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I have stuck on problem. I have to draw 11-sided polygon. 10 sides of polygon must be equal, while one side must be longer (at least twice longer than any other side).

To make things more complicated it must fit circle as close as possible.

Is there anyway to determine at least close approximation of side lenghts ?

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A hint: Inscribe a long side to your liking, compute the corresponding central angle $ \alpha$, and divide $360^\circ-\alpha$ into $10$ equal pieces.

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Consider a regular $n$-gon for any $n$ greater than 11. You will certainly be able to inscribe this in a circle and it will have at least 10 equal sides. A little change and you will have an $11$ sided polygon matching your requirements.

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To make it fit as close as possible, the long side must be exactly twice as long as the others. If the ten small angles are $2\beta$, the large angle is $2\alpha$, where $\sin\alpha=2\sin\beta$ and $\alpha+10\beta=\pi$. So try to solve $\arcsin(2\sin\beta)+10\beta=\pi$

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