Question 155, The edge length of the regular tetrahedron $\mathrm{ABCD}$ is 2, the angle $\theta$ between edge $\mathrm{AD}$ and plane $\alpha$ is $\theta \in\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$, and vertex $\mathrm{A}$ is on plane $\alpha$, while B, C, D are all outside of $\alpha$. Find the range of the distance from the midpoint $\mathrm{E}$ of edge $\mathrm{BC}$ to plane $\alpha$.
Question 155, Solution: We only need to consider $\triangle \mathrm{ADE}$. Clearly, $\triangle \mathrm{ADE}$ is an isosceles triangle, with $DE = AE = \sqrt{3}$, and $AD = 2$.

As shown in the figure, take the midpoint of $AD$ as $F$, then $EF \perp AD$, and $EF = \sqrt{2}$. Rotating $\triangle ADE$ around the line $AD$ for a full circle, we get all possible positions of point $E$. When the angle $\theta$ between $AD$ and plane $\alpha$ is fixed, let the projection of $E$ on plane $\alpha$ be $G$. Clearly, the minimum and maximum distances from $E$ to plane $\alpha$ are achieved when $E$, $G$, $A$, and $F$ are coplanar (the maximum distance is achieved when $E$ is at the position $E'$ shown in the figure, and $F$ is the midpoint of $EE'$). At this time, the minimum distance is $AE \cdot \sin \angle EAG = \sqrt{3} \sin (\theta - \angle DAE) = \sin \theta - \sqrt{2} \cos \theta$; the maximum distance is $AE \cdot \sin \angle E'AG = \sqrt{3} \sin (\theta + \angle DAE) = \sin \theta + \sqrt{2} \cos \theta$.

Noting that $\theta \in \left[\frac{\pi}{3}, \frac{\pi}{2}\right]$, the minimum distance is $\sin \frac{\pi}{3} - \sqrt{2} \cos \frac{\pi}{3} = \frac{\sqrt{3} - \sqrt{2}}{2}$, and the maximum distance is $\sqrt{3} \sin (\theta + \angle DAE) = \sqrt{3} \sin \left(\frac{\pi}{3} + \angle DAE\right) = \sin \frac{\pi}{3} + \sqrt{2} \cos \frac{\pi}{3} = \frac{\sqrt{3} + \sqrt{2}}{2}$.

In summary, the range of distances from the midpoint $E$ of edge $\mathrm{BC}$ to plane $\alpha$ is $\left[\frac{\sqrt{3} - \sqrt{2}}{2}, \frac{\sqrt{3} + \sqrt{2}}{2}\right]$.