Finding the middle ground bisectors in \(p\)-geometry (Q2820326)

The setting of the paper is the real plane, namely the two dimensional vector space of ordered pairs \(\overrightarrow{v}=(x,y)\in \mathbb R^2\). The plane is equiped with a \(p\)-norm, \(1\leq p \leq \infty\), which is so defined: \(\|(x,y)\|_p = (|x|^p + |y|^p)^{1/p}\) for \(1\leq p < \infty\) and, for \(p=\infty\), \(\|(x,y)\|_{\infty}=\max\{|x|, |y|\}\). For the case \(p=1,2\) and \(\infty\), the corresponding norms are the well known taxicab norm, Euclidean norm and Chebyshev norm. The paper focus on bisectors of a line segment with respect to the \(p\)-norm and discusses some geometric properties. In particular, denote by \(l_2^p\) the pair \((\mathbb R^2, \|\cdot \|_p)\) and let \(M_p(\overrightarrow{v}_1,\overrightarrow{v}_2)=\{\overrightarrow{w}\in l_2^p \;: \;\|\overrightarrow{w}-\overrightarrow{v}_1\|_p= \|\overrightarrow{w}-\overrightarrow{v}_2\|_p\}\). The authors prove that the intersection:NEWLINENEWLINENEWLINE\[NEWLINE\bigcap_{1\leq p \leq \infty}M_p(\overrightarrow{v}_1,\overrightarrow{v}_2)NEWLINE\]NEWLINENEWLINENEWLINE\noindent consists exactly of three points as long as the line segment \(\overrightarrow{v}_1\overrightarrow{v}_2\) is not parallel to a coordinate axis or to a side of the unit taxicab circle. When the segment is parallel to a coordinate axis, then all \(p\)-bisectors coincide with the Euclidean bisector. When the line segment is parallel to a side of the taxicab unit circle, then all \(p\)-bisectors, \(1< p \leq \infty\), are the same.