Hellenic Mathematical Olympiad

Let $Z$ be the intersection of $c_2$ with $AB$ and $M$ the intersection of $KZ$ with $BE$. The angle $A\hat{B}\Delta$ (and so the angle $Z\hat{B}K$) is right since it sees the diameter $A\Delta$ of the circle $c(O,R)$. Since $Z\hat{B}K=90^{\circ }$, $ZK$ is a diameter of $c_1$ and thus $ZK$ is the midperpendicular of the common chord $BE$ of the circles $c_1$ and $c_2$. The line $AB$ is tangent to the circle $c_1$, so ${\hat{B}}_1=\hat{\Gamma }$ and from the right-angled triangle $BMZ$ we have: $${\hat{Z}}_1=90^{\circ }-{\hat{B}}_1=90^{\circ }-\hat{\Gamma }(1)$$ Figure 3 The angles $\hat{\Delta },\hat{\Gamma }$ see the same arc of $c$, so $\hat{\Delta }=\hat{\Gamma }$. From the right-angled triangle $AB\Delta$, we have: $${\hat{A}}_1=90^{\circ }-\hat{\Delta }=90^{\circ }-\hat{\Gamma }(2)$$

From (1) and (2) we obtain that $\widehat{A_1}=\widehat{Z_1}$, so $\text{οποˊτε}\Delta A\Delta$ is parallel to $KZ$ and as a result $A\Delta KZ$ is a trapezium. Moreover, since $O$ is the midpoint of $A\Delta$, we conclude that $OB$ passes through the midpoint of $KZ$ which is the center of $c_2$. Therefore the centers of $c,c_2$ and the point $B$ are collinear.

We will prove that the circle $c_2$ is tangent to $c(O,R)$ at the point $B$. To this end, we will prove that these circles have common tangent at the point $B$. Let $BP$ be the tangent of $c(O,R)$ at the point $B$ and let $\widehat{KBP}=\omega$. In order to prove that $BP$ is tangent to $c_2$ at $B$, it suffices to prove that $$\widehat{BEK}=\omega .(1)$$ Since $BP$ is tangent to $c(O,R)$, we have $\widehat{\Delta \Gamma B}=\omega$. Moreover, $A\Delta$ is a diameter, so $\widehat{A\Gamma \Delta }=90^{\circ }$, and as a result $\widehat{E\Gamma B}=90^{\circ }-\omega$. However, $$\widehat{BKE}=2\widehat{B\Gamma E}=2(90^{\circ }-\omega )=180^{\circ }-2\omega ,(2)$$ and from the isosceles triangle $BKE$ we have $\widehat{KBE}=\widehat{KEB}=\omega$, so (1) holds, which is the desired result.