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

Latest revision as of 22:44, 22 December 2016

Problem

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 $\frac{\frac{135}{2}}{360}\pi = \frac{3}{16}\pi$ .

The area of the entire triangle is $\frac{1}{2}bh = \frac{\sqrt{2}}{2}$ . Thus, the answer is $\frac{\sqrt{2}}{2} - \frac{3}{16}\pi = \frac{8\sqrt{2} - 3\pi}{16} \Longrightarrow \mathrm{(C)}$ .

@@ Line 1: / Line 1: @@
 ==Problem==
-<div style="float:right">
+[[Image:2006 CyMO-19.PNG|250px|right]]
-[[Image:2006 CyMO-19.PNG|250px]]
-</div>
-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
+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
-A. <math>\frac{4\sqrt3-\pi}{6}</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>
-B. <math>4\left(\sqrt2-\frac{\pi}{3}\right)</math>
+==Solution==
+<math>A  \Delta B</math> is a [[right triangle]] with an angle of <math>45^{\circ}</math>, so it is a <math>45-45-90</math> triangle with <math>B\Delta = \frac{AB}{\sqrt{2}} = 1</math>.
-C. <math>\frac{8\sqrt2-3\pi}{16}</math>
+The area of the entire circle is <math>(1)^2\pi = \pi</math>. The central angle of the sector is <math>\frac{180-45}{2} = \frac{135}{2}</math>, so the area is <math>\frac{\frac{135}{2}}{360}\pi = \frac{3}{16}\pi</math>.
-D. <math>\frac{\pi}{8}</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>.
-E. None of these
-==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>.
 ==See also==
 {{CYMO box|year=2006|l=Lyceum|num-b=18|num-a=20}}
 [[Category:Introductory Geometry Problems]]

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

Latest revision as of 22:44, 22 December 2016

Problem

Solution

See also