\section*{Problem 5 - 310935}

Determine and plot in an \(x, y\)-coordinate system all points whose coordinates \((x ; y)\) satisfy the equation \(|x+y|+|x-y|=4\).
}

Through a point reflection at the coordinate origin or equivalently a rotation around it by \(180^{\circ}\), every point \(\left(x_{P} ; y_{P}\right)\) is transformed into the point \(\left(-x_{P} ;-y_{P}\right)\). If the first point satisfies the condition, then due to the absolute values, the second point also satisfies it, and vice versa. Therefore, it suffices to construct the figure in the region \(x \geq 0\) and then continue it by this point reflection or rotation, so we can assume \(x \geq 0\) without loss of generality.

Furthermore, through reflection at the \(x\)-axis, every point \(\left(x_{P} ; y_{P}\right)\) is transformed into the point \(\left(x_{P} ;-y_{P}\right)\). If the first point satisfies the condition \(\left|x_{P}+y_{P}\right|+\left|x_{P}-y_{P}\right|=4\), then due to

\[
\left|x_{P}-y_{P}\right|+\left|x_{P}+y_{P}\right|=\left|x_{P}+y_{P}\right|+\left|x_{P}-y_{P}\right|=4
\]

the second point also satisfies it, and vice versa. Therefore, we can assume \(y \geq 0\) without loss of generality and then continue the figure by reflection at the \(x\)-axis.

Finally, through reflection at the angle bisector of the first quadrant, every point \(\left(x_{P} ; y_{P}\right)\) is transformed into the point \(\left(y_{P} ; x_{P}\right)\). If the first point satisfies the condition, then due to

\[
\left|y_{P}+x_{P}\right|+\left|y_{P}-x_{P}\right|=\left|x_{P}+y_{P}\right|+\left|x_{P}-y_{P}\right|
\]

the second point also satisfies it, and vice versa. Therefore, we can assume \(x \leq y\) without loss of generality and then continue the figure by reflection at the angle bisector of the first quadrant.

Together, we now have the following assumptions, which we could make without loss of generality: \(0 \leq x \leq y\). Then, however, \(x+y\) and \(-(x-y)\) are always non-negative, so the equation simplifies to \((x+y)+(y-x)=4\) or \(y=2\). To be drawn is thus the segment of the line parallel to the \(x\)-axis through \((0 ; 2)\), for which the \(x\)-coordinates of the points lying on it are non-negative and \(\leq 2\), i.e., the segment from the point \((0 ; 2)\) to the point \((2 ; 2)\).

By reflection at the angle bisector of the first quadrant, one obtains the additional segment from the point \((2 ; 2)\) to the point \((2 ; 0)\). By reflection at the \(x\)-axis, this segment extends to the point \((2 ;-2)\) and one obtains as the reflection of the first segment the segment between the points \((2 ;-2)\) and \((0 ;-2)\). The point reflection/rotation around the coordinate origin finally completes the figure, so that one ultimately obtains exactly the square (or more precisely: its boundary) with the vertices \(( \pm 2 ; \pm 2)\) as the solution set.