# Napoleon’s theorem 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

 $L=\frac{1+X+C}{3}\qquad M=\frac{C+Y+0}{3}\qquad N=\frac{0+1+Z}{3}\;.$

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

 $\displaystyle 3(M-L)$ $\displaystyle=$ $\displaystyle Y-1-X$ $\displaystyle=$ $\displaystyle(2\alpha-1)C-1-\alpha$ $\displaystyle 3(N-L)$ $\displaystyle=$ $\displaystyle Z-X-C$ $\displaystyle=$ $\displaystyle(\alpha-2)C+1-2\alpha$ $\displaystyle=$ $\displaystyle(2\alpha-2-\alpha)C-\alpha+1-\alpha$ $\displaystyle=$ $\displaystyle(2\alpha^{2}-\alpha)C-\alpha-\alpha^{2}$ $\displaystyle=$ $\displaystyle 3(M-L)\alpha$

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 NapoleonsTheorem 2013-03-22 13:48:50 2013-03-22 13:48:50 drini (3) drini (3) 7 drini (3) Theorem msc 51M04