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

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In the figure <math>ABC</math> is isosceles triangle with<math> AB=AC=\sqrt2</math> and <math>\ang A=45^\circ</math>. If <math>BD</math> is altitude of the triangle and the sector <math>BLDKB</math> belongs to the circle <math>(B,BD)</math>, the area of the shaded region is
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In the figure <math>AB\Delta</math> is isosceles triangle with<math> AB=A\Delta=\sqrt2</math> and <math>\ang A=45^\circ</math>. If <math>BD</math> is altitude of the triangle and the sector <math>BLDKB</math> belongs to the circle <math>(B,BD)</math>, the area of the shaded region is
  
 
A. <math>\frac{4\sqrt3-\pi}{6}</math>
 
A. <math>\frac{4\sqrt3-\pi}{6}</math>

Revision as of 08:16, 16 October 2007

Problem

2006 CyMO-19.PNG

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

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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