Triangles $ ABC$ and $ ADE$ have areas $ 2007$ and $ 7002,$ respectively, with $ B \equal{} (0,0),$ $ C \equal{} (223,0),$ $ D \equal{} (680,380),$ and $ E \equal{} (689,389).$ What is the sum of all possible x-coordinates of $ A?$ $ \textbf{(A)}\ 282 \qquad \textbf{(B)}\ 300 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 900 \qquad \textbf{(E)}\ 1200$

Let $ABCD$ be a trapezoid with $AB$ parallel to $CD$ and perpendicular to $BC$. Let $M$ be a point on $BC$ such that $\angle AMB = \angle DMC$. If $AB = 3$, $BC = 24$, and $CD = 4$, what is the value of $AM + MD$?

In trapezoid $ABCD$, $AD$ is parallel to $BC$. Knowing that $AB=AD+BC$, prove that the bisector of $\angle A$ also bisects $CD$.

Given a convex, $n$-sided polygon $P$, form a $2n$-sided polygon $\text{clip}(P)$ by cutting off each corner of $P$ at the edges' trisection points. In other words, $\text{clip}(P)$ is the polygon whose vertices are the $2n$ edge trisection points of $P$, connected in order around the boundary of $P$. Let $P_1$ be an isosceles trapezoid with side lengths $13,13,13,$ and $3$, and for each $i\geq 2$, let $P_i=\text{clip}(P_{i-1}).$ This iterative clipping process approaches a limiting shape $P_\infty=\lim_{i\to\infty}P_i$. If the difference of the areas of $P_{10}$ and $P_\infty$ is written as a fraction $\tfrac xy$ in lowest terms, calculate the number of positive integer factors of $x\cdot y$.

Let $K$, in square units, be the area of a trapezoid such that the shorter base, the altitude, and the longer base, in that order, are in arithmetic progression. Then: $\textbf{(A)}\ K \; \text{must be an integer} \qquad \textbf{(B)}\ K \; \text{must be a rational fraction} \\ \textbf{(C)}\ K \; \text{must be an irrational number} \qquad \textbf{(D)}\ K\; \text{must be an integer or a rational fraction} \qquad$ $\textbf{(E)}\ \text{taken alone neither} \; \textbf{(A)} \; \text{nor} \; \textbf{(B)} \; \text{nor} \; \textbf{(C)} \; \text{nor} \; \textbf{(D)} \; \text{is true}$

It is known that a circle can be inscribed in a trapezium $ABCD$. Prove that the two circles, constructed on its oblique sides as diameters, touch each other. (D. Fomin, Leningrad)

Let $PQ$ be a line segment of constant length $\lambda$ taken on the side $BC$ of a triangle $ABC$ with the order $B,P,Q,C$, and let the lines through $P$ and $Q$ parallel to the lateral sides meet $AC$ at $P_1$ and $Q_1$ and $AB$ at $P_2$ and $Q_2$ respectively. Prove that the sum of the areas of the trapezoids $PQQ_1P_1$ and $PQQ_2P_2$ is independent of the position of $PQ$ on $BC.$

Trapezoid $ABCD$ is an isosceles trapezoid with $AD=BC$. Point $P$ is the intersection of the diagonals $AC$ and $BD$. If the area of $\triangle ABP$ is $50$ and the area of $\triangle CDP$ is $72$, what is the area of the entire trapezoid? [i]Proposed by David Altizio

Given $\triangle{ABC}$ with $\angle{B}=60^{\circ}$ and $\angle{C}=30^{\circ}$, let $P,Q,R$ be points on the sides $BA,AC,CB$ respectively such that $BPQR$ is an isosceles trapezium with $PQ \parallel BR$ and $BP=QR$.\\ Find the maximum possible value of $\frac{2[ABC]}{[BPQR]}$ where $[S]$ denotes the area of any polygon $S$.

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".

The triangle $ABC$ acute with gravity center $M$ is such that $\angle AMB = 2 \angle ACB$. Prove that: [b](a)[/b] $AB^4=AC^4+BC^4-AC^2 \cdot BC^2,$ [b](b)[/b] $\angle ACB \geq 60^o$.

$ABCD$ is a trapezoid with $AB$ parallel to $CD$. The external bisectors of the angles at $ B$ and $C$ intersect at $ P$. The external bisectors of the angles at $ A$ and $D$ intersect at $Q$. Show that the length of $PQ$ is equal to half the perimeter of the trapezoid $ABCD$.

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)

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} $

The bases of a trapezoid have lengths 10 and 21, and the legs have lengths $\sqrt{34}$ and $3 \sqrt{5}$. What is the area of the trapezoid?

A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly 1/1985. [asy] size(200); pair A=(0,1), B=(1,1), C=(1,0), D=origin; draw(A--B--C--D--A--(1,1/6)); draw(C--(0,5/6)^^B--(1/6,0)^^D--(5/6,1)); pair point=( 0.5 , 0.5 ); //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("$1/n$", (11/12,1), N, fontsize(9));[/asy]

In a trapezoid $ABCD$, the legs $AB$ and $CD$ meet in $M$ and the diagonals $AC$ and $BD$ meet in $N.$ Let $AC=a$ and $BC=b.$ Find the area of triangles $AMD$ and $AND$ in terms of $a$ and $b.$

An isosceles trapezoid is divided by each diagonal into two isosceles triangles. Determine the angles of the trapezoid.

Let $ABCD$ be a right-angled trapezoid and $M$ be the midpoint of its greater lateral side $CD$. Circumcircles $\omega_{1}$ and $\omega_{2}$ of triangles $BCM$ and $AMD$ meet for the second time at point $E$. Let $ED$ meet $\omega_{1}$ at point $F$, and $FB$ meet $AD$ at point $G$. Prove that $GM$ bisects angle $BGD$.

The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$

Let $ K$ be the circumscribed circle of the trapezoid $ ABCD$ . In this trapezoid the diagonals $ AC$ and $ BD$ are perpendicular. The parallel sides $ AB\equal{}a$ and $ CD\equal{}c$ are diameters of the circles $ K_{a}$ and $ K_{b}$ respectively. Find the perimeter and the area of the part inside the circle $ K$, that is outside circles $ K_{a}$ and $ K_{b}$.

The consecutive angles of a trapezoid form an arithmetic sequence. If the smallest angle is $75^\circ$, then the largest angle is $\textbf{(A) }95^\circ\qquad \textbf{(B) }100^\circ\qquad \textbf{(C) }105^\circ\qquad \textbf{(D) }110^\circ\qquad \textbf{(E) }115^\circ$

Let $ABC$ be an isosceles triangle with $AB=AC$. Consider a variable point $P$ on the extension of the segment $BC$ beyound $B$ (in other words, $P$ lies on the line $BC$ such that the point $B$ lies inside the segment $PC$). Let $r_{1}$ be the radius of the incircle of the triangle $APB$, and let $r_{2}$ be the radius of the $P$-excircle of the triangle $APC$. Prove that the sum $r_{1}+r_{2}$ of these two radii remains constant when the point $P$ varies. [i]Remark.[/i] The $P$-excircle of the triangle $APC$ is defined as the circle which touches the side $AC$ and the [i]extensions[/i] of the sides $AP$ and $CP$.

The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD \equal{} \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP \equal{} \angle DAQ$. [i]Author: Vyacheslav Yasinskiy, Ukraine[/i]

Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} \equal{} \angle{BCD}\qquad\text{and}\qquad \angle{CQD} \equal{} \angle{ABC}.\] Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]