## Task A-3.3.

Given is a cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge length 1. On its edges $\overline{A B}, \overline{A D}, \overline{C^{\prime} D^{\prime}}$ and $\overline{B^{\prime} C^{\prime}}$, points $P, Q, R$ and $S$ are chosen respectively such that $P Q R S$ is a square with its center at the center of the given cube. What is the length of the side of this square?
## Solution.

Points $A, P, Q$ are centrally symmetric to points $C^{\prime}$, $R, S$ with respect to the center of the cube, respectively. Therefore, $|A P|=\left|C^{\prime} R\right|,|A Q|=\left|C^{\prime} S\right|$.

Let $|A P|=\left|C^{\prime} R\right|=x,|A Q|=\left|C^{\prime} S\right|=y,|P Q|=|Q R|=|R S|=|S P|=a$.

![](https://cdn.mathpix.com/cropped/2024_05_30_44b5caa35f0d2fab7d89g-12.jpg?height=796&width=779&top_left_y=750&top_left_x=570)

In the right triangle $A P Q$, we have

$$
a^{2}=x^{2}+y^{2}
$$

Let $Q_{1}$ and $S_{1}$ be the orthogonal projections of points $Q$ and $S$ onto the edge $\overline{B C}$, respectively.

Then $P S$ is the diagonal of a cuboid with $P, B$ and $S_{1}$ as three vertices of the lower base, and the height is $\overline{S S_{1}}$. Therefore, we have $a^{2}=|P S|^{2}=|P B|^{2}+\left|B B^{\prime}\right|^{2}+\left|B^{\prime} S\right|^{2}$, i.e.,

$$
a^{2}=(1-x)^{2}+1+(1-y)^{2}
$$

Furthermore, $\overline{Q S}$ is the diagonal of a cuboid with $Q, Q_{1}$ and $S_{1}$ as three vertices of the lower base, and the height is $\overline{S S_{1}}$. Since $\left|Q_{1} S_{1}\right|=1-(1-y)-(1-y)=2 y-1$,

we have

$$
2 a^{2}=|Q S|^{2}=\left|Q Q_{1}\right|^{2}+\left|Q_{1} S_{1}\right|^{2}+\left|S_{1} S\right|^{2}=1+(2 y-1)^{2}+1
$$

Equating $a^{2}$ in (1) and (2), we get $2 x+2 y=3$.

Therefore, $a^{2}=\left(\frac{3}{2}-y\right)^{2}+y^{2}$, and substituting into (3) we get

$$
2\left(\frac{3}{2}-y\right)^{2}+2 y^{2}=2+(2 y-1)^{2}
$$

After simplification, we get $4 y=3$, i.e., $y=\frac{3}{4}$.

Then $x=\frac{3}{2}-y=\frac{3}{4}$, and finally $a=\frac{3 \sqrt{2}}{4}$.

1 point

Note: If the student does not correctly justify why $x=y$, but solves the problem under this assumption, they should be awarded 6 points.