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

Home > Shapes > Geometry
Eight Points (Posted on 2008-02-01) Difficulty: 3 of 5
Eight points are placed on the surface of a sphere with a radius of 1. The shortest distance between any two points is greater than 1.2. How can the points be arranged?

Hint: They are not arranged as a cube. The cube would have an edge length of only 2/sqrt(3) = 1.1547.

See The Solution Submitted by Brian Smith    
Rating: 4.4000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Summary? | Comment 19 of 28 |
(In reply to re: Summary? by Charlie)

Define two adjacent vertices of the square
in the northern hemisphere,
   A (a,0,sqrt(1-a^2))
   B (0,a,sqrt(1-a^2))
Define a vertice of the square in the
southern hemisphere adjacent to A and B,
   C (a*sqrt(2)/2,a*sqrt(2)/2,-sqrt(1-a^2))
   |AB|^2 = 2a^2
   |AC|^2 = 4 - a^2*(2+sqrt(2))
Setting |AB|^2 = |AC|^2 gives
   a^2 = 4/(4+sqrt(2))
Therefore, for the twisted squares, 
the maximum minimum is
   |AB| = |AC| = sqrt(2a^2)
        = sqrt(8/(4+sqrt(2))) ~= 1.215562524

Edited on February 3, 2008, 12:08 pm
  Posted by Bractals on 2008-02-03 12:06:04

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 (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information