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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 $\Delta AXY, \Delta XZB$ and $\Delta ABC$ 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

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

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

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

We now show that the lengths $PQ$, $QR$ and $RP$ are equal. $$\begin{array}{rl}P{Q}^2& =(x+\frac{1}{2}y-\frac{1}{2}x{)}^2+(\frac{\sqrt{3}}{6}y-\frac{\sqrt{3}}{6}x{)}^2\\ & =\frac{x^2+2xy+y^2}{4}+\frac{y^2-2xy+x^2}{12}\\ & =\frac{x^2+xy+y^2}{3}.\end{array}$$ $$\begin{array}{rl}Q{R}^2& =(\frac{1}{2}(x+y)-(x+\frac{1}{2}y{)}^2+(-\frac{\sqrt{3}}{6}(x+y)-\frac{\sqrt{3}}{6}y{)}^2\\ & =\frac{1}{4}{x}^2+\frac{1}{12}(x^2+4xy+4y^2)\\ & =\frac{x^2+xy+y^2}{3}.\end{array}$$ $$\begin{array}{rl}R{P}^2& =(\frac{1}{2}(x+y)-\frac{1}{2}x{)}^2+(-\frac{\sqrt{3}}{6}(x+y)-\frac{\sqrt{3}}{6}x{)}^2\\ & =\frac{1}{4}{y}^2+\frac{1}{12}(4x^2+4xy+y^2)\\ & =\frac{x^2+xy+y^2}{3}.\end{array}$$ As $$PQ=QR=RP=\sqrt{\frac{x^2+xy+y^2}{3}}$$ for any $x$ it follows that $\Delta PQR$ is equilateral whatever the position of $X$.