Question

A circle of radius $1$ is internally tangent to two circles of radius $2$ at points $A$ and $B,$ where $AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the figure, that is outside the smaller circle and inside each of the two larger circles?

$\dfrac{5}{3}\pi - 3\sqrt{2}$

$\dfrac{5}{3}\pi - 2\sqrt{3}$

$\dfrac{8}{3}\pi - 3\sqrt{3}$

$\dfrac{8}{3}\pi - 3\sqrt{2}$

$\dfrac{8}{3}\pi - 2\sqrt{3}$

Answer: B

Concepts:sector area decomposition special right triangle symmetry

Difficulty rating: 2270

Solution:

Let the large circles have centers $A$ and $B,$ let $C$ be the center of the small circle, and let $D$ be a point where the two large circles meet.

Then $\triangle ACD$ is right with $AC = 1$ and $AD = 2,$ so $CD = \sqrt3,$ $\angle CAD = 60^\circ,$ and its area is $\dfrac{\sqrt3}{2}.$

One quarter of the shaded region equals the $60^\circ$ sector of the radius- $2$ circle (area $\dfrac{2\pi}{3}$ ) minus $\triangle ACD$ (area $\dfrac{\sqrt3}{2}$ ) minus a quarter of the small circle (area $\dfrac{\pi}{4}$ ), giving $\dfrac{2\pi}{3} - \dfrac{\sqrt3}{2} - \dfrac{\pi}{4} = \dfrac{5\pi}{12} - \dfrac{\sqrt3}{2}.$

Multiplying by $4,$ the shaded area is $\dfrac{5\pi}{3} - 2\sqrt3.$

Thus, the correct answer is B.

2004 AMC 10B Problem 25

Answer: B

Difficulty rating: 2270

Solution:

Problem 25 in Other Years