PRMO 2015 question 10

10. What is the greatest possible perimeter of a right-angled triangle with integer side lengths if one of the sides has length 12?

Solution:
If we have to maximize the perimeter, we should not keep 12 as the hypotenuse. That way hypotenuse will be longer than 12 and the perimeter will be more as compared to keeping 12 as hypotenuse.

Let hypotenuse be y and the other leg be x.
x^2 + 12^2 = y^2
y^2 - x^2 = 144
(y + x)(y - x) = 144

Since x,y are both integers the problem now reduces to finding integer factors of 144.
Let's try them one by one.
y + x = 144
y - x = 1
=> 2y = 145 => y  = 72.5, x = 71.5, not integer move on.

y + x = 72
y - x = 2
=> 2y = 74 => y = 37, x = 35 valid solution

y + x = 48
y - x = 3
=> 2y = 51 => y = 25.5, x = 22.5, not integer move on.

Should we try more factors now?
Not really.
Even if x,y solve for integer values, their values are constantly decreasing as the factors are becoming smaller. So y = 37, x = 35 is the right answer.
Perimeter = 37 + 35 + 12 = 84 Answer.

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