Nine lines are given such that every one of them intersects given square $ABCD$ on two trapezoids, which area ratio is $2 : 3$. Prove that at least $3$ of those $9$ lines pass through the same point

$ABCD$ is a trapezoid with $AB || CD$. There are two circles $\omega_1$ and $\omega_2$ is the trapezoid such that $\omega_1$ is tangent to $DA$, $AB$, $BC$ and $\omega_2$ is tangent to $BC$, $CD$, $DA$. Let $l_1$ be a line passing through $A$ and tangent to $\omega_2$(other than $AD$), Let $l_2$ be a line passing through $C$ and tangent to $\omega_1$ (other than $CB$). Prove that $l_1 || l_2$.

Let $A_0A_1A_2A_3A_4A_5$ be a convex hexagon inscribed in a circle. Define the points $A_0'$, $A_2'$, $A_4'$ on the circle, such that \[ A_0A_0' \parallel A_2A_4, \quad A_2A_2' \parallel A_4A_0, \quad A_4A_4' \parallel A_2A_0 . \] Let the lines $A_0'A_3$ and $A_2A_4$ intersect in $A_3'$, the lines $A_2'A_5$ and $A_0A_4$ intersect in $A_5'$ and the lines $A_4'A_1$ and $A_0A_2$ intersect in $A_1'$. Prove that if the lines $A_0A_3$, $A_1A_4$ and $A_2A_5$ are concurrent then the lines $A_0A_3'$, $A_4A_1'$ and $A_2A_5'$ are also concurrent.

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]

There are two triangles $ABC$ and $BKL$ on the plane so that the segment $AK$ is divided into three equal parts by the point of intersection of the medians $\vartriangle ABC$ and the point of intersection of the bisectors $ \vartriangle BKL $ ($AK $ - median $ \vartriangle ABC$, $KA$ - bisector $\vartriangle BKL $) and quadrilateral $KALC $ is trapezoid. Find the angles of the triangle $BKL$. (Bogdan Rublev)

$BC$ is a diameter of a circle and the points $X,Y$ are on the circle such that $XY\perp BC$. The points $P,M$ are on $XY,CY$ (or their stretches), respectively, such that $CY||PB$ and $CX||PM$. Let $K$ be the meet point of the lines $XC,BP$. Prove that $PB\perp MK$.

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 $|BC| = |AD|/2$

Let $ ABCD$ be an isosceles trapezoid with $ AD \parallel BC$. Its diagonals $ AC$ and $ BD$ intersect at point $ M$. Points $ X$ and $ Y$ on the segment $ AB$ are such that $ AX \equal{} AM$, $ BY \equal{} BM$. Let $ Z$ be the midpoint of $ XY$ and $ N$ is the point of intersection of the segments $ XD$ and $ YC$. Prove that the line $ ZN$ is parallel to the bases of the trapezoid. [i]Author: A. Akopyan, A. Myakishev[/i]

Let $ABC$, $AC \not= BC$, be an acute triangle with orthocenter $H$ and incenter $I$. The lines $CH$ and $CI$ meet the circumcircle of $\bigtriangleup ABC$ at points $D$ and $L$, respectively. Prove that $\angle CIH = 90^{\circ}$ if and only if $\angle IDL = 90^{\circ}$

On square $ABCD,$ points $E,F,G,$ and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34.$ Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P,$ and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$. [asy] size(200); defaultpen(linewidth(0.8)+fontsize(10.6)); pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(120))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", (H+E)/2,fontsize(15)); label("$x$", (E+F)/2,fontsize(15)); label("$y$", (G+F)/2,fontsize(15)); label("$z$", (H+G)/2,fontsize(15)); label("$w:x:y:z=269:275:405:411$",(sqrt(850)/2,-4.5),fontsize(11)); [/asy]

Point $C_{1}$ of hypothenuse $AC$ of a right-angled triangle $ABC$ is such that $BC = CC_{1}$. Point $C_{2}$ on cathetus $AB$ is such that $AC_{2} = AC_{1}$; point $A_{2}$ is defined similarly. Find angle $AMC$, where $M$ is the midpoint of $A_{2}C_{2}$.

In a trapezoid $ABCD$, the side $DA$ is perpendicular to the bases $AB$ and $CD$. Also $AB=45$, $CD =20$, $BC =65$. Let $P$ be a point on the side $BC$ such that $BP=45$ and let $M$ be the midpoint of $DA$. Calculate the length of $PM$ .

Find the necessary and sufficient condition that a trapezoid can be formed out of a given four-bar linkage.

Circles $ S$ and $ T$ intersect at $ P$ and $ Q$, with $ S$ passing through the centre of $ T$. Distinct points $ A$ and $ B$ lie on $ S$, inside $ T$, and are equidistant from the centre of $ T$. The line $ PA$ meets $ T$ again at $ D$. Prove that $ |AD| \equal{} |PB|$.

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$

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 $

In an acute-angled triangle $ABC, CD$ is the altitude. A line through the midpoint $M$ of side $AB$ meets the rays $CA$ and $CB$ at $K$ and $L$ respectively such that $CK = CL.$ Point $S$ is the circumcenter of the triangle $CKL.$ Prove that $SD = SM.$

Let $s=\frac{AB+BC+AC}{2}$ be half-perimeter of triangle $ABC$. Let $L$ and $N$be a point's on ray's $AB$ and $CB$, for which $AL=CN=s$. Let $K$ is point, symmetric of point $B$ by circumcenter of $ABC$. Prove, that perpendicular from $K$ to $NL$ passes through incenter of $ABC$. Solution for problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=365714&p=2011659#p2011659]here[/url]

Prove the following inequality. \[ \frac{2}{2\plus{}e^{\frac 12}}<\int_0^1 \frac{dx}{1\plus{}xe^{x}}<\frac{2\plus{}e}{2(1\plus{}e)}\]

Points $A$, $B$, $C$, and $D$ lie on a circle such that chords $\overline{AC}$ and $\overline{BD}$ intersect at a point $E$ inside the circle. Suppose that $\angle ADE =\angle CBE = 75^\circ$, $BE=4$, and $DE=8$. The value of $AB^2$ can be written in the form $a+b\sqrt{c}$ for positive integers $a$, $b$, and $c$ such that $c$ is not divisible by the square of any prime. Find $a+b+c$. [i]Proposed by Tony Kim[/i]

Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$. Segments $AY$ and $BX$ meet at $P$. Point $Z$ is the foot of the perpendicular from $P$ to line $XY$. Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$. Let $Q$ be the intersection of segments $AY$ and $XC$. Prove that \[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]

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?

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

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

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 $C.$