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Find three points along a circle with radius r such that the sum of distance between any pair of the three points is maximised.

Intuition is pointing to a equilateral triangle. How do I prove that?

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    $\begingroup$ This is a special case of the isodiametric problem for triangles: among all inscribed triangles find one with maximal perimeter. The solution is indeed the equilateral triangle. More generally, regular polygons have maximal perimeter among inscribed polygons with odd number of sides. There is a geometric proof on p. 390 of Mossinghoff's paper. But what methods are you expected to use? $\endgroup$
    – Conifold
    Commented Dec 12, 2019 at 11:35

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HINT Consider three points on a circle $(1,0)$, $(\cos\theta,\sin\theta)$ and $(\cos\phi,\sin\phi)$ with $\phi>\theta$

The expression to be maximized using calculus is $$S=2\sin(\frac{\theta}2)+2\sin(\frac{\phi-\theta}2)+2\sin(\frac{\phi}2)$$

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  • $\begingroup$ I think you need a factor of 2 on the last term as well, the inequality should be non-strict (there is no guarantee that the maximal shape does not degenerate), and I am not sure how he would handle the inequality constraint using calculus. $\endgroup$
    – Conifold
    Commented Dec 12, 2019 at 11:27

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