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.
|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,
|α''||= γ + λ(β - γ),|
|β''||= α + λ(γ - α),|
|γ''||= β + λ(α - β).|
|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
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]|
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.
- B. Grünbaum, A RELATIVE OF "NAPOLEON'S THEOREM", Geombinatorics 10(2001), 116 - 121 (available online)
Copyright © 1996-2018 Alexander Bogomolny