BrainDen.com - Brain Teasers

## Question

If we spread out 24 points on the

surface of a unit sphere (i.e., a

sphere with radius 1) so that each

point is a distance D away from its

nearest neighbour, what would be

the value of D?

## Recommended Posts

• 0 Here is a guess.

You have an area of the sphere of 4pi, you have 24 points to cover the sphere. Each point can cover pi/6 area. Therefore, pi d^2 = pi/6 or d = sqrt(6).

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• 0 With this description, you can take any value for D between 0 and 0.2.

Choose a positive real number D up to 0.2 and put pairs of points with distance D to each other anywhere on the square, as long as you make sure the distance between points of two differnent couples is higher than or equal to D.

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• 0 Here is a guess.

You have an area of the sphere of 4pi, you have 24 points to cover the sphere. Each point can cover pi/6 area. Therefore, pi d^2 = pi/6 or d = sqrt(6).

Typo, d=sqrt(1./6)

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• 0 oops it was a sphere....lol I guess i'd better read next time

Edited by Doo
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• 0

I'm late but I'll take a crack at it...

I believe the area of the sphere is 4*pi. The dots are evenly spaced, so each dot is covering pi/6 area. The distance "d" then should be (pi/6)^0.5, or 0.7236012545582676593630146272908

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• 0 I could be completely wrong here but here's my quick guess from my high school math education...

If I read this correctly, the questions is essentially stating that there are 24 evenly spaced out points along the lining of the sphere, thus making D equal to the (Circumference/24). Since Circumference = (Pi)Diameter and Diameter = 2*Radius --> Circumference = (Pi)/(2*1) = (Pi/2) --> D = (Pi/2)/24 --> D = (Pi/48).

Am I missing something here?

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• 0

I could be completely wrong here but here's my quick guess from my high school math education...

If I read this correctly, the questions is essentially stating that there are 24 evenly spaced out points along the lining of the sphere, thus making D equal to the (Circumference/24). Since Circumference = (Pi)Diameter and Diameter = 2*Radius --> Circumference = (Pi)/(2*1) = (Pi/2) --> D = (Pi/2)/24 --> D = (Pi/48).

Am I missing something here?

basekick, welcome to Brain Den! I do think you're missing something ...

The question was about the area of a sphere (3D). You start by working from the formula for the crcumfrence of a circle (2D). Also it seems to me that you switched from multiplication to division here: "Circumference = (Pi)Diameter and Diameter = 2*Radius --> Circumference = (Pi)/(2*1)".

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• 0

Woops! I left out an important word

in the OP. I should have asked for

the maximum value of D. Here's the

corrected problem:

If we spread out 24 points on the

surface of a unit sphere (i.e., a

sphere with radius 1) so that each

point is a distance D away from its

nearest neighbour, what would be

the maximum value of D?

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• 0

well drats, a little late. dont know if this is even correct and now, even if the math pans out, certainly can't prove it's a max.

envisioned a series of squares with 24 total vertexes on the sphere. basically there would be a ring of eight squares about the middle of the sphere each sharing one side and two on each pole in such a way that the distance from the two on the poles to the ones in the ring about the middle of the sphere would be the same as the edge length of each square. my sketch is no where near scan worthy. right or wrong the following is probably easily reduced but with my limited trig skills -

sqrt(2)sin{tan-1(sqrt(2)tan(pi/8))} = .7148

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• 0

I could be completely wrong here but here's my quick guess from my high school math education...

If I read this correctly, the questions is essentially stating that there are 24 evenly spaced out points along the lining of the sphere, thus making D equal to the (Circumference/24). Since Circumference = (Pi)Diameter and Diameter = 2*Radius --> Circumference = (Pi)/(2*1) = (Pi/2) --> D = (Pi/2)/24 --> D = (Pi/48).

Am I missing something here?

I'm sorry, I goofed. You could do this with any circle on the surface of the sphere.

A zig-zag pattern encircling the sphere would work as well. Alas! I neglected to add

a crucial word to the OP. I corrected it in post #9.

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• 0

well drats, a little late. dont know if this is even correct and now, even if the math pans out, certainly can't prove it's a max.

envisioned a series of squares with 24 total vertexes on the sphere. basically there would be a ring of eight squares about the middle of the sphere each sharing one side and two on each pole in such a way that the distance from the two on the poles to the ones in the ring about the middle of the sphere would be the same as the edge length of each square. my sketch is no where near scan worthy. right or wrong the following is probably easily reduced but with my limited trig skills -

sqrt(2)sin{tan-1(sqrt(2)tan(pi/8))} = .7148

you measured distance between points as

being the length of arc of a great circle

between two points. Is that correct?

