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$.

$(BEL 2) (a)$ Find the equations of regular hyperbolas passing through the points $A(\alpha, 0), B(\beta, 0),$ and $C(0, \gamma).$ $(b)$ Prove that all such hyperbolas pass through the orthocenter $H$ of the triangle $ABC.$ $(c)$ Find the locus of the centers of these hyperbolas. $(d)$ Check whether this locus coincides with the nine-point circle of the triangle $ABC.$

The diagonals of a convex quadrilateral divide it into four triangles. The radii of the circles circumscribed around these triangles are equal. Can such a property have a quadrilateral other than: a) parallelogram, b) rhombus? (Sharygin Igor)

$ AA_{3}$ and $ BB_{3}$ are altitudes of acute-angled $ \triangle ABC$. Points $ A_{1}$ and $ B_{1}$ are second points of intersection lines $ AA_{3}$ and $ BB_{3}$ with circumcircle of $ \triangle ABC$ respectively. $ A_{2}$ and $ B_{2}$ are points on $ BC$ and $ AC$ respectively. $ A_{1}A_{2}\parallel AC$, $ B_{1}B_{2}\parallel BC$. Point $ M$ is midpoint of $ A_{2}B_{2}$. $ \angle BCA \equal{} x$. Find $ \angle A_{3}MB_{3}$.

Let $ABCD$ be a parallelogram with $\angle ABC=120^\circ$, $AB=16$ and $BC=10$. Extend $\overline{CD}$ through $D$ to $E$ so that $DE=4$. If $\overline{BE}$ intersects $\overline{AD}$ at $F$, then $FD$ is closest to $\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }5$ [asy] size(200); defaultpen(linewidth(0.8)); pair A=origin,B=(16,0),C=(26,10*sqrt(3)),D=(10,10*sqrt(3)),E=(0,10*sqrt(3)); draw(A--B--C--E--B--A--D); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("$E$",E,N); label("$F$",extension(A,D,B,E),W); label("$4$",(D+E)/2,N); label("$16$",(8,0),S); label("$10$",(B+C)/2,SE); [/asy]

In the parallelogram $ABCD$, a line through $C$ intersects the diagonal $BD$ at $E$ and $AB$ at $F$. If $F$ is the midpoint of $AB$ and the area of $\vartriangle BEC$ is $100$, find the area of the quadrilateral $AFED$.

Prove that, for all convex pentagons $P_1 P_2 P_3 P_4 P_5$ with area 1, there are indices $i$ and $j$ (assume $P_7 = P_2$ and $P_6 = P_1$) such that: \[ \text{Area of} \ \triangle P_i P_{i+1} P_{i+2} \le \frac{5 - \sqrt 5}{10} \le \text{Area of} \ \triangle P_j P_{j+1} P_{j+2}\]

Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ are perpendicular. Let $O$ be the circumcenter of $ABCD$, $K$ the intersection of the diagonals, $ L\neq O $ the intersection of the circles circumscribed to $OAC$ and $OBD$, and $G$ the intersection of the diagonals of the quadrilateral whose vertices are the midpoints of the sides of $ABCD$. Prove that $O, K, L$ and $G$ are collinear

Let $ABCD$ be a parallelogram of center $O$. Prove that for any point $M\in (AB)$, there exist unique points $N\in (OC)$ and $P\in (OD)$ such that $O$ is the center of mass of $\triangle MNP$.

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]

The area of parallelogram $ABCD$ is $51\sqrt{55}$ and $\angle{DAC}$ is a right angle. If the side lengths of the parallelogram are integers, what is the perimeter of the parallelogram?

Let $\Gamma_1$ be a circle and $P$ a point outside of $\Gamma_1$. The tangents from $P$ to $\Gamma_1$ touch the circle at $A$ and $B$. Let $M$ be the midpoint of $PA$ and $\Gamma_2$ the circle through $P$, $A$ and $B$. Line $BM$ cuts $\Gamma_2$ at $C$, line $CA$ cuts $\Gamma_1$ at $D$, segment $DB$ cuts $\Gamma_2$ at $E$ and line $PE$ cuts $\Gamma_1$ at $F$, with $E$ in segment $PF$. Prove lines $AF$, $BP$, and $CE$ are concurrent.

Let $ ABC$, $ l$ and $ P$ be arbitrary triangle, line and point. $ A',B',C'$ are reflections of $ A,B,C$ in point $ P$. $ A''$ is a point on $ B'C'$ such that $ AA''\parallel l$. $ B'',C''$ are defined similarly. Prove that $ A'',B'',C''$ are collinear.

Equilateral triangles $ABE$ and $BCF$ are erected externally onthe sidess $AB$ and $BC$ of a parallelogram $ABCD$. Prove that $\vartriangle DEF$ is equilateral.

