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# 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

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

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 MathPages.

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## Recent Activity

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new correction: Error in proof of Proposition 2 by alex2907

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