Napoleon’s theorem

Theorem.

If equilateral triangles are erected externally on the three sides of any given triangle, then their centres are the vertices of an equilateral triangle.

If we embed the statement in the complex plane, the proof is a mere calculation. In the notation of the figure, we can assume that $A=0$, $B=1$, and $C$ is in the upper half plane. The hypotheses are

$$\frac{1-0}{Z-0}=\frac{C-1}{X-1}=\frac{0-C}{Y-C}=\alpha$$ (1)

where $\alpha =\exp \pi i/3$, and the conclusion we want is

$$\frac{N-L}{M-L}=\alpha$$ (2)

where

$$L=\frac{1+X+C}{3}\mathit{\hspace{1em}\hspace{1em}}M=\frac{C+Y+0}{3}\mathit{\hspace{1em}\hspace{1em}}N=\frac{0+1+Z}{3}.$$

From (1) and the relation ${\alpha }^2=\alpha -1$, we get $X,Y,Z$:

$$X=\frac{C-1}{\alpha }+1=(1-\alpha )C+\alpha$$

$$Y=-\frac{C}{\alpha }+C=\alpha C$$

$$Z=1/\alpha =1-\alpha$$

and so

$3(M-L)$ $=$ $Y-1-X$

$=$ $(2\alpha -1)C-1-\alpha$

$3(N-L)$ $=$ $Z-X-C$

$=$ $(\alpha -2)C+1-2\alpha$

$=$ $(2\alpha -2-\alpha )C-\alpha +1-\alpha$

$=$ $(2{\alpha }^2-\alpha )C-\alpha -{\alpha }^2$

$=$ $3(M-L)\alpha$

proving (2).

Remarks: The attribution to Napoléon Bonaparte (1769-1821) is traditional, but dubious. For more on the story, see http://www.mathpages.com/home/kmath270/kmath270.htmMathPages.

Title	Napoleon’s theorem
Canonical name	NapoleonsTheorem
Date of creation	2013-03-22 13:48:50
Last modified on	2013-03-22 13:48:50
Owner	drini (3)
Last modified by	drini (3)
Numerical id	7
Author	drini (3)
Entry type	Theorem
Classification	msc 51M04