The diagonals of a trapezoid $ABCD$ with $AD \parallel BC$ intersect at point $P$. Point $Q$ lies between the parallel lines $AD$ and $BC$ such that the line $CD$ separates points $P$ and $Q$, and $\angle AQD=\angle CQB$. Prove that $\angle BQP = \angle DAQ$.

The point $E$ lies on the base $[AD]$ of the trapezoid $ABCD$. The perimeters of the triangles $ABE, BCE$ and $CDE$ are equal. Prove that $|BC| = |AD|/2$

An isosceles trapezoid $ABCD$ with bases $AB$ and $CD$ has $AB=13$, $CD=17$, and height $3$. Let $E$ be the intersection of $AC$ and $BD$. Circles $\Omega$ and $\omega$ are circumscribed about triangles $ABE$ and $CDE$. Compute the sum of the radii of $\Omega$ and $\omega$.

Find the smallest integer $n$ satisfying the following condition: regardless of how one colour the vertices of a regular $n$-gon with either red, yellow or blue, one can always find an isosceles trapezoid whose vertices are of the same colour.

The lateral sidelines $AB$ and $CD$ of trapezoid $ABCD$ meet at point $S$. The bisector of angle $ASC$ meets the bases of the trapezoid at points $K$ and $L$ ($K$ lies inside segment $SL$). Point $X$ is chosen on segment $SK$, and point $Y$ is selected on the extension of $SL$ beyond $L$ such a way that $\angle AXC - \angle AYC = \angle ASC$. Prove that $\angle BXD - \angle BYD = \angle BSD$.

Let $ ABCDE$ be a convex pentagon such that $ AB\plus{}CD\equal{}BC\plus{}DE$ and $ k$ a circle with center on side $ AE$ that touches the sides $ AB$, $ BC$, $ CD$ and $ DE$ at points $ P$, $ Q$, $ R$ and $ S$ (different from vertices of the pentagon) respectively. Prove that lines $ PS$ and $ AE$ are parallel.

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 $AY$.

Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let $t$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Let the trapezoids $A_iB_iC_iD_i$ ($i = 1, 2, 3$) be similar and have the same clockwise direction. Their angles at $A_i$ and $B_i$ are $60^o$ and the sides $A_1B_1$, $B_2C_2$ and $A_3D_3$ are parallel. The lines $B_iD_{i+1}$ and $C_iA_{i+1}$ intersect at the point $P_i$ (the indices are understood cyclically, i.e. $A_4 = A_1$ and $D_4 = D_1$). Prove that the points $P_1$, $P_2$ and $P_3$ lie on a line.

In $\triangle ABC$ we have $BC>CA>AB$. The nine point circle is tangent to the incircle, $A$-excircle, $B$-excircle and $C$-excircle at the points $T,T_A,T_B,T_C$ respectively. Prove that the segments $TT_B$ and lines $T_AT_C$ intersect each other.

Two trapezoid angles and diagonals are respectively equal. Is it true that such are the trapezoid equal?

A point $ P$ is selected at random from the interior of the pentagon with vertices $ A \equal{} (0,2)$, $B \equal{} (4,0)$, $C \equal{} (2 \pi \plus{} 1, 0)$, $D \equal{} (2 \pi \plus{} 1,4)$, and $ E \equal{} (0,4)$. What is the probability that $ \angle APB$ is obtuse? [asy] size(150); pair A, B, C, D, E; A = (0,1.5); B = (3,0); C = (2 *pi + 1, 0); D = (2 * pi + 1,4); E = (0,4); draw(A--B--C--D--E--cycle); label("$A$", A, dir(180)); label("$B$", B, dir(270)); label("$C$", C, dir(0)); label("$D$", D, dir(0)); label("$E$", E, dir(180)); [/asy] $ \displaystyle \textbf{(A)} \ \frac {1}{5} \qquad \textbf{(B)} \ \frac {1}{4} \qquad \textbf{(C)} \ \frac {5}{16} \qquad \textbf{(D)} \ \frac {3}{8} \qquad \textbf{(E)} \ \frac {1}{2}$

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

Let $ ABCD$ be a trapezoid with $ AB\parallel{}CD$. Let $ a \equal{} AB$ and $ b \equal{} CD$. For $ MN\parallel{}AB$ such that $ M$ lies on $ AD,$ $ N$ lies on $ BC$, and the trapezoids $ ABNM$ and $ MNCD$ have the same area, the length of $ MN$ equals [img]http://i250.photobucket.com/albums/gg265/geometry101/NielsHenrikAbel1997Number6.jpg[/img] A. $ \sqrt{ab}$ B. $ \frac{a\plus{}b}{2}$ C. $ \frac{a^2 \plus{} b^2}{a\plus{}b}$ D. $ \sqrt{\frac{a^2 \plus{} b^2}{2}}$ E. $ \frac{a^2 \plus{} (2 \sqrt{2} \minus{} 2)ab \plus{} b^2}{\sqrt{2} (a\plus{}b)}$

