AMC 10 · 2025 · #3
Grade 7 geometry-2dHow many isosceles triangles are there with positive area whose side lengths are all positive integers and whose longest side has length 2025?
Pick an answer.
AMC 10 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: Count the whole-number-sided isosceles triangles that have positive area and whose longest side is exactly 2025.
Givens: The triangle is isosceles, so at least two sides are equal; its sides can be written as A, A, B with A and B positive integers.; All side lengths are positive integers.; The longest side has length 2025.; The triangle has positive area, so the three sides must satisfy the triangle inequality (no flat, degenerate triangle).
Unknowns: How many such triangles exist.
Understand
Restated: Count the whole-number-sided isosceles triangles that have positive area and whose longest side is exactly 2025.
Givens: The triangle is isosceles, so at least two sides are equal; its sides can be written as A, A, B with A and B positive integers.; All side lengths are positive integers.; The longest side has length 2025.; The triangle has positive area, so the three sides must satisfy the triangle inequality (no flat, degenerate triangle).
Plan
Primary tool: #7 Identify Subproblems
Secondary: #4 Introduce a Variable, #14 Extreme Principle, #2 Make a Systematic List
An isosceles triangle has an equal pair A, A and a third side B. The number 2025 is the longest side, and it can play one of two roles: it is one of the equal sides, or it is the single unequal side. Those two roles never happen at the same time, so splitting the count into these two separate subproblems lets us count each cleanly and add. Inside each case, naming the sides with a variable and pushing to the triangle-inequality boundary turns the count into counting a run of consecutive integers.
Execute — Answer: D
7.G.A.2 Step 1 Name the sides
- Write the isosceles sides as A, A, B with A and B positive integers.
- Because 2025 is the longest side, one of these values equals 2025 and no side is larger than 2025.
💡 Every isosceles triangle is just an equal pair plus a third side, so two letters capture all of them.
7.G.A.2 Step 2 Split by the role of 2025
- The longest side 2025 is either one of the two equal sides, or it is the single unequal side.
- These are mutually exclusive, so count each case on its own and add the totals.
💡 The biggest side has to be somewhere in the pattern A, A, B, and there are only two spots it can sit.
6.EE.B.8 Step 3 Case 1: equal sides are 2025
- The sides are 2025, 2025, B.
- To keep 2025 the longest side we need B at most 2025.
- The triangle inequality only requires 2025 + B > 2025, i.e.
- B is at least 1, and the other inequality 2025 + 2025 > B holds automatically since B is at most 2025.
- So B runs over every integer from 1 to 2025, giving 2025 triangles (B = 2025 is the equilateral one, still isosceles).
💡 With two long equal sides, any positive base up to 2025 still closes into a real triangle.
6.EE.B.5 Step 4 Case 2: base is 2025
- The sides are A, A, 2025 with A less than 2025 (so 2025 stays strictly longest).
- The triangle inequality needs the two equal sides to reach across the base: A + A > 2025, so A > 1012.5, meaning A is at least 1013.
- Combined with A at most 2024, the integer values are 1013 through 2024.
- That is 2024 - 1013 + 1 = 1012 triangles.
💡 The two equal legs must together stretch past the base, which forces each leg to be more than half of 2025.
4.OA.A.3 Step 5 Add the two cases
- The cases share no triangle: Case 1 always has an equal pair of 2025, while Case 2 has its equal pair below 2025.
- Adding the counts gives 2025 + 1012 = 3037, so the answer is (D).
💡 Two non-overlapping piles of triangles just add together.
7.G.A.2 Write the isosceles sides as A, A, B with A and B positive integers. Because 202 7.G.A.2 The longest side 2025 is either one of the two equal sides, or it is the single 6.EE.B.8 The sides are 2025, 2025, B. To keep 2025 the longest side we need B at most 202 6.EE.B.5 The sides are A, A, 2025 with A less than 2025 (so 2025 stays strictly longest). 4.OA.A.3 The cases share no triangle: Case 1 always has an equal pair of 2025, while Case Review
Reasonableness: The two cases are genuinely disjoint (the equal pair is either 2025 or strictly below 2025), so no triangle is double counted, and the boundary A = 1013 barely satisfies 2A = 2026 > 2025, confirming none of the Case 2 triangles are flat. The total 3037 matches choice (D). The tempting wrong answers line up with common slips: 2025 forgets Case 2, and 3012 = 2000 + 1012 comes from mistakenly starting Case 1 at B = 26 or dropping the equilateral endpoint.
Alternative: Instead of splitting by role, list by the equal side length s. If s = 2025 the base B ranges over 1..2025 (2025 triangles); if s < 2025 then 2025 is the base and the inequality 2s > 2025 forces s in 1013..2024 (1012 triangles). Same partition, same total 3037.
CCSS standards used (min grade 7)
7.G.A.2Draw geometric shapes with given conditions including triangles (Using the triangle inequality to decide which integer side triples actually form a real (positive-area) triangle and which side is longest.)6.EE.B.8Write an inequality of the form x > c or x < c and graph on a number line (Turning 'stays the longest side' and 'the triangle closes up' into the bounds 1 <= B <= 2025 in Case 1.)6.EE.B.5Understand solving an equation or inequality as finding values that make it true (Finding and counting the integer values of A that satisfy 2A > 2025 with A <= 2024 in Case 2.)4.OA.A.3Solve multi-step word problems using the four operations with whole numbers (Counting each run of consecutive integers and adding the two case totals, 2025 + 1012 = 3037.)
⭐ Ask where the longest side can sit in the pattern A, A, B, count each spot with the triangle inequality, and add.
⭐ Ask where the longest side can sit in the pattern A, A, B, count each spot with the triangle inequality, and add.
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