I just use the usual euclidean distance

in 3-space. Either one is fine -- I just

wanted to know which you used.

couldn't see how to get there. I would

need a bit more of an explanation, including

which distance measure you used. If you are

a bit (how much depends on the metric you

used).

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• 0

If you position the points in the vertices of a

snub cube then the straight (euclidian) distance between points will be approximately 0.744206.
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• 0

you measured distance between points as

being the length of arc of a great circle

between two points. Is that correct?

I just use the usual euclidean distance

in 3-space. Either one is fine -- I just

wanted to know which you used.

couldn't see how to get there. I would

need a bit more of an explanation, including

which distance measure you used. If you are

a bit (how much depends on the metric you

used).

my answer was supposed to be a straight line/Euclidean distance. See k-man may have us both beat. Checked out his link. Quite the shape. Can't visualize if the vertices lie on a shere? Do now have the name of the polyhedron I was trying to describe - rhombicuboctahedron. For simplicity, I started by defining the edge length of the rhombicuboctahedron as 1.

tan(pi/8) = (1/2) / L (the length from center of sphere to center of square when the square edge length is 1) or L = 1/2tan(pi/8)

tan-1{(sqrt(2)/2)/L} = the angle B formed by L and the radius of the sphere (the length from the corner of a square to the center of the sphere)

so when the radius of the sphere is 1, half the diagonal of a square = sin(B).

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• 0 I Imagined a structure that would have it's vertexes on the sphere.

First draft of this would be to imagine 3 square prisms placed along the Ox, Oy,Oz axes with the vertexes of each prism placed on the sphere surface.

Now, variate the height of the prisms until you get the Distance between one vertex from the prism placed on one axe and the closest vertex from the prisms placed on the other 2 axes to be equal to the side of the square that is base to your prisms( all 3 prisms have the same dimensions).

Another way to see this structure is to take an regulate octagonal prism on the horizontal plane and merge it with another octagonal prism placed in the vertical plane, with the condition that the height of the prism equals the length of one of the octagon's sides.

Did some calcules and i got distance D to be

D= sqrt(17(6+sqrt(2) ) / 17.

In numeric terms, this is D = 06604, approximately.

My calculus might be erroneous, but i think this structure can best approximate the positioning of the points in your problem.

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• 0

Here's my wild guess. I know working is a bit concise. I can explain the working in detail later

4(pi)r2= 12.5714286 now straighten the surface area. and put it in 7x5. so 7d x 5d=12.571428 d= 0.599319342

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• 0

I thought my solution was good -- the order 4

permutahedron. It has D=.632456. In short

order, mine was beaten by plainglazed with

his rhombicuboctahedron and D=.714813.

Then came k-man's snub cube (best so far)

with D=.744206. I've checked each of these

myself. In case anyone's interested, here

are coordinates for snub cube vertices on

the surface of the unit sphere:

distance to
nearest
X Y Z neighbour
0.462321 -0.251359 -0.850340 0.744206
-0.462321 0.251359 -0.850340 0.744206
-0.462321 -0.251359 0.850340 0.744206
0.462321 0.251359 0.850340 0.744206
0.462321 0.850340 -0.251359 0.744206
0.462321 -0.850340 0.251359 0.744206
-0.462321 0.850340 0.251359 0.744206
-0.462321 -0.850340 -0.251359 0.744206
0.251359 0.462321 -0.850340 0.744206
0.251359 -0.462321 0.850340 0.744206
-0.251359 0.462321 0.850340 0.744206
-0.251359 -0.462321 -0.850340 0.744206
0.251359 -0.850340 -0.462321 0.744206
-0.251359 0.850340 -0.462321 0.744206
-0.251359 -0.850340 0.462321 0.744206
0.251359 0.850340 0.462321 0.744206
0.850340 -0.251359 0.462321 0.744206
0.850340 0.251359 -0.462321 0.744206
-0.850340 0.251359 0.462321 0.744206
-0.850340 -0.251359 -0.462321 0.744206
0.850340 -0.462321 -0.251359 0.744206
-0.850340 0.462321 -0.251359 0.744206
-0.850340 -0.462321 0.251359 0.744206
0.850340 0.462321 0.251359 0.744206

Can anyone do better than k-man?

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• 0 It never said anything about the points having to try to be as far apart as they can, or that the last point has to touch the first point. just that the distance of one is D away from another. The farthest distance the 24 can be spread then would be a line on its circumference right? Im sure there are other ways but either way all of these can be correct answers.

