Given a triangle ABC and a straight line L.
Find the point P on L such that
is the smallest.
(In reply to one messy method
by Steven Lord)
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