Show how to distribute nine points around the surface of a sphere in such a way that each point is equidistant from its four nearest neighbors.
Place the first three points equally spaced about a great circle (equator) of the sphere.
Place the next three in the top hemisphere equally spaced along *some* small circle (latitude) but rotated 60 degrees (so as to be on different longitudes)
Place the final three in the bottom hemisphere in the same fashion.
If the small circle is chosen correctly, each of the upper points is equidistant from the other two upper points and two of the equator points. Each equator point will be equidistant from two of the upper and two of the lower points.
How do we know _which_ small circle to use?
Let the sphere have radius 1 and the height above the plane of the equator x.
The distance between a point on the equator to an upper point must be the same as the distance between two upper points.
This leads to the equation:
(1-sqrt(1-x^2)/2)^2+x^2+(sqrt(3)/2*(sqrt(1-x^2))^2=(sqrt(3)(sqrt(1-x^2))
Which has solution x=3/5
This corresponds to the arcsin(3/5) = 36.9 degree north/south line of latitude small circles.
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Posted by Jer
on 2021-06-03 14:47:02 |
Solution
