2011 AMC 10A Problem 18

Attempt Problem 18 of the 2011 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10A solutions, or check the answer key.

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18.

Circles A,A, B,B, and CC each have radius 1.1. Circles AA and BB share one point of tangency. Circle CC has a point of tangency with the midpoint of AB‾.\overline{AB}. What is the area inside circle CC but outside circle AA and circle B?B?
3−π23 - \dfrac{\pi}{2}
π2\dfrac{\pi}{2}
22
3π4\dfrac{3\pi}{4}
1+π21+\dfrac{\pi}{2}
Answer: C
Concepts:circle areasectorarea decomposition
Small hint:
Use the unit squares formed by the centers and intersection points
Big hint:
Two quarter-circle subtractions plus a semicircle leave a simple area
Solution:
The area of this region is the area of circle CC minus the area of the overlapping regions with AA and B.B.
From the diagram, we can find the area of half of one of the overlapping regions by finding the area of the sector and subtracting the area of the triangle.
This area is then 14π⋅12−12⋅1⋅1=π4−12. \dfrac{1}{4} \pi \cdot 1^2 - \dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{\pi}{4} - \dfrac{1}{2}.
There are four of these that we must subtract, which leaves us with a final answer of π⋅12−4(π4−12)=2. \pi \cdot 1^2 - 4(\dfrac{\pi}{4} - \dfrac{1}{2}) = 2.
Thus, C is the correct answer.
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