If we increase the sides of an isosceles right triangle by $4 \mathrm{~cm}$, then the area of the triangle increases by $112 \mathrm{~cm}^{2}$. What are the sides of the original triangle?
Let $x$ be the leg of the right triangle, then according to the problem

$$
\frac{(x+4)^{2}}{2}=\frac{x^{2}}{2}+112
$$

from which

$$
x=26
$$

The leg of the triangle is $26 \mathrm{~cm}$, so the hypotenuse is $26 \sqrt{2} \mathrm{~cm}$.

(László Szántó, Pécs.)

Number of solutions: 30.