2011 AMC 10B Problem 22

The whole problem comes down to one flat picture, so first find how tall the pyramid is, then slice it.

Step A: the height of the pyramid.

1. Every edge of the pyramid is $1$ (the base sides are $1$ and the side faces are equilateral).
2. The tip sits straight above the center of the base. The center is half a diagonal away from each base corner. The diagonal of a $1 \times 1$ square is $\sqrt{2}$, so that distance is $\frac{\sqrt{2}}{2}$.
3. The tip, the center and a corner make a right triangle with long side $1$ (a slanted edge). So the height $h$ satisfies
$$h^2 + \left(\tfrac{\sqrt{2}}{2}\right)^2 = 1^2, \quad h^2 = 1 - \tfrac{1}{2} = \tfrac{1}{2}, \quad h = \tfrac{\sqrt{2}}{2}.$$

Step B: slice the pyramid at the cube's top.

4. Let the cube have side $s$. Its top face is at height $s$, and its four top edges touch the four slanted faces. So the top face fills the horizontal slice of the pyramid at height $s$ exactly.
5. A horizontal slice of the pyramid is a smaller square. At height $0$ its side is $1$; at the tip (height $h$) its side is $0$; in between it shrinks evenly. At height $s$ the side is
$$1 \times \frac{h - s}{h} = 1 - \frac{s}{h} = 1 - s\sqrt{2}.$$
(Here $\frac{1}{h} = \frac{2}{\sqrt{2}} = \sqrt{2}$.)
6. That slice is the cube's top face, so its side is $s$:
$$s = 1 - s\sqrt{2}, \quad s(1 + \sqrt{2}) = 1, \quad s = \frac{1}{1 + \sqrt{2}}.$$
7. Multiply top and bottom by $\sqrt{2} - 1$, using $(\sqrt{2} + 1)(\sqrt{2} - 1) = 1$: $s = \sqrt{2} - 1 \approx 0.414$.

Step C: the volume.

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

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

A pyramid gets smaller evenly as you go up, so a slice at any height is a scaled-down copy of the base. "The cube's top edges touch the side faces" just says the cube's top face is one of those slices. Once you know the pyramid's true height, that gives a single simple equation for $s$.

Cut the pyramid straight down through its tip, parallel to one pair of base edges. The cut shows a triangle with base $1$ and height $\frac{\sqrt{2}}{2}$, and the cube shows up as an $s$ by $s$ square standing on the base, with its top corners on the two slanted sides. The small triangle above the square has the same shape as the big one, so its base divided by its height matches: $\frac{s}{\frac{\sqrt{2}}{2} - s} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$. Multiplying out, $s = \sqrt{2}\left(\frac{\sqrt{2}}{2} - s\right) = 1 - \sqrt{2}\,s$, the same equation as before, so $s = \sqrt{2} - 1$.

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