AMC 12 B, 2024

Answer (B): Let $O$ be the center of the octagon, and let $r = AO$. As can be seen from the figure below, $AD = 1 + \sqrt{2}$.

Because $\triangle AVD$ is an isosceles right triangle,
$$AV = \frac{\sqrt{2}}{2} \cdot AD = \frac{2+\sqrt{2}}{2}.$$
Applying the Law of Cosines to $\triangle AOH$ yields
$$r^2 + r^2 = 1^2 + 2r^2 \cos 45^\circ = 1 + r^2 \sqrt{2}.$$
Thus $r^2(2 - \sqrt{2}) = 1$ and
$$r^2 = \frac{1}{2 - \sqrt{2}} = \frac{2 + \sqrt{2}}{2}.$$
The Pythagorean Theorem applied to $\triangle VOA$ gives the requested square of the height of the pyramid:
$$VO^2 = AV^2 - r^2 = \left(\frac{2+\sqrt{2}}{2}\right)^2 - \frac{2+\sqrt{2}}{2} = \frac{1+\sqrt{2}}{2}.$$

Place the figure in a three-dimensional coordinate system with the center $O$ of the base of the pyramid at the origin, the octagon in the $x$-$y$ plane with positive $x$ coordinates for $A$, $B$, $C$, and $D$, the $y$-axis parallel to $\overline{AD}$, and apex $V(0, 0, h)$ on the positive $z$-axis. See the figure.

Then consider vectors
$$\overrightarrow{OV} = \langle 0, 0, h \rangle, \quad \overrightarrow{OA} = \left\langle \frac{1}{2}, -\frac{1+\sqrt{2}}{2}, 0 \right\rangle, \quad \text{and} \quad \overrightarrow{OD} = \left\langle \frac{1}{2}, \frac{1+\sqrt{2}}{2}, 0 \right\rangle.$$
It follows that
$$\overrightarrow{AV} = \overrightarrow{OV} - \overrightarrow{OA} = \left\langle -\frac{1}{2}, \frac{1+\sqrt{2}}{2}, h \right\rangle$$
and
$$\overrightarrow{DV} = \overrightarrow{OV} - \overrightarrow{OD} = \left\langle -\frac{1}{2}, -\frac{1+\sqrt{2}}{2}, h \right\rangle.$$
Because $\overrightarrow{AV}$ and $\overrightarrow{DV}$ are perpendicular, their dot product is zero. Therefore
$$0 = \left\langle -\frac{1}{2}, \frac{1+\sqrt{2}}{2}, h \right\rangle \cdot \left\langle -\frac{1}{2}, -\frac{1+\sqrt{2}}{2}, h \right\rangle = \frac{1}{4} - \frac{3+2\sqrt{2}}{4} + h^2,$$
from which $h^2 = \frac{1+\sqrt{2}}{2}$.