2006 Cyprus MO/Lyceum/Problem 19

Problem

In the figure $AB\Delta$ is isosceles triangle with $AB=A\Delta=\sqrt2$ and $\ang A=45^\circ$ (Error compiling LaTeX. Unknown error_msg). If $BD$ is altitude of the triangle and the sector $BLDKB$ belongs to the circle $(B,BD)$ , the area of the shaded region is

A. $\frac{4\sqrt3-\pi}{6}$

B. $4\left(\sqrt2-\frac{\pi}{3}\right)$

C. $\frac{8\sqrt2-3\pi}{16}$

D. $\frac{\pi}{8}$

E. None of these

Solution

$ADB$ is a right triangle with an angle of $45^{\circ}$ , so it is a $45-45-90 \triangle$ and $BD = \frac{AB}{\sqrt{2}} = 1$ . The area of the entire circle is $(1)^2\pi = \pi$ . To find the area of the sector, we find the central angle is $\frac{180-45}{2} = \frac{135}{2}$ , and 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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