# 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.

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

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)