Napoleon’s theorem

Theorem.

If equilateral trianglesare erected externally on the three sides of any given triangle, thentheir centres are the vertices of an equilateral triangle.

If we embed the statement in the complex plane, the proofis a mere calculation. In the notationof 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) istraditional, but dubious. For more on the story, seehttp://www.mathpages.com/home/kmath270/kmath270.htmMathPages.