Determine the sides of a right triangle if the difference between the legs is $\alpha$, and the difference between the hypotenuse and the altitude to it is $\beta$. For which values of $\alpha$ and $\beta$ is the problem possible.
Let the larger leg of the right triangle be $x$, the smaller leg $y$, the hypotenuse $z$, and the height to the hypotenuse $h$.

According to the problem,

$$
x-y=\alpha
$$

$$
z-h=\beta
$$

Furthermore,

$$
\begin{gathered}
x^{2}+y^{2}=z^{2} \\
x y=h z
\end{gathered}
$$

From equations (1) and (3), it follows that:

$$
2 x y=z^{2}-\alpha^{2}
$$

From equations (2) and (4):

$$
x y=z(z-\beta)
$$

Finally, from equations (5) and (6):

$$
2 z(z-\beta)=z^{2}-\alpha^{2}
$$

or, rearranging in descending powers of $z$:

$$
z^{2}-2 \beta z+\alpha^{2}=0
$$

The roots of this equation are:

$$
z=\beta \pm \sqrt{\beta^{2}-\alpha^{2}}
$$

Thus, $x$ and $-y$, as it follows from equations (1) and (4), are the roots of the following quadratic equation:

$$
\lambda^{2}-x \lambda-h z=0
$$

where

$$
h=z-\beta
$$

For $z$ to satisfy the problem from the values obtained from equation (8), it must be real, positive, and, as it follows from (10), greater than $\beta$. The condition for reality is

$$
\beta>\alpha
$$

and in this case, the larger root of equation (8) satisfies the problem. At the same time, the roots of equation (9) are also suitable, as the constant term being negative ensures that the roots are real and of opposite signs.

(Friedmann Bernát, 7th grade gymnasium, S.-A.-Ujhely.)

The problem was also solved by: Grünhut Béla, Visnya Aladár, and Weisz Lipót, from Pécsett; Mayer Miksa, from Budapest, and Szabó István, from Debrecen.