Problem 24 · 2025 AMC 8
Hard
Geometry & Measurement
perimetercaseworkarea-decomposition
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Answer: E — 4 trapezoids.
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Hint 1 of 2
Two 60° base angles are begging for an equilateral triangle. Slice one off: draw a line through A parallel to the slanted side CD — what shape gets cut out, and what's left?
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Hint 2 of 2
That cut makes an equilateral triangle ABE (so AB = BE = AE) plus a parallelogram ADCE (so AD = EC). Relabel the equal lengths x and y, and the whole perimeter collapses to 3x + 2y = 30 — now it's just counting integer solutions.
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Approach: auxiliary parallel line: equilateral triangle + parallelogram, then count integer solutions
- The two 60° angles are the cue. Draw a segment through A parallel to CD, meeting BC at E. Because ∠B = 60° and AB = DC, triangle ABE has all 60° angles — it's equilateral, so AB = BE = AE = x.
- What remains, ADCE, is a parallelogram (AE ∥ DC and AD ∥ EC), so the two parallel sides match: AD = EC = y.
- Now the perimeter rewrites entirely in x and y: AB + BC + CD + DA = x + (x + y) + x + y = 3x + 2y = 30. The geometry is gone — it's a counting problem.
- Need positive integers: y = (30 − 3x)/2 must be positive (so x < 10) and an integer (so 30 − 3x even, i.e. x even). That leaves x ∈ {2, 4, 6, 8} — 4 trapezoids.
- Why this transfers: a parallel auxiliary line through a vertex peels a trapezoid into a triangle + parallelogram, and a 60° angle makes that triangle equilateral. The payoff is converting a shape question into a tidy linear equation to count.
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