A triangle's sides are: $a=15 \mathrm{~cm}, b=20 \mathrm{~cm}, c=25 \mathrm{~cm}$. What is the length of the angle bisector $A A_{1}$?
Let $A B C$ be the given triangle. Since

$$
a^{2}+b^{2}=15^{2}+20^{2}=25^{2}=c^{2} \text {, therefore } C \angle=90^{\circ} \text {. }
$$

Thus

$$
A A_{1}=\frac{b}{\cos \frac{A}{2}}=b \sqrt{\frac{2}{1+\cos A}}
$$

But

$$
\cos A=\frac{b}{c}=\frac{4}{5}
$$

and thus

$$
A A_{1}=b \frac{\sqrt{10}}{3}=\frac{20 \sqrt{10}}{3}=21.08 \text{ cm}
$$

(Kornél Miklóssy, Arad.)

The problem was also solved by: L. Bánó, J. Dévai, E. Dévai, F. Eckhart, P. Ehrenstein, M. Fekete, E. Freund, P. Füstös, D. Goldstein, E. Tandlich, Z. Harsányi, P. Heimlich, Gy. Jánosy, J. Kepes, J. Kiss, S. Petrik, S. Pichler, S. Sárközy, J. Szécsi, J. Szőlőssy, B. Tóth, J. Winkler, R. Wottitz.