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February 24th, 2007, 12:32 AM  #1 
Senior Member Joined: Jan 2007 From: India Posts: 161 Thanks: 0  shortest distance
..can any1 help A geodesic is (in mathematics) the shortest line between two points on a mathematically defined surface (as a straight line on a plane or an arc of a great circle on a sphere). You are given a 1x1x2 cuboid. You are required to determine 2 points on the surface of the cuboid such that they have the maximum possible geodesic. Hint: The required distance is not the geodesic between the mid points of the two faces as commonly misunderstood. 
February 24th, 2007, 03:50 AM  #2 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
I have no proof of this, but I would imagine that the two points are opposite corners. The distance between them is âˆš10â‰ˆ3.1623 as opposed to 3, the distance between the centers of 1x1 faces.

February 24th, 2007, 10:34 AM  #3  
Senior Member Joined: Jan 2007 From: India Posts: 161 Thanks: 0  Quote:
 
February 24th, 2007, 01:52 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 16,942 Thanks: 1253 
Cut the surface and "unfold" it to a flat surface. There are various ways of doing that. Now join the opposite corners with a straight line. Check whether any different way of unfolding results in a shorter line. Obviously, reject any such line that leaves the surface. Calculate the length of the shortest line using Pythagoras's theorem.

February 24th, 2007, 06:09 PM  #5 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
Obviously, calculating the length of the geodesic between two opposite corners does not show that that is indeed the longest geodesic. The next step would be to rigorously prove that it is.

February 25th, 2007, 07:54 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 16,942 Thanks: 1253 
There are a limited number of ways to flatten out the surface that need to be considered. Even so, a rigorous proof seems a bit tedious.

March 4th, 2007, 10:01 AM  #7 
Member Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0 
would it be less boring of a proof, if one shows that if any point has length longer than the diagonal corners, one would lie outside of the cube?


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