A regular quadrilateral truncated pyramid has a height and a top edge that are equal. The volume of the truncated pyramid is twice the volume of the largest cube that can fit inside it. What angle do the side faces form with the base plane?
Our intuition suggests that the largest cube that can be inscribed in a frustum is the one whose one face coincides with the frustum's top face. In this case, since the height of the frustum is \(a\), we indeed get a cube inscribed in the frustum. A cube with a larger volume cannot be inscribed in the frustum. See the 2nd remark.

To calculate the angle between the lateral face and the base, let's take a plane that passes through the center of the top face, is perpendicular to the base, and is parallel to one edge of the square face. This plane cuts out an isosceles trapezoid \(ABCD\) from the frustum, with parallel sides \(a\) and \(b\), height \(a\), and the angle between the lateral edge and the base edge, \(\alpha\), is the same as the angle between the two planes we are looking for. From this:

\[
\operatorname{tg} \alpha = \frac{a}{\frac{(b-a)}{2}} = \frac{2a}{b-a}
\]

The volume of the cube is \(a^3\); the volume of the frustum is \(\frac{a}{3}(a^2 + b^2 + ab)\), and according to the condition:

\[
2a^3 = \frac{a}{3}(a^2 + b^2 + ab)
\]

Rearranging the equation, we get the quadratic equation:

\[
-5a^2 + ab + b^2 = 0
\]

Solving for \(a\):

\[
a = \frac{-b \pm \sqrt{b^2 + 20b^2}}{-10} = \frac{1 + \sqrt{21}}{10} b
\]

(We only consider the positive value, since \(a > 0\).)

Substitute the value of \(a\) obtained into (1):

\[
\operatorname{tg} \alpha = \frac{\frac{1 + \sqrt{21}}{5} b}{b - \frac{1 + \sqrt{21}}{10} b} = \frac{2 + 2\sqrt{21}}{9 - \sqrt{21}} \approx 2.5275
\]

From which \(\alpha \approx 68.4^\circ\).

Remarks. 1. Most solvers naturally assumed that the edge length of the cube can be at most as large as the edge length of the frustum's top face, and only performed the calculations. These solutions received 3 points.

2. We will show that a cube with an edge length greater than \(a\) cannot be inscribed in the frustum. Suppose we inscribed a cube with edge length \(b\) into the frustum. If it does not touch the base, move it in the direction of the base, perpendicular to it, until it has a common point with the base. During this movement, it remains inside the frustum because any point inside it remains inside with such a movement. (We only drew the base and top planes of the frustum.)

The face of the cube obtained now, or one of its edges, or one of its vertices, will be on the base; one of its vertices will definitely be on the base, let this be \(A\), and the adjacent vertices \(B_1, B_2, B_3\), and their common neighbors \(C_1, C_2, C_3\) as shown in the third figure. Finally, let the eighth vertex of the cube be \(D\). Project the cube onto the base plane. The contour of the cube's projection will be \(B_1', C_3', B_2', C_1', B_3', C_2'\). Therefore, \(D'\) will fall into this polygon. Since this polygon is the union of three (possibly degenerate) quadrilaterals (which are the projections of the faces of the cube containing \(A\)), \(D'\) will fall into one of them, for example, the quadrilateral \(A, B_1', C_3', B_2'\). Thus, the distance \(\left|A D'\right|\) is either no greater than the distance \(\left|A C_3'\right|\), which cannot be greater than \(\left|A C_3\right| = b \cdot \sqrt{2}\), or no greater than the distances \(\left|A B_1'\right|\) and \(\left|A B_2'\right|\), which cannot be greater than \(\left|A B_1\right| = \left|A B_2\right| = b\). Therefore, \(\left|A D'\right| \leq b \cdot \sqrt{2}\) always holds. We also know that \(|A D| = b \cdot \sqrt{3}\). From this, by the Pythagorean theorem, it follows that \(\left|D D'\right|^2 > 3 \cdot b^2 - 2 \cdot b^2 = b^2\), since the angle at \(D'\) in the triangle \(D D' A\) is a right angle. Thus, \(\left|D D'\right| \geq b\), since we are dealing with positive numbers. Considering that the segment \(D D'\) is perpendicular to the base (and top) plane, and both points lie between these two planes (or on them), the distance between them is at most \(a\). Therefore, \(a \geq \left|D D'\right| \geq b\).
![](https://cdn.mathpix.com/cropped/2024_05_02_5bd7841a704be2a077cag-2.jpg?height=1460&width=816&top_left_y=115&top_left_x=644)