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2006 Cyprus MO/Lyceum/Problem 19

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Problem

In the figure, $AB\Gamma$ is an isosceles triangle with $AB=A\Gamma=\sqrt2$ and $\ang A=45^\circ$ . 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

$\mathrm{(A)}\ \frac{4\sqrt3-\pi}{6}\qquad\mathrm{(B)}\ 4\left(\sqrt2-\frac{\pi}{3}\right)\qquad\mathrm{(C)}\ \frac{8\sqrt2-3\...$

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

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_Cyprus_MO/Lyceum/Problem_19"

Category: Introductory Geometry Problems

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