Béla Bollobás’s proof of Napoleon’s triangle theorem

Recently I came across a super cute proof of Napoleon’s triangle theorem using complex numbers in Béla Bollobás’s book “The Art of Mathematics: Coffee time in Memphis”. Realizing the fact that I never knew it till now, and the fact that it is neither discussed in the standard textbooks on Euclidean geometry I read till day nor any online reading source (for instance Wikipedia doesn’t discuss this proof, however the Cut the Knot page on Napolean’s triangle discusses this proof), I think that despite being so beautiful this proof is not very popular (especially among math Olympians and other mathematics enthusiasts at the undergraduate level), and thus I decided to dedicate this blogpost for discussing the proof.

Proposition 1 (Napoleon’s triangle theorem)
Given any triangle ABC, we construct equilateral triangles \Delta A'BC, \Delta B'CA, \Delta C'AB erected outwards (i.e., the points A',B',C' lie in the exterior of \Delta ABC) on the sides BC, CA, AB respectively. Let A'', B'', C'' be the centroids of triangles A'BC, B'CA, C'AB respectively. Then \Delta A''B''C'' must be an equilateral triangle.

Here is an illustration of how the setup of Napoleon’s triangle theorem looks like.

Figure 1: Pictorial description of the setup of Napoleon’s triangle which says that the dotted triangle must be equilateral.

Here A'BC, B'CA, C'AB are equilateral triangles erected outwards on the sides BC, CA, AB of $\Delta ABC,$ and $A”, B”, C”$ are the centroids of $\Delta A’BC, B’CA, C’AB$ respectively; we wish to show that the dotted triangle (i.e., the triangle with its edges marked by dotted lines

Proof: We notice that a triangle \Delta PQR is equilateral if and only if the angle between the sides PQ and PR is 60^{\circ} and |PQ| = |PR|; in other words \Delta PQR is equilateral if and only if rotating the side PQ by an angle of 60^{\circ} gives the side PR of this triangle.

So, it suffices to show that rotation of side A''B'' of triangle \Delta A''B''C'' gives the side A''C'' of this triangle; in terms of complex numbers a rotation by an angle of 60^{\circ} basically corresponds to multiplication by the complex number e^{i \pi / 3}.

The idea is to write each side of triangle \Delta A''B''C'' in terms of complex numbers and then verify the fact that multiplication by e^{i \pi / 3} rotates the side A''B'' to yield A''C''.

Let \alpha = CA'', \beta = AB'', \gamma = BC'', and \tau = e^{i \pi / 3}. Then one has \tau ^3 = -1, and \tau = 1+\tau^2. Also one obtains A''B = \tau \alpha, B''C = \tau \beta, and C''A = \tau \gamma. Thus we immediately get

\displaystyle (1+\tau)(\alpha + \beta + \gamma) = \alpha + \tau \alpha + \beta + \tau \beta + \gamma + \tau \gamma = 0 ~~~~~ (1).

whence \alpha + \beta + \gamma = 0 (since 1+ \tau \ne 0 and the product of two non-zero complex numbers must necessarily be non-zero).

Now, notice that \tau A''C'' = \tau (A''B + BC'') = \tau (\tau \alpha + \gamma).
Hence, \tau A''C'' - A''B'' = \tau(\tau \alpha + \gamma) + (\tau \beta + \alpha) = (\tau^2 +1) \alpha + \tau \beta + \tau \gamma = 0 thus proving the claim. \square

EDIT: (Last updated on 7 June, 2021)

Remark: Some sources believe that the proof discussed above was known much before Béla Bollobás’s discovery of the proof.

#complex-numbers, #easy-beautiful-proofs, #geometry