2007 AMC 12B Problem 23

Below is the professionally curated solution for Problem 23 of the 2007 AMC 12B, 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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Concepts:right triangle Diophantine Equation Simon’s Favorite Factoring Trick

Difficulty rating: 2140

23.

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?

$6$

$7$

$8$

$10$

$12$

Solution:

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

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

The pair $(3,4)$ is extraneous (its area $6$ does not equal $3$ times its perimeter $12$ ), so exactly $6$ triangles work.

Thus, the correct answer is A.

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