Let XY denote the vector from point X to point Y.

Let P be a point in the plane of triangle ABC.

Let PQ = PA + PB + PC.

What is the locus of points Q as P traces the triangle ABC ?

You're quite right Charlie, I forgot to add back the vector

AP(x, x* yb /xb) as the point P moves from A to B

The locus of point Q is therefore

PQ + AP = PB + PC

=xb+xc-2*x, yb+yc-2*x*yb/xb

which does give a line twice the length of AB which passes throught the point (xc, yc) when x=xb/2 so the mid point of each side of the resulting triangle is the corner opposite the original side in the original triangle.

Posted by Adrian on 2008-05-21 05:34:47