## Task A-4.5. (4 points)

A regular tetrahedron $A B X Y$ is placed inside a cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with side length 1 such that the point $X$ lies in the plane $A B C D$. Determine the distance between the points $Y$ and $A^{\prime}$.
## First Solution.

Let $Z$ be the foot of the perpendicular from point $Y$ to the plane $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, and $W$ be the foot of the perpendicular from point $Y$ to the plane $A B C D$. It holds that $\left|Z A^{\prime}\right|=|W A|$.

The point $W$ is the centroid of the equilateral triangle $A B X$, so $|W A|=\frac{2}{3} \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}$.

Triangle $A W Y$ is a right triangle with a right angle at vertex $W$, so by the Pythagorean theorem, we have:

$$
|W Y|=\sqrt{|A Y|^{2}-|A W|^{2}}=\sqrt{1^{2}-\left(\frac{\sqrt{3}}{3}\right)^{2}}=\frac{\sqrt{6}}{3}
$$

From this, we get:

$$
|Z Y|=|Z W|-|Y W|=1-\frac{\sqrt{6}}{3}=\frac{3-\sqrt{6}}{3}
$$

Finally, from the fact that triangle $A^{\prime} Y Z$ is a right triangle with a right angle at vertex $Z$, we have:

$$
\begin{aligned}
\left|A^{\prime} Y\right| & =\sqrt{\left|A^{\prime} Z\right|^{2}+|Y Z|^{2}} \\
& =\sqrt{|A W|^{2}+|Y Z|^{2}} \\
& =\sqrt{\left(\frac{\sqrt{3}}{3}\right)^{2}+\left(\frac{3-\sqrt{6}}{3}\right)^{2}} \\
& =\sqrt{2-\frac{2 \sqrt{6}}{3}}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_d6d90a155ebaa296cbbeg-27.jpg?height=830&width=808&top_left_y=1778&top_left_x=630)