Napoleon's Theorem: Third Proof with Complex Numbers

Napoleon's theorem claims that the centers A', B', C', of the equilateral triangles A''BC, AB''C, ABC'', erected on the sides (either all inwardly or all outwardly) of a given triangle ABC form an equilateral triangle. In the proof we are going to use complex numbers. The proof comes from [Bollobás, pp. 124-125] where the author makes an observation that after the slogan 'let's use vectors and complex numbers' no more thinking is needed. While this is true that one of algebra's purposes and uses is to mechanize solving problems, this is a third proof of Napoleon's theorem that makes use of complex numbers. So that, perhaps, some deliberation as to which road to choose might follow a conscious decision to base a proof on complex numbers.

The applet below serves to illustrate the proof.

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Let a = CA', b = AB', c = BC', j = e^iπ/3, the counterclockwise rotation through 60° so that j³ = -1 and j = 1 + j². We have A'B = ja, B'C = jb, and C'A = jc, and conclude that

0 = a + ja + b + jb + c + jc = (1 + j)(a + b + c)

so that

a + b + c = 0.

For ΔA'B'C' to be equilateral suffice it to have, say, A'B' = jA'C'. Let's see that this is indeed so:

	jA'C' - A'B'	= j(A'B + BC') + (B'C + CA')
		= j(ja + c) + (jb + a)
		= (j² + 1)a + jc + jb
		= j(a + b + c)
		= 0

and we are done.

References

B. Bollobás, The Art of Mathematics: Coffee Time in Memphis, Cambridge University Press, 2006

Napoleon's Theorem

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