I have to say, simplependulum - your solution is very inspiring!

So to sum it up, in the general case we have:

$\displaystyle

\sum_{k=1}^{n} D_k = \frac{a}{2 \sin^2{\frac{\pi}{2n}}} = 2R\cot{\frac{\pi}{2n}}

$

$\displaystyle

\prod_{\stackrel{k=1}{\scriptstyle{k}\ne i}}^{n} D_k = n \left( \frac{a}{2\sin{\frac{\pi}{n}}} \right)^{n-1} = n R^{n-1}

$

Where $\displaystyle R$ is the radius of the circumscribed circle.

Thank you Drexel28 for this interesting problem with surprisingly simple solution formulas.

Edit:

After noticing that the last expression for the product of the distances does not include trigonometry, an interesting question came into my head: can you find this formula without the use of trigonometry?