AMC 8 · 2025 · #24

Grade 7 geometry-2dnumber-theory

Archetype Small-Domain Constrained Search

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

📏 Long solution 💡 3 insights 📊 Diagram

Problem

In trapezoid $ABCD$ , angles $B$ and $C$ measure $60^\circ$ and $AB = DC$ . The side lengths are all positive integers, and the perimeter of $ABCD$ is 30 units. How many non-congruent trapezoids satisfy all of these conditions?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: In trapezoid $ABCD$, the base angles $\angle B$ and $\angle C$ are both $60^\circ$ 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?

Givens: Trapezoid $ABCD$ with $\angle B = \angle C = 60^\circ$; Legs are equal: $AB = DC$ (isosceles trapezoid); All four side lengths are positive integers; Perimeter $= AB + BC + CD + DA = 30$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$

Unknowns: The number of non-congruent trapezoids (i.e., the number of distinct side-length multisets) satisfying all the conditions

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #2 Make a Systematic List, #7 Identify Subproblems

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.

Execute — Answer: E

#1 Draw a Diagram 4.G.A.2 Step 1

Name the sides.
Call the shorter base $AD = a$, the longer base $BC = b$, and each equal leg $AB = DC = x$, all positive integers.
Drop perpendiculars from $A$ and $D$ to $BC$, hitting it at $E$ and $F$.
The middle piece $AEFD$ is a rectangle (so $EF = a$), and the two side pieces $\triangle ABE$ and $\triangle DCF$ are congruent right triangles with a $60^\circ$ angle at the base.

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

💡 Classifying the trapezoid by its parallel sides and labeling the matching legs is exactly the Grade 4 "identify parallel sides" skill.

#7 Identify Subproblems 7.G.B.5 Step 2

Use the $30$–$60$–$90$ ratio inside $\triangle ABE$.
The angle at $B$ is $60^\circ$ and the angle at $E$ is $90^\circ$, so the angle at $A$ inside the triangle is $30^\circ$.
In a $30$–$60$–$90$ triangle with hypotenuse $x$, the side opposite the $30^\circ$ angle is $\tfrac{x}{2}$.
That side is $BE$, so $BE = \tfrac{x}{2}$.
By the same argument $CF = \tfrac{x}{2}$.

$$BE = CF = \dfrac{x}{2}$$

💡 Reading off the two complementary acute angles of a right triangle (here $30^\circ$ and $60^\circ$) is a Grade 7 angle-relationship move.

#7 Identify Subproblems 4.MD.C.7 Step 3

Express $b$ in terms of $a$ and $x$.
The long base splits as $BC = BE + EF + FC = \tfrac{x}{2} + a + \tfrac{x}{2} = a + x$.
So the relationship between the three side lengths is just $b = a + x$.

$$b = \dfrac{x}{2} + a + \dfrac{x}{2} = a + x$$

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

#7 Identify Subproblems 6.EE.B.6 Step 4

Use the perimeter to get one equation.
The perimeter is $AD + BC + AB + DC = a + b + 2x = 30$.
Substitute $b = a + x$ from the previous step to get $a + (a + x) + 2x = 30$, i.e.
$2a + 3x = 30$.

$$2a + 3x = 30$$

💡 Turning the word "perimeter $= 30$" into a single equation with two letters is Grade 6 variable-expression work.

#2 Make a Systematic List 6.EE.B.7 Step 5

List all positive-integer solutions.
From $2a = 30 - 3x$, we need $30 - 3x$ to be positive and even.
Positive forces $x \le 9$.
Even forces $3x$ even, hence $x$ even.
So $x \in \{2, 4, 6, 8\}$.
Walk through each in order: $x=2 \Rightarrow a = 12,\ b = 14$; $x=4 \Rightarrow a = 9,\ b = 13$; $x=6 \Rightarrow a = 6,\ b = 12$; $x=8 \Rightarrow a = 3,\ b = 11$.
All four give $a \ge 1$ and $b > a$, so all four are honest trapezoids — none collapses into a rectangle or a triangle.

$$\begin{array}{c|c|c|c} x & a & b & \text{sides}\\ \hline 2 & 12 & 14 & \{2,2,12,14\}\\ 4 & 9 & 13 & \{4,4,9,13\}\\ 6 & 6 & 12 & \{6,6,6,12\}\\ 8 & 3 & 11 & \{8,8,3,11\} \end{array}$$

💡 Solving $2a + 3x = 30$ in positive integers by walking $x$ through its legal values is the Grade 6 "solve an equation for the unknown" skill plus a parity check.

#2 Make a Systematic List 6.EE.B.7 Step 6

Count the non-congruent trapezoids.
Each row above has a different multiset of side lengths, so the four trapezoids are pairwise non-congruent.
The count is $4$, matching choice (E).

$$\text{Count} = 4 \Rightarrow \textbf{(E)}$$

💡 Each distinct $(x, a, b)$ from the systematic list is a distinct trapezoid, so counting rows is the answer.

[1] #1 4.G.A.2 Name the sides. Call the shorter base $AD = a$, the longer base $BC = b$, and ea

[2] #7 7.G.B.5 Use the $30$–$60$–$90$ ratio inside $\triangle ABE$. The angle at $B$ is $60^\ci

[3] #7 4.MD.C.7 Express $b$ in terms of $a$ and $x$. The long base splits as $BC = BE + EF + FC

[4] #7 6.EE.B.6 Use the perimeter to get one equation. The perimeter is $AD + BC + AB + DC = a +

[5] #2 6.EE.B.7 List all positive-integer solutions. From $2a = 30 - 3x$, we need $30 - 3x$ to b

[6] #2 6.EE.B.7 Count the non-congruent trapezoids. Each row above has a different multiset of s

Review

Reasonableness: The leg $x$ has to be even and less than $10$, leaving only $x \in \{2,4,6,8\}$ — 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).

Alternative: Tool #6 (Guess and Check) on $x$ directly: just try $x = 1, 2, 3, \dots, 9$ and compute $a = (30 - 3x)/2$. Discard non-integer or non-positive $a$. Only $x = 2, 4, 6, 8$ survive, giving the same four trapezoids — fewer algebra steps, same answer.

CCSS standards used (min grade 7)

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines (Identifying $ABCD$ as a trapezoid with parallel bases $AD$ and $BC$ and labeling the equal legs $AB = DC$.)
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems (Adding the three sub-segments of $BC$ — $BE + EF + FC = \tfrac{x}{2} + a + \tfrac{x}{2}$ — to express the long base as $b = a + x$.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Letting $a$, $b$, $x$ stand for the side lengths and writing the perimeter condition as a single equation $2a + 3x = 30$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving $2a + 3x = 30$ for $a$ in terms of $x$ and enumerating the positive-integer solutions.)
7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles (Using the $30^\circ$–$60^\circ$–$90^\circ$ angle relationships in $\triangle ABE$ to get the leg-overhang ratio $BE = \tfrac{x}{2}$.)

⭐ 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 7)

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