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| − | == Problem ==
| + | #REDIRECT [[2003 AMC 10B Problems/Problem 19]] |
| − | Three semicircles of radius <math>1</math> are constructed on diameter <math>AB</math> of a semicircle of
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| − | radius <math>2</math>. The centers of the small semicircles divide <math>AB</math> into four line segments
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| − | of equal length, as shown. What is the area of the shaded region that lies within
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| − | the large semicircle but outside the smaller semicircles?
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| − | <math>\textbf{(A) } \pi - \sqrt{3} \qquad\textbf{(B) } \pi - \sqrt{2} \qquad\textbf{(C) } \frac{\pi + \sqrt{2}}{2} \qquad\textbf{(D) } \frac{\pi +\sqrt{3}}{2} \qquad\textbf{(E) } \frac{7}{6}\pi - \frac{\sqrt{3}}{2}</math>
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| − | == Solution ==
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| − | Each small semicircle shares a radius with an adjacent circle. Therefore, the radii drawn to the points of intersection will create equilateral triangles. Bisect the <math>120^\circ</math> angles on the sides and complete the incomplete triangles with lines so that the unshaded region is broken into two types of pieces: a circular segment from a <math>60^\circ</math> sector of radius <math>1</math>, and an equilateral triangle with side length <math>1</math>.
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| − | There are <math>7</math> equilateral triangles and <math>5</math> circular segments. One equilateral triangle has area <math>\frac{1}{2}(1)\left( \frac{\sqrt{3}}{2} \right)=\frac{\sqrt{3}}{4}</math>, and one segment as area <math>\frac{60}{360}\pi (1^2)-\frac{\sqrt{3}}{4}=\frac{\pi}{6}-\frac{\sqrt{3}}{4}</math>.
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| − | So the shaded region is the area of the large semicircle minus the area of the unshaded parts, which is <math>\frac{1}{2}\pi (2^2) - (7)\frac{\sqrt{3}}{4} - 5\left( \frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) = 2\pi - \frac{7\sqrt{3}}{4} - \frac{5\pi}{6} - \frac{5\sqrt{3}}{4}= \boxed{\textbf{(E) } \frac{7}{6}\pi - \frac{\sqrt{3}}{2}}</math>.
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