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 Eight Points (Posted on 2008-02-01)
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)

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 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 thesouthern hemisphere adjacent to A and B,`
`   C (a*sqrt(2)/2,a*sqrt(2)/2,-sqrt(1-a^2))`
`Then`
`   |AB|^2 = 2a^2`
`       and`
`   |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

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