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 ?

I think the locus is another triangle, three times the size of the original, rotated by 180 degrees

Say A is 0, 0 : B is xb, yb and C is xc, yc

As P goes from A to B let x go from 0 to xb

<o:p> </o:p>

P is thus x, x* yb /xb

<o:p> </o:p>

PA = A-P = -x, -x*yb/xb

PB = B- P= xb-x, yb – x*yb/xb

PC= C-P = xc-x, yc– x*yb/xb

PQ= xb+xc-3*x, yb+yc-3*x*yb/xb

So we have a straight line 3 times the length of AB drawn in the opposite direction

similar reasoning for othr two lines

Posted by Adrian on 2008-05-15 14:23:40