Given a trapezoid, divide it by a line into two quadrilaterals in such a way that both of them are cyclic with the same circumradius. Discuss conditions of solvability.

$A, B, C$, and $D$ are the four different points on the circle $O$ in the order. Let the centre of the scribed circle of triangle $ABC$, which is tangent to $BC$, be $O_1$. Let the centre of the scribed circle of triangle $ACD$, which is tangent to $CD$, be $O_2$. (1) Show that the circumcentre of triangle $ABO_1$ is on the circle $O$. (2) Show that the circumcircle of triangle $CO_1O_2$ always pass through a fixed point on the circle $O$, when $C$ is moving along arc $BD$.

Construct the $ \triangle ABC $, with $ AC <BC $, if the circumcircle is known, and the points $ D, E, F $ in it, where they intersect, respectively, the altitude, the median and the angle bisector that they start from the vertex $ C $.

The triangle $ABC$ has circumcentre $O$ and circumcircle $\Gamma$. Let $AI$ be a diameter of $\Gamma$. The ray $AI$ extends to intersect the circumcircle $\omega$ of $\triangle BOC$ for the second time at a point $P$. Let $AD$ and $IQ$ be perpendicular to $BC$, with $D$ and $Q$ on $BC$. Let $M$ be the midpoint of $BC$. (a) Prove that $|AD| \cdot |QI| = |CD| \cdot |CQ| = |BD| \cdot |BQ|$. (b) Prove that $IM$ is parallel to $PD$.

It is given a triangle $ABC$ whose circumcenter is $O$ and orthocenter $H$. If $AO=AH$ find the angle $\hat{BAC}$ of that triangle.

Points $M$, $N$, $K$ lie on the sides $BC$, $CA$, $AB$ of a triangle $ABC$, respectively, and are different from its vertices. The triangle $MNK$ is called[i] beautiful[/i] if $\angle BAC=\angle KMN$ and $\angle ABC=\angle KNM$. If in the triangle $ABC$ there are two beautiful triangles with a common vertex, prove that the triangle $ABC$ is right-angled. [i]Proposed by Nairi M. Sedrakyan, Armenia[/i]

In an acute angled triangle $ABC$, $\angle A = 30^{\circ}$, $H$ is the orthocenter, and $M$ is the midpoint of $BC$. On the line $HM$, take a point $T$ such that $HM = MT$. Show that $AT = 2 BC$.

Given an acute triangle $ABC$ and $(O)$ be its circumcircle. Let $G$ be the point on arc $BC$ that doesn't contain $O$ of the circumcircle $(I)$ of triangle $OBC$. The circumcircle of $ABG$ intersects $AC$ at $E$ and circumcircle of $ACG$ intersects $AB$ at $F$ ($E\ne A, F\ne A$). a) Let $K$ be the intersection of $BE$ and $CF$. Prove that $AK,BC,OG$ are concurrent. b) Let $D$ be a point on arc $BOC$ (arc $BC$ containing $O$) of $(I)$. $GB$ meets $CD$ at $M$ , $GC$ meets $BD$ at $N$. Assume that $MN$ intersects $(O)$ at $P$ nad $Q$. Prove that when $G$ moves on the arc $BC$ that doesn't contain $O$ of $(I)$, the circumcircle $(GPQ)$ always passes through two fixed points.

Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$. The points $D,E$ and $F$ on the sides $BC,CA$ and $AB$ respectively are such that $BD+BF=CA$ and $CD+CE=AB$. The circumcircles of the triangles $BFD$ and $CDE$ intersect at $P \neq D$. Prove that $OP=OI$.

Let $ ABC $ be a triangle in which $ O,I, $ are the circumcenter, respectively, incenter. The mediators of $ IA,IB,IC, $ form a triangle $ A_1B_1C_1. $ Show that $ \overrightarrow{OI}=\overrightarrow{OA_1} +\overrightarrow{OA_2} +\overrightarrow{OA_3} . $

Let $ABCD$ be a isosceles trapezoid with $AB \| CD $ , $AD=BC$, $ AC \cap BD = $ { $O$ }. $ M $ is the midpoint of the side $AD$ . The circumcircle of triangle $ BCM $ intersects again the side $AD$ in $K$. Prove that $OK \| AB $ .

Let $ P$ be a point in triangle $ ABC$, and define $ \alpha,\beta,\gamma$ as follows: \[ \alpha\equal{}\angle BPC\minus{}\angle BAC, \quad \beta\equal{}\angle CPA\minus{}\angle \angle CBA, \quad \gamma\equal{}\angle APB\minus{}\angle ACB.\] Prove that \[ PA\dfrac{\sin \angle BAC}{\sin \alpha}\equal{}PB\dfrac{\sin \angle CBA}{\sin \beta}\equal{}PC\dfrac{\sin \angle ACB}{\sin \gamma}.\]

Let $\triangle{ABC}$ be a triangle, $\omega$ its circumcircle and $O$ the center of $\omega$. Let $P$ be a point on the segment $BC$. We denote by $Q$ the second intersection point of the circumcircles of triangles $\triangle{AOB}$ and $\triangle{APC}$. Prove that the line $PQ$ and the tangent to $\omega$ at point $A$ intersect on the circumcircle of triangle $\triangle AOB$.

