- MacTutor History of Mathematics

Solution: Day 4, problem 1

This is not really geometry at all, but algebraic number theory.
Without loss of generality, assume that the sides of the polygon have length 1, and place it in the complex plane with

B

and

C

at 0 and 1, respectively, where

A, B, C, D, ...

are the vertices of the polygon. Then the rationality of the angles at

C, D, ...

implies by induction that each of the vertices in turn is a sum of roots of unity. Hence the number

A

is a sum of roots of unity and has absolute value 1.
Then A is an algebraic unit in an abelian number field and has absolute value 1. But such a number has all its conjugates of absolute value 1 (since if

c

denotes complex conjugation and

s

any other element of the Galois group then we have

|A^{s}|^{2} = A^{s} (A^{s})^{c} = A^{s} (A^{c})^{s}) = (A A^{c})^{s}= 1^{s} = 1

) and hence is a root of unity by Kronecker's theorem.

For the counterexample, consider the (3, 4, 5) right triangle. By marking off unit distances on all sides, starting at the vertices, we can think of it as an equilateral dodecagon with ten rational angles (nine equal to 180° and one equal to 90° ) and two non-adjacent irrational ones (arctan(3 /4 ) is irrational because 3 /4 is not an algebraic integer and hence not a root of unity).

It is also easy to give a counterexample with only 5 sides.