A quadrilateral $ABCD$ without parallel sides is circumscribed around a circle with centre $O$. Prove that $O$ is a point of intersection of middle lines of quadrilateral $ABCD$ (i.e. barycentre of points $A,\,B,\,C,\,D$) iff $OA\cdot OC=OB\cdot OD$.

Let $ABCD$ be a parallelogram with $\angle ABC = 105^\circ$. Inside the parallelogram, there is a point $E$ such that triangle $BEC$ is equilateral and $\angle CED = 135^\circ$. Let $K$ be the midpoint of side $AB$. Determine the measure of angle $\angle BKC$.

The parallelogram $ABCD$ has $AB=a,AD=1,$ $\angle BAD=A$, and the triangle $ABD$ has all angles acute. Prove that circles radius $1$ and center $A,B,C,D$ cover the parallelogram if and only \[a\le\cos A+\sqrt3\sin A.\]

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 $BL$.

Let $ABCD$ be a parallelogram such that $AB = 35$ and $BC = 28$. Suppose that $BD \perp BC$. Let $\ell_1$ be the reflection of $AC$ across the angle bisector of $\angle BAD$, and let $\ell_2$ be the line through $B$ perpendicular to $CD$. $\ell_1$ and $\ell_2$ intersect at a point $P$. If $PD$ can be expressed in simplest form as $\frac{m}{n}$, find $m + n$.

Circles $\omega_1$ and $\omega_2$ with centers at points $O_1$ and $O_2$ intersect at points $A$ and $B$. Let point $C$ be such that $AO_2CO_1$ is a parallelogram. An arbitrary line is drawn through point $A$, which intersects the circles $\omega_1$ and $\omega_2$ at points $X$ and $Y$, respectively. Prove that $CX = CY$. [i]Proposed by Oleksii Masalitin[/i]

A point $B$ lies on a chord $AC$ of circle $\omega.$ Segments $AB$ and $BC$ are diameters of circles $\omega_1$ and $\omega_2$ centered at $O_1$ and $O_2$ respectively. These circles intersect $\omega$ for the second time in points $D$ and $E$ respectively. The rays $O_1D$ and $O_2E$ meet in a point $F,$ and the rays $AD$ and $CE$ do in a point $G.$ Prove that the line $FG$ passes through the midpoint of the segment $AC.$

A $k\times \ell$ 'parallelogram' is drawn on a paper with hexagonal cells (it consists of $k$ horizontal rows of $\ell$ cells each). In this parallelogram a set of non-intersecting sides of hexagons is chosen; it divides all the vertices into pairs. Juniors) How many vertical sides can there be in this set? Seniors) How many ways are there to do that? [asy] size(120); defaultpen(linewidth(0.8)); path hex = dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--cycle; for(int i=0;i<=3;i=i+1) { for(int j=0;j<=2;j=j+1) { real shiftx=j*sqrt(3)/2+i*sqrt(3),shifty=j*3/2; draw(shift(shiftx,shifty)*hex); } } [/asy] [i](T. Doslic)[/i]

Let $ABCD$ be a parallelogram of center $O$. Points $M$ and $N$ are the midpoints of $BO$ and $CD$, respectively. Prove that if the triangles $ABC$ and $AMN$ are similar, then $ABCD$ is a square.

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.

Three $ \Delta $'s and a $ \diamondsuit $ will balance nine $ \bullet $'s. One $ \Delta $ will balance a $ \diamondsuit $ and a $ \bullet $. [asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,2)--(-12,0)--(12,0)--(12,2)); draw(ellipse((-12,5),8,3)); draw(ellipse((12,5),8,3)); label("$\Delta \hspace{2 mm}\Delta \hspace{2 mm}\Delta \hspace{2 mm}\diamondsuit $",(-12,6.5),S); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm} \bullet $",(12,5.2),N); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet $",(12,5.2),S); fill((44,0)--(40,-2)--(48,-2)--cycle,black); draw((34,2)--(34,0)--(54,0)--(54,2)); draw(ellipse((34,5),6,3)); draw(ellipse((54,5),6,3)); label("$\Delta $",(34,6.5),S); label("$\bullet \hspace{2 mm}\diamondsuit $",(54,6.5),S);[/asy] How many $ \bullet $'s will balance the two $ \diamondsuit $'s in this balance? [asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,4)--(-12,2)--(12,-2)--(12,0)); draw(ellipse((-12,7),6.5,3)); draw(ellipse((12,3),6.5,3)); label("$?$",(-12,8.5),S); label("$\diamondsuit \hspace{2 mm}\diamondsuit $",(12,4.5),S);[/asy] $ \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5 $