All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
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.)
one messy method | Comment 1 of 4
If we call L the x-axis we can arbitrarily represent any triangle as three points above this axis, at heights y1, y2, and y3. Let us put the first point on the y-axis (x1=0) and place the other two at x2 and x3 so the three points of the triangle are (0,y1) (x2,y2), (x3,y3). Placing point P on the x-axis at x, that is (x,0). Note that this arrangement maintains the generality of the problem. (Con't)

Edited on June 24, 2018, 2:07 pm
  Posted by Steven Lord on 2018-06-04 23:48:05

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information