AMC 8 · 2005 · #15

Grade 5 geometry-2d
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
isosceles-triangleperimetersystematic-enumerationlinear-diophantine bound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: isosceles-triangleperimeter
📏 Medium solution 💡 3 insights
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Problem
Count the isosceles triangles whose three side lengths are positive integers and whose perimeter is 23.

Pick an answer.

(A)
2
(B)
4
(C)
6
(D)
9
(E)
11

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

The problem asks "how many," the candidate set is small and finite, and each candidate is easy to test — three classic signs that Tool #2 (Systematic List) is the right primary. Order the candidates by the repeated side a: once a is fixed, the perimeter forces the third side b = 23 - 2a, so there is at most one triangle per value of a. Tool #6 (Guess and Check) then handles each candidate the same way — write the three sides and check the triangle inequality a + a ＞ b. No algebra is needed; the longest case to handle is 11 × 1 = 11.

1STEP 1

Set up the list: an isosceles triangle has sides (a, a, b) with a + a + b = 23, so b = 23 - 2a. List candidates by increasing a.

a + a + b = 23 → b = 23 - 2a
2STEP 2

Bound a: since b = 23 - 2a ≥ 1, we get a ≤ 11, and a ≥ 1, so a runs over 1 to 11 before the triangle test.

b ≥ 1 → 23 - 2a ≥ 1 → a ≤ 11
3STEP 3

Test the triangle inequality a + a ＞ b for each a from 1 to 11: it fails for a ≤ 5 and holds for a ≥ 6.

a & b = 23 - 2a & sides & a+a ＞ b ? ; 1 & 21 & (1,1,21) & 2 ＞ 21? no ; 2 & 19 & (2,2,19) & 4 ＞ 19? no ; 3 & 17 & (3,3,17) & 6 ＞ 17? no ; 4 & 15 & (4,4,15) & 8 ＞ 15? no ; 5 & 13 & (5,5,13) & 10 ＞ 13? no ; 6 & 11 & (6,6,11) & 12 ＞ 11? yes ; 7 & 9 & (7,7,9) & 14 ＞ 9? yes ; 8 & 7 & (8,8,7) & 16 ＞ 7? yes ; 9 & 5 & (9,9,5) & 18 ＞ 5? yes ; 10 & 3 & (10,10,3) & 20 ＞ 3? yes ; 11 & 1 & (11,11,1) & 22 ＞ 1? yes ;
4STEP 4

Count the passing rows a = 6 to 11: six triangles in all.

valid a ∈ {6, 7, 8, 9, 10, 11} → count = 6 → (C)
Answer
6
Spot-check the boundary a = 6: sides (6, 6, 11) give 6 + 6 = 12 ＞ 11, just barely a triangle — and dropping to a = 5 gives (5, 5, 13) with 5 + 5 = 10 ＜ 13, which fails. So the cutoff a ≥ 6 is sharp. On the other end, a = 11 gives (11, 11, 1), a very thin triangle but legal since 11 + 11 ＞ 1. The six valid triangles are (6,6,11), (7,7,9), (8,8,7), (9,9,5), (10,10,3), (11,11,1) — all have perimeter 23 and all satisfy 2a ＞ b, confirming answer (C).
💡Key takeaway

When a problem asks "how many," make a systematic list — once the perimeter pins the third side to the repeated side, the only question per row is whether the two equal sides are long enough to close the triangle.

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