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 *P*_{0}, *P*_{1}, …, *P*_{n-1}. Next, let us denote by *L*_{k}, *k*=0,1,2,…,*n*-1, the distance from *P*_{0} to *P*_{k}.

Then, first we see that *L*_{0}=0, and that *L*_{k}=*L*_{n–k}.

Next, let us denote by *d*_{k} the ratio

.

Now, consider the triangle with vertices *P*_{0}, *P*_{1}, and *P*_{k}. We name the angle at *P*_{k} as *α* and the angle at *P*_{1} 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 .

Now, *α* subtends exactly such an arc (the one from *P*_{0} to *P*_{1}). Hence, by the inscribed angle theorem, we find .

Next, we note that *β* subtends *n*–*k* sides of the polygon, and thus an arc of angle . 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:

.

Rearranging,

.

so

,

now, since , we see

(this is equivalent to using our earlier property that *L*_{k}=*L*_{n–k}, and thus *d*_{k}=*d*_{n–k}).

Note that *L*_{1}=*L*_{n-1} is the length of the side of the *n*-gon. Thus, for a regular polygon with sides of unit length, *L*_{1}=1, and then the *d*_{k} 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 *d*_{k}=*d*_{n–k}.

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Tags: Geometry, Inscribed Angle Theorem, Law of Sines, Math, Monday Math, Regular Polygon, Trigonometry

This entry was posted on October 25, 2010 at 12:44 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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November 1, 2010 at 12:21 am |

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