Given a triangle ABC and a straight line L.
Find the point P on L such that
PA2+PB2+PC2
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.