Given a triangle ABC and a straight line L.

Find the point P on L such that

** PA**^{2}+PB^{2}+PC^{2}
is the smallest.

(In reply to

re(2): one messy method by Steven Lord)

The answer makes sense. When A B and C are expressed in a coordinate system where L is the x-axis, their y distance contributions,

where d^2 = x^2 + y^2, being orthogonal to the line, are non-negotiable. Therefore their x distances alone are to be minimized, and this is done by choosing P to be located at their mean.