AMC 10 · 2021 · #14

Grade 10 geometry-2d

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

law-of-cosinessine-area-formulaequilateral-trianglearea-triangles identify-subproblemsconvert-to-algebra ↑ Prerequisites: law-of-cosinesarea-triangles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A hexagon has all six sides the same length. Three of its interior angles, no two of them at neighbouring vertices, each measure 30 degrees. The region it encloses has area six root three. Find the hexagon's perimeter.

Pick an answer.

(A)

(B)

$4\sqrt3$

(C)

(D)

(E)

$12\sqrt3$

How to solve

Strategy Identify Subproblems

A spiky six-sided shape has no area formula, but the three sharp corners are exactly where it can be cut. Joining the three blunt vertices slices the hexagon into three corner triangles, each with two sides of the common length and the 30° angle between them, plus one triangle in the middle. The corner triangles are forced to be congruent, which makes the middle triangle equilateral, so every piece has an area that depends on the single unknown side. Adding the four areas gives one equation in one unknown, and the given area solves it.

1STEP 1

Name the side length

Give the side a name.

AB = BC = CD = DE = EF = FA = s, perimeter = 6s

2STEP 2

Cut off the three spikes

Cut off the three spikes.

[ABCDEF] = [FAB] + [BCD] + [DEF] + [BDF]

3STEP 3

The middle triangle is equilateral

The middle is an equilateral triangle.

△ FAB ≅ △ BCD ≅ △ DEF (SAS) → BF = BD = DF = b

4STEP 4

Area of one corner triangle

Compute one corner triangle's area.

BH = s/2, AH = s√(3)/2, [FAB] = 1/2 · s · s/2 = s²/4

5STEP 5

Square the middle triangle's side

Find the middle triangle's side squared.

b² = (s/2)² + (s - s√(3)/2)² = s²/4 + (7/4 - √(3))s² = (2 - √(3))s²

6STEP 6

Area of the middle triangle

Compute the middle triangle's area.

[BDF] = √(3)/4b² = √(3)/4(2 - √(3))s² = (√(3)/2 - 3/4)s²

7STEP 7

Add the pieces and solve

Adding and solving gives a perimeter of twelve root three.

3/4s² + (√(3)/2 - 3/4)s² = √(3)/2s² = 6√(3) → s² = 12 → s = 2√(3) → 6s = 12√(3)

Answer

12√3

First check that the figure is consistent. Each corner triangle is isosceles with apex 30°, so its base angles are (180° - 30°)/2 = 75°. At a blunt vertex such as B the interior angle is 75° + 60° + 75° = 210°, using the 60° of the equilateral middle triangle. The six angles then total 3 · 30° + 3 · 210° = 720°, exactly the hexagon angle sum, and 210° ＞ 180° confirms the reflex corners drawn in the figure. Next check the area: with s = 2√(3), √(3)/2s² = √(3)/2 · 12 = 6√(3), as required. A compact second look: √(3)/2s² = 2·√(3)/4s², so this hexagon always has exactly twice the area of an equilateral triangle with the same side; that triangle has area 3√(3) here, and doubling gives 6√(3). Finally, the other choices really do fail. Writing the area in terms of the perimeter P gives √(3)/2(P/6)² = √(3)P²/72, which turns the five choices into areas 2√(3)/9, 2√(3)/3, 2√(3), 9√(3)/2, and 6√(3). Only the last one matches.

💡Key takeaway

Slice the three sharp corners off the hexagon: the corners are congruent, what is left in the middle is equilateral, and the four areas add to √(3)/2s², so the given area pins down the side.

Name the side length
Cut off the three spikes
The middle triangle is equilateral
Area of one corner triangle
Square the middle triangle's side
Area of the middle triangle
Add the pieces and solve