Problem 15 · 2005 AMC 8
Medium
Counting & Probability
triangle-inequalityisosceles
How many different isosceles triangles have integer side lengths and perimeter 23?
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Answer: C — 6 triangles.
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Hint 1 of 2
An isosceles triangle is pinned down by just one number — the leg length y — because the base is then forced to be 23 − 2y. So you're really counting valid leg lengths, not triples.
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Hint 2 of 2
Two fences bound y: the base must be positive (so y isn't too big) and the triangle inequality 2·leg > base must hold (so y isn't too small). Count the integers between.
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Approach: one variable (the leg), bounded both ways
- Let the two legs be y and the base x, so 2y + x = 23. The base is whatever's left: x = 23 − 2y — one number controls everything.
- Lower fence (triangle inequality): the two legs must beat the base, 2y > x = 23 − 2y, giving y > 5.75, so y ≥ 6.
- Upper fence (base positive): x = 23 − 2y ≥ 1 gives y ≤ 11.
- So y = 6, 7, 8, 9, 10, 11 — 6 triangles.
- Why this transfers: reduce a 'count the shapes' problem to one free variable, then squeeze it between two inequalities and count integers. Here, conveniently, the odd perimeter 23 makes the base 23 − 2y always odd, so it can never tie a leg — no equilateral or degenerate cases to worry about.
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