2008 AMC 10B Problem 15

Attempt Problem 15 of the 2008 AMC 10B 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 2008 AMC 10B solutions, or check the answer key.

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

15.

How many right triangles have integer leg lengths aa and bb and a hypotenuse of length b+1,b+1, where b<100?b\lt 100?
66
77
88
99
1010
Answer: A
Concepts:Pythagorean Tripleperfect square
Small hint:
Set a2+b2=(b+1)2a^2+b^2=(b+1)^2 and simplify
Big hint:
This gives a2=2b+1,a^2=2b+1, so a2a^2 is an odd perfect square with a2<201a^2\lt 201
Solution:
From a2+b2=(b+1)2a^2+b^2=(b+1)^2 we get a2=2b+1,a^2=2b+1, so aa is odd and a2a^2 is an odd perfect square.
Since b<100,b\lt 100, we need a2=2b+1<201,a^2=2b+1\lt 201, and a2≥9a^2\ge 9 for b≥4.b\ge 4. The odd squares 9,25,49,81,121,1699,25,49,81,121,169 give a=3,5,7,9,11,13,a=3,5,7,9,11,13, which is 66 triangles.
Thus, the correct answer is A.
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