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Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 19"

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[[Image:2006 CyMO-19.PNG|250px|right]]
 
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In the figure, <math>AB\Gamma</math> is an [[isosceles triangle]] with<math> AB=A\Gamma=\sqrt2</math> and <math>\ang A=45^\circ</math>. If <math>B\Delta</math> is an [[altitude]] of the [[triangle]] and the [[sector]] <math>B\Lambda \Delta KB</math> belongs to the circle <math>(B,B\Delta )</math>, the [[area]] of the shaded region is
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In the figure, <math>AB\Gamma</math> is an [[isosceles triangle]] with<math> AB=A\Gamma=\sqrt2</math> and <math>\angle A=45^\circ</math>. If <math>B\Delta</math> is an [[altitude]] of the [[triangle]] and the [[sector]] <math>B\Lambda \Delta KB</math> belongs to the circle <math>(B,B\Delta )</math>, the [[area]] of the shaded region is
  
 
<math>\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}</math>
 
<math>\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}</math>

Revision as of 10:11, 27 November 2016

Problem

2006 CyMO-19.PNG

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

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
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