[bonanova]

1. Simple problem
   .
   Imagine a 3-dimensional extension of the Thanks, now I don't have to draw it.
   Three spheres, of 3" radius lie on a table top, each sphere touching the other two.
   What is the largest sphere that fits in the space bounded by these four objects?
   That is, the largest sphere that lies on the table and between and under the other three spheres.
   .
   
2. Harder problem
   .
   What is the greatest number of spheres that have all pairs tangent at distinct single points?
   How many configurations are there?
   Can you find a relationship among the radii?
   

[Guest]

Is "r" 2/5. If Its not I have no idea how to get to the answer

[Guest]

1)Connect the centres (A,B,C) of the three spheres to form an equilateral triangle with sides of 6” each. I’ll refer to this as the base.

2)Connect the centre (O) of the new smaller sphere (of radius r) to the three large spheres. These lines, along with the base, create a triangular pyramid (ABCO). This pyramid now has a base of 6”, a height (measured perpendicular to the base) of 3 – r. The length of the sloping sides is 3 + r.

3)As the slopes are all the same, they meet at a point perpendicularly above the centre of the base. From this, we can deduce the length of the other two measurements needed to write an equation for the length of the sloping line:

(3+r)² = (3-r)² + 3² + (3*tan(30))²

9 + r² + 6r = 9 + r² - 6r + 9 + 3

12r = 12

r = 1

[Guest]

Using the same thinking as the "Intimate Circle" problem, it would seem as if the answer is 5, in two configurations analogous to the solution there? Don't know how to go about proving that though - just seems like the reasonable answer.

Edited October 4, 2010 by rajat_magic

[k-man]

The only possible solutions that I can think of are the same as with the "flat" problem with the answers being:

5 spheres (4 spheres making up a pyramid and the 5th sphere is either inside or around the other 4)

radius of the inner sphere is r*(sqrt(3/8)-1) and the circumsphere is r*(sqrt(3/8)+1)

[Guest]

The only possible solutions that I can think of are the same as with the "flat" problem

... with three spheres of radius R, and two of radius R/6.

[Guest]

brb

Edited October 4, 2010 by Spabbarter

[bonanova]

... with three spheres of radius R, and two of radius R/6.

Nice!

[bonanova]

The table top is a portion of a sphere of infinite radius.

Just as the line segment is a portion of a circle of infinite radius.

[Guest]

have 3 spheres in a triangular shape as before. place two addtional spheres of the same radius on top and bottom on the middle space. question, is it possible to construct a 6th sphere in the middle space that is tangent to all 5 spheres? i would think yes, but not sure.

[EventHorizon]

The solutions so far all seem to be part of the same family of configurations.

If you start with the pyramid k-man suggested and "pull" the middle sphere down (while only changing the radius of the middle sphere and the top sphere) you eventually reach d3k3's configuration. Pull further, the radius of the middle sphere eventually grows to infinity (table top mentioned by bonanova). Pull further still, and the middle sphere inverts and the configuration is a pyramid inside a larger sphere (a configuration also mentioned by k-man).

And each of these is similar to answers for the circle problem as well.

Is there a fundamentally different configuration that cannot be reached this way? (ignoring changing the radii of the other 3 circles)

Also, do you think that if you extend the problem to 4d spheres (aka glomes, hyperspheres, or 3-spheres), would the answer be 6? Would the answer generally be n+2 for n dimensions (n>1)?

[bonanova]

The solutions so far all seem to be part of the same family of configurations.

If you start with the pyramid k-man suggested and "pull" the middle sphere down (while only changing the radius of the middle sphere and the top sphere) you eventually reach d3k3's configuration. Pull further, the radius of the middle sphere eventually grows to infinity (table top mentioned by bonanova). Pull further still, and the middle sphere inverts and the configuration is a pyramid inside a larger sphere (a configuration also mentioned by k-man).

And each of these is similar to answers for the circle problem as well.

