2006 Cyprus MO/Lyceum/Problem 8

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 $\frac{1}{2}\mathrm{(B)}$ .

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 7	Followed by Problem 9
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2006 Cyprus MO/Lyceum/Problem 8

Problem

Solution

See also