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 $\frac{\frac{135}{2}}{360} = \frac{3}{16}\pi$ .

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

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 18	Followed by Problem 20
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2006 Cyprus MO/Lyceum/Problem 19

Problem

Solution

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