$X$ -ed Community, But Not Circles!

The area of a triangle $A = rs$ and $A = \sqrt{s(s - a)(s - b)(s - c)}$ , and in a right triangle, the inradius is $r = \frac{1}{2}(a + b - c) = s - c$ .

The exradii have equations $r_a = \cfrac{A}{s - a}$ , $r_b = \cfrac{A}{s - b}$ , and $r_c = \cfrac{A}{s - c} = \cfrac{A}{r} = s$ .

By substitution, $A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{s \cdot \cfrac{A}{r_a} \cdot \cfrac{A}{r_b} \cdot r} = \sqrt{\cfrac{A^3}{r_a r_b}}$ , which solves to $A = r_a r_b$ .

Substituting $A = r_a r_b$ into $r_a = \cfrac{A}{s - a}$ solves to $r_b = s - a$ , and similarly, $r_b = \cfrac{A}{s - b}$ solves to $r_a = s - b$ .

Adding $r_b = s - a$ and $r_a = s - b$ gives $r_a + r_b = 2s - a - b = c$ , so $c = r_a + r_b$ .

Substituting $r = s - c$ into $A = rs$ gives $A = (s - c)s$ , which solves to $s = \frac{1}{2}(c + \sqrt{c^2 + 4A})$ .

And substituting $A = r_a r_b$ and $c = r_a + r_b$ into $r_c = s = \frac{1}{2}(c + \sqrt{c^2 + 4A})$ gives $r_c = \frac{1}{2}(r_a + r_b + \sqrt{(r_a + r_b)^2 + 4r_a r_b})$ .

Therefore, if $r_a = 5$ and $r_b = 12$ , then $r_c = \frac{1}{2}(5 + 12 + \sqrt{(5 + 12)^2 + 4 \cdot 5 \cdot 12}) = \boxed{20}$ .