Let n>6 positive integer and $A_1A_2...A_n$ a convex polygon. Prove that there exist i and j such that $$|\cos(A_i)-\cos(A_j)|<{1\over 2(n-6)}$$

My trials: I tried by contradiction but it didn’t work. I also tried supposing WLOG that angles A1>A2>A3>...>An. I also tried factoring cos(Ai)-cos(Aj) but it didn’t work for me either.

Would a pigheon principle or anything be useful here? Please help.

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    $\begingroup$ @MariaMazur I suppose $\cos\angle A_{i-1}A_iA_{i+1}$? $\endgroup$ – Hagen von Eitzen May 30 at 17:17
  • $\begingroup$ Yeah. It is about the interior angle. $\endgroup$ – furfur May 30 at 17:20

The angles $A_i$ are between $0$ and $\pi$ and we know that $\sum(\pi-A_i)=2\pi$. If the claim is false, then for each integer $k$, there is at most one $i$ with $$1-\frac{k+1}{2(n-6)}<\cos A_i\le 1-\frac{k}{2(n-6)}.$$ It follows that the $k$th largest angle has cosine $\le 1-\frac{k-1}{2(n-6)}$. Then the six smallest angles have cosine $\le 1-\frac{n-1}{2(n-6)}$, $\le1-\frac{n-2}{2(n-6)}$, $\le1-\frac{n-3}{2(n-6)}$, $\le1-\frac{n-4}{2(n-6)}$, $\le1-\frac{n-5}{2(n-6)}$, and $\le1-\frac{n-6}{2(n-6)}=\frac 12$, respectively. We conclude that the $6$th smallest angle is $\le \frac23\pi$ and the five smaller angles are $<\frac23\pi$. Therefore, $$ \sum(\pi-A_i)>6\cdot \frac\pi3=2\pi,$$ contradiction

  • $\begingroup$ Could you please explain this statement: “It follows that the $k$th largest angle has cosine $\le 1-\frac{k-1}{2(n-6)}$. ” $\endgroup$ – furfur May 30 at 17:48

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