Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
Line 13: Line 13:
  
 
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>.
 
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>.
 +
{{MAA Notice}}

Revision as of 09:26, 4 July 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 has 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 $\frac{1}{2}\pi (2^2) - 7\left( \frac{\sqrt{3}}{4}\right) - 5\left( \frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) = 2\pi - \frac{7\sqrt{3}}{4} - \frac{5\pi}{6} + \frac{5\sqrt{3}}{4}= \boxed{\textbf{(E) } \frac{7}{6}\pi - \frac{\sqrt{3}}{2}}$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12B_Problems/Problem_16&oldid=53990"