$\begin{aligned} & {\left[\begin{array}{l}\text { Regular tetrahedron }\end{array}\right]} \\ & \text { [ Pythagorean theorem in space ] }\end{aligned}$

Each edge of the triangular pyramid $P A B C$ is equal to $1 ; B D$ is the height of triangle $A B C$. The equilateral triangle $B D E$ lies in a plane forming an angle $\phi$ with the edge $A C$, and points $P$ and $E$ lie on the same side of the plane $A B C$. Find the distance between points $P$ and $E$.
Since all edges of the pyramid PABC are equal, it is a regular tetrahedron. Let \( M \) be the center of the base \( ABC \), \( N \) be the orthogonal projection of the vertex \( E \) of the equilateral triangle \( BDE \) onto the plane \( ABC \), \( K \) be the midpoint of \( BD \), and \( F \) be the foot of the perpendicular dropped from point \( E \) to the height \( PM \) of the tetrahedron \( PABC \). Since \( EK \perp BD \), by the theorem of three perpendiculars, \( NK \perp BD \), so \( EKN \) is the linear angle of the dihedral angle formed by the planes \( ABC \) and \( BDE \). Since \( NK \parallel AC \), we have \( \angle EKN = \phi \). Further, we have:

\[
\begin{gathered}
BD = \frac{\sqrt{3}}{2}, \quad MD = \frac{\sqrt{3}}{6}, \quad KD = \frac{1}{2} BD = \frac{\sqrt{3}}{4}, \quad PM = \sqrt{\frac{2}{3}}, \\
KM = KD - MD = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{12}, \\
EK = BD \cdot \frac{\sqrt{3}}{2} = \frac{3}{4}, \quad EN = EK \sin \phi = \frac{3}{4} \sin \phi \\
NK = EK \cos \phi = \frac{3}{4} \cos \phi, \quad MN^2 = NK^2 + KM^2 = \frac{9}{16} \cos^2 \phi + \frac{1}{45} \\
PE^2 = EF^2 + PF^2 = MN^2 + (PM - MF)^2 = MN^2 + (PM - EN)^2 = \\
= \frac{9}{16} \cos^2 \phi + \frac{1}{45} + \left( \sqrt{\frac{2}{3}} - \frac{3}{4} \sin \phi \right)^2 = \\
= \frac{9}{16} \cos^2 \phi + \frac{1}{45} + \frac{2}{3} - \sqrt{\frac{3}{2}} \sin \phi + \frac{9}{16} \sin^2 \phi = \\
= \frac{9}{16} + \frac{1}{45} + \frac{2}{3} - \sqrt{\frac{3}{2}} \sin \phi = \frac{5}{4} - \sqrt{\frac{3}{2}} \sin \phi = \frac{5}{4} - \frac{\sqrt{6}}{2} \sin \phi
\end{gathered}
\]

Therefore,

\[
PE = \sqrt{\frac{5}{4} - \frac{\sqrt{6}}{2} \sin \varphi} = \frac{1}{2} \sqrt{5 - 2 \sqrt{6} \sin \varphi}
\]

## Answer

\(\frac{1}{2} \sqrt{5 - 2 \sqrt{6} \sin \varphi}\)

## Problem