2025 AMC 10A Problem 3
How many isosceles triangles are there with positive area whose side lengths are all positive integers and whose longest side has length ?
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Key insight
Split by which sides are equal. Two sides of 2025 works with any third side up to 2025. Two equal shorter sides must add to more than 2025.
Solution
- The longest side is , so every side is a whole number from to . Isosceles means two sides are equal. Either the equal pair is the long side , or the equal pair is shorter. Count each case.
- "Positive area" means the triangle is not flat, so the two shorter sides together must be longer than the longest side (equal is not allowed).
Case 1: two sides are . The triangle is with from to .
- Is it always a real triangle? The two short sides are and , and is bigger than for any . So every works. That is triangles (including the equilateral one, ).
Case 2: the two equal sides are shorter. The triangle is with smaller than .
- Now the rule really matters. Try small numbers first: gives , too short to reach across . So must be big enough that .
- means , so is at least . Check the edge: works, while fails.
- So can be . That is triangles.
- No triangle is counted twice: in Case 1 the repeated side is , in Case 2 it is less than .
- Total: .
The answer is .
Why this works
Fixing the longest side makes the list finite, and then each isosceles triangle is decided by one other number. The triangle rule (the two shorter sides must add to more than the longest) is easy in one case and does real work in the other, which is why the two cases have different sizes. When counting triangles, always ask which case each triangle falls in and apply that rule to it.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Positive area means the triangle inequality must be strict. The longest side is , so every side is at most and at least one side equals . An isosceles triangle has a pair of equal sides, and the two cases are decided by whether that pair is the longest side or not.
The equal pair is the longest side. The triangle is with . The inequality is automatic and holds for every such , so all values work. This case includes the equilateral triangle .
The equal pair is shorter. The triangle is with , and now the inequality bites: the two short sides must beat the long one, so
Hence runs from to , giving triangles.
The cases are disjoint, since the repeated side equals in the first and is smaller in the second, so the total is
The answer is .
Review in the AoPS books
- Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
- Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
- Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
- Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.