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2012 AMC 10A Problem 18

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2012 AMC 10A Problem 18AMC 10 Step by Step
Name:                             Date: September 13, 2026
Circles · difficulty 3 of 5 · about 5 min · position P16-20

The closed curve in the figure is made up of 99 congruent circular arcs each of length 2π3\frac{2\pi}{3} , where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 22 . What is the area enclosed by the curve?

figure

  1. A)

    2π+62\pi+6

  2. B)

    2π+432\pi+4\sqrt{3}

  3. C)

    3π+43\pi+4

  4. D)

    2π+33+22\pi+3\sqrt{3}+2

  5. E)

    π+63\pi+6\sqrt{3}

Answer

E

Key insight

Each arc has radius 1 and spans 120 degrees; three vertices carry 240-degree outward bulges and three carry 120-degree dents, so hexagon plus 2pi minus pi.

Solution

Read off the arcs first. Each arc is centered at a hexagon vertex and passes through the midpoints of the two sides at that vertex, so its radius is half the side, r=1r = 1. An arc of length 2π3\frac{2\pi}{3} with radius 11 subtends 2π3\frac{2\pi}{3} radians, i.e. 120∘120^\circ.

At three alternate vertices the curve bulges outward: the two side-midpoints are joined by an arc going around the outside of the vertex. Since the hexagon's interior angle is 120∘120^\circ, the outside sweep is 360∘−120∘=240∘360^\circ - 120^\circ = 240^\circ, which is two of the nine arcs. At the other three vertices the curve dips inward through a single 120∘120^\circ arc, exactly filling the interior angle. That accounts for 3⋅2+3⋅1=93 \cdot 2 + 3 \cdot 1 = 9 arcs.

Now compute the area as hexagon plus bulges minus dents:

  • Hexagon of side 22: 332⋅22=63\frac{3\sqrt3}{2} \cdot 2^2 = 6\sqrt3.
  • Each outward bulge is a 240∘240^\circ sector of radius 11, lying entirely outside the hexagon: 240360π=2π3\frac{240}{360}\pi = \frac{2\pi}{3}. Three of them add 2π2\pi.
  • Each inward dent removes a 120∘120^\circ sector of radius 11: π3\frac{\pi}{3}. Three of them remove π\pi.

Total: 63+2π−π=63+π6\sqrt3 + 2\pi - \pi = 6\sqrt3 + \pi.

The answer is (E) π+63\boxed{\textbf{(E)}\ \pi+6\sqrt{3}}.

Why this works

Curves made of arcs centered at polygon vertices are handled by starting from the polygon and adjusting with sectors: bulges add a sector, dents subtract one. The angles come for free from the polygon's interior and exterior angles (120∘120^\circ and 240∘240^\circ here), and the given arc length pins down the radius. Always convert "arc length" into "radius and central angle" before touching areas.

Alternative approach

Sanity check with the answer form: the region is a hexagon-like shape of area about 63≈10.46\sqrt3 \approx 10.4 with net circular adjustment +π+\pi; only choice (E) has the 636\sqrt3 term that a side-22 hexagon produces.

The trap

Treating all nine arcs as bulging outward (or all inward), or using the hexagon area with side 1 instead of side 2.

Common mistakes

  • Treating all nine arcs as bulging outward (or all inward), or using the hexagon area with side 1 instead of side 2.
  • Using radius 22 (the side) instead of 11 (half the side) for the sectors, which quadruples the π\pi terms.

Techniques

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