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 DAXY, DXZB and DABC 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 2 x, Ö3 6 x),

Q = (x + 1 2 y, Ö3 6 y), and

R = ( 1 2 (x + y), -Ö3 6 (x +y)).

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

PQ2 = (x + 1 2 y - 1 2 x)2 + ( Ö3 6 y - Ö3 6 x)2 = x2 + 2xy + y2 4 + y2 - 2xy + x2 12 = x2 +xy +y2 3 .

QR2 = ( 1 2 (x + y) - (x + 1 2 y)2 + (- Ö3 6 (x + y) - Ö3 6 y)2 = 1 4 x2 + 1 12 (x2 + 4xy + 4y2) = x2 +xy +y2 3 .

RP2 = ( 1 2 (x + y) - 1 2 x)2 + (- Ö3 6 (x + y) - Ö3 6 x)2 = 1 4 y2 + 1 12 (4x2 + 4xy + y2) = x2 +xy +y2 3 .

As

PQ = QR = RP =   æ  ú Ö x2 + xy + y2 3

for any x it follows that DPQR is equilateral whatever the position of X.