2021 AMC 10B Fall Problem 11

Attempt Problem 11 of the 2021 AMC 10B Fall 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 2021 AMC 10B Fall solutions, or check the answer key.

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11.

A regular hexagon of side length 11 is inscribed in a circle. Each minor arc of the circle determined by a side of the hexagon is reflected over that side. What is the area of the region bounded by these 66 reflected arcs?

532π \frac{5\sqrt{3}}{2} - \pi

33π 3\sqrt{3}-\pi

433π2 4\sqrt{3}-\frac{3\pi}{2}

π32 \pi - \frac{\sqrt{3}}{2}

π+32 \frac{\pi + \sqrt{3}}{2}

Answer: B
Concepts:area decompositionregular polygoncircle area
Difficulty rating: 1630
Solution:

The original circle is made from the regular hexagon plus 66 equal circular segments. Reflecting each minor arc over its side puts those same 66 segments inside the hexagon instead.

Therefore the average of the circle's area and the reflected-arc region's area is the area of the regular hexagon. The hexagon has area 634=332,6\cdot\frac{\sqrt3}{4}=\frac{3\sqrt3}{2}, and the circle has radius 1,1, so its area is π.\pi.

If the desired area is A,A, then A+π2=332,\frac{A+\pi}{2}=\frac{3\sqrt3}{2}, so A=33π.A=3\sqrt3-\pi.

Thus, the answer is B .

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