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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)

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re: For Charlie's thoughtful consideration | Comment 23 of 28 |
(In reply to For Charlie's thoughtful consideration by FrankM)

Let's see:

My original solution (after correcting the program) has points at 30.73586710236 degrees north and south latitudes. The ones in the north have longitudes, say, of 0, 90, -90 and 180 (same as -180). The ones in the south have longitudes of 45, -45, 135, -135.

Each point is 74.85849218562 degrees of great circle arc, or chord distance of 1.21556252413 radii from each of its four nearest neighbors. For the point at 30.73586710236 north on the 0 meridian, its nearest neighbors are those at that latitude at longitudes 90 and -90, and the ones at 30.73586710236 south with longitudes +45 and -45. All four are at that same distance: 74.85849218562 degrees of great circle arc, or chord distance of 1.21556252413 radii. I assumed the 0-degree, +45 degree was the latitude-preserving twist you were referring to. The point in the northern hemisphere does indeed have two nearest neighbors in the south.

Now perform the longitude-preserving shear. The point at 30.7+ degrees north and zero longitude, for sake of argument(it could be another, but they all work equally well), moves to a longitude that is closer to the equator than 30.7 degrees. Meanwhile, either the point at longitude +45 or the point at longitude -45 has moved further north than its 30.7-degree latitude, while the other has moved south.  The one that has moved north will be closer than the 74.85849218562-degree separation that it had before from the point on the prime meridian.

Adding this to the end of my program:

   lat1 = 25: lat2 = -25: lonDiff = 45: lat = lat1
   sinLat = SIN(lat * dr): cosLat = COS(lat * dr)
   sinLat2 = -sinLat
   distOnTri = acos(sinLat * sinLat2 + cosLat * cosLat * COS(lonDiff * dr))
   angular = distOnTri
   distOnTri = 2 * SIN(distOnTri * dr / 2)' convert arc degrees to chord length
   PRINT USING "##.########### "; lat; distOnTri; angular


25.00000000000  1.09342861288 66.28378271595

indicating if the point on the prime meridian is moved down to 25 north, while either the point at +45 or -45 is moved north to 25 south, that the distance between them will be 1.093 instead of 1.215.


  Posted by Charlie on 2008-02-03 23:46:25
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