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If a regular polygon inscribed in a circle of radius $1$ has side lengths of

$$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}$$

then how many sides does it have? We know that the polygon is constructible (so the number of his sides can be written as the product of a power of 2 times a combination of Fermat primes).

(original problem image)

Community
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Display maths
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    You've been here a while, so you should know that the Math.SE community prefers/expects questions to include something of what the asker knows about a problem. (What have you tried? Where did you get stuck? What's the source? etc) This information helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining things you already know, duplicating your effort, or using techniques beyond your skill level. As comments are easily overlooked, please edit the question to include details. – Blue May 21 '20 at 18:56
  • Hint: The angle subtended by two adjacent vertices at the center is $\frac{2\pi}{n}. Find the other two angles in the isosceles triangle so formed and then use the Law of Sines. – Vishu May 21 '20 at 18:58
  • Is this a contest problem? In any case, please cite the source of the problem. – amWhy May 21 '20 at 19:09
  • @amWhy I may be wrong, but IMHO this is too easy to be a contest problem. Falls under the same umbrella as $\sqrt{2+\sqrt{2+\sqrt2}}$ and friends. Only the starting point of the half-angle formula is different. My favorite among them is this. – Jyrki Lahtonen May 21 '20 at 19:26
  • I'm puzzled by this users Problem statements from this user, @Jyrki (and I'm not necessarily wondering about high level math competition, and I believe such questions ought to be accompanied by more context, at the very least, citing the source of the question). But I respect your take on it. – amWhy May 21 '20 at 19:28
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    Alex, I recommend that you study our guide for new askers. Listing a few things we expect, and gives pointers to avoiding negative attention :-) – Jyrki Lahtonen May 21 '20 at 19:32
  • Thanks for the link I should have provided in the first place, @Jyrki! – amWhy May 21 '20 at 19:34
  • I'm still new enough that idk if answering such questions is poor form. I agree with Jyrki that the question is easy enough to not be a contest problem. – Integrand May 21 '20 at 19:39
  • It’s a problem that I found in my 12th grade math book. – Display maths May 22 '20 at 01:23

1 Answers1

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The side length is the distance from $(1,0)$ to $(\cos(2\pi/n),\sin(2\pi/n))$. Then we have $$ 2-2\cos(2\pi/n) = {2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}} $$ $$ 2\cos(2\pi/n) = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} $$Square and simplify. $$ 4\cos^2(2\pi/n) = 2+\sqrt{2+\sqrt{2+\sqrt{3}}} $$ $$ 2(2\cos^2(2\pi/n) -1)= \sqrt{2+\sqrt{2+\sqrt{3}}} $$Use the double-angle formula. $$ 2\cos(4\pi/n)= \sqrt{2+\sqrt{2+\sqrt{3}}} $$Repeat. $$ 2\cos(8\pi/n)= \sqrt{2+\sqrt{3}} $$ $$ 2\cos(16\pi/n)= \sqrt{3} $$ $$ \cos(16\pi/n)= \sqrt{3}/2 $$Then $n=96$.

Integrand
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