Competition · AMC preparation · step 4 of 4

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

Name the side lengths

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

Use the 30-60-90 ratio

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

Express b using a and x

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

Adding the three pieces of BC to get its total length is the Grade 4 "angle/segment measure is additive" idea applied to lengths.

The longer base of the trapezoid is exactly the shorter base plus one leg length.

▸ Why?

The two perpendiculars dropped from the top corners cut the longer base into a left piece, a middle piece, and a right piece, and those three pieces together make up the whole longer base.

▸ Why?

The feet of the perpendiculars land on the longer base and divide it with no gaps and no overlaps, so the three pieces add back to the full base.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

The middle piece has the same length as the shorter base, because the shorter base and the middle piece are the two horizontal sides of the rectangle formed between the perpendiculars.

▸ Why?

Sliding the shorter base straight down the two equal perpendiculars lays it exactly onto the middle piece, so the two segments must have equal length.

Rigid motion preserves length and angleSliding, turning, or flipping a shape lays it exactly onto a copy, so every length and angle stays the same (folding is the flip case).

▸ Why?

Each outer piece is half of a leg, so the left and right pieces together add exactly one whole leg length to the middle piece.

▸ Why?

Each corner triangle has a right angle at the foot and a 60 degree angle at the base, which forces its remaining angle to be 30 degrees.

A triangle's angles sum to 180 degreesThe three corners of any flat triangle always add up to one straight angle, 180 degrees.

▸ Why?

Flipping a corner triangle across its vertical height copies the leg and the outer piece onto the other side, building a triangle with all equal sides whose base is a full leg long, so the outer piece is half of that leg.

Rigid motion preserves length and angleSliding, turning, or flipping a shape lays it exactly onto a copy, so every length and angle stays the same (folding is the flip case).

4.MD.C.7Identify Subproblems

4STEP 4

Turn the perimeter into an equation

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

2a + 3x = 30

5STEP 5

List the integer solutions

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

Count the distinct trapezoids

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!

Name the side lengths
Use the 30-60-90 ratio
Express b using a and x
Turn the perimeter into an equation
List the integer solutions
Count the distinct trapezoids

A parent dashboard for the family lives at sensimlab.com.

Name the side lengths

Use the 30-60-90 ratio

Express b using a and x

Turn the perimeter into an equation

List the integer solutions

Count the distinct trapezoids

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