In the triangle $ABC$, $\angle C$ is obtuse and $D$ is a fixed point on the side $BC$, different from $B$ and $C$. For any point $M$ on the side $BC$, different from $D$, the ray $AM$ intersects the circumcircle $S$ of $ABC$ at $N$. The circle through $M, D$ and $N$ meets $S$ again at $P$, different from $N$. Find the location of the point $M$ which minimises $MP$.

As shown in the figure, quadrilateral $ ABCD$ is inscribed in a circle with $ AC$ as its diameter, $ BD \perp AC,$ and $ E$ the intersection of $ AC$ and $ BD.$ Extend line segment $ DA$ and $ BA$ through $ A$ to $ F$ and $ G$ respectively, such that $ DG \parallel{} BF.$ Extend $ GF$ to $ H$ such that $ CH \perp GH.$ Prove that points $ B, E, F$ and $ H$ lie on one circle. [asy] defaultpen(linewidth(0.8)+fontsize(10));size(150); real a=4, b=6.5, c=9, d=a*c/b, g=14, f=sqrt(a^2+b^2)*sqrt(a^2+d^2)/g; pair E=origin, A=(0,a), B=(-b,0), C=(0,-c), D=(d,0), G=A+g*dir(B--A), F=A+f*dir(D--A), M=midpoint(G--C); path c1=circumcircle(A,B,C), c2=Circle(M, abs(M-G)); pair Hf=F+10*dir(G--F), H=intersectionpoint(F--Hf, c2); dot(A^^B^^C^^D^^E^^F^^G^^H); draw(c1^^c2^^G--D--C--A--G--F--D--B--A^^F--H--C--B--F); draw(H--B^^F--E^^G--C, linetype("2 2")); pair point= E; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$F$", F, dir(point--F)); label("$G$", G, dir(point--G)); label("$H$", H, dir(point--H)); label("$E$", E, NE);[/asy]

The point $O$ is the centre of the circumscribed circle of the acute-angled triangle $ABC$. The line $AO$ cuts the side $BC$ in point $N$, and the line $BO$ cuts the side $AC$ at point $M$. Prove that if $CM=CN$, then $AC=BC$.

Let $AD$ be the base of trapezoid $ABCD$. It is known that the circumcenter of triangle $ABC$ lies on $BD$. Prove that the circumcenter of triangle $ABD$ lies on $AC$. [i]Proposed by Ye.Bakayev[/i]

In-circle of $ABC$ with center $I$ touch $AB$ and $AC$ at $P$ and $Q$ respectively. $BI$ and $CI$ intersect $PQ$ at $K$ and $L$ respectively. Prove, that circumcircle of $ILK$ touch incircle of $ABC$ iff $|AB|+|AC|=3|BC|$.

In acute triangle $ABC$, $AB > AC$. $D$ and $E$ are the midpoints of $AB$, $AC$ respectively. The circumcircle of $ADE$ intersects the circumcircle of $BCE$ again at $P$. The circumcircle of $ADE$ intersects the circumcircle $BCD$ again at $Q$. Prove that $AP = AQ$.

A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$.

Let $P_0$, $P_1$, $P_2$ be three points on the circumference of a circle with radius $1$, where $P_1P_2 = t < 2$. For each $i \ge 3$, define $P_i$ to be the centre of the circumcircle of $\triangle P_{i-1} P_{i-2} P_{i-3}$. (1) Prove that the points $P_1, P_5, P_9, P_{13},\cdots$ are collinear. (2) Let $x$ be the distance from $P_1$ to $P_{1001}$, and let $y$ be the distance from $P_{1001}$ to $P_{2001}$. Determine all values of $t$ for which $\sqrt[500]{ \frac xy}$ is an integer.

The vertices of a right-angled triangle are on a circle of radius $R$ and the sides of the triangle are tangent to another circle of radius $r$ (this is the circle that is inside triangle). If the lengths of the sides about the right angles are 16 and 30, determine the value of $R+r$.

Let $ P$ be an interior point of an acute angled triangle $ ABC$. The lines $ AP,BP,CP$ meet $ BC,CA,AB$ at points $ D,E,F$ respectively. Given that triangle $ \triangle DEF$ and $ \triangle ABC$ are similar, prove that $ P$ is the centroid of $ \triangle ABC$.

Let $ABCD$ be a cyclic quadrilateral and let $H_{A}, H_{B}, H_{C}, H_{D}$ be the orthocenters of the triangles $BCD$, $CDA$, $DAB$ and $ABC$ respectively. Show that the quadrilaterals $ABCD$ and $H_{A}H_{B}H_{C}H_{D}$ are congruent.

Let $ABC$ be an acute scalene triangle. Let $D$ and $E$ be points on the sides $AB$ and $AC$, respectively, such that $BD=CE$. Denote by $O_1$ and $O_2$ the circumcentres of the triangles $ABE$ and $ACD$, respectively. Prove that the circumcircles of the triangles $ABC, ADE$, and $AO_1O_2$ have a common point different from $A$. [i]Proposed by Patrik Bak, Slovakia[/i]