AMC 10 · 2021 · #21

Grade 8 geometry-2d

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

area-trianglessimilar-figuresangle-sum-triangleperimeter identify-subproblemsconvert-to-algebra ↑ Prerequisites: area-triangles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

An equiangular hexagon ABCDEF has every interior angle equal to 120°. Extending alternate sides — first AB, CD, EF, then BC, DE, FA — produces two triangles whose areas are 192√(3) and 324√(3). Find the perimeter of the hexagon in the form m + n√(p) (with p square-free) and compute m + n + p.

Pick an answer.

(A)

~47

(B)

~52

(C)

~55

(D)

~58

(E)

~63

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram) — sketch the hexagon and extend the two triples of alternate sides until they cross; the picture immediately shows two triangles, each with three small "corner" triangles snipped at the hexagon's vertices. Tool #9 (Easier Problem) — a regular hexagon is the easy case (192 = 324), so we expect the answer for the general equiangular hexagon to still hinge on the side-length-to-area relation s² √(3)/4 for equilateral triangles. Tool #7 (Subproblems) — recover the two big triangle side lengths separately from the two areas; then read off the hexagon side lengths from the diagram. Tool #13 (Algebra) — the perimeter is just the sum of the two big triangle sides, since every hexagon side sits on one of the two big triangles.

1STEP 1

Every interior angle is 120°, so each extended corner snips a 60° chip — making both big outer triangles equilateral.

each interior angle=120° → exterior=60° → both big triangles are equilateral

2STEP 2

Invert the equilateral area formula area = s²· $\frac{\sqrt{3}}{4}$ : from the two areas the triangles have sides s₁ = 16√(3) and s₂ = 36.

s₁ = 16√(3), s₂ = 36

3STEP 3

Each big-triangle side = one hexagon edge + two chip edges; summing the three sides gives P₁ + 2 P₂ = 108 and 2 P₁ + P₂ = 48√(3).

P₁ + 2 P₂ = 108, 2 P₁ + P₂ = 48√(3)

4STEP 4

Sanity-check on a regular hexagon: P₁ = P₂ makes the two relations collapse to 3 P₁ = 3 s, confirming the formulas hold.

regular hexagon: P₁ = P₂ → 3 P₁ = 3 s₁ = 3 s₂ ✓

5STEP 5

Eliminate to solve the pair: P₁ = 32√(3) - 36 and P₂ = 72 - 16√(3).

P₁ = 32√(3) - 36, P₂ = 72 - 16√(3)

6STEP 6

Add the two subsums for the full perimeter P = 36 + 16√(3), so m = 36, n = 16, p = 3.

P = 36 + 16√(3) → m + n + p = 36 + 16 + 3 = 55

Answer

~55

Each P_i should be positive. P₁ = 32√(3) - 36 ≈ 55.4 - 36 = 19.4 ＞ 0 and P₂ = 72 - 16√(3) ≈ 72 - 27.7 = 44.3 ＞ 0, so the hexagon is geometrically realizable. The two big triangle side lengths 16√(3) ≈ 27.7 and 36 are different, which matches the unequal areas 192√(3) vs 324√(3) (ratio

\frac{324}{192}

\frac{27}{16}

, square root

\frac{√(27)}{4}

= 3√(3)/4 ≈ 1.30 =

\frac{36}{27.7}

— consistent). Final answer 55 matches choice (C) exactly.

💡Key takeaway

This AMC 10 problem only needs Grade 8 angle reasoning and a 2 × 2 linear system you already know — extend the alternate sides, notice every corner chip and both big triangles are equilateral (since 180° - 120° = 60°), recover side lengths s₁ = 16√(3) and s₂ = 36 from areas, then solve the two linear equations P₁ + 2 P₂ = 108 and 2 P₁ + P₂ = 48√(3) to get perimeter = 36 + 16√(3), so m + n + p = 55.

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