# 2006 Cyprus MO/Lyceum/Problem 8

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## Problem

In the figure $AB\Gamma \Delta E$ is a regular 5-sided polygon and $Z$, $H$, $\Theta$, $I$, $K$ are the points of intersections of the extensions of the sides. If the area of the "star" $AHB\Theta \Gamma I\Delta KEZA$ is 1, then the area of the shaded quadrilateral $A\Gamma IZ$ is

$\mathrm{(A)}\ \frac{2}{3}\qquad\mathrm{(B)}\ \frac{1}{2}\qquad\mathrm{(C)}\ \frac{3}{7}\qquad\mathrm{(D)}\ \frac{3}{10}\qquad\mathrm{(E)}\ \text{None of these}$

## Solution

In the quadrilateral $A\Gamma IZ$, we have three isosceles triangles $A\Gamma\Delta$, $AZE$, and $\Gamma \Delta I$. Those are congruent to each other, as well as $HAB$, $B\Gamma\Theta$, and $EK\Delta$. Also, $AE\Delta$ is congruent to $AB\Gamma$. Thus we have two figures of equal area: $A\Gamma IZ$ and a combination of two figures: $HB\Theta\Gamma A$ and $EK\Delta$. Since the area of the whole star is 1, the area of $AZI\Gamma$ is $\frac{1}{2}\mathrm{(B)}$.