Napoleon's RelativesNapoleon'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.
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, We have
Further
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
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
Napoleon's Theorem
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