Problem no 1.160 on my previous question.

Let $ABCD...PQ$ represent a regular polygon of $n$ sides inscribed in a circle of unit radius. Prove that the product of the lengths of the diagonals $AC, AD, ... , AP$ is $\frac14 n \csc^2 \left( \frac{\pi}{n} \right )$.

How do I proceed? I am thinking $\displaystyle \prod_{k=1}^{n-2} \left | 1 - e^{i\frac{2\pi k }{n}} \right |$, am I in right direction? And how do I simplify it?

  • $\begingroup$ what is the source of the problems? $\endgroup$ – lab bhattacharjee Feb 21 '13 at 15:57
  • $\begingroup$ @labbhattacharjee it's Schaums Series Complex Variables ... want it?? $\endgroup$ – hasExams Feb 21 '13 at 15:57
  • $\begingroup$ Well, $k$ should go from $2$ to $n-2$ in that expression. $\endgroup$ – Thomas Andrews Feb 21 '13 at 16:01
  • $\begingroup$ Related: math.stackexchange.com/questions/287976/… $\endgroup$ – lab bhattacharjee Feb 21 '13 at 16:01
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    $\begingroup$ I think the answer should be $\frac{1}{4}n\csc^2 \pi/n$. For example, when $n=4$, there is one diagonal of length $2$, and $\csc \pi/4 = \sqrt 2$ $\endgroup$ – Thomas Andrews Feb 21 '13 at 16:13

Define $p(z)=\prod_{k=1}^{n-1} \left(z-e^{2\pi ik/n}\right)=1+z+z^2+... + z^{n-1}$.

Then note that the value you are looking for is:

$$\left|\frac{p(1)} {(1-e^{2\pi i/n})(1-e^{-2\pi i/n})}\right|$$

But $p(1)=n$. And the denominator is $2-2\cos(2\pi/n)$. But $1-\cos 2x = 2\sin^2 x$, so we get that the value you are looking for is $$\frac{n}{4\sin^2 \frac{\pi}{n}}$$

  • $\begingroup$ why am i looking for that form?? $\endgroup$ – hasExams Feb 21 '13 at 16:23
  • $\begingroup$ Form was a typo, meant "for." But you are looking for that value because $|p(1)|$ is the product of all the lengths from $AB\cdot AC\cdot AD\dots AQ$ and the denominator is dividing by the lengths that are not diagonals, $AB\cdot AQ$. $\endgroup$ – Thomas Andrews Feb 21 '13 at 16:25
  • $\begingroup$ Oh!! I see .. also I understand the comment you made. I was reading improperly. $\endgroup$ – hasExams Feb 21 '13 at 16:28

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