $\angle BEC$. Prove that $EQ$ also bisects $\angle AEP$. [i]Proposed by Li4.[/i]

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 trapezium $ ABCD$, in which $ AB$ is parallel to $ CD$, is inscribed in a circle with centre $ O$. Suppose the diagonals $ AC$ and $ BD$ of the trapezium intersect at $ M$, and $ OM \equal{} 2$. [b](a)[/b] If $ \angle AMB$ is $ 60^\circ ,$ find, with proof, the difference between the lengths of the parallel sides. [b](b)[/b] If $ \angle AMD$ is $ 60^\circ ,$ find, with proof, the difference between the lengths of the parallel sides. [b][Weightage 17/100][/b]

Let $ABCD$ be a trapezoid with bases $BC \parallel AD$, where $AD > BC$, and non-parallel legs $AB$ and $CD$. Let $M$ be the intersection of $AC$ and $BD$. Let $\Gamma_1$ be a circumference that passes through $M$ and is tangent to $AD$ at point $A$; let $\Gamma_2$ be a circumference that passes through $M$ and is tangent to $AD$ at point $D$. Let $S$ be the intersection of the lines $AB$ and $CD$, $X$ the intersection of $\Gamma_1$ with the line $AS$, $Y$ the intesection of $\Gamma_2$ with the line $DS$, and $O$ the circumcenter of triangle $ASD$. Show that $SO \perp XY$.

Let $ABCD$ be a trapezium with $AB\parallel CD$. Let $P$ be a point on $AC$ such that $C$ is between $A$ and $P$; and let $X, Y$ be the midpoints of $AB, CD$ respectively. Let $PX$ intersect $BC$ in $N$ and $PY$ intersect $AD$ in $M$. Prove that $MN\parallel AB$.