2019 AMC 10A Problem 16

Attempt Problem 16 of the 2019 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 2019 AMC 10A solutions, or check the answer key.

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16.

The figure below shows 1313 circles of radius 11 within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius 1?1?

4π34 \pi \sqrt{3}

7π7 \pi

π(33+2)\pi\left(3\sqrt{3} +2\right)

10π(31)10 \pi \left(\sqrt{3} - 1\right)

π(3+6)\pi\left(\sqrt{3} + 6\right)

Answer: A
Concepts:circle areaequilateral triangletangent circles
Difficulty rating: 1540
Solution:

We know ABC\triangle ABC and ABO\triangle ABO are equilateral triangles.

We get that OC=23OC = 2\sqrt{3} using special right triangles to find the altitudes of the triangles.

The radius of the larger circle is therefore 23+1,2\sqrt{3} + 1, since there is the extra unit radius after OC.\overline{OC}.

The area of the larger circle is (23+1)2π=(13+43)π. (2\sqrt{3} + 1)^2\pi = (13 + 4\sqrt{3})\pi.

The area of all the inner circles is 13π.13\pi.

The area of the shaded region is (13+43)π13π=4π3. (13 + 4\sqrt{3})\pi - 13\pi = 4\pi\sqrt{3}.

Thus, A is the correct answer.

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