Product of Diagonals in Regular N-gon

  Let ε0 = 1, ..., εN-1 be the vertices of a regular N-gon inscribed on the unit circle. Show that the product of the lengths of all diagonals of the N-gon emanating from ε0 is N.

Solution

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Copyright © 1996-2018 Alexander Bogomolny

  Let ε0 = 1, ..., εN-1 be the vertices of a regular N-gon inscribed on the unit circle. Show that the product of the lengths of all diagonals of the N-gon emanating from ε0 is N.

Solution

Let's shift the N-gon one unit to the left so that the vertices will become ω = ε - 1. In particular, ω0 = ε0 - 1 = 0 falls onto the origin. We are thus interested in the product Πωk, where k ranges from 1 to N-1.

Vertices ε are the Nth roots of unity: εN = 1. Except for ε0, all other ε's are the roots of the equation

  εN-1 + εN-2 + ... + ε + 1 = 0.

This means that ω's satisfy the equation

  (ω + 1)N-1 + (ω + 1)N-2 + ... + (ω + 1) + 1 = 0.

This is a polynomial equation of degree N-1 with the free coefficient equal to N and the roots ω1, ω2, ..., ωN-1. As is well known, the product of all roots of a polynomial equation is given by its free coefficient (up to a sign), in particular

  Πωk = (-1)N-1N,

which shows that

  Π|ωk| = |Πωk| = N.

Exactly what we set out to prove.

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Copyright © 1996-2018 Alexander Bogomolny

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