# 2006 Cyprus MO/Lyceum/Problem 19

## Problem

In the figure, $AB\Gamma$ is an isosceles triangle with $AB=A\Gamma=\sqrt2$ and $\angle A=45^\circ$. If $B\Delta$ is an altitude of the triangle and the sector $B\Lambda \Delta KB$ belongs to the circle $(B,B\Delta )$, the area of the shaded region is $\mathrm{(A)}\ \frac{4\sqrt3-\pi}{6}\qquad\mathrm{(B)}\ 4\left(\sqrt2-\frac{\pi}{3}\right)\qquad\mathrm{(C)}\ \frac{8\sqrt2-3\pi}{16}\qquad\mathrm{(D)}\ \frac{\pi}{8}\qquad\mathrm{(E)}\ \text{None of these}$

## Solution $A \Delta B$ is a right triangle with an angle of $45^{\circ}$, so it is a $45-45-90$ triangle with $B\Delta = \frac{AB}{\sqrt{2}} = 1$.

The area of the entire circle is $(1)^2\pi = \pi$. The central angle of the sector is $\frac{180-45}{2} = \frac{135}{2}$, so the area is $\frac{\frac{135}{2}}{360}\pi = \frac{3}{16}\pi$.

The area of the entire triangle is $\frac{1}{2}bh = \frac{\sqrt{2}}{2}$. Thus, the answer is $\frac{\sqrt{2}}{2} - \frac{3}{16}\pi = \frac{8\sqrt{2} - 3\pi}{16} \Longrightarrow \mathrm{(C)}$.