[2(pi)r]/24 = D

since D(1) and d(24) dont have to be the same distance as D1 and D2, the distance D is from 0 (non inclusive) and [2(pi)r]/24

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• 0

It never said anything about the points having to try to be as far apart as they can, or that the last point has to touch the first point. just that the distance of one is D away from another. The farthest distance the 24 can be spread then would be a line on its circumference right? Im sure there are other ways but either way all of these can be correct answers.

[2(pi)r]/24 = D

since D(1) and d(24) dont have to be the same distance as D1 and D2, the distance D is from 0 (non inclusive) and [2(pi)r]/24

novah,

Why say [2(pi)r]/24 is the largest? By your reasoning, I would think [2(pi)r]/2 would be the greatest distance. This places all but one dot on one location, and the one extra dot on the far side of the sphere. Then the distance D is half the circumfrence. If your value of [2(pi)r]/24 is pertinent, it seems you are spacing the dots evenly along one equatorial circle, but the OP seems to indicate they are spaced across theentire surface of the sphere.

And here's a question for Superprismatic...

Some of the earlier discussion seems to be about actual distance, point-to-point, like the chord length of an arc (in 2D terms) rather than the arc length. I assumed you were measurig distance along the surface of the sphere rather than through the sphere. If the points are evenly spaced, would not my original response be correct? I took the area ofthe sphere, divided by 24 (the area covered by each dot, so to say), and then took the square root of that value. If the dots are evenly spaced in all directions, would not the square root of one dot's area yield the maximum distance between dots?

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• 0

novah,

Why say [2(pi)r]/24 is the largest? By your reasoning, I would think [2(pi)r]/2 would be the greatest distance. This places all but one dot on one location, and the one extra dot on the far side of the sphere. Then the distance D is half the circumfrence. If your value of [2(pi)r]/24 is pertinent, it seems you are spacing the dots evenly along one equatorial circle, but the OP seems to indicate they are spaced across theentire surface of the sphere.

And here's a question for Superprismatic...

Some of the earlier discussion seems to be about actual distance, point-to-point, like the chord length of an arc (in 2D terms) rather than the arc length. I assumed you were measurig distance along the surface of the sphere rather than through the sphere. If the points are evenly spaced, would not my original response be correct? I took the area ofthe sphere, divided by 24 (the area covered by each dot, so to say), and then took the square root of that value. If the dots are evenly spaced in all directions, would not the square root of one dot's area yield the maximum distance between dots?

It really doesn't matter what distance metric is used, but I want to know so that I can compare them.

On the circle (2 dimensional), "evenly spaced" is well defined. That is not true about the 3 dimensional sphere. In any case, your solution of D=.7236 (no matter if it is arc-length or euclidean distance) is still smaller than k-man's D=.7442. I don't really know why you take the square root of "the area covered by each dot" to get your distance. I suppose that the "area covered by each dot" is the area of the largest circle which doesn't include other dots which is centered on the dot in question.

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• 0

It really doesn't matter what distance metric is used, but I want to know so that I can compare them.

On the circle (2 dimensional), "evenly spaced" is well defined. That is not true about the 3 dimensional sphere. In any case, your solution of D=.7236 (no matter if it is arc-length or euclidean distance) is still smaller than k-man's D=.7442. I don't really know why you take the square root of "the area covered by each dot" to get your distance. I suppose that the "area covered by each dot" is the area of the largest circle which doesn't include other dots which is centered on the dot in question.

the square root was used based on the concept that any area can be expressed as 'length x width', and if the points were equidistant, then the length and width would be equal. Hence if LxW=A and L=W, then L^2=A, then A^0.5=L. So, the square root of the area would be the maximum distance D. Without attempting to think through the actual shape of the 1/24 of the surface of a sphere, I just divided the sphere's area by 24 then applied my simple algebra against that value. But, in reading some of your other posts, and other people's posts on this subject, the thinking goes way deeper. I can't verify k-man's value of D, but I do notice that the snub cube is not comprised of truly regular faces. There are two variations of surface geometry.

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• 0

the square root was used based on the concept that any area can be expressed as 'length x width', and if the points were equidistant, then the length and width would be equal. Hence if LxW=A and L=W, then L^2=A, then A^0.5=L. So, the square root of the area would be the maximum distance D. Without attempting to think through the actual shape of the 1/24 of the surface of a sphere, I just divided the sphere's area by 24 then applied my simple algebra against that value. But, in reading some of your other posts, and other people's posts on this subject, the thinking goes way deeper. I can't verify k-man's value of D, but I do notice that the snub cube is not comprised of truly regular faces. There are two variations of surface geometry.

It's not the area of the faces that count, it's the length of the edges. The snub cube (with vertices on the unit circle) has the closest neighbour of every point at a distance of 0.7442.

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