A regular tetrahedron lies with its four vertices at respective distances from the origin as 1, 2, 3 and 4 units. What is the length of each edge of the tetrahedron?
I think I have proven there is no such tetrahedron.
Call the verticies at distances 1,2,3,4 by A,B,C,D respectively.
There are limits on the possible lengths of each edge.
The distance from A to B cannot be less than 1 or greater than 3, etc:
AB = [1,3]
AC = [2,4]
AD = [3,5]
BC = [1,5]
BD = [2,6]
CD = [1,7]
The only number contained in all these intervals is 3, which would be have to be the solution.
The only way AB = 3 is if the origin is between A and B and the three are collinear.
The only way AD = 3 is if A is between D and the origin amd the three are collinear.
But this would make A, B and D collinear so there's no tetrahedron (also BD = 6 not 3)

Posted by Jer
on 20120308 11:31:16 