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2021 AMC Fall 10B Problem 11

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2021 AMC Fall 10B Problem 11AMC 10 Step by Step
Name:                             Date: September 13, 2026
Circles · difficulty 3 of 5 · about 4 min · position P11-15

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?

  1. A)

    532−π\frac{5\sqrt{3}}{2} - \pi

  2. B)

    33−π3\sqrt{3}-\pi

  3. C)

    43−3π24\sqrt{3}-\frac{3\pi}{2}

  4. D)

    π−32\pi - \frac{\sqrt{3}}{2}

  5. E)

    π+32\frac{\pi + \sqrt{3}}{2}

Answer

B

Key insight

Each reflected arc bulges into the hexagon, so the region is the hexagon minus six circular segments, each a 60-degree sector minus an equilateral triangle.

Solution

A regular hexagon with side 11 inscribed in a circle has circumradius 11, and each side subtends a 60∘60^\circ arc.

The region between a side and its minor arc is a circular segment. Its area is a 60∘60^\circ sector minus an equilateral triangle of side 11:
60360π⋅12−34=π6−34.\frac{60}{360}\pi \cdot 1^2 - \frac{\sqrt3}{4} = \frac{\pi}{6} - \frac{\sqrt3}{4}.

Reflecting the arc over the side flips this segment to the inside of the hexagon. The region bounded by the six reflected arcs is therefore the hexagon with six such segments removed.

The hexagon consists of six equilateral triangles of side 11, area 6⋅34=3326 \cdot \frac{\sqrt3}{4} = \frac{3\sqrt3}{2}. So the required area is
332−6(π6−34)=332+332−π=33−π.\frac{3\sqrt3}{2} - 6\left(\frac{\pi}{6} - \frac{\sqrt3}{4}\right) = \frac{3\sqrt3}{2} + \frac{3\sqrt3}{2} - \pi = 3\sqrt3 - \pi.

The answer is (B) 33−π\boxed{\textbf{(B)}\ 3\sqrt{3}-\pi}.

Why this works

Reflection preserves area, so the reflected arcs cut out of the hexagon exactly what the original arcs added outside it. Any region bounded by arcs is best handled by decomposing into sectors and triangles; "segment = sector minus triangle" is the standard building block.

Alternative approach

Sanity check with sizes: the hexagon has area about 2.62.6, and the answer must be smaller since the arcs curve inward. Choice (B) is about 5.20−3.14=2.065.20 - 3.14 = 2.06, and (A) is about 1.191.19; the removed area is six thin segments of about 0.090.09 each, so (B) fits.

The trap

Adding the six segments to the hexagon (as if the arcs bulged outward) and getting an answer larger than the hexagon.

Common mistakes

  • Adding the six segments to the hexagon (as if the arcs bulged outward) and getting an answer larger than the hexagon.
  • Using a 30∘30^\circ sector or radius 32\frac{\sqrt3}{2} for the segments; the arc is 60∘60^\circ on a circle of radius 11.

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors) · Exploit symmetry to reduce work or pair up objects