AMC 10 · 2012 · #14

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
circular-sectorarea-regular-hexagonarea-difference complementary-countingidentify-subproblems ↑ Prerequisites: area-regular-hexagonarea-circles
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A closed curve is built from equal arcs centred on the corners of a regular hexagon. Find the area it encloses.

Pick an answer.

(A)
$2\pi+6$
(B)
$2\pi+4\sqrt{3}$
(C)
$3\pi+4$
(D)
$2\pi+3\sqrt{3}+2$
(E)
$\pi+6\sqrt{3}$
How to solve
Strategy Identify Subproblems

The curve is irregular, but its pieces are all circular arcs centered at the hexagon's vertices, so the honest move is to build the area out of pieces I can compute (Tool #7). Draw the hexagon that connects the centers (Tool #1) and use it as a baseline area. Then, instead of chasing the wavy boundary directly, reframe the curve as the hexagon with three outward bulges added and three inward bites removed (Tool #16). Every added or removed piece is a circular sector, so the whole answer becomes hexagon area plus outer sectors minus inner sectors.

1STEP 1

Find the arc radius

The arc length gives a radius of 1.

2π r = 3·2π/3=2π → r = 1
2STEP 2

Area of the center hexagon

The hexagon's own area is 6√3.

hexagon = 6·√(3) = 6√(3)
3STEP 3

Reframe curve vs. hexagon

The curve is the hexagon plus bulges minus bites.

Area = 6√(3) + (outer) - (inner)
4STEP 4

Compute the sector areas

Both are simple sectors.

inner=3·π/3=π, outer=3·2π/3=2π
5STEP 5

Add it all up

Adding gives π+6√3, choice (E).

6√(3) - π + 2π = π + 6√(3) → (E) π+6√(3)
Answer
π+6√(3)
Count the arcs the framing uses: three inward 120° arcs (3 arcs) plus three outward 240° bulges, each of which is two 120° arcs (6 arcs), totals 3+6=9 arcs of length 2π/3 each — exactly the 9 congruent arcs the problem names. The area π+6√(3)≈ 3.14+10.39=13.53, a bit larger than the hexagon's 6√(3)≈ 10.39, which fits a shape whose outward bulges beat its inward dents. Among the choices, (E) is the only one carrying a lone π with the hexagon's 6√(3), matching our net of one extra circle.
💡Key takeaway

Draw the hexagon through the arc centers (6√(3)), then treat the curve as that hexagon with three outward bulges added (2π) and three inward bites removed (π); the bulges win by one circle, giving (E) π+6√(3).

  • Find the arc radius
  • Area of the center hexagon
  • Reframe curve vs. hexagon
  • Compute the sector areas
  • Add it all up

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