**Problem**: (2007 AMC12B - #23) How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $ 3 $ times their perimeters?

**Solution**: Well, basically, you should know the Pythagorean triple generating formula, i.e. $ a = k(u^2-v^2) $, $ b = 2kuv $, $ c = k(u^2+v^2) $. Substitute accordingly and we have to solve the diophantine equation

$ \frac{1}{2} \cdot k(u^2-v^2) \cdot 2kuv = 3 \cdot (k(u^2-v^2)+2kuv+k(u^2+v^2)) $

which conveniently simplifies to

$ kv(u-v) = 6 $.

Obviously then $ k = 1, 2, 3, 6 $ so look at these cases:

$ k = 1 $: We can take $ v = 1, 2, 3, 6 $ to get the triples $ (14, 48, 50); (20, 21, 29); (16, 30, 34); (13, 84, 85) $.

$ k = 2 $: We can take $ v = 1, 3 $ to get the triples $ (16, 30, 34); (14, 48, 50) $ both of which are already counted.

$ k = 3 $: We can take $ v = 1, 2 $ to get the triples $ (18, 24, 30); (15, 24, 39) $.

$ k = 6 $: We can take $ v = 1 $ to get the triple $ (18, 24, 30) $, which is already counted.

So we have $ 4+2 = 6 $ triangles. QED.

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Comment: Not too hard if you knew the generating formula for Pythagorean triples. It was a little annoying having to check for repeated triples, but at least there weren't that many.

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Practice Problem: (2007 AMC 12B - #24) How many pairs of positive integers $ (a, b) $ are there such that $ \text{gcd}(a, b) = 1 $ and

$ \frac{a}{b}+\frac{14b}{9a} $

is an integer?

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=135187

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