## Task 24/63

Four non-adjacent corners of a cuboid $V=abc$ are beveled by planar cuts such that a segment $x$ is removed from each edge. The remaining corners are treated similarly, with the cuts being made such that at each corner, exactly the remaining edge segments that were left after the first beveling are removed.

How large must the segment $x$ be chosen so that the resulting fourteen-faced polyhedron, bounded by eight triangles and six parallelograms, has an extreme volume?
When the first truncation occurs, four pyramids with the volume $\frac{1}{6} x^{3}$ are removed, followed by four more (also triangular) pyramidal volumes of $\frac{1}{6}(a-x)(b-x)(c-x)$.

What remains is (as shown in the figure)

$$
V(x)=a b c-\frac{2}{3} x^{3}-\frac{2}{3}(a-x)(b-x)(c-x)=\frac{1}{3} a b c+\frac{2}{3}(a b+b c+a c) x-\frac{2}{3}(a+b+c) x^{2}
$$

From this, we get

$$
V^{\prime}(x)=\frac{2}{3}(a b+b c+a c)-\frac{4}{3}(a+b+c) x ; \quad V^{\prime \prime}(x)=-\frac{4}{3}(a+b+c)
$$

If $b>c$, then in the extreme case, $x_{e} \leq c$ must hold. This leads to

$$
c^{2}+\frac{a+b}{2} x-\frac{a b}{2} \geq 0
$$

and finally, by solving the quadratic inequality (negative root values are not considered),

$$
c \geq \frac{1}{4}\left[\sqrt{(a+b)^{2}+8 a b}-(a+b)\right]
$$

b) For the cube $a^{3}$, we get $x_{e}=\frac{a}{2}, V_{\max }=\frac{5}{6} a^{3}$.

The truncated body is the solid bounded by eight equilateral triangles and six squares, one of the 15 semi-regular Archimedean solids.