37th Hellenic Mathematical Olympiad 2020

If $A\Gamma E=\omega$, from the right angled triangle $\Gamma \Theta B$ we have: $$\Gamma \hat{B}\Theta =90^{\circ }-B\hat{\Gamma }\Theta =90^{\circ }-\omega ,(1)$$ since $B\Gamma \Theta =A\Gamma E=\omega$, and moreover $B\hat{E}K=90^{\circ }$, and so $$\Gamma \hat{B}\Theta +B\hat{H}K=90^{\circ }-\omega +90^{\circ }=180^{\circ }-\omega <180^{\circ },$$ Which gives that the lines $B\Theta$ and $\Delta H$ intersect at a point, say $K$. Therefore, it is enough to prove that the points $Z$, $E$ and $K$ are collinear. The quadrilateral $E\Delta \Gamma Z$ is isosceles trapezium, because $E\Delta \parallel Z\Gamma$ and $ZE=AE=\Gamma \Delta$. Indeed, the triangles $AEZ$ and $A\Gamma E$ have the angle $A$ in common and $A\hat{E}Z=A\Gamma E=\omega$, therefore they have equal angles. Therefore the triangle $AZE$ is isosceles with $A\hat{E}Z=E\hat{A}Z$. It follows that: $AE=ZE$. figure 2 From the isosceles trapezium $E\Delta \Gamma Z$ we have that: $$E\hat{Z}\Gamma =Z\hat{\Gamma }\Delta =180^{\circ }-\Gamma \hat{A}E=180^{\circ }-\left(\frac{180^{\circ }-\omega }{2}\right)=90^{\circ }+\frac{\omega }{2}.(2)$$ From the orthogonal triangle $\Delta HZ$ we have $$Z\hat{\Delta }H=90^{\circ }-\Delta \hat{Z}H=90^{\circ }-\omega .(3)$$ From (1) and (3) it follows that the quadrilateral $B\Delta ZK$ is cyclic. If $\Lambda$ is the midpoint of the segment $\Gamma B$ we observe that the quadrilateral $E\Delta \Lambda \Gamma$ has $E\Delta \parallel \Gamma \Lambda$, $\Gamma E=\Delta E=A\Gamma =\Gamma \Lambda$, since $A\Gamma =\frac{1}{3}AB$ and $\Lambda$ is the midpoint of $\Gamma B$. Hence $E\Delta \Lambda \Gamma$ is rhombus. Hence, we have $\Lambda \Gamma \Delta =\omega$ and moreover $\Delta \Lambda =\Lambda \Gamma =\Lambda B=\frac{\Gamma B}{2}$. It means that the triangle $\Gamma \Delta B$ is orthogonal at $\Delta$ and from the isosceles triangle $\Lambda \Delta B$ we have : $\Delta \hat{B}\Lambda =\frac{\omega }{2}$. From the cyclic quadrilateral $B\Delta ZK$ we have: $$Z\hat{K}\Delta =\Delta \hat{B}Z=\Delta \hat{B}\Lambda =\frac{\omega }{2},(4)$$ οπότε από το ορθογώνιο τρίγωνο $HZK$ και τη σχέση (4) έπεται ότι: $$\Gamma \hat{Z}K=H\hat{Z}K=90^{\circ }-\frac{\omega }{2}(5)$$ From (2) and (5) it follows that: $$E\hat{Z}\Gamma +\Gamma \hat{Z}K=90^{\circ }+\frac{\omega }{2}+90^{\circ }-\frac{\omega }{2}=180^{\circ },$$ and hence the points $A$, $E$ and $K$ are collinear.