In the figure, $AB \perp BC$, $BC \perp CD$, and $BC$ is tangent to the circle with center $O$ and diameter $AD$. In which one of the following cases is the area of $ABCD$ an integer? [asy] size(170); defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=(-1/sqrt(2),1/sqrt(2)), B=(-1/sqrt(2),-1), C=(1/sqrt(2),-1), D=(1/sqrt(2),-1/sqrt(2)); draw(unitcircle); dot(O); draw(A--B--C--D--A); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$O$",O,N); [/asy] $ \textbf{(A)}\ AB=3, CD=1\qquad\textbf{(B)}\ AB=5, CD=2\qquad\textbf{(C)}\ AB=7, CD=3\qquad\textbf{(D)}\ AB=9, CD=4\qquad\textbf{(E)}\ AB=11, CD=5 $

A trapezium is given with parallel bases having lengths $1$ and $4$. Split it into two trapeziums by a cut, parallel to the bases, of length $3$. We now want to divide the two new trapeziums, always by means of cuts parallel to the bases, in $m$ and $n$ trapeziums, respectively, so that all the $m + n$ trapezoids obtained have the same area. Determine the minimum possible value for $m + n$ and the lengths of the cuts to be made to achieve this minimum value.

Let $ A_{1}A_{2}A_{3}A_{4}A_{5}$ be a convex pentagon, such that \[ [A_{1}A_{2}A_{3}] \equal{} [A_{2}A_{3}A_{4}] \equal{} [A_{3}A_{4}A_{5}] \equal{} [A_{4}A_{5}A_{1}] \equal{} [A_{5}A_{1}A_{2}].\] Prove that there exists a point $ M$ in the plane of the pentagon such that \[ [A_{1}MA_{2}] \equal{} [A_{2}MA_{3}] \equal{} [A_{3}MA_{4}] \equal{} [A_{4}MA_{5}] \equal{} [A_{5}MA_{1}].\] Here $ [XYZ]$ stands for the area of the triangle $ \Delta XYZ$.

Let $ABCD$ be a trapezoid with bases $AB$ and $CD$. Bisectors of $AD$ and $BC$ intersect line segments $BC$ and $AD$ respectively in points $P$ and $Q$. Show that $\angle APD = \angle BQC$.

Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE:EB=CF:FD$. Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$. Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.

Let $ABCD$ be a trapezium (trapezoid) with $AD$ parallel to $BC$ and $J$ be the intersection of the diagonals $AC$ and $BD$. Point $P$ a chosen on the side $BC$ such that the distance from $C$ to the line $AP$ is equal to the distance from $B$ to the line $DP$. [i]The following three questions 1, 2 and 3 are independent, so that a condition in one question does not apply in another question.[/i] 1.Suppose that $Area( \vartriangle AJB) =6$ and that $Area(\vartriangle BJC) = 9$. Determine $Area(\vartriangle APD)$. 2. Find all points $Q$ on the plane of the trapezium such that $Area(\vartriangle AQB) = Area(\vartriangle DQC)$. 3. Prove that $PJ$ is the angle bisector of $\angle APD$.

The following isosceles trapezoid consists of equal equilateral triangles with side length $1$. The side $A_1E$ has length $3$ while the larger base $A_1A_n$ has length $n-1$. Starting from the point $A_1$ we move along the segments which are oriented to the right and up(obliquely right or left). Calculate (in terms of $n$ or not) the number of all possible paths we can follow, in order to arrive at points $B,\Gamma,\Delta, E$, if $n$ is an integer greater than $3$. [color=#00CCA7][Need image][/color]

In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD = BC = 5, AB = 4,$ and $DC = 10$. The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in the right triangle $DEF$. Then $CF =$ [asy] size(200); defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5,3); label("$A$", A, N); label("$B$", B, N); label("$C$", C, S); label("$D$", D, S); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); markscalefactor=0.05; draw(rightanglemark(E,F,D), linewidth(0.7));[/asy] $\text{(A)} \ 3.25 \qquad \text{(B)} \ 3.5 \qquad \text{(C)} \ 3.75 \qquad \text{(D)} \ 4.0 \qquad \text{(E)} \ 4.25$

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]

Trapezium $ABCD$ and parallelogram $MBDK$ are located so that the sides of the parallelogram are parallel to the diagonals of the trapezoid (see fig.). Prove that the area of the gray part is equal to the sum of the areas of the black part. (Yu. Blinkov) [img]https://cdn.artofproblemsolving.com/attachments/b/9/cfff83b1b85aea16b603995d4f3d327256b60b.png[/img]

