SELECTION EXAMINATION 2019

We will prove that the point $T$ belongs to the circle $c_1(O,\Gamma ,M)$ and the point $N$ belongs to the circle $c_2(O,A,\Delta )$, that is the quadrilaterals $NAO\Delta$ and $MO\Gamma T$ are cyclic. Then we conclude the following: the line $A\Delta$ is the radical axis of the circles $C$ and $c_2$, the line $\Gamma T$ is the radical axis of the circles $C$ and $c_1$ and the line $O\Sigma$ is the radical axis $C_1$ and $C_2$, and therefore the lines $A\Delta$, $\Gamma T$ and $O\Sigma$ are passing through the radical center of the three circles.

Since $B\Delta$ is diameter of the circle $c$, we have $B\hat{A}\Delta =90^{\circ }$ and so $N\hat{A}\Delta =180^{\circ }-B\hat{A}\Delta =90^{\circ }$. Also, $ON$ is the perpendicular bisector of the segment $B\Delta$, and so: $N\hat{O}\Delta =90^{\circ }$. From the last two angle equalities we conclude that the quadrilateral NAOΔ is cyclic.

The triangle $NB\Delta$ is isosceles with $NB=N\Delta$ (because $N$ belongs to the perpendicular bisector $B\Delta$), and so $N\hat{B}\Delta =N\hat{A}B$. Therefore taking in mind that $OA=OB=O\Delta =OT=R$ we conclude that the isosceles triangles $OAB$ and $O\Delta T$ are equal. Therefore $$(\alpha )OA=OT\text{ and }(\beta )AB=\Delta T\Rightarrow NA=NT.$$ Hence the triangles $OAN$ and $OTN$ are equal, because they have their sides equal one to one and so $$N\hat{O}A=N\hat{O}T=\frac{1}{2}\cdot A\hat{O}T,(1)$$ Moreover we have that $$M\hat{\Gamma }T=A\hat{\Gamma }T=\frac{1}{2}\cdot A\hat{O}T(2)$$ From relations (1) and (2) it follows that $M\hat{\Gamma }T=N\hat{O}T=M\hat{O}T$, from which we conclude that the quadrilateral MOΓT is cyclic.