Problem 23 · 2012 AMC 8
Hard
Geometry & Measurement
hexagon-decompositionscaling-area
An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?
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Answer: C — Area 6.
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Hint 1 of 2
A regular hexagon is secretly 6 little equilateral triangles meeting at its center — so the whole problem is about comparing those small triangles to the big one. First find the hexagon's side from the equal-perimeter clue.
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Hint 2 of 2
Equal perimeters: the triangle's 3 sides equal the hexagon's 6 sides, so each hexagon side is half the triangle's side. Key fact: when you halve a length, area shrinks by the square — to 1/4, not 1/2.
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Approach: hexagon = six equilateral triangles; area scales as (length)²
- Equal perimeters means 3 × (triangle side) = 6 × (hexagon side), so the hexagon's side is half the triangle's side.
- Cut the hexagon into its 6 natural equilateral triangles (from center to the corners) — each has that half-length side.
- Halving the side scales area by (1/2)² = 1/4 (area always scales as the square of length), so each mini-triangle has area 4 × 1/4 = 1.
- Six of them: hexagon area = 6 × 1 = 6.
- The big takeaway: area scales like the square of the length ratio — double the side, 4× the area; half the side, 1/4 the area. This single fact handles most "similar figures" comparisons without any formula for the area itself.
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