\section*{Problem 4 - 021244}

Given is a rectangle with sides \(2 a\) and \(2 b\), where \(a > b\). From this rectangle, four congruent right triangles (one at each corner, with the legs lying on the sides of the rectangle) are to be cut off such that the remaining figure forms an octagon with equal sides.

The side of the octagon is to be expressed in terms of \(a\) and \(b\) and constructed from \(a\) and \(b\). Additionally, the conditions under which the problem is solvable are to be specified.
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The lengths of the catheti of the cut-off right-angled triangular areas are denoted by \(x\) and \(y\), and the length of the hypotenuse, which is equal to the side length of the octagon to be calculated, is denoted by \(c\). Then we have:

\[
\begin{aligned}
& 2 x+c=2 a, \text { i.e., } \quad x=\frac{2 a-c}{2} \\
& 2 y+c=2 b, \text { i.e., } \quad y=\frac{2 b-c}{2} \\
& x^{2}+y^{2}=c^{2}
\end{aligned}
\]

Thus, due to (1), (2), and (3),

\[
\frac{(2 a-c)^{2}}{4}+\frac{(2 b-c)^{2}}{4}=c^{2} \Rightarrow c^{2}+2(a+b) c-2\left(a^{2}+b^{2}\right)=0
\]

It follows that either

\[
\begin{gathered}
c=-(a+b)+\sqrt{(a+b)^{2}+2\left(a^{2}+b^{2}\right)} \text { or } \\
c=-(a+b)-\sqrt{(a+b)^{2}+2\left(a^{2}+b^{2}\right)}
\end{gathered}
\]

must hold. Since \(a\), \(b\), and \(c\) are positive, \(c\) cannot satisfy relation (5). If the construction is possible at all, \(c\) must satisfy condition (4).

Remark: The construction is possible if and only if \(c<2 \min (a, b)\), i.e., if \(\max (a, b)<\) \(3 \min (a, b)\).

Taken from [2]