What is the height of the center of a regular pentagon $ABCDE$ with a unit side above the point $M$, for which it is true that the line $AM$ is perpendicular to the plane $CDM$?
Let the midpoint of side $CD$ be denoted by $F$. If $AM$ is perpendicular to the plane $CDM$, it is also perpendicular to the line $FM$ within that plane. Conversely, if $M$ is chosen such that the triangle $AFM$ is a right triangle, then $AM$ will be perpendicular to the entire plane $CDM$, since it is perpendicular to both $FM$ and the line $CD$. The triangle $AFM$ is a right triangle at $M$ if and only if the altitude from $M$ to the hypotenuse is the geometric mean of the two segments it creates on the hypotenuse. If $O$ is the center of the pentagon $ABCDE$, then the desired height is $\sqrt{\overline{AO} \cdot \overline{OF}}$.
![](https://cdn.mathpix.com/cropped/2024_05_02_f64cf02cc4311262ae13g-1.jpg?height=522&width=1160&top_left_y=392&top_left_x=470)

Let the intersection of the angle bisector of $\angle C$ in triangle $ACD$ with the opposite side be denoted by $G$. Then

$$
\angle CDG = \angle DGC = 72^\circ, \quad \angle ACG = \angle CAG = 36^\circ,
$$

so $AG = GC = CD = 1$, and the length $c$ of the diagonal $AD$ can be expressed using the golden ratio due to the similarity of triangles $ACD$ and $CDG$:

$$
c: 1 = 1: (c - 1)
$$

This quadratic equation yields $c > 1$, so $c = (\sqrt{5} + 1) / 2$.

Let the point opposite to $A$ on the circumcircle of the pentagon $ABCDE$ be denoted by $H$, the radius of the circle by $r$, and the length of $AF$ by $f$. Using the theorem about the segments of the hypotenuse, we get $AH \cdot AF = AD^2$, or $r = c^2 / 2f$. By the Pythagorean theorem, $f^2 = c^2 - 1/4$, so

$$
\begin{gathered}
AO \cdot OF = r(f - r) = \frac{c^2}{2} - \frac{c^4}{4c^2 - 1} = \frac{1}{2} \frac{c^2(2c^2 - 1)}{4c^2 - 1} = \\
= \frac{1}{2} \frac{(c + 1)(2c + 1)}{4c + 3} = \frac{1}{2} \frac{5c + 3}{4c + 3} = \frac{1}{2} + \frac{1}{4} \frac{\sqrt{5} + 1}{2\sqrt{5} + 5} = \frac{3\sqrt{5} + 5}{20}
\end{gathered}
$$

(We used multiple times that $c^2 = c + 1$.)

Thus, the desired height is $\frac{1}{2} \sqrt{\frac{3\sqrt{5} + 5}{5}} = 0.76512$.