Playing with Pythagorean Triplets!

It's known that a PPT has hypotenuse of length $m^2 + n^2$ and legs of length $2mn$ and $m^2 - n^2$ , where $m,n \in \mathbb Z$ of different parity with $gcd(m,n) =1$ .

Let $ABC$ be a PPT, and let $r$ be its inradius. Let $s= \frac12 (a+b+c)$ . (s = semiperimeter of $\Delta ABC$ ). On one hand, the area of $\Delta ABC$ ) is $\frac12(2mn)(m^2 - n^2) = mn(m^2 - n^2)$ ; on the other hand, the area is $rs$ .

Thus,

$r = \dfrac{mn(m^2-n^2)}{s} = \dfrac{mn(m^2-n^2)}{\frac12 (2mn + m^2 - n^2 + m^2 + n^2)}$ $= \dfrac{mn(m^2 - n^2)}{mn + m^2} = n(m-n)$

Searching for a pair of PPTs with minimal $r$ , we find that:

(i) If $r=1$ , then $n(m-n) = 1$ , which implies that $n=1, m=2$ . Therefore, we have only one PPT when $r=1$ .

(ii) If $r=2$ , then $n(m-n) = 2$ , which implies that $(m,n) = (3,1) \text{ or } (m,n) = (3,2)$ . Since 1 and 3 have the same parity, we again have only one PPT when $r=2$ .

(iii) If $r=3$ , then $n(m-n) = 3$ , which implies that $(m,n) = (=4,1) \text{ or } (m,n) = (4,3)$ . These lead to the required pair of PPTs.

Therefore the pair of PPTs having congruent incircles with minimal integer radius are the triangles with side lengths $(8,15,17)$ and the triangle with side lengths $(7,24,25)$ .

So $8+7+15+24+17+25=\boxed{96}$ .