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

Home > Shapes > Geometry
Angry Friends (Posted on 2004-08-17) Difficulty: 4 of 5
Six friends were playing a game of indoor soccer together, but got into a fight during their game. Upset with each other, they decided to position themselves in the (square) 300x300 foot gymnasium so as to maximize the distance between the closest pair. Where should they each stand?

No Solution Yet Submitted by ThoughtProvoker    
Rating: 3.6667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution New Solution | Comment 5 of 15 |

I don't know if this is THE maximized distance, but I think it is larger than what has been found so far.

I placed two players on two opposite corners of the gym.  Then I placed the other 4 players such that one was on each side of the gym.

To maximize the distances, I placed them such that they were all equidistant to each other.  So I am basically dealing with a hexagon with equal sides, but NOT equal angles.  Almost like chopped off two opposite corners of the square.  By symmetry, I am chopping off a 45 degree angle.

My basic equation to solve is X (the distance a long a side of the square between a corner and side player) + Y (the leg of a 45 degree right triangle with hypotenuse = X) = 300 ft = X + X*cos(45) = X * (1 + 1/sqrt(2))

So X = 300ft / (1 + 1/sqrt(2)) = 175.736 ft.

  Posted by nikki on 2004-08-17 09:57:31
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