Triangle ABC has equilateral triangles drawn on its edges. Points P, Q and R are the centres of the equilateral triangles. Experimentation with the interactive diagram leads to the conjecture that PQR is an equilateral triangle.

There are many ways to prove this result. Here we have chosen one which uses only the cosine rule and one which uses complex numbers to represent vectors, and multiplication by complex numbers to rotate the vectors by 60 degrees.

Another proof using a tessellation of the plane is discussed on the 'Cut-the-knot' website.

First the proof using the Cosine Rule. The sides of triangle ABC are written as

a, b

and

c

. Centroids of equilateral triangles are at the intersection of the altitudes so

∠ PAB

and

∠ RAC

are both 30 degrees. Hence

AP = \frac{2}{3} . \frac{\sqrt{3} c}{2} = \frac{c}{\sqrt{3}}

and

AR = \frac{2}{3} . \frac{\sqrt{3} b}{2} = \frac{b}{\sqrt{3}} .

It follows that

∠ PAR = (∠ A + 60)

degrees. By the cosine rule

{PR}^{2} = {AP}^{2} + {AR}^{2} - 2 AP . AR \cos (∠ A + 60) = \frac{1}{3} (c^{2} + b^{2} - 2 bc \cos (∠ A + 60) (1) .

Now

\cos (∠ A + 60) = \frac{1}{2} \cos A - \frac{\sqrt{3}}{2} \sin A

and, from

▵ ABC

\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 bc}

and

\sin A = \frac{2 Area ▵ ABC}{bc}

. Substituting for

\cos (∠ A + 60)

in (1) and simplifying the expression gives:

{PR}^{2} = \frac{1}{3} [\frac{a^{2} + b^{2} + c^{2}}{2} + 2 \sqrt{3} Area ▵ ABC] .

This formula is completely symmetric in

a, b

and

c

and it follows that

{RQ}^{2}

and

{QP}^{2}

have the same value and that

▵ PQR

is equilateral.

Next the proof using complex numbers as vectors. We use $λ = e^{π i / 3}$ so that $λ^{2} = λ - 1$ .

Also multiplying a complex number by $λ$ rotates it by 60 degrees.

Referring to the given diagram let $A, B, C$ be represented by the complex numbers $a, b, c$ . The third vertex of the equilateral triangle drawn on $AB$ is represented by the complex number $a + λ (b - a)$ . Therefore the centre of this triangle P is represented by $p$ where

$p = \frac{1}{3} ([2 - λ] a + [1 + λ] b) .$

Similarly

$q = \frac{1}{3} (2 - λ] b + [1 + λ] c),$

and

$r = \frac{1}{3} (2 - λ] c + [1 + λ] a) .$

To show that PQR is equilateral it is sufficient to show that $r - q = λ [p - q]$ and this follows using simple algebra and $λ^{2} = λ - 1$ .