In a right angled triangle with integer sides, the radius of the inscribed circle is 12. Compute the largest possible length of the hypotenuse.

For a right triangle with legs $$a, b$$ and hypotenuse $$c$$ the inradius is $$r = \frac{a+b-c}{2}$$, so $$a+b-c = 24$$. Substituting $$c = a+b-24$$ into $$a^2+b^2 = c^2$$ gives $$(a-24)(b-24) = 288$$, and then $$c = 24 + d + \frac{288}{d}$$ where $$d = a - 24$$. This is largest when $$d = 1$$, giving $$a = 25$$, $$b = 312$$ and $$c = 313$$.