2007 AMC 12B Problem 23

Attempt Problem 23 of the 2007 AMC 12B 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 2007 AMC 12B solutions, or check the answer key.

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23.

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to 33 times their perimeters?

66

77

88

1010

1212

Answer: A
Concepts:right triangleDiophantine EquationSimon’s Favorite Factoring Trick
Difficulty rating: 2140
Solution:

Let the legs be ab.a\le b. The condition is 12ab=3(a+b+a2+b2),\tfrac12 ab=3\left(a+b+\sqrt{a^2+b^2}\right), so ab6a6b=6a2+b2. ab-6a-6b=6\sqrt{a^2+b^2}.

Squaring and simplifying gives ab(ab12a12b+72)=0,ab(ab-12a-12b+72)=0, hence (a12)(b12)=72.(a-12)(b-12)=72. The positive integer solutions are (a,b)=(3,4),(a,b)=(3,4), (13,84),(13,84), (14,48),(14,48), (15,36),(15,36), (16,30),(16,30), (18,24),(18,24), (20,21).(20,21).

The pair (3,4)(3,4) is extraneous: its area is 6,6, while its perimeter is 1212 and three times that is 36.36. So exactly 66 triangles work.

Thus, the correct answer is A.

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