# Equilateral triangle in $l_p^2$

I was trying to calculate the side of an equilateral triangle with the vertices on the unit sphere of $l_p^2$, when $1<p<\infty$. When I say "equilateral", I mean with respect to the distance in $l_p^2$. With the exception of $p=2$, where is a trivial problem of Euclidean geometry, I am having problems finding the coordinates of the vertices, as I get some equations I don't seem to be able to solve.

What I tried was fixing one vertex at $(0,1)$ and the other two having coordinates $(a,b)$ and $(-a,b)$. I am unable to solve for $a$ and $b$.

1. Any ideas on how I solve this, or perhaps a different approach?
2. Are there any known results for the regular $n$-gon with vertices on the unit sphere of $l_p^n$. Again, I can calculate this pretty easily for $p=2$.

Edit: In view of Christian Blatter below, I will change the question to finding the largest $\lambda$ such that there exist $3$ points on the unit sphere of $l_p^2$ such that the distance between any $2$ is equal to $\lambda$. Similarly for $l_p^n$. I think such largest distance should exist by a simple compactness argument.

-
What is $l_p^2\$? – Christian Blatter Mar 5 '13 at 14:52
@ChristianBlatter $\mathbb{R}^2$ with the $p$-norm, $||(x,y)||=(|x|^p+|y|^p)^{1/p}$. – Theo Mar 5 '13 at 18:09

## 1 Answer

The norm $$\|(x,y)\|:=(|x|^p+|y|^p)^{1/p}\qquad(p\ne2)$$ does not admit a continuous group of isometries keeping the origin fixed. As a consequence the side-length of equilateral triangles inscribed in $S_{(p)}^1$ is not uniquely determined.

A hint: Try to solve your problem in the cases $p=1$ and $p=\infty$. In these cases $S_{(p)}^1$ is a square with respect to $x$ and $y$, and you might be able to find equilateral triangles inscribed in $S_{(p)}^1$ explicitly.

-
I understand. I modified the question to reflect what i was really looking for. – Theo Mar 5 '13 at 19:15