AMC 8 · 2025 · #24
Grade 7 geometry-2dnumber-theoryProblem
In trapezoid ABCD, angles B and C measure 60∘ 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.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $4$
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
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.
💡 Classifying the trapezoid by its parallel sides and labeling the matching legs is exactly the Grade 4 "identify parallel sides" skill.
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 $CF = \tfrac{x}{2}$.
💡 Reading off the two complementary acute angles of a right triangle (here $30^\circ$ and $60^\circ$) is a Grade 7 angle-relationship move.
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 $b = a + x$.
💡 Adding the three pieces of $BC$ to get its total length is the Grade 4 "angle/segment measure is additive" idea applied to lengths.
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.
- $2a + 3x = 30$.
💡 Turning the word "perimeter $= 30$" into a single equation with two letters is Grade 6 variable-expression work.
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.
💡 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.
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).
💡 Each distinct $(x, a, b)$ from the systematic list is a distinct trapezoid, so counting rows is the answer.
4.G.A.2 Name the sides. Call the shorter base $AD = a$, the longer base $BC = b$, and ea 7.G.B.5 Use the $30$–$60$–$90$ ratio inside $\triangle ABE$. The angle at $B$ is $60^\ci 4.MD.C.7 Express $b$ in terms of $a$ and $x$. The long base splits as $BC = BE + EF + FC 6.EE.B.6 Use the perimeter to get one equation. The perimeter is $AD + BC + AB + DC = a + 6.EE.B.7 List all positive-integer solutions. From $2a = 30 - 3x$, we need $30 - 3x$ to b 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.2Classify 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.7Recognize 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.6Use 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.7Solve 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.5Use 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!
⭐ 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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