3 triangles

This is Sue Liu's solution proving that the triangle $PQR$ is equilateral whatever the position of the point $X$ . Many congratulations Sue on all your excellent work.

Let $AX = x$ and $XB = y$ where we know that $AX + XB = AB$ (constant). Let the points $P, Q$ and $R$ be the centres (centroids) of the triangles $Δ AXY, Δ XZB$ and $Δ ABC$ respectively.

We use the fact that the medians of a triangle intersect at the centroid and this point divides the medians in the ratio one third to two thirds.

If we set the point $A$ as the origin, then the points $P$ , $Q$ and $R$ , being the centroids of the equilateral triangles $AXY$ , $XZB$ and $ABC$ , have coordinates

$P = (\frac{1}{2} x, \frac{\sqrt{3}}{6} x),$

$Q = (x + \frac{1}{2} y, \frac{\sqrt{3}}{6} y), and$

$R = (\frac{1}{2} (x + y), \frac{- \sqrt{3}}{6} (x + y)) .$

We now show that the lengths $PQ$ , $QR$ and $RP$ are equal.

$\begin{matrix} {PQ}^{2} & = (x + \frac{1}{2} y - \frac{1}{2} x)^{2} + (\frac{\sqrt{3}}{6} y - \frac{\sqrt{3}}{6} x)^{2} \\ = \frac{x^{2} + 2 xy + y^{2}}{4} + \frac{y^{2} - 2 xy + x^{2}}{12} \\ = \frac{x^{2} + xy + y^{2}}{3} . \end{matrix}$

$\begin{matrix} {QR}^{2} & = (\frac{1}{2} (x + y) - (x + \frac{1}{2} y)^{2} + (- \frac{\sqrt{3}}{6} (x + y) - \frac{\sqrt{3}}{6} y)^{2} \\ = \frac{1}{4} x^{2} + \frac{1}{12} (x^{2} + 4 xy + 4 y^{2}) \\ = \frac{x^{2} + xy + y^{2}}{3} . \end{matrix}$

$\begin{matrix} {RP}^{2} & = (\frac{1}{2} (x + y) - \frac{1}{2} x)^{2} + (- \frac{\sqrt{3}}{6} (x + y) - \frac{\sqrt{3}}{6} x)^{2} \\ = \frac{1}{4} y^{2} + \frac{1}{12} (4 x^{2} + 4 xy + y^{2}) \\ = \frac{x^{2} + xy + y^{2}}{3} . \end{matrix}$

As

$PQ = QR = RP = \sqrt{\frac{x^{2} + xy + y^{2}}{3}}$

for any $x$ it follows that $Δ PQR$ is equilateral whatever the position of $X$ .