Folklore

In the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge 1, points $T, P$, and $Q$ are the centers of the faces $A A^{\prime} B^{\prime} B, A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, and $B B^{\prime} C^{\prime} C$ respectively. Find the distance from point $P$ to the plane $A T Q$.
Vertices $B^{\prime}$ and $C$ of the cube lie in the plane $A T Q$, so the planes $A T Q$ and $A B^{\prime} C$ coincide (see figure). By symmetry, the base $H$ of the perpendicular $P H$ dropped onto the plane $A B^{\prime} C$ lies on the line $B^{\prime} O$. Thus, $P H$ is the height of the right triangle $B^{\prime} R O$ with legs 1 and $\frac{\sqrt{2}}{2}$. Therefore,

$P H=\frac{1 \cdot \frac{\sqrt{2}}{2}}{\sqrt{1+\frac{1}{2}}}=\frac{\sqrt{3}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_23a8a0dc60f629d15d84g-01.jpg?height=426&width=463&top_left_y=1016&top_left_x=803)

Answer

$\frac{\sqrt{3}}{3}$

Problem