Solve the right triangle if the hypotenuse (a) and the angle bisector of the right angle ( $l$ ) are given.
First, let's solve problem 634. Let the angle formed by the angle bisector $AD$ and side $a$ be $\delta$, or $180^{\circ}-\delta$; further, let $BD=x$ and $DC=a-x$. From triangles $ABD$ and $ADC$:

$$
x: l=\sin \frac{\alpha}{2}: \sin \left(\delta+\frac{\alpha}{2}\right)
$$

and

$$
(a-x): l=\sin \frac{\alpha}{2}: \sin \left(\delta-\frac{\alpha}{2}\right)
$$

Adding these two equations:

$$
\begin{gathered}
\frac{a}{l}=\sin \frac{\alpha}{2} \cdot \frac{\sin \left(\delta-\frac{\alpha}{2}\right)+\sin \left(\delta+\frac{\alpha}{2}\right)}{\sin \left(\delta-\frac{\alpha}{2}\right) \cdot \sin \left(\delta+\frac{\alpha}{2}\right)} \\
\frac{a}{l}=\sin \frac{\alpha}{2} \cdot \frac{4 \sin \delta \cos \frac{\alpha}{2}}{2 \sin ^{2} \delta-1+\cos \alpha} \\
\frac{a}{l}=\frac{2 \sin \alpha \sin \delta}{2 \sin ^{2} \delta-1+\cos \alpha}
\end{gathered}
$$

Rearranging the equation, we get:

$$
\sin ^{2} \delta-\frac{l \sin \alpha}{\alpha} \sin \delta+\frac{1}{2}(\cos \alpha-1)=0
$$

From this equation:

$$
\sin \delta=\frac{l \sin \alpha \pm \sqrt{(1+\cos \alpha)\left[l^{2}\left(1+\cos \alpha+2 a^{2}\right]\right.}}{2 a}
$$

The components of the triangle are therefore:

$$
\begin{gathered}
\beta=\delta-\frac{\alpha}{2}, \gamma=180^{\circ}-\left(\delta+\frac{\alpha}{2}\right) \\
b=\frac{a \sin \beta}{\sin \alpha}, c=\frac{a \sin \gamma}{\sin \alpha}, \quad t=\frac{a l}{2} \sin \delta
\end{gathered}
$$

For $\alpha=90^{\circ}$, then

$$
\sin \delta=\frac{1}{2 a}\left(l \pm \sqrt{l^{2}+2 a^{2}}\right)
$$

In (1) and (2), the radical must be taken with a positive sign, because $\delta<180^{\circ}$, and thus $\sin \delta$ cannot be negative.

(Jenő Filkorn, Nyitra.)

Problem 568 was also solved by: M. Dolowschiák, B. Eisenberg, E. Freibauer, B. Krausz, Gy. Krisztián, A. Kürth, Gy. Lukhaub, E. Miletits, G. Petrogalli, Gy. Perl, P. Rozlosnik, G. Scholtz, Ö. Spitzer, J. Tinyó, Á. Weisz, J. Weisz.

Problem 634 was also solved by: M. Dolowschiák, F. Faith, E. Freibauer, Ö. Kornis, B. Krausz, Gy. Krisztián, Gy. Lukhaub, R. Obláth, F. Pálfy, Gy. Perl, G. Sasvári, J. Weisz.