There is a unit-radius circle and you must form a polygon all of whose vertices are located on the circle, such as below:
What is the biggest possible value of the sum of squares of side lengths of such a polygon?
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$\begingroup$ All the edges or all the vertices? I see no edges on the circle, and I believe what you describe is impossible without infinite edges. $\endgroup$– jpmc26Oct 1, 2018 at 23:34
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$\begingroup$ I have submitted an edit that's currently pending to clarify this issue. Note that the author considers the circle a filled-in solid object, hence the edges are "on" the circle. Hopefully the edit will go through soon to avoid future confusion. $\endgroup$– Apollys supports MonicaOct 1, 2018 at 23:57
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$\begingroup$ @Apollys Approved ;) $\endgroup$– QuintecOct 2, 2018 at 0:37
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2 Answers
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The greatest possible sum of the squares of the side lengths is
9, constructed using an equilateral triangle.
Details below; I have no spoilered them; read on at your peril.
Lemma If our polygon has any obtuse angle, we may remove an obtuse-angled vertex without reducing the sum of squared side lengths. (Nitpick: Not if it has only three vertices and we aren't prepared to consider degenerate two-sided "polygons".)
Proof Cosine rule.
Corollary A sum-of-squares-maximizing polygon either has only non-obtuse angles or is a triangle.
Corollary A sum-of-squares-maximizing polygon is a triangle or rectangle.
Thanks to Pythagoras, any rectangle has the same sum of squared side-lengths — namely, 8. Let us now consider triangles:
If the triangle's angles are $\alpha,\beta,\gamma$ then the sum of squares is $2(\sin^2\alpha+\sin^2\beta+\sin^2\gamma)$. Maximizing this is the same as minimizing $S=\cos a+\cos b+\cos c$ where $a=2\alpha$ etc., so therefore $a+b+c=2\pi$.
We can't change $a,b,c$ independently because their sum is fixed. What happens if we increase $a$ a bit and decrease $b$ a bit? Well, $\frac{dS}{da}-\frac{dS}{db}=\sin b-\sin a$, so in an optimal triangle all of $a,b,c$ have the same sine; that is, any two of them differ by an even multiple of $\pi$ or add up to an odd multiple of $\pi$. Since they're all positive and add up to $2\pi$, this means that any pair are equal or add up to $\pi$.
If two of them add up to $\pi$ then the third must equal $\pi$. Therefore we have a right-angled triangle (remember that $a,b,$ and $c$ are double the angles of the triangle), and thanks to Pythagoras any such has the same sum of squared side-lengths, namely 8.
Otherwise, all the angles are equal, and we have an equilateral triangle, where all the sides have length $\sqrt{3}$, for a sum-of-squares of 9. This is clearly better than 8.
answered Oct 1, 2018 at 18:58
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$\begingroup$ How do you know that making one of the angles of the triangle a little bigger and another of the angles a little smaller doesn't increase your total value? For me, this is the interesting part of the question (but maybe I missed something obvious). $\endgroup$ Oct 1, 2018 at 23:50
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$\begingroup$ Ooo! I read it as specifying a regular polygon but now you mention it it doesn't say anything of the kind -- it's just the diagrams that give that impression. So my answer is extremely incomplete. Let's see whether I can fix it. $\endgroup$– Gareth McCaughan ♦Oct 2, 2018 at 0:31
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$\begingroup$ OK, fixed. Thanks @Apollys for pointing out the error! $\endgroup$– Gareth McCaughan ♦Oct 2, 2018 at 0:48
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$\begingroup$ @GarethMcCaughan I think that makes sense... but I found it a bit difficult to understand fully so I suggested an edit. $\endgroup$ Oct 2, 2018 at 4:31
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$\begingroup$ Thanks. I have to confess that for my taste the edits make it harder rather than easier to read, but I don't feel strongly enough about it to revert them :-). $\endgroup$– Gareth McCaughan ♦Oct 2, 2018 at 10:31
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The maximum value you can get is
9, from a triangle
Because
As you increase the number of sides, the square of the sides gets smaller faster than the increase in the number
