2007 AMC 12B Problem 23

Question

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

With legs

a\le b,

write

\tfrac12 ab=3\left(a+b+\sqrt{a^2+b^2}\right)

and isolate the radical

Squaring and simplifying yields

(a-12)(b-12)=72;

check the factor-pair candidates in the original equation to remove extraneous ones

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 is

6,

while its perimeter is

12

and three times that is

36.

So exactly

6

triangles work.

Thus, the correct answer is A.

Problem 23 in other years