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For a regular n-gon of side length s, draw all the diagonals from a given vertex, i.e. pick a vertex, and draw line segments from it to every other vertex. Since there are n vertices, there will be n-3 diagonals, or n-1 when also including the sides from the vertex to the adjacent vertices.

I want an expression for the tuple of all side lengths, for a given n. For example, for n=3, the tuple is just the two adjacent sides, so it is $(s,s)$. For a square, the tuple is $(s,s\sqrt{2},s)$. For n=5, using the law of cosines, and the expression for the interior angle of a regular polygon ($\theta=\frac{\pi(n-2)}{n}$), the tuple is $(s,s\sqrt{1-2cos(\frac{3\pi}{5})},s\sqrt{1-2cos(\frac{3\pi}{5})},s)$. The tuple will be symmetric and have n-1 elements.

I'm looking for a geometric derivation of the expression for each element of the tuple.


My question is related to the following Math SE post, but is different: Regular polygon diagonal lengths. It seems like the accepted solution here defined the diagonal as the distance from the center of the polygon/circle to any and all vertices. But I'm interested in the length of the line segment from a single vertex to al the other vertices.

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Place the regular polygon with $v_0=(1,0)$ as the first vertex, then the other vertices are given by $v_k=(\cos2\pi k/n, \sin2\pi k/n)$, for $k=1,2,\ldots,n-1$. Use the distance formula to calculate distance from $v_0$ to $v_k$.

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  • $\begingroup$ Shouldn't it be $v_k=(\cos2\pi k/n, \sin2\pi k/n)$? $\endgroup$
    – PattuX
    Jan 26, 2018 at 3:30
  • $\begingroup$ @PattuX: Yes, thats right. I am correcting it. $\endgroup$ Jan 26, 2018 at 5:47

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