We have a compact cube. We want to cut a cone from it such that its axis coincides with one of the cube's space diagonals, and its volume is as large as possible. Express the dimensions of the cone in terms of the edge length of the cube.
Let the endpoints of the body diagonal of the cube extracted along the axis be denoted by $A$ and $B$, its center by $C$, the vertices adjacent to $A$ by $A_{1}$, $A_{2}$, and $A_{3}$, and the reflection of $A_{i}$ about $C$ by $B_{i}$ $(i=1,2,3)$. First, determine the geometric locus of the points of the cube that, when rotated by any angle around the $A B$ axis, result in a point that still belongs to the cube. Let this geometric locus be denoted by $M$, and the frustum of the cone that can be carved out of the cube around the $A B$ axis by any rotation is clearly a part of $M$.

![](https://cdn.mathpix.com/cropped/2024_05_02_77b108928da6552c1479g-1.jpg?height=1082&width=1027&top_left_y=386&top_left_x=533)

Figure 1

First, examine the intersection $M_{0}$ of $M$ with the plane $A B_{1} B A_{1}$. This is clearly part of the rectangle $A B_{1} B A_{1}$ and the rectangle $A B_{1}^{*} B A_{1}^{*}$ obtained by a $180^{\circ}$ rotation of $A B_{1} B A_{1}$ about the $A B$ axis. We will show that the intersection $M_{0}$ is the common part of these two rectangles.

![](https://cdn.mathpix.com/cropped/2024_05_02_77b108928da6552c1479g-1.jpg?height=781&width=848&top_left_y=1721&top_left_x=625)

From this, it follows that $M$ is obtained by rotating this common part around the $A B$ axis.

Let the intersection of $A B_{1}^{*}$ and $A_{1} B$ be denoted by $D$. It suffices to show that $D$ is in $M_{0}$, because if we have already shown this, then the fact that any point $E$ on the segment $C D$ and any point $F$ on the segment $A E$ belongs to $M_{0}$ follows from the convexity of $M_{0}$. The other points of the intersection can be obtained from the points of the triangle $A C D$ by reflection about the $A B$ axis and the center $C$ (both $M_{0}$ and $M$ are clearly symmetric about the $A B$ axis and the center $C$). The convexity of $M_{0}$ follows from the fact that the intersection of convex shapes is also convex, and $M_{0}$ is the intersection of the convex body $M$ formed by the rotation of the convex cube about the $A B$ axis and the convex plane $A B_{1} B A_{1}$. Indeed, if a set of convex shapes in space and the points $P, Q$ are in each of these shapes, then any point on the segment $P Q$ is also in each of these shapes (since the shapes are convex), so any point on the segment $P Q$ is also in the intersection of these shapes.

Let us see why $D$ is in $M$. $D$ is always in the plane $S$ that is the perpendicular bisector of the segment $A B$, so it is the same as the point on the line $A_{1} B$ in the intersection of the plane $S$ and the cube. It is well known that this intersection is a regular hexagon, whose vertices are the midpoints of the segments $A_{1} B_{2}, B_{2} A_{3}, A_{3} B_{1}, B_{1} A_{2}, A_{2} B_{3}, B_{3} A_{1}$, and $D$ is the midpoint of one of the sides of this hexagon. Therefore, when $D$ is rotated about the $A B$ axis, we get the circle inscribed in the hexagon, and every point of this circle is in the hexagon, and thus in the cube.

Returning to our original goal, let us now examine the maximal volume cone that can be inscribed in the double cone $M$ formed by the rotation of the triangle $A B D$ about the $A B$ axis. The center of the base circle of the cone we are looking for is on the segment $A B$, denoted by $O$. Without loss of generality, we can assume that $O$ is on the $B C$ part of the segment $A B$. For a fixed $O$, the base circle of the cone is already determined, so among the cones with the same $O$, the one with the maximum volume is the one with the maximum height, i.e., the one whose apex is farthest from $O$. Since this apex is also on $A B$, it can only be $A$ (if $O$ is the same as $C$, $B$ could also be, but this does not represent a significantly new possibility). Let the edge of the cube be denoted by $a$, the height of the cone by $m$, and the radius of its base circle by $r$. Then $A B=a \sqrt{3}$, and from similar triangles we get

$$
r:(a \sqrt{3}-m)=A A_{1}: A_{1} B, \quad r=\frac{a \sqrt{3}-m}{\sqrt{2}}
$$

so the volume of the cone is

$$
V=\frac{m r^{2} \pi}{3}=\frac{\pi}{6} m(a \sqrt{3}-m)^{2}
$$

where $\frac{a \sqrt{3}}{2} \leq m<a \sqrt{3}$. Now, the derivative of $V$ with respect to $m$ is

$$
\frac{d V}{d m}=\frac{\pi}{6}\left[(a \sqrt{3}-m)^{2}-2 m(a \sqrt{3}-m)\right]=\frac{\pi}{6}(a \sqrt{3}-m)(a \sqrt{3}-2 m)
$$

which is negative on the given interval, so $V$ decreases monotonically as $m$ increases, and $V$ is maximal when $m$ is minimal, i.e., $m=\frac{a \sqrt{3}}{2}$, or $O$ is the same as $C$. Then $r=\frac{a \sqrt{3}}{2 \sqrt{2}}$, and $V=\frac{\pi}{6} m^{3}=\frac{\pi \sqrt{3}}{16} a^{3}=0.34 a^{3}$, which seems like a significant waste, but if we consider that we can carve out two such cones from the cube, this waste becomes understandable.