A3. Triangle $P R T$ is an isosceles right triangle because $P T=R T=1$ and $P R=P Q \sqrt{2}=\sqrt{2}$. Let the length of the edge of the cube be $r$, then:

$r+r \sqrt{2}+r=E A+A C+C G=P A+A C+C R=P R=\sqrt{2}$.

Thus, $r=\frac{\sqrt{2}}{2+\sqrt{2}}=\sqrt{2}-1$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f71f49936c57e5c73f63g-1.jpg?height=305&width=460&top_left_y=1458&top_left_x=1432)
A3. In a regular quadrilateral pyramid $\begin{gathered}T \\ P Q R S\end{gathered}$ where all edges have a length of 1, there is a cube $\begin{gathered}E F G H \\ A B C D\end{gathered}$.

The base $A B C D$ of the cube lies in the base $P Q R S$ of the pyramid, and the vertices $E, F, G$ and $H$ lie on the edges $P T, Q T, R T$ and $S T$.

Calculate the length of the edge of the cube.

![](https://cdn.mathpix.com/cropped/2024_04_17_0b81ff2ef64511bfe8eeg-1.jpg?height=470&width=487&top_left_y=2038&top_left_x=1410)