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Greets to all. Recently I came across the following question:

Find the value of n such that a regular n-gon satisfies the property $\frac{1}{A_1A_2} = \frac{1}{A_1A_3} + \frac{1}{A_1A_4}.$

Here $A_1, A_2, A_3, A_4$ are any four consecutive vertices of the relevant figure.

I have been able to find that it is 7 through both trigonometric manipulation and Ptolemy’s theorem but the first thought that sprung to my mind when I read the words “regular n-gon” we’re to try the approach of complex roots (generally of unity). However, I have been unable to get very far with this approach.

I have gone so far as to take the $z$ to be the complex $n^{th}$ root of unity and hence rewrite the problem to find said $z$ given that $\frac{1}{|z-1|} = \frac1{|z^2-1|} + \frac1{|z^3-1|}$

Further, the idea struck to multiply both sides of the equation with |z-1| and then with |z| (which is 1) to get $1 = \frac1{|1+\frac1z|} + \frac1{|z+\frac1z+1|}$

The advantage of this is that $z + \frac{1}{z} + 1$ is purely real, and so we can lift the modulus. However I am unable to get past this stage and fear I have hit a dead end. Replacing $\frac1z$ with $z^{n-1}$ does not help me much either.

Thank you for your help.

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    $\begingroup$ Fun fact: Proving the identity for $n = 7$ was a problem in a German Math Olympiad in 2014. As far as I can remember, Ptolemy was by far the nicest solution but there indeed also was an approach via complex numbers. $\endgroup$
    – Qi Zhu
    Commented Jul 26, 2019 at 5:11
  • $\begingroup$ Oh, really. That’s interesting. The idea of complex numbers is surely the first to strike in a question with the word “regular n-gon”. Perhaps it is something of a red herring with respect to elegance. $\endgroup$ Commented Jul 26, 2019 at 5:15

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It's $$\frac{1}{\sin\frac{180^{\circ}}{n}}=\frac{1}{\sin\frac{360^{\circ}}{n}}+\frac{1}{\sin\frac{540^{\circ}}{n}}$$ or $$\cos\frac{180^{\circ}}{n}-\cos\frac{900^{\circ}}{n}=\cos\frac{360^{\circ}}{n}-\cos\frac{720^{\circ}}{n}+\cos\frac{180^{\circ}}{n}-\cos\frac{540^{\circ}}{n}$$ or $$\cos\frac{360^{\circ}}{n}+\cos\frac{900^{\circ}}{n}=\cos\frac{720^{\circ}}{n}+\cos\frac{540^{\circ}}{n}$$ or $$\cos\frac{630^{\circ}}{n}\cos\frac{270^{\circ}}{n}=\cos\frac{630^{\circ}}{n}\cos\frac{90^{\circ}}{n}$$ and since $$\cos\frac{270^{\circ}}{n}=\cos\frac{90^{\circ}}{n}$$ is impossible, we obtain: $$\cos\frac{630^{\circ}}{n}=0.$$ Can you end it now?

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    $\begingroup$ Hey, thanks for your answer. This is the trigonometric proof which I have indeed figured myself. However, I am looking for a proof involving complex numbers. Thank you again though. $\endgroup$ Commented Jul 26, 2019 at 5:19
  • $\begingroup$ If so, see here: math.stackexchange.com/questions/1390471 the Seven's solution. $\endgroup$ Commented Jul 26, 2019 at 5:22

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