Let $A B C$ be a triangle with sides $a, b, c$. The altitudes from vertices $A, B$, and $C$ intersect the opposite sides at points $A_{1}, B_{1}, C_{1}$, respectively. What are the lengths of the sides of triangle $A_{1} B_{1} C_{1}$?
From triangle $A A_{1} B$:

$$
A_{1} B=c \cos B
$$

From triangle $B C_{1} C$:

$$
B C_{1}=a \cos B
$$

From triangle $A_{1} C_{1} B$:

$$
{\overline{A_{1} C_{1}}}^{2}={\overline{A_{1} B}}^{2}+{\overline{B C_{1}}}^{2}-2 \overline{A_{1} B} \cdot \overline{B C_{1}} \cdot \cos B
$$

Substituting (1) and (2) into (3):

$$
\begin{gathered}
{\overline{A_{1} C_{1}}}^{2}=c^{2} \cos ^{2} B+a^{2} \cos ^{2} B-2 a c \cos ^{3} B \\
{\overline{A_{1} C_{1}}}^{2}=\cos ^{2} B\left(c^{2}+a^{2}-2 c a \cos B\right)
\end{gathered}
$$

$$
{\overline{A_{1} C_{1}}}^{2}=b^{2} \cos ^{2} B
$$

from which

$$
A_{1} C_{1}=b \cos B
$$

or

$$
A_{1} C_{1}=b \frac{a^{2}+c^{2}-b^{2}}{2 a c}
$$

Similarly, we obtain:

$$
A_{1} B_{1}=c \cos C=c \frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$

and

$$
B_{1} C_{1}=a \cos A=a \frac{b^{2}+c^{2}-a^{2}}{2 b c}
$$

(Hofbauer Ervin, 7th grade, Budapest, Evangelical Gymnasium)

The problem was also solved by: Grünhut Béla, 7th grade, Pécs; Kántor Nándor and Langheim Pál István, 7th grade, Budapest, Evangelical Gymnasium; Kiss Béla, Ludovika-Academy student, Budapest; Visnya Aladár, 8th grade, Pécs; Friedmann Bernát, S.A.-Ujhely; Suschnik József, Kecskemét.