Hellenic Mathematical Olympiad

We will prove that the circle $c_3$ passes from the center $O$ of $c(O,R)$. From the inscribed quadrilateral $O\Delta \Gamma E$ (in the circle $c_3$) we get: ${\hat{O}}_1=\hat{\Gamma }$. From the inscribed quadrilateral $O\Delta BZ$ (in the circle $c_1$) we get: ${\hat{O}}_2=\hat{B}$. Summing up the above equations we find: ${\hat{O}}_1+{\hat{O}}_2=\hat{B}+\hat{\Gamma }\Rightarrow EOZ=\hat{B}+\hat{\Gamma }=180^{\circ }-\hat{A}$, and therefore the quadrilateral $AEOZ$ is cyclic.

Now we are going to prove that the circles $c_1,c_2,c_3$ are equal. From the cyclic quadrilateral $O\Delta BZ$ we have: ${\hat{\Delta }}_2={\hat{Z}}_2$. Figure 1 From the cyclic quadrilateral $O\Delta \Gamma E$ we have: ${\hat{\Delta }}_2={\hat{E}}_1$ and hence: $${\hat{\Delta }}_2={\hat{Z}}_2={\hat{E}}_1.$$ These three angles go in the equal chords $OB$, $O\Gamma$ and $OA$ of the circles $c_1,c_2$ and $c_3$, respectively. Therefore these three circles are equal.

Now in the equal circles $c_1$ and $c_2$, the angles ${\hat{Z}}_1$ and ${\hat{\Delta }}_1$ go in the equal chords $OK$ and $OM$ ($OK=OM=R$), and so: ${\hat{Z}}_1={\hat{\Delta }}_1$. From the last equality we conclude that the points $K$, $\Delta$, $M$ are collinear. Similarly we prove that the points $M$, $E$, $N$ and $N$, $Z$, $K$ are collinear.

From the equalities of angles $B\hat{\Delta }K=\Gamma \hat{\Delta }M$ and $\Gamma \hat{E}M=A\hat{E}N$ we get the equality of segments $AN=BK=\Gamma M$ (chords of the circle $c(O,R)$).

The triangles $AB\Gamma$ and $KMN$ have common circumcenter $O$ and the triangle $KMN$ is the image of $AB\Gamma$ in the rotation $R(O,\omega )$, where $$A\hat{O}N=B\hat{O}K=\Gamma \hat{O}M=\hat{\omega }.\text{ Hence the triangles are equal.}$$