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 ?
(In reply to
Solution? by Adrian)
When P moves from (0,0) to (xb,yb), PQ does indeed go from (xb+xc,yb+yc) to (xc2*xb,yc2*yb).
However, PQ is a vector; it isn't the point Q. The vector takes the point P, now at (xb, yb), to the point Q at (xcxb,ycyb), and so the traced line of Q for this segment is from (xb+xc,yb+yc) to (xcxb,ycyb), and thus is twice (not three times) the length of the segment traced by P.
Symmetry shows that if it were three times, the opposite vertex would fall 1/3 of the way along the generated segment; but 1/3 from which direction? ... there's no asymmetry to favor one or the other.

Posted by Charlie
on 20080515 17:16:20 