A cube has two parallel faces $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$. Mark the point that divides the edges $A B, B C, C D, D A ; A_{1} D_{1}, D_{1} C_{1}, C_{1} B_{1}$, $B_{1} A_{1} ; A A_{1}, C C_{1} ; B_{1} B, D_{1} D$ in the ratio $\lambda$. (The $\lambda$-ratio point $Z$ on the edge $X Y$ is the internal point such that $X Z: Z Y=\lambda$.) Cut off each vertex of the cube with the plane passing through the 3 division points on the edges emanating from that vertex. For which value of $\lambda$ will the volume of the remaining solid be the smallest?
We can work more simply if we phrase the question like this: "At which value of $\lambda$ is the combined volume of the cut-off parts of the cube the largest?" - furthermore, if we do not relate the two parts of the edges to each other, but each to the original edge of the cube. Thus, for the point $Z$ dividing the edge $XY$ in the ratio $\lambda$,

$$
X Z=p=\frac{\lambda}{\lambda+1} \cdot X Y, \quad Z Y=q=\frac{1}{\lambda+1} \cdot X Y \quad \text { and } \quad p+q=a
$$

where $a$ denotes the length of the edges of the cube.

The cut-off body in each of the 8 cube vertices is a triangular pyramid, and its volume is one-sixth of the product of the edges (of length $p$ or $q$) converging at the vertex common with the cube. In the first 8 listed edges, each of the 8 cube vertices once takes the role of $X$ and once that of $Y$, so far, therefore, each of the 8 vertices is a common point of one $p$ and one $q$ length edge segment. Thus, in the starting points of the last 4 edges: at $A$, $B_{1}$, $C$, and $D_{1}$, two $p$ and one $q$ length edge segments meet, while in the other 4 vertices, one $p$ and two $q$ length edge segments meet. According to this, the combined volume of the cut-off pyramids is

![](https://cdn.mathpix.com/cropped/2024_05_02_1fb6a29cee1540d74393g-1.jpg?height=781&width=523&top_left_y=688&top_left_x=779)

and it is obvious that we only need to find the value of $\lambda$ for which the product of the $p$ and $q$ segments, whose sum is the constant $a$, is the largest. The quadratic function $p q=p(a-p)$ reaches its maximum at $p=\frac{a}{2}$, in which case $p=q$, and according to (1), the sought ratio $\lambda=1$, that is, we take the midpoint as the division point on every edge of the cube. - With this, we have solved the problem as outlined. (The volume of the remaining body is $\frac{5}{6}$ of the cube's volume.)

Remark. In the case of $\lambda=1$, the remaining body is bounded by 6 squares and 8 equilateral triangles. If we reflect the vertices of the triangles onto the midpoint of the opposite side, we get the vertices of a regular octahedron. (2nd figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_1fb6a29cee1540d74393g-1.jpg?height=528&width=507&top_left_y=1785&top_left_x=793)

2nd figure