2011 AMC 10B Problem 22

Height of the pyramid. Every edge has length $1$. The apex sits above the center of the base, and the distance from the center to a base vertex is half the diagonal, $\frac{\sqrt{2}}{2}$. So the height is
$$h = \sqrt{1^2 - \left(\tfrac{\sqrt{2}}{2}\right)^2} = \sqrt{\tfrac{1}{2}} = \tfrac{\sqrt{2}}{2}.$$

Cross-section at the cube's top. Let the cube have side $s$. Slice the pyramid horizontally at height $s$. The slice is a square similar to the base, scaled by $\frac{h - s}{h}$, so its side is
$$1 \cdot \frac{h - s}{h} = 1 - \frac{s}{h} = 1 - s\sqrt{2}.$$

Matching the cube. The top face of the cube has all four edges on the lateral faces, so the top face is this cross-sectional square (its four sides lie on the four slanted faces). Hence
$$s = 1 - s\sqrt{2} \quad\Longrightarrow\quad s = \frac{1}{1 + \sqrt{2}} = \sqrt{2} - 1.$$

Volume. $(\sqrt{2} - 1)^2 = 3 - 2\sqrt{2}$, so
$$s^3 = (3 - 2\sqrt{2})(\sqrt{2} - 1) = 3\sqrt{2} - 3 - 4 + 2\sqrt{2} = 5\sqrt{2} - 7.$$

The answer is $\boxed{\textbf{(A)}\ 5\sqrt{2} - 7}$.

A solid inscribed in a pyramid is controlled by one horizontal cross-section: the section at height $s$ is a scaled copy of the base with factor $(h - s)/h$, a similar-triangles fact seen in the vertical slice through the apex and the midpoints of two opposite base edges. Once the height is known, "top face touches all lateral faces" becomes a single linear equation in $s$.

Work in the vertical slice through the apex and the midpoints of two opposite base edges. That slice is an isosceles triangle with base $1$ and height $\frac{\sqrt{2}}{2}$; the cube appears as an $s \times s$ square standing on the base with its top corners on the slanted sides. The little triangle above the square is similar to the whole: $\frac{s}{\frac{\sqrt{2}}{2} - s} = \frac{1}{\frac{\sqrt{2}}{2}}$, giving $s = \sqrt{2} - 1$ again.

• Using height $1$ or $\frac{\sqrt{3}}{2}$ for the pyramid (the slant height) instead of computing it from the apex to the center of the base, which is $\frac{\sqrt{2}}{2}$.
• Reporting the side $\sqrt{2} - 1$ or its square $3 - 2\sqrt{2}$ rather than cubing to get the volume.
• Slicing through the apex and a base diagonal and then matching the cube's top face to that slice; the cube's faces are parallel to the base edges, so the relevant slice is through the edge midpoints.