A right-angled isosceles hexahedron (a rectangular prism) has faces that meet at a vertex in the ratio of $16: 21: 28$. What are the lengths of the edges if the space diagonal is $29 \mathrm{~cm}$?
Let the edges meeting at one vertex be denoted by $a, b, c$. For these 3 unknowns, our task provides the following 3 equations:

\[
\begin{gathered}
\frac{a b}{b c}=\frac{16}{21} \\
\frac{a b}{a c}=\frac{16}{28}=\frac{4}{7}
\end{gathered}
\]

and

\[
a^{2}+b^{2}+c^{2}=29^{2}
\]

From (1), $a=\frac{16 c}{21}, \quad$ from (2), $b=\frac{4 c}{7}$. Substituting these values into (3),

\[
\frac{256 c^{2}}{441}+\frac{16 c^{2}}{49}+c^{2}=29^{2}
\]

which simplifies to

\[
256 c^{2}+144 c^{2}+441 c^{2}=29^{2} \cdot 21^{2}
\]

or

\[
841 c^{2}=29^{2} \cdot 21^{2}
\]

from which

\[
29 c=29 \cdot 21
\]

and thus

\[
c=21 \text{ cm}, \quad a=\frac{16 \cdot 21}{21}=16 \text{ cm} \text{ and } b=\frac{4 \cdot 21}{7}=12 \text{ cm}
\]

Pék István (Nagykőrös, State Vocational School Boarding School)