\section*{Problem 3 - 201033}

Given a point \(P\) in a plane \(\epsilon\). Investigate whether there exists a square \(A B C D\) in \(\epsilon\) such that \(P A=\sqrt{2} \text{ cm}, P B=\sqrt{5} \text{ cm}, P C=\sqrt{8} \text{ cm}\)!

If this is the case, determine the side length of each such square!
}

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1433.jpg?height=363&width=334&top_left_y=1412&top_left_x=264)

Let the sought side length be \(a\). Using the Law of Cosines in triangle \(A B P\), we get:

\[
2=a^{2}+5-2 \sqrt{5} a \cos \beta \quad \rightarrow \quad \text { (1) } \quad 2 \sqrt{5} a \cos \beta=a^{2}+3
\]

and in triangle \(P B C\):

\[
8=a^{2}+5-2 \sqrt{5} a \cos \left(\frac{\pi}{2}-\beta\right) \quad \rightarrow \quad 8=a^{2}+5-2 \sqrt{5} a \sin \beta
\]

\[
2 \sqrt{5} a \sin \beta=a^{2}-3
\]

If we square equations (1) and (2) and add them, we get:

\[
\begin{gathered}
20 a^{2}=a^{4}+6 a^{2}+9+a^{4}-6 a^{2}+9 \\
a^{4}-10 a^{2}+9=0 \\
a^{2}=5 \pm \sqrt{25-9} \quad \rightarrow \quad a_{1}^{2}=9 \quad a_{2}^{2}=1
\end{gathered}
\]

Assuming positive side lengths, the side length of the triangle can either be \(a=3\) or \(a=1\) (in the latter case, \(P\) would of course lie outside the square).