2003 AMC 12B Problem 16

Below is the professionally curated solution for Problem 16 of the 2003 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12B solutions, or check the answer key.

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Concepts:sectorequilateral trianglearea decomposition

Difficulty rating: 1680

16.

Three semicircles of radius 11 are constructed on diameter AB\overline{AB} of a semicircle of radius 2.2. The centers of the small semicircles divide AB\overline{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?

π3\pi - \sqrt{3}

π2\pi - \sqrt{2}

π+22\dfrac{\pi + \sqrt{2}}{2}

π+32\dfrac{\pi + \sqrt{3}}{2}

76π32\dfrac{7}{6}\pi - \dfrac{\sqrt{3}}{2}

Solution:

The large semicircle has area 12π(2)2=2π.\dfrac{1}{2}\pi(2)^2 = 2\pi.

Where the small semicircles overlap, adjacent ones meet at points a distance 11 from two centers, forming equilateral triangles. The region removed from the large semicircle consists of five congruent 6060^\circ sectors of radius 1,1, each of area π6,\dfrac{\pi}{6}, together with two equilateral triangles of side 1,1, each of area 34.\dfrac{\sqrt{3}}{4}.

The shaded area is 2π5π6234=76π32. 2\pi - 5\cdot\frac{\pi}{6} - 2\cdot\frac{\sqrt{3}}{4} = \frac{7}{6}\pi - \frac{\sqrt{3}}{2}.

Thus, the correct answer is E.

Problem 16 in Other Years