AMC 10 · 2020 · #11

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

area-regular-hexagonarea-circlesequilateral-trianglesymmetry-argument identify-subproblemsarea-differencesymmetry-argument ↑ Prerequisites: area-regular-hexagonarea-circles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A regular hexagon has side length 2. Six semicircles lie inside it, one on each side, each with the side as its diameter so each bulges into the hexagon. Find the area of the region inside the hexagon but outside all six semicircles.

Pick an answer.

(A)

$6\sqrt3 - 3\pi$

(B)

$\frac{9\sqrt3}{2} - 2\pi$

(C)

$\frac{3\sqrt3}{2} - \frac{\pi}{3}$

(D)

$3\sqrt3 - \pi$

(E)

$\frac{9\sqrt3}{2} - \pi$

How to solve

Strategy Draw a Diagram

Tool #1 (Draw): subdivide the hexagon into a grid of small equilateral triangles of side 1 — 24 of them — so every region (shaded or white) is a clean union of these tiles or circular sectors. Tool #9 (Easier Problem): by 6-fold rotational symmetry, the shaded region is 6 congruent pieces; find the area of one piece and multiply. Tool #7 (Subproblems): each piece is (rhombus of 2 small triangles) minus (one 60° sector of radius 1). Tool #3 (Eliminate): the simplified expression 3√(3) - π matches exactly one answer choice.

1STEP 1

Cut the hexagon into triangles

Cut it into small equilateral triangles.

hexagon = 24 small triangles of side 1

2STEP 2

Use the six-fold symmetry

Measuring one piece is enough.

shaded = 6 · (one piece)

3STEP 3

Name the piece

A piece is a rhombus minus a sector.

one piece = (rhombus of 2 unit triangles) - (60° sector of radius 1)

4STEP 4

Measure both parts

Compute the rhombus and the sector.

rhombus = √(3)/2, sector = π/6

5STEP 5

Multiply by six

Take six of that piece.

shaded = 6 ( √(3)/2 - π/6 ) = 3√(3) - π

6STEP 6

Match the choice

The area is three root three minus pi.

3√(3) - π → (D)

Answer

3√3 - π

The hexagon's area is 3√(3)/2 · 4 = 6√(3) ≈ 10.39. The total semicircle area (with overlap) is 6 · π/2 = 3π ≈ 9.42 — so the white region cannot be that large; some semicircle area is double-counted. The shaded answer 3√(3) - π ≈ 5.196 - 3.14 ≈ 2.06 is positive and well below the hexagon area, exactly as expected for the leftover slivers near the vertices.

💡Key takeaway

This AMC 12 problem only needs Grade 7 area formulas you already know! Cut the side-2 hexagon into 24 small triangles of side 1. By 6-fold symmetry, the shaded region is 6 identical pieces near the vertices — each piece is a 2-triangle rhombus (area √(3)/2) minus a 60° sector of radius 1 (area π/6). Multiply by 6: 6(√(3)/2 - π/6) = 3√(3) - π, answer (D).