Draw two squares in a right-angled triangle. One such that one of its sides lies on the hypotenuse of the triangle, and the other such that two of its sides lie on the legs of the triangle. Which square is larger?
Let the legs of the right triangle be $a$ and $b$, the hypotenuse be $c$, the height be $m$, the side of the first square be $x_{1}$, and the side of the second square be $x_{2}$. Then, from similar triangles:

$$
x_{1}: m-x_{1}=c: m
$$

from which

$$
x_{1}=\frac{m c}{m+c}=\frac{a b}{m+c}
$$

furthermore

$$
x_{2}: a-x_{2}=b: a
$$

from which

$$
x_{2}=\frac{a b}{a+b}
$$

But (247. ex.) $m+c>a+b$ and thus

$$
x_{1}<x_{2}
$$

Therefore, the square is larger whose two sides lie on the legs of the triangle.

(József Kiss, Pápa.)

The problem was also solved by: A. Fuchs, P. Füstös, Z. Harsányi, P. Heimlich, Gy. Jánosy, M. Martini, Gy. Meleghy, S. Pichler, E. Sárközy, Gy. Schuster, E. Tandlich, B. Tóth.