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

one messy method by Steven Lord)

continued...

sum = y_1^2 + x^2 + (x_2-x)^2 + y_2^2 + (x_3-x)^2 + y_3^2

This sum is minimum when d(sum)/dx = 0

This occurs when

2x - 2 x_2 + 2x - 2 x_3 + 2x = 0

x = (x2 + x3) / 3

which is now a remarkably simple result, independent of the y's

In the language of the problem, drop 3 perpendiculars from A B and C to L, add the lengths the that second and third are from the first and find the P that is 1/3 that distance from the 1st.

The result is the average of x1, 2, and x3, with x1=0. Had we left x1 as non-zero, the answer is:

x = (x1+x2+x3)/3

*Edited on ***June 24, 2018, 2:08 pm**

*Edited on ***June 24, 2018, 4:56 pm**