AMC 8 · 2025 · #24

Grade 7 geometry-2dnumber-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

perimetersimilar-triangleslinear-diophantineangle-sum-triangle convert-to-algebrasystematic-enumerationcasework ↑ Prerequisites: perimeterpythagorean-theoremlinear-equations-two-var

📏 Long solution 💡 3 insights 📊 Diagram

Problem

In trapezoid ABCD, the base angles ∠ B and ∠ C are both 60° and the legs are equal (AB = DC), so the trapezoid is isosceles with parallel bases AD and BC. Every side length is a positive integer and the perimeter is 30. How many non-congruent trapezoids fit all of these conditions?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The asy figure is a starting point, but the key move is Tool #1 (Draw a Diagram): add height segments from A and D down to base BC. That splits the trapezoid into a rectangle in the middle and two congruent 30–60–90 right triangles on the sides — a Tool #7 (Identify Subproblems) decomposition. The 30–60–90 ratio forces each leg overhang to be exactly half the leg, which produces the simple relation BC = AD + AB. Combined with perimeter = 30, this gives one linear equation in two unknowns, so Tool #2 (Make a Systematic List) — running through the small set of legal leg lengths in order — finishes the count without algebra heavy lifting.

1STEP 1

Bases AD = a, BC = b and legs = x; dropping perpendiculars splits BC into a rectangle and two 30–60–90 triangles.

AD = a, BC = b, AB = DC = x

2STEP 2

In each 30–60–90 triangle the hypotenuse is the leg x, so the side opposite 30° — the overhang BE = CF — is x/2.

BE = CF = x/2

3STEP 3

The long base splits as BC = x/2 + a + x/2, which collapses to the clean relation b = a + x.

b = x/2 + a + x/2 = a + x

4STEP 4

Perimeter a + b + 2x = 30; substitute b = a + x to get 2a + 3x = 30.

2a + 3x = 30

5STEP 5

Since 30 − 3x must be positive and even, x must be even and under 10, giving x ∈ {2, 4, 6, 8} — each a real trapezoid with b ＞ a.

x & a & b & sides ; 2 & 12 & 14 & {2,2,12,14} ; 4 & 9 & 13 & {4,4,9,13} ; 6 & 6 & 12 & {6,6,6,12} ; 8 & 3 & 11 & {8,8,3,11}

6STEP 6

Each row is a different side-length multiset, so all four are non-congruent: the count is 4, choice (E).

Count = 4 → (E)

Answer

The leg x has to be even and less than 10, leaving only x ∈ {2,4,6,8} — a small finite set, so an answer of 4 is the natural upper bound. Sanity check the extremes: at x=2, the trapezoid is short and wide (2,2,12,14); at x=8, it is tall and narrow (8,8,3,11). Both still have b = a + x ＞ a, so both keep BC as the longer base. No case is degenerate (no a = 0 triangle, no a = b rectangle), so the count 4 is correct and matches answer (E).

💡Key takeaway

This AMC 8 problem only needs Grade 7 angle-relationship facts (the 30–60–90 triangle) plus simple Grade 6 equation-listing you already know!

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