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Difference between revisions of "2003 AMC 12B Problems/Problem 16"

(Created page with "== Problem == Three semicircles of radius <math>1</math> are constructed on diameter <math>AB</math> of a semicircle of radius <math>2</math>. The centers of the small semicircle...")
 
(→‎Solution)
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== Solution ==
 
== 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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Each small semicircle shares a radius with an adjacent circle. Therefore, the radii to the points of intersection will create equilateral triangles. Draw these triangles, and then bisect the <math>120^\circ</math> angles on the sides and complete the incomplete triangles so that the unshaded region is broken into two types of pieces: circular segments from a <math>60^\circ</math> sector of a circle with radius <math>1</math>, and equilateral triangles with side length <math>1</math>.
  
 
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>.
 
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>.
+
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\left( \frac{\sqrt{3}}{4}\right) - 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>.

Revision as of 00:10, 21 January 2013

Problem

Three semicircles of radius $1$ are constructed on diameter $AB$ of a semicircle of radius $2$. The centers of the small semicircles divide $AB$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

$\textbf{(A) } \pi - \sqrt{3} \qquad\textbf{(B) } \pi - \sqrt{2} \qquad\textbf{(C) } \frac{\pi + \sqrt{2}}{2} \qquad\textbf{(D) } \frac{\pi +\sqrt{3}}{2} \qquad\textbf{(E) } \frac{7}{6}\pi - \frac{\sqrt{3}}{2}$

Solution

Each small semicircle shares a radius with an adjacent circle. Therefore, the radii to the points of intersection will create equilateral triangles. Draw these triangles, and then bisect the $120^\circ$ angles on the sides and complete the incomplete triangles so that the unshaded region is broken into two types of pieces: circular segments from a $60^\circ$ sector of a circle with radius $1$, and equilateral triangles with side length $1$.

There are $7$ equilateral triangles and $5$ circular segments. One equilateral triangle has area $\frac{1}{2}(1)\left( \frac{\sqrt{3}}{2} \right)=\frac{\sqrt{3}}{4}$, and one segment as area $\frac{60}{360}\pi (1^2)-\frac{\sqrt{3}}{4}=\frac{\pi}{6}-\frac{\sqrt{3}}{4}$.

So the shaded region is the area of the large semicircle minus the area of the unshaded parts, which is $\frac{1}{2}\pi (2^2) - 7\left( \frac{\sqrt{3}}{4}\right) - 5\left( \frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) = 2\pi - \frac{7\sqrt{3}}{4} - \frac{5\pi}{6} - \frac{5\sqrt{3}}{4}= \boxed{\textbf{(E) } \frac{7}{6}\pi - \frac{\sqrt{3}}{2}}$.

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