Let $ABCD$ be a parallelogram. Half-circles $\omega_1,\omega_2,\omega_3$ and $\omega_4$ with diameters $AB,BC,CD$ and $DA$, respectively, are erected on the exterior of $ABCD$. Line $l_1$ is parallel to $BC$ and cuts $\omega_1$ at $X$, segment $AB$ at $P$, segment $CD$ at $R$ and $\omega_3$ at $Z$. Line $l_2$ is parallel to $AB$ and cuts $\omega_2$ at $Y$, segment $BC$ at $Q$, segment $DA$ at $S$ and $\omega_4$ at $W$. If $XP\cdot RZ=YQ\cdot SW$, prove that $PQRS$ is cyclic. [i]Proposed by José Alejandro Reyes González[/i]

A quadrilateral $ABCD$ is inscribed in a circle $\omega$. The tangent to $\omega$ at $A$ intersects the ray $CB$ at $K$, and the tangent to $\omega$ at $B$ intersects the ray $DA$ at $M$. Prove that if $AM=AD$ and $BK=BC$, then $ABCD$ is a trapezoid.

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 $AFED$.

On the side $BC$ of the triangle $ABC$ there are two points $D$ and $E$ such that $BD < BE$. Denote by $p_1$ and $p_2$ the perimeters of triangles $ABC$ and $ADE$ respectively. Prove that \[p_1 > p_2 + 2\cdot \min\{BD, EC\}.\]

This is the one with the 8 circles? I made each circle into the square in which the circle is inscribed, then calculated it with that. It got the right answer but I don't think that my method is truly valid...

Let $w$-circumcircle of triangle $ABC$ with an obtuse angle $C$ and $C '$symmetric point of point $C$ with respect to $AB$. $M$ midpoint of $AB$. $C'M$ intersects $w$ at $N$ ($C '$ between $M$ and $N$). Let $BC'$ second crossing point $w$ in $F$, and $AC'$ again crosses the $w$ at point $E$. $K$-midpoint $EF$. Prove that the lines $AB, CN$ and$ KC'$are concurrent.

For $n>1$ consider an $n\times n$ chessboard and place identical pieces at the centers of different squares. [list=i] [*] Show that no matter how $2n$ identical pieces are placed on the board, that one can always find $4$ pieces among them that are the vertices of a parallelogram. [*] Show that there is a way to place $(2n-1)$ identical chess pieces so that no $4$ of them are the vertices of a parallelogram. [/list]

Let $ABC$ be a scalene triangle with circumcircle $\omega$. The tangent to $\omega$ at $A$ meets $BC$ at $D$. The $A$-median of triangle $ABC$ intersects $BC$ and $\omega$ at $M$ and $N$, respectively. Suppose that $K$ is a point such that $ADMK$ is a parallelogram. Prove that $KA = KN$. [i]Proposed by Alexandru Lopotenco (Moldova)[/i]

Let an acute-angled triangle $ABC$ with $AB<AC<BC$, inscribed in circle $c(O,R)$. The angle bisector $AD$ meets $c(O,R)$ at $K$. The circle $c_1(O_1,R_1)$(which passes from $A,D$ and has its center $O_1$ on $OA$) meets $AB$ at $E$ and $AC$ at $Z$. If $M,N$ are the midpoints of $ZC$ and $BE$ respectively, prove that: [b]a)[/b]the lines $ZE,DM,KC$ are concurrent at one point $T$. [b]b)[/b]the lines $ZE,DN,KB$ are concurrent at one point $X$. [b]c)[/b]$OK$ is the perpendicular bisector of $TX$.

A circumcircle of triangle $ABC$ meets the sides $AD$ and $CD$ of a parallelogram $ABCD$ at points $K$ and $L$ respectively. Let $M$ be the midpoint of arc $KL$ not containing $B$. Prove that $DM \perp AC$. by E.Bakaev

Let $ABCD$ be a parallelogram, such that the point $M$ is the midpoint of the side $CD$ and lies on the bisector of the angle $\angle BAD$. Prove that $\angle AMB = 90^o$.

Given a non-isoceles triangle $ABC$ inscribes a circle $(O,R)$ (center $O$, radius $R$). Consider a varying line $l$ such that $l\perp OA$ and $l$ always intersects the rays $AB,AC$ and these intersectional points are called $M,N$. Suppose that the lines $BN$ and $CM$ intersect, and if the intersectional point is called $K$ then the lines $AK$ and $BC$ intersect. $1$, Assume that $P$ is the intersectional point of $AK$ and $BC$. Show that the circumcircle of the triangle $MNP$ is always through a fixed point. $2$, Assume that $H$ is the orthocentre of the triangle $AMN$. Denote $BC=a$, and $d$ is the distance between $A$ and the line $HK$. Prove that $d\leq\sqrt{4R^2-a^2}$ and the equality occurs iff the line $l$ is through the intersectional point of two lines $AO$ and $BC$.