On the trapezoid $ABCD$ , side $DA$ is perpendicular to the bases $AB$ and $CD$ . The base $AB$ measures $45$, the base $CD$ measures $20$ and the $BC$ side measures $65$. Let $P$ on the $BC$ side such that $BP$ measures $45$ and $M$ is the midpoint of $DA$. Calculate the measure of the $PM$ segment.

A trapezoid $ABCD$ is given with $BC // AD$. Assume that the bisectors of the angles $BAD$ and $CDA$ intersect on the perpendicular bisector of the line segment $BC$. Prove that $|AB|= |CD|$ or $|AB| +|CD| =|AD|$.

Let $ A,B,C$ be collinear points on the plane with $ B$ between $ A$ and $ C$. Equilateral triangles $ ABD,BCE,CAF$ are constructed with $ D,E$ on one side of the line $ AC$ and $ F$ on the other side. Prove that the centroids of the triangles are the vertices of an equilateral triangle, and that the centroid of this triangle lies on the line $ AC$.

A trapezoid $ABCD$ with bases $BC$ and $AD$ is given. The points $K$ and $L$ are chosen on the sides $AB$ and $CD$, respectively, so that $KL \parallel AD$. It turned out that the areas of the quadrilaterals $AKLD$ and $KBCL$ are equal. Find the length $KL$ if $BC = 3, AD = 5$.

Let $ABCD$ be a trapezoid and a tangential quadrilateral such that $AD || BC$ and $|AB|=|CD|$. The incircle touches $[CD]$ at $N$. $[AN]$ and $[BN]$ meet the incircle again at $K$ and $L$, respectively. What is $\dfrac {|AN|}{|AK|} + \dfrac {|BN|}{|BL|}$? $ \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16 $

A circle center $O$ is inscribed in the quadrilateral $ABCD$. $AB$ is parallel to and longer than $CD$ and has midpoint $M$. The line $OM$ meets $CD$ at $F$. $CD$ touches the circle at $E$. Show that $DE = CF$ iff $AB = 2CD$.

In $\triangle ABC$ with incenter $I$, $AB = 61$, $AC = 51$, and $BC=71$. The circumcircles of triangles $AIB$ and $AIC$ meet line $BC$ at points $D$ ($D \neq B$) and $E$ ($E \neq C$), respectively. Determine the length of segment $DE$. [i]James Tao[/i]

Consider a trapezium $ ABCD $ in which $ AB\parallel CD. $ Show that $$ (AC^2+AB^2-BC^2)(BD^2-BC^2+CD^2) =(AC^2-AD^2+CD^2)(BD^2+AB^2-AD^2) . $$

Two parallel chords in a circle have lengths $10$ and $14$, and the distance between them is $6$. The chord parallel to these chords and midway between them is of length $\sqrt{a}$ where $a$ is [asy] // note: diagram deliberately not to scale -- azjps void htick(pair A, pair B, real r){ D(A--B); D(A-(r,0)--A+(r,0)); D(B-(r,0)--B+(r,0)); } size(120); pathpen = linewidth(0.7); pointpen = black+linewidth(3); real min = -0.6, step = 0.5; pair[] A, B; D(unitcircle); for(int i = 0; i < 3; ++i) { A.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)[0]); B.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)[1]); D(D(A[i])--D(B[i])); } MP("10",(A[0]+B[0])/2,N); MP("\sqrt{a}",(A[1]+B[1])/2,N); MP("14",(A[2]+B[2])/2,N); htick((B[1].x+0.1,B[0].y),(B[1].x+0.1,B[2].y),0.06); MP("6",(B[1].x+0.1,B[0].y/2+B[2].y/2),E);[/asy] $\textbf{(A)}\ 144 \qquad \textbf{(B)}\ 156 \qquad \textbf{(C)}\ 168 \qquad \textbf{(D)}\ 176 \qquad \textbf{(E)}\ 184$

$ABC$ is an acute triangle with orthocenter $H$. Points $D$ and $E$ lie on segment $BC$. Circumcircle of $\triangle BHC$ instersects with segments $AD$,$AE$ at $P$ and $Q$, respectively. Prove that if $BD^2+CD^2=2DP\cdot DA$ and $BE^2+CE^2=2EQ\cdot EA$, then $BP=CQ$.

Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$. [color=red][Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .][/color]