Which polygon is the one?

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?

• 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. Oct 1 '18 at 23:34
• 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. Oct 1 '18 at 23:57
• @Apollys Approved ;) Oct 2 '18 at 0:37

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.

• 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). Oct 1 '18 at 23:50
• 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. Oct 2 '18 at 0:31
• OK, fixed. Thanks @Apollys for pointing out the error! Oct 2 '18 at 0:48
• @GarethMcCaughan I think that makes sense... but I found it a bit difficult to understand fully so I suggested an edit. Oct 2 '18 at 4:31
• 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 :-). Oct 2 '18 at 10:31

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