Let $ABCD$ be a isosceles trapezoid such that $AB || CD$ and all of its sides are tangent to a circle. $[AD]$ touches this circle at $N$. $NC$ and $NB$ meet the circle again at $K$ and $L$, respectively. What is $\dfrac {|BN|}{|BL|} + \dfrac {|CN|}{|CK|}$? $ \textbf{(A)}\ 4 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 9 \qquad\textbf{(E)}\ 10 $

The diagonals of a trapezoid $ ABCD $ whose bases are $ [AB] $ and $ [CD] $ intersect at $P.$ Prove that \[S_{PAB} + S_{PCD} > S_{PBC} + S_{PDA},\] Where $S_{XYZ} $ denotes the area of $\triangle XYZ $.

A trapezoid with bases $a,b$ and altitude $h$ is circumscribed around a circl.. Prove that $h^2\le ab$.

In trapezoid $ABCD$, diagonal $AC$ is the bisector of angle $A$. Point $K$ is the midpoint of diagonal $AC$. It is known that $DC = DK$. Find the ratio of the bases $AD: BC$.

Let $ABCD$ be a cyclic quadrilateral. The bisectors of angles $BAD$ and $BCD$ intersect in point $K$ such that $K \in BD$. Let $M$ be the midpoint of $BD$. A line passing through point $C$ and parallel to $AD$ intersects $AM$ in point $P$. Prove that triangle $\triangle DPC$ is isosceles.

A square $ABCD$ is given. A point $P$ is chosen inside the triangle $ABC$ such that $\angle CAP = 15^\circ = \angle BCP$. A point $Q$ is chosen such that $APCQ$ is an isosceles trapezoid: $PC \parallel AQ$, and $AP=CQ, AP\nparallel CQ$. Denote by $N$ the midpoint of $PQ$. Find the angles of the triangle $CAN$.

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]

Let $ ABCD$ be a trapezoid inscribed in a circle $ \mathcal{C}$ with $ AB\parallel CD$, $ AB\equal{}2CD$. Let $ \{Q\}\equal{}AD\cap BC$ and let $ P$ be the intersection of tangents to $ \mathcal{C}$ at $ B$ and $ D$. Calculate the area of the quadrilateral $ ABPQ$ in terms of the area of the triangle $ PDQ$.

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

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. $b) a\cdot AP=b\cdot BP=c\cdot PC.$

Let $c$ be a real number, and let $z_1, z_2$ be the two complex numbers satisfying the quadratic $z^2 - cz + 10 = 0$. Points $z_1, z_2, \frac{1}{z_1}$, and $\frac{1}{z_2}$ are the vertices of a (convex) quadrilateral $Q$ in the complex plane. When the area of $Q$ obtains its maximum value, $c$ is the closest to which of the following? $\textbf{(A)}~4.5\qquad\textbf{(B)}~5\qquad\textbf{(C)}~5.5\qquad\textbf{(D)}~6\qquad\textbf{(E)}~6.5$

Let $ABCD$ be an isosceles trapezoid with bases $AB = 92$ and $CD = 19$. Suppose $AD = BC = x$ and a circle with center on $\overline{AB}$ is tangent to segments $\overline{AD}$ and $\overline{BC}$. If $m$ is the smallest possible value of $x$, then $m^2 = $ $ \textbf{(A)}\ 1369\qquad\textbf{(B)}\ 1679\qquad\textbf{(C)}\ 1748\qquad\textbf{(D)}\ 2109\qquad\textbf{(E)}\ 8825 $

Let $ABC$ be a scalene triangle with circumcircle $\Gamma$, and let $D$,$E$,$F$ be the points where its incircle meets $BC$, $AC$, $AB$ respectively. Let the circumcircles of $\triangle AEF$, $\triangle BFD$, and $\triangle CDE$ meet $\Gamma$ a second time at $X,Y,Z$ respectively. Prove that the perpendiculars from $A,B,C$ to $AX,BY,CZ$ respectively are concurrent. [i]Proposed by Michael Kural[/i]

A point $ P$ is randomly selected from the rectangular region with vertices $ (0, 0)$, $ (2, 0)$, $ (2, 1)$, $ (0, 1)$. What is the probability that $ P$ is closer to the origin than it is to the point $ (3, 1)$? $ \textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{4}{5} \qquad \textbf{(E)}\ 1$

In triangle $ABC$ with $AB>BC$, $BM$ is a median and $BL$ is an angle bisector. The line through $M$ and parallel to $AB$ intersects $BL$ at point $D$, and the line through $L$ and parallel to $BC$ intersects $BM$ at point $E$. Prove that $ED$ is perpendicular to $BL$.