Is there a fundamentally different configuration that cannot be reached this way? (ignoring changing the radii of the other 3 circles)

Also, do you think that if you extend the problem to 4d spheres (aka glomes, hyperspheres, or 3-spheres), would the answer be 6? Would the answer generally be n+2 for n dimensions (n>1)?

Bingo! http://brainden.com/forum/uploads/emoticons/default_thumbsup.gif' alt=':thumbsup:'>

... that three intimate circles disjoin the remaining area of the plane so that, banning intersections, which preclude tangencies anyway, only one more circle can have access to the first three. Either of two, inside or outside, but not both, since they'd reside in areas inaccessible to each other.

Similarly, four spheres [tetrahedral centers] divide 3-space into two regions that are mutually inaccessible [by spheres, at least], so that only one, or another, sphere can be added.

So the max = n+2 formula does hold:

In 4-space, and higher n- spaces, analogous partitions are formed by the first 5, or n+1, hyperspheres, permitting one but only one additional hypersphere to join the group intimately.

For there to be a fundamentally different configuration, you'd need a fundamentally different arrangement of three intimate circles on the plane, four intimate spheres in space, etc. Hard to imagine what they might be.

Can you imagine an intimate configuration for n=1?

[bonanova]

have 3 spheres in a triangular shape as before. place two addtional spheres of the same radius on top and bottom on the middle space. question, is it possible to construct a 6th sphere in the middle space that is tangent to all 5 spheres? i would think yes, but not sure.

A 6th sphere would keep the 4th and 5th from touching.

[bonanova]

Can anyone imagine an intimate configuration for n=1?

i.e. 1-dimensional space.

[EventHorizon]

Can anyone imagine an intimate configuration for n=1?

i.e. 1-dimensional space.

This would require connecting 3 1D circles, which would be line segments if you think of the pattern of removing slices or using shadows/projections from higher order circles to lower the dimensionality.

If you have two line segments that touch at 1 point on a line, the only line segment on that line that touches both and doesn't intersect either is the line that starts and ends at that point the other two lines share (eg, [-1,0], [0,0], and [0,1]). I don't think this necessarily works due to the third "line" degenerating to a 0D point, but it's the only solution I see that follows the equation, has non-overlapping and unique line segments, and stays in the right number of dimensions.

Edited October 27, 2010 by EventHorizon

[bonanova]

This would require connecting 3 1D circles, which would be line segments if you think of the pattern of removing slices or using shadows/projections from higher order circles to lower the dimensionality.

If you have two line segments that touch at 1 point on a line, the only line segment on that line that touches both and doesn't intersect either is the line that starts and ends at that point the other two lines share (eg, [-1,0], [0,0], and [0,1]). I don't think this necessarily works due to the third "line" degenerating to a 0D point, but it's the only solution I see that follows the equation, has non-overlapping and unique line segments, and stays in the right number of dimensions.

I agree with your point [npi] that a line segment includes its interior; a degenerate triangle therefore does not comprise a 1-dimensional case of three objects that touch only at three distinct points.

But does a circle consistently collapse to a line segment? Consider:

A sphere is a surface - the locus of points in 3-space equidistant from a center point. A circle is a curved line - the locus of points in 2-space equidistant from a center point. It is consistent, then, for a circle to collapse to the locus of points in 1-space equidistant from a center point. That is, a circle collapses to a point pair.

Taking this view, collinear point pairs AB and BC, even when point B is the leftmost or rightmost of the three, touch only at point B, and can be mutually intimate at three distinct points with point pair AC.

The interior volume of a sphere [which does not belong to the sphere, rather to a ball] collapses to the interior area of a circle [which does not belong to the circle, rather to a disk] which collapses to the interior length of a point pair [which does not belong to the point pair, rather to a line segment.

A sphere's exterior, a 2-dimensional [in spherical coordinates] surface area, becomes a circle's exterior, a 1-dimensional [in radial coordinates] perimeter line, which then becomes a point pair's exterior, the 0-dimensional [in any coordinates] points.

[EventHorizon]

Yeah, projections aren't the same as slices as they include the interior (guess I should have thought about it more). With point pairs it makes sense how it would work.