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Equidistant point (Posted on 2018-06-03) Difficulty: 3 of 5
Given a triangle ABC and a straight line L.

Find the point P on L such that
PA2+PB2+PC2 is the smallest.

No Solution Yet Submitted by Ady TZIDON    
Rating: 2.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: one messy method | Comment 2 of 4 |
(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
  Posted by Steven Lord on 2018-06-05 02:22:13

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