Monday Math 140

What are the lengths of the diagonals of a regular n-sided polygon with sides of unit length?

Consider any regular n-gon. First, let us label the vertices with P0, P1, …, Pn-1. Next, let us denote by Lk, k=0,1,2,…,n-1, the distance from P0 to Pk.

Then, first we see that L0=0, and that Lk=Lnk.
Next, let us denote by dk the ratio
Now, consider the triangle with vertices P0, P1, and Pk. We name the angle at Pk as α and the angle at P1 as β.

Now, we note that since any regular polygon may be inscribed in a circle, the angles α and β are inscribed angles in the circle. Note that each side of the n-gon is the chord of an arc of angle \frac{2\pi}n.

Now, α subtends exactly such an arc (the one from P0 to P1). Hence, by the inscribed angle theorem, we find \alpha=\frac12\frac{2\pi}n=\frac{\pi}n.
Next, we note that β subtends nk sides of the polygon, and thus an arc of angle (n-k)\frac{2\pi}n=2\pi-\frac{2k\pi}n. So, we again apply the inscribed angle theorem to get
Now, we use the law of sines for these two angles and their opposite sides:
now, since \sin(\pi-\theta)=\sin\theta, we see
(this is equivalent to using our earlier property that Lk=Lnk, and thus dk=dnk).
Note that L1=Ln-1 is the length of the side of the n-gon. Thus, for a regular polygon with sides of unit length, L1=1, and then the dk are thus the lengths of the diagonals, and we have our answer; for a regular n-sided polygon with sides of unit length, the diagonals have lengths
k=2,3,…,n-2, with dk=dnk.


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3 Responses to “Monday Math 140”

  1. Monday Math 141 « Twisted One 151's Weblog Says:

    […] us combine the results from the previous two weeks (here and here). We found that for a regular n-gon with unit sides, the diagonals have lengths , k=2,3,…,n-2, […]

  2. jackaljim Says:

    Some of the pictures are missing… (can’t be seen)

    • twistedone151 Says:

      Fixed. I’m in the process of switching posts over to WordPress’ LaTeX from the no-longer working server I previously used.

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