33rd Hellenic Mathematical Olympiad

a. From the right angled triangle $B\Gamma \Delta$ we have: ${\hat{\Gamma }}_1=90^{\circ }-\hat{B}$. The line joining the centers of the circles $c_1$ and $c_2$ is the perpendicular bisector of their common chord $\Delta E$. If $T$ is the point of intersection of the lines $B\Gamma$ and $\Delta E$, then: $${\hat{\Gamma }}_1+{\hat{\Delta }}_1+{\hat{\Delta }}_2=90^{\circ },\text{ that is }{\hat{\Delta }}_1+{\hat{\Delta }}_2=\hat{B}=\hat{\Gamma }.(1)$$ From the isosceles triangle $\Gamma \Delta Z$ we have: ${\hat{\Delta }}_1={\hat{Z}}_1$ and ${\hat{\Gamma }}_2=2{\hat{\Delta }}_1=90^{\circ }-\hat{A}$, and hence: $${\hat{\Delta }}_1=45^{\circ }-\frac{\hat{A}}{2}.(2)$$ From the isosceles triangle $AB\Gamma$ we have: $\hat{B}=90^{\circ }-\frac{\hat{A}}{2}$ \quad (3) From (1), (2), (3) we get: $${\hat{\Delta }}_1+{\hat{\Delta }}_2=\hat{B}=\hat{\Gamma }\iff {\hat{\Delta }}_2=\hat{B}-{\hat{\Delta }}_1\stackrel{(2),(3)}{\iff }{\hat{\Delta }}_2=90^{\circ }-\frac{\hat{A}}{2}-\left(45^{\circ }-\frac{\hat{A}}{2}\right)=45^{\circ }.$$

b. The angle ${\hat{\Delta }}_1$ is created from the chord $\Delta M$ and the tangent $\Delta \Gamma$ of the circle $c_1$, and hence $\Delta \hat{E}M={\hat{\Delta }}_1$. Also we have $\Delta \hat{E}K=\Delta \hat{Z}K={\hat{Z}}_1$. Since ${\hat{\Delta }}_1={\hat{Z}}_1$, we get $\Delta \hat{E}M=\Delta \hat{E}K$, and the points $E$, $M$, $K$ are collinear.

c. We will prove that $\Delta \hat{B}M=90^{\circ }-\hat{A}$. We have $\Delta \hat{B}M=2\Delta \hat{E}M=2{\hat{\Lambda }}_1={\hat{\Gamma }}_2=90^{\circ }-\hat{A}$, and hence $BM\perp A\Gamma$. Moreover $E\hat{K}Z=E\hat{\Delta }Z={\hat{\Lambda }}_2=45^{\circ }$, and so the orthogonal triangle $EZK$ isosceles. Then its median $E\Gamma$ is also an altitude, that is $E\Gamma \perp KZ$. Hence $BM\parallel E\Gamma$ (both are perpendicular to the line $AZ$).

Figure 2