Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 19"
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==Problem== | ==Problem== | ||
− | {{ | + | <div style="float:right"> |
+ | [[Image:2006 CyMO-19.PNG|250px]] | ||
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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 | ||
+ | |||
+ | A. <math>\frac{4\sqrt3-\pi}{6}</math> | ||
+ | |||
+ | B. <math>4\left(\sqrt2-\frac{\pi}{3}\right)</math> | ||
+ | |||
+ | C. <math>\frac{8\sqrt2-3\pi}{16}</math> | ||
+ | |||
+ | D. <math>\frac{\pi}{8}</math> | ||
+ | |||
+ | E. None of these | ||
==Solution== | ==Solution== | ||
− | {{ | + | <math>ADB</math> is a [[right triangle]] with an angle of <math>45^{\circ}</math>, so it is a <math>45-45-90 \triangle</math> and <math>BD = \frac{AB}{\sqrt{2}} = 1</math>. The area of the entire circle is <math>(1)^2\pi = \pi</math>. To find the area of the sector, we find the central angle is <math>\frac{180-45}{2} = \frac{135}{2}</math>, and the area is <math>\frac{\frac{135}{2}}{360} = \frac{3}{16}\pi</math>. The area of the entire triangle is <math>\frac{1}{2}bh = \frac{\sqrt{2}}{2}</math>. Thus the answer is <math>\frac{\sqrt{2}}{2} - \frac{3}{16}\pi = \frac{8\sqrt{2} - 3\pi}{16} \Longrightarrow \mathrm{(C)}</math>. |
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==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=18|num-a=20}} | {{CYMO box|year=2006|l=Lyceum|num-b=18|num-a=20}} | ||
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+ | [[Category:Introductory Geometry Problems]] |
Revision as of 18:07, 15 October 2007
Problem
In the figure is isosceles triangle with and $\ang A=45^\circ$ (Error compiling LaTeX. Unknown error_msg). If is altitude of the triangle and the sector belongs to the circle , the area of the shaded region is
A.
B.
C.
D.
E. None of these
Solution
is a right triangle with an angle of , so it is a and . The area of the entire circle is . To find the area of the sector, we find the central angle is , and the area is . The area of the entire triangle is . Thus the answer is .
See also
2006 Cyprus MO, Lyceum (Problems) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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