Question

Circles $A,$ $B,$ and $C$ each have radius $1.$ Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$

$3 - \dfrac{\pi}{2}$

$\dfrac{\pi}{2}$

$2$

$\dfrac{3\pi}{4}$

$1 + \dfrac{\pi}{2}$

Solution:

Place $A = (-1, 0),$ $B = (1, 0),$ so their tangency point is the origin, the midpoint of $\overline{AB}.$ Then $C = (0, 1),$ since $C$ passes through the origin.

The distance from $C$ to $A$ (and to $B$ ) is $\sqrt2.$ Two unit circles whose centers are $\sqrt2$ apart overlap in a lens of area $2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) - \tfrac{\sqrt2}{2}\sqrt{4 - 2} = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1.$

Circles $A$ and $B$ meet only at the origin, so the two lenses do not overlap. The wanted area is $\pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2.$

Thus, the correct answer is C.

2011 AMC 12A Problem 11

Solution:

Problem 11 in Other Years