In the parallelogram $ABCD$, $\angle BAD =60 ^ \circ$. Let $E $ be the intersection point of the diagonals. The circle circumscribed to the triangle $ACD$ intersects the line $AB$ at the point $K$ (different from $A$), the line $BD$ at the point $P$ (different from $D$), and to the line $BC$ in $L$ (different from $C$). The line $EP$ intersects the circumscribed circle of the triangle $CEL$ at the points $E$ and $M$. Show that the triangles $KLM$ and $CAP$ are congruent.

A parallelopiped has vertices $A_1, A_2, ... , A_8$ and center $O$. Show that: \[ 4 \sum_{i=1}^8 OA_i ^2 \leq \left(\sum_{i=1}^8 OA_i \right) ^2 \]

Let $ABC$ be a triangle, $O$ its circumcenter, $S$ its centroid, and $H$ its orthocenter. Denote by $A_1, B_1$, and $C_1$ the centers of the circles circumscribed about the triangles $CHB, CHA$, and $AHB$, respectively. Prove that the triangle $ABC$ is congruent to the triangle $A_1B_1C_1$ and that the nine-point circle of $\triangle ABC$ is also the nine-point circle of $\triangle A_1B_1C_1$.

Given acute triangle $ABC$ with $AB>AC$, let $M$ be the midpoint of $BC$. $P$ is a point in triangle $AMC$ such that $\angle MAB=\angle PAC$. Let $O,O_1,O_2$ be the circumcenters of $\triangle ABC,\triangle ABP,\triangle ACP$ respectively. Prove that line $AO$ passes through the midpoint of $O_1 O_2$.

Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$

Let $ABCD$ be a tetragonal. [b](a)[/b] If the plane $(P)$ cuts $ABCD,$ find the necessary and sufficient condition such that the area formed from the intersection of the plane $(P)$ and the tetragonal be a parallelogram. Prove that the problem has three solutions in this case. [b](b)[/b] Consider one of the solutions of [b](a)[/b]. Find the situation of the plane $(P)$ for which the parallelogram has maximum area. [b](c)[/b] Find a plane $(P)$ for which the parallelogram be a lozenge and then find the length side of his lozenge in terms of the length of the edges of $ABCD.$

Let $I$ be the center of the incircle of non-isosceles triangle $ABC,A_{1}=AI\cap BC$ and $B_{1}=BI\cap AC.$ Let $l_{a}$ be the line through $A_{1}$ which is parallel to $AC$ and $l_{b}$ be the line through $B_{1}$ parallel to $BC.$ Let $l_{a}\cap CI=A_{2}$ and $l_{b}\cap CI=B_{2}.$ Also $N=AA_{2}\cap BB_{2}$ and $M$ is the midpoint of $AB.$ If $CN\parallel IM$ find $\frac{CN}{IM}$.

The point P lies on the side $CD$ of the parallelogram $ABCD$ with $\angle DBA = \angle CBP$. Point $O$ is the center of the circle passing through the points $D$ and $P$ and tangent to the straight line $AD$ at point $D$. Prove that $AO = OC$.

Let $ABCD$ be a convex quadrilateral. Assume line $AB$ and $CD$ intersect at $E$, and $B$ lies between $A$ and $E$. Assume line $AD$ and $BC$ intersect at $F$, and $D$ lies between $A$ and $F$. Assume the circumcircles of $\triangle BEC$ and $\triangle CFD$ intersect at $C$ and $P$. Prove that $\angle BAP=\angle CAD$ if and only if $BD\parallel EF$.

Consider a point $G$ in the interior of a parallelogram $ABCD$. A circle $\Gamma$ through $A$ and $G$ intersects the sides $AB$ and $AD$ for the second time at the points $E$ and $F$ respectively. The line $FG$ extended intersects the side $BC$ at $H$ and the line $EG$ extended intersects the side $CD$ at $I$. The circumcircle of triangle $HGI$ intersects the circle $\Gamma$ for the second time at $M \ne G$. Prove that $M$ lies on the diagonal $AC$.

Let $ABC$ be an acute triangle with $AB \neq AC$. Let $D$ be the foot of the altitude from $A$ and $\omega$ the circumcircle of the triangle. Let $\omega_1$ be the circle tangent to $AD$, $BD$ and $\omega$. Let $\omega_2$ be the circle tangent to $AD$, $CD$ and $\omega$. Let $\ell$ be the interior common tangent to both $\omega_1$ and $\omega_2$, different from $AD$. Prove that $\ell$ passes through the midpoint of $BC$ if and only if $2BC = AB + AC$.

Given a parallelogram $ABCD$, in which the angle $\angle ABC$ is obtuse. Line $AD$ intersects the circle a second time $\omega$ circumscribed around triangle $ABC$, at the point $E$. Line $CD$ intersects second time circle $\omega$ at point $F$. Prove that the circumcenter of triangle $DEF$ lies on the circle $\omega$.

Inside the parallelogram $ABCD$, point $M$ is chosen, and inside the triangle $AMD$, point $N$ is chosen in such a way that $$\angle MNA + \angle MCB =\angle MND + \angle MBC = 180^o.$$ Prove that lines $MN$ and $AB$ are parallel.