# Napoleon's Relatives

Napoleon's theorem claims that the centers A', B', C', of the equilateral triangles A''BC, AB''C, ABC'', erected on the sides (either all inwardly or all outwardly) of a given triangle ABC form an equilateral triangle. There is a small family of triangles - related to Napoleon's - that were discovered relatively recently [Grünbaum].

The applet below serves to illustrate the construction.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Let Am, Bm, Cm be the midpoints of  B''C'', A''C'', and A''B'', respectively. Then triangles ABmCm, AmBCm, AmBmC are equilateral. If A*, B*, C* are their respective centers then triangle A*B*C* is also equilateral.

Complex numbers come in handy for proving this results. Let α, β, γ correspond to points A, B, C; α'', β'', γ'' to A'', B'', C''; αm, βm, γm to Am, Bm, Cm. Leta λ be a rotation by 60°: λ³ = -1, λ² - λ + 1 = 0.

We have

 α'' = γ + λ(β - γ), β'' = α + λ(γ - α), γ'' = β + λ(α - β).

Further

 αm = (β'' + γ'')/2 = (α + β)/2 + λ(γ - β)/2, βm = (α'' + γ'')/2 = (β + γ)/2 + λ(α - γ)/2, γm = (α'' + β'')/2 = (γ + α)/2 + λ(β - α)/2.

We wish to establish that &alpha, βm, γm form an equilateral triangle. Suffice it to show that α = A, where A = βm + λ(γm - βm).

Well, γm - βm = (γ - β)/2 + λ(2α - β - γ)/2. Thus

 A = βm + λ(γm - βm) = (β + γ)/2 + λ(α - γ)/2 + λ[(γ - β)/2 + λ(2α - β - γ)/2] = [(β + γ)/2 + (2α - β - γ)/2] + λ[(α - γ)/2 + (γ - β)/2 + (β - α)/2] = α,

as needed. Note that triangles ABC and ABmCm have the same orientation. A proof for the triangle A*B*C* is not very much different, however its orientation is with a natural consequence for the proof.

All the above assertions hold when triangles ABC'', AB''C, and A''BC are drawn with a reversed orientation. The triangle A*B*C* will be different, but the sum of the areas of the two (such triangles formed with different orientations) is one fourth the area of ΔABC.

### References

1. B. Grünbaum, A RELATIVE OF "NAPOLEON'S THEOREM", Geombinatorics 10(2001), 116 - 121 (available online)