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We know that an ellipse is the locus of all point $z$ in the plane with $$|z-a|+|z-b|=\lambda$$

where $a,b$ are two given points in the plane and $\lambda$ is a constant.

Now we consider the following generalization:

For three given points $a,b,c \in \mathbb{R}^{2}$ define $$A_{\lambda}=\{z\in \mathbb{R}^{2} \;\text{with}\;\; |z-a|+|z-b|+|z-c|=\lambda\}$$

How is the geometric description of $A_{\lambda}$? Is it a closed curve, at least when $\lambda$ is very large?

More generally, assume that $G$ is a compact topological group with Haar measure $\mu$. Assume that $G$ is topologically embedded in the plane. Define:

$$A_{\lambda}=\{z\in \mathbb{R}^{2} \;\text{with}\;\; \int_{G} |z-g|d\mu=\lambda\}$$

Is the topology of $A_{\lambda}$ independent of choosing sufficiently large $\lambda$?

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    $\begingroup$ You might look at the $n$-ellipse, e.g., images at this MSE posting. $\endgroup$ Commented Feb 9, 2016 at 15:35
  • $\begingroup$ @JosephO'Rourke Thank you very much for the link. $\endgroup$ Commented Feb 9, 2016 at 15:38
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    $\begingroup$ Interesting historical footnote: James Clerk Maxwell, On the description of oval curves, and those having a plurality of foci; with remarks by Professor Forbes, Proceedings of the Royal Society of Edinburgh 2 #28 (1845-466), 89-91. Forbes actually read the paper before the Royal Society of Edinburgh (on 6 April 1846) because Maxwell was not allowed due to his age. Maxwell was only 14 years old at the time. $\endgroup$ Commented Feb 9, 2016 at 19:41
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    $\begingroup$ As explained in the paper cited in said MSE post, all $n$-ellipses are boundaries of compact and convex sets (though their algebraic closures aren't necessarily). The argument carries over to the case where you have any compactly supported finite measure $\mu$ on $\mathbb R^n$ and want to keep $f_\mu(z)=\int_{\mathbb R^n}|z-x|\,d\mu(x)$ constant. Simply regard $f_\mu$ along a straight line as a superposition of convex functions. Being the boundary of a compact convex set, the topology is clear (except for one degenerate case for each $\mu$ ...). $\endgroup$ Commented Feb 9, 2016 at 20:12
  • $\begingroup$ @DaveLRenfro thank you very much for your very interesting historical remark. $\endgroup$ Commented Mar 8, 2016 at 17:18

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