2006 Cyprus MO/Lyceum/Problem 19

Revision as of 07:22, 16 October 2007 by 1=2 (talk | contribs) (Solution)

Problem

2006 CyMO-19.PNG

In the figure $AB \Gamma$ is isosceles triangle with$AB=A\Gamma=\sqrt2$ and $\ang A=45^\circ$ (Error compiling LaTeX. Unknown error_msg). If $B\Delta$ is altitude of the triangle and the sector $B\Lambda \Delta K B$ 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

$A  \Delta B$ 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 $\frac{\sqrt{2}}{2} - \frac{3}{16}\pi = \frac{8\sqrt{2} - 3\pi}{16} \Longrightarrow \mathrm{(C)}$.

See also

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