34th Hellenic Mathematical Olympiad

Figure 1 We draw the altitude $H\Lambda$ of the triangle $B\Gamma H$. Let it intersect $A\Gamma$ at $K$. We put $EK=x$, $KZ=y$ and $KH=z$. Then $H\Lambda =\alpha +z$ and $$E_{B\Gamma H}=\frac{1}{2}\alpha (\alpha +z)(1)$$ The triangles $\Gamma \Delta E$ and $EHK$ are similar. Hence $$\frac{KH}{\Gamma \Delta }=\frac{KE}{\Delta E}\iff \frac{z}{\alpha }=\frac{x}{\frac{\alpha }{3}}\iff z=3x(2)$$ Moreover, the triangles $ABZ$ and $ZKH$ are similar and hence $$\frac{KH}{AB}=\frac{KZ}{AZ}\iff \frac{z}{\alpha }=\frac{y}{\frac{\alpha }{4}}\iff z=4y(3)$$ Since $$x+y=A\Gamma -AZ-\Delta E=\alpha -\frac{\alpha }{4}-\frac{\alpha }{3}=\frac{5\alpha }{12}(4)$$ from (2), (3), and (4) we have $x+y=\frac{5\alpha }{12}\iff \frac{z}{3}+\frac{z}{4}=\frac{5\alpha }{12}\iff z=\frac{5\alpha }{7}$, and $$E=\frac{1}{2}\alpha \left(\alpha +\frac{5\alpha }{7}\right)=\frac{12{\alpha }^2}{14}=\frac{6{\alpha }^2}{7}.$$