A right pyramid has regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V.$ Segments $\overline{AV}$ and $\overline{DV}$ are perpendicular. What is the square of the height of the pyramid? $\textbf{(A) }1 \qquad$ $ \textbf{(B) }\frac{1+\sqrt2}{2} \qquad$ $ \textbf{(C) }\sqrt2 \qquad$ $ \textbf{(D) }\frac32 \qquad$ $ \textbf{(E) }\frac{2+\sqrt2}{3} \qquad$
Solution
Let $O$ be the center of the base, $R$ the circumradius of the octagon, and $h=VO$ the height. The side $AB=1$ subtends $45^\circ$ at $O$, so $1=2R^2(1-\cos45^\circ)=R^2(2-\sqrt2)$, giving $R^2=\dfrac1{2-\sqrt2}=\dfrac{2+\sqrt2}2$.
$A$ and $D$ are three sides apart, so $\angle AOD=135^\circ$ and $AD^2=2R^2(1-\cos135^\circ)=R^2(2+\sqrt2)$.
By symmetry $VA=VD=\sqrt{R^2+h^2}$. Since $\overline{AV}\perp\overline{DV}$, $AD^2=VA^2+VD^2=2(R^2+h^2)$.
So $R^2(2+\sqrt2)=2R^2+2h^2$, and $h^2=\dfrac{\sqrt2}2R^2=\dfrac{\sqrt2}{2}\cdot\dfrac{2+\sqrt2}2=\dfrac{1+\sqrt2}2$.
A right pyramid has regular hexagon $ABCDEF$ with side length $1$ as its base and apex $V$. Segments $\overline{AV}$ and $\overline{CV}$ are perpendicular. What is the volume of the pyramid?
