A square-based frustum has a base edge and every side edge of 4. The edge of its top face is 2. What is the maximum distance between two vertices of the frustum?
Solution. We label the vertices of the frustum as shown in Figure 1. We calculate the lengths of the base and lateral diagonals of the pyramid, as well as the length of the space diagonal. The bases are squares, so the lengths of the diagonals are

$$
A C=4 \sqrt{2}=\sqrt{32}, \quad \text { and } \quad A^{\prime} C^{\prime}=2 \sqrt{2}=\sqrt{8}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_487f030785868e6ae950g-1.jpg?height=415&width=428&top_left_y=367&top_left_x=838)

Figure 1

The lateral faces of the pyramid are isosceles trapezoids. For example, the vertices of one such trapezoid are $A, B, A^{\prime}, B^{\prime}$ (Figure 2). The projection of $B^{\prime}$ onto the side $A B$ is $E$.

$$
\begin{aligned}
E B & =\frac{4-2}{2}=1, \quad B^{\prime} E=\sqrt{4^{2}-1^{2}}=\sqrt{15} \\
A B^{\prime 2} & =3^{2}+(\sqrt{15})^{2}=24, \quad A B^{\prime}=\sqrt{24}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_487f030785868e6ae950g-1.jpg?height=458&width=377&top_left_y=1224&top_left_x=863)

Finally, we project a plane perpendicular to the base onto the diagonal of the base. This plane cuts an isosceles trapezoid from the pyramid. This plane contains the space diagonal $A C^{\prime}$ of the pyramid (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_487f030785868e6ae950g-1.jpg?height=453&width=496&top_left_y=1844&top_left_x=804)

Similarly to the previous calculations, we can find the length of the space diagonal using the Pythagorean theorem:

$$
\begin{aligned}
C^{\prime} F & =\sqrt{16-2}=\sqrt{14} \\
A{C^{\prime}}^{2} & =(\sqrt{14})^{2}+(3 \sqrt{2})^{2}=14+18=32 \\
A C^{\prime} & =\sqrt{32}
\end{aligned}
$$

Comparing the results, we see that the longest distance between any two vertices is $\sqrt{32}$, which is the length of the space diagonal and the diagonal of the base.