PlanetMath: Napoleon's theorem

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Napoleon's theorem

(Theorem)

Theorem 1 If equilateral triangles are erected externally on the three sides of any given triangle, then their centres are the vertices of an equilateral triangle.

$\includegraphics{napoleon}$

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

, and

is in the upper half plane. The hypotheses are

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

(1)

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

$\displaystyle \frac{N-L}{M-L}=\alpha$

(2)

where

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

From (1) and the relation $\alpha^2=\alpha-1$ , we get

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

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

$\displaystyle Z=1/{\alpha}=1-\alpha$

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.

"Napoleon's theorem" is owned by drini. [ full author list (3) | owner history (2) ]

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Cross-references: conclusion, upper half plane, complex plane, centres, triangle, sides, equilateral triangles

This is version 4 of Napoleon's theorem, born on 2003-07-31, modified 2007-06-17.
Object id is 4538, canonical name is NapoleonsTheorem.
Accessed 3712 times total.

Classification:

AMS MSC:

51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries)

Pending Errata and Addenda

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