2005 AMC 8 Problem 15

Attempt Problem 15 of the 2005 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

How many different isosceles triangles have integer side lengths and perimeter 23?23?
22
44
66
99
1111
Answer: C
Concepts:triangle inequalityparitycasework
Small hint:
Let the equal side length be aa and the base be bb.
Big hint:
Use 2a+b=232a+b=23 and the triangle inequality.
Solution:
Let the equal side length be aa and the base be bb. Then 2a+b=232a+b=23, so bb must be odd.
The triangle inequality requires b<2ab<2a. Since b=23−2ab=23-2a, this gives 23−2a<2a23-2a<2a, so a>234a>\frac{23}{4}. Thus a≥6a\ge6.
Also b≥1b\ge1, so 2a≤222a\le22 and a≤11a\le11. The possible values a=6,a=6, 7,7, 8,8, 9,9, 10,10, and 1111 all work, giving 66 triangles.
Thus, C is the correct answer.
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