\section*{Problem 6 - 081046}

The figure shows a cube \(W = A B C D E F G H\) with edge length \(a\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1269.jpg?height=583&width=580&top_left_y=388&top_left_x=744)

In the side faces \(A B C D, A B F E, A D H E, B C G F, D C G H, E F G H\) of \(W\), there are edge-parallel squares

\[
A_{1} B_{1} C_{1} D_{1}, A_{2} B_{2} F_{2} E_{2}, A_{3} D_{3} H_{3} E_{3}, B_{4} C_{4} G_{4} F_{4}, D_{5} C_{5} G_{5} H_{5}, E_{6} F_{6} G_{6} H_{6}
\]

with edge length \(x < a\) and with the midpoints \(M_{1}, \ldots, M_{6}\), such that the three lines \(g_{M_{1} M_{6}}, g_{M_{2} M_{5}}, g_{M_{3} M_{4}}\) are edge-parallel and intersect at the same point. From \(W\), the three cuboids

\[
A_{1} B_{1} C_{1} D_{1} E_{6} F_{6} G_{6} H_{6}, A_{2} B_{2} F_{2} E_{2} D_{5} C_{5} G_{5} H_{5}, A_{3} D_{3} H_{3} E_{3} B_{4} C_{4} G_{4} F_{4}
\]

are cut out. For what value of \(x\) does the resulting residual body have half the volume of the original cube?
}

Each of the three cut-out cuboids has a volume of \(x^{2} \cdot a\). However, they overlap in a cube of edge length \(x\) (with the center as the intersection point of the three lines \(g_{M_{1} M_{6}}, g_{M_{2} M_{5}}\) and \(g_{M_{3} M_{4}}\)), so that this part is only removed by one cuboid, but not again by the other two.

Thus, the remaining body has a volume of \(a^{3}-3 a x^{2}+2 x^{3}\). If this is to equal half the volume of the original cube, then

\[
\frac{1}{2} a^{3}=a^{3}-3 a x^{2}+2 x^{3} \quad \text { or } \quad 4 x^{3}-6 a x^{2}+a^{3}=0
\]

must hold. Clearly, \(x=\frac{a}{2}\) satisfies this equation, since

\[
4 \cdot\left(\frac{a}{2}\right)^{3}-6 \cdot a \cdot\left(\frac{a}{2}\right)^{2}+a^{3}=\left(\frac{4}{8}-\frac{6}{4}+1\right) \cdot a^{3}=0
\]

is true. Therefore, \(4 x^{3}-6 a x^{2}+a^{3}\) can be represented as \(4 x^{3}-6 a x^{2}+a^{3}=(2 x-a) \cdot\left(2 x^{2}-2 a x-a^{2}\right)\). If \(x \neq \frac{a}{2}\), then the second factor \(2 x^{2}-2 a x-a^{2}\) must be zero. The solutions to the resulting equation in \(x\) are

\[
\frac{a}{2} \pm \sqrt{\frac{a^{2}}{4}+\frac{a^{2}}{2}}=\frac{1 \pm \sqrt{3}}{2} \cdot a
\]

However, one solution \(\frac{1-\sqrt{3}}{2} \cdot a\) is negative (for \(a>0\)) and the other \(\frac{1+\sqrt{3}}{2} \cdot a\) is greater than \(a\) and thus also not valid.

The only possible choice for \(x\) in the interval \([0, a]\) is \(x=\frac{1}{2} a\) as the only solution.