superprismatic Posted July 16, 2009 Report Share Posted July 16, 2009 If I ask you to place N equally space points on a circle, you would have no particular trouble doing so. The same is not true for equally spaced points on a sphere, as is evidenced by the irregular-looking pattern of dimples on a golf ball. But what does equally spaced points on a sphere mean? I have come up with a working definition which is probably not original to me: The points are equally spaced if and only if the minimum distance from a point to any other point is maximized. That is, find the closest neighbor to every point, then find the smallest such distance, and make sure that it is as large as possible. So, for a 2-point example, the points would be antipodal -- that's as far away from each other than they can get. For three points, they will all lie on a great circle. For four points, they form the vertices of a regular tetrahedron (a tetrahedron of which all edges have the same length). Suppose one places 8 equally-spaces points (according to the definition above) on a unit sphere (a sphere of radius 1). What would be the shortest euclidean distance between any pair of points? Euclidean distance is the straight-line distance (not the distance over the surface of the sphere) between the points. How would you describe the figure whose vertices are those points? It is NOT a cube. Quote Link to comment Share on other sites More sharing options...
0 roolstar Posted July 17, 2009 Report Share Posted July 17, 2009 The points will be placed based on the following visualization: imagine 2 symetrical regular tetrahedrons. Now flip one of them vertically (Top summit becomes down). Now merge them together as if one's summit pierces the other's bse triangle. Now the final object (after we join all neighboring points) will be like 4 tetrahedrons put together and each 2 of them will be sharing 1 side between them... These tetrahedrons are not regular, in fact, the base is still an equilateral triangle, but the distance from the summit to the plane containing that base will be half the diatnce in a regular tetrahedron with that same base. WHAT?? In simple terms, the height of the new tetrahedrons will be half of the original ones before the merger. I am not sure if my explanation is clear, but I'm pretty sure I got the visual solution of this puzzle. Now for the distance, It would be calculated like so: Based on my figure on the left, the distance we are looking for is simply half of the hypotenuse of any right triangle with 2 antipodal summits (Hypoteneuse = 2) Therefore distance = 1 = radius of the sphere It's a bit late for me to put more details in my solution (4:40 AM where I live) So i'm gonna settle for this and maybe "superprismatic" can tell me if I'm on the right track at least. Quote Link to comment Share on other sites More sharing options...
0 unreality Posted July 17, 2009 Report Share Posted July 17, 2009 if it's not a cube, then the spacing distance is greater than the cube's spacing distance. Two adjacent points on a cube differ by the side of the cube, call that s s*sqrt(3) = length of the diagonal of the cube = diameter of circumscribed sphere = 2r s = r*(2/sqrt3) s is the side of the cube and also the spacing distance for the cube fitting. In other words, the Euclidean distance if you use a cube is the radius multiplied by 2/sqrt3 which is 1.15470054 so 1.15470054*r is the spacing distance to beat For that reason, we know that roolstar's procedure (if I understood it correctly) is too close together Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 17, 2009 Report Share Posted July 17, 2009 (edited) 1) ~1.216 2)a square antiprism (basically a cube with 2 opposing faces twisted 45 degrees relative to each other) It would have 10 faces; 8 of them triangles, 2 of them squares Edited July 17, 2009 by Matou Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted July 17, 2009 Report Share Posted July 17, 2009 1) ~1.216 2)a square antiprism (basically a cube with 2 opposing faces twisted 45 degrees relative to each other) It would have 10 faces; 8 of them triangles, 2 of them squares Quick back of the envelope sketch convinces me. Nice. Quote Link to comment Share on other sites More sharing options...
0 roolstar Posted July 17, 2009 Report Share Posted July 17, 2009 unreality, you are right in your reasoning and this made me review my calculations. As I expected I made a mistake in calculating the distance: After review, distance = R*2/sqrt(3) = 2/sqrt(3) = 1.155 Now I'm pretty sure about my calculation, but I'm not sure if we can maximize that distance even more. Maybe Matou's method is closer? Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 17, 2009 Author Report Share Posted July 17, 2009 1) ~1.216 2)a square antiprism (basically a cube with 2 opposing faces twisted 45 degrees relative to each other) It would have 10 faces; 8 of them triangles, 2 of them squares Matou's got it! Quote Link to comment Share on other sites More sharing options...
0 roolstar Posted July 17, 2009 Report Share Posted July 17, 2009 Matou's got it! Nice... So my method is better than the cube, but there's an even better answer (MATOU's). Love that kind of puzzles... If someonw would take the time on rendering it on some 3D modeling software, I would love to see a 3D model of Matou's method V/s mine and even V/s cube. I'm pretty sure it would become easier to see why Matou's structure maximizes more than mine. Nice puzzle superprismatic... Cheers Quote Link to comment Share on other sites More sharing options...
Question
superprismatic
If I ask you to place N equally space points on a circle,
you would have no particular trouble doing so. The same
is not true for equally spaced points on a sphere, as is
evidenced by the irregular-looking pattern of dimples
on a golf ball.
But what does equally spaced points on a sphere mean?
I have come up with a working definition which is probably
not original to me: The points are equally spaced if and
only if the minimum distance from a point to any other
point is maximized. That is, find the closest neighbor
to every point, then find the smallest such distance,
and make sure that it is as large as possible. So, for
a 2-point example, the points would be antipodal -- that's
as far away from each other than they can get. For three
points, they will all lie on a great circle. For four
points, they form the vertices of a regular tetrahedron
(a tetrahedron of which all edges have the same length).
Suppose one places 8 equally-spaces points (according to
the definition above) on a unit sphere (a sphere of radius
1). What would be the shortest euclidean distance between
any pair of points? Euclidean distance is the straight-line
distance (not the distance over the surface of the sphere)
between the points. How would you describe the figure
whose vertices are those points?
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