2024 AMC 12B Problem 23

Let $R$ be the circumradius of the octagon and $L$ the length of each lateral edge, so $L^2 = h^2 + R^2.$ Since $\angle AVD = 90^\circ,$ $AD^2 = 2L^2.$ Vertices $A$ and $D$ are three steps apart, a central angle of $135^\circ,$ so $AD^2$ $= 2R^2(1 - \cos 135^\circ)$ $= R^2(2 + \sqrt2).$ Setting $R^2(2 + \sqrt2) = 2(h^2 + R^2)$ gives $2h^2 = R^2\sqrt2.$ For a regular octagon of side $1,$ $R^2 = \dfrac{1}{2\sin^2(22.5^\circ)} = \dfrac{2 + \sqrt2}{2}.$ Therefore $h^2$ $= \dfrac{R^2\sqrt2}{2}$ $= \dfrac{(2 + \sqrt2)\sqrt2}{4}$ $= \dfrac{2\sqrt2 + 2}{4}$ $= \dfrac{1 + \sqrt2}{2}.$ Thus, the correct answer is B.