How many ways can 2 1x2s be rotated to fit into a sphere?

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In summary, the conversation is discussing how many cubes of side 1 centimeter can fit into a sphere of diameter 3 centimeters. There are different opinions on the number, with some saying 4, 5, 6, or 8. The conversation also touches on mathematical concepts such as volume, the Pythagorean theorem, Cartesian coordinates, and the equation for a circle. Ultimately, it is concluded that the maximum number of cubes that can fit inside the sphere is 4, arranged in a tetrahedral or square prism formation.
  • #1
Loren Booda
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How many cubes of side 1 centimeter can fit into a sphere of diameter 3 centimeters?
 
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  • #2
I don't know. How many?
 
  • #3
I'll say 8.

Since the volume of such a sphere is ~ 9.42 cm3, no more than 9 unbroken cubes could possibly fit into this sphere
 
  • #4
Mark44 said:
I'll say 8.

Since the volume of such a sphere is ~ 9.42 cm3, no more than 9 unbroken cubes could possibly fit into this sphere

Try again.
 
  • #5
I certainly agree. If you can't bend or break the cubes or the sphere you can't fit 8.

It is natural to think so though. You can fit four squares of side length 1 into a circle of diameter 3. You can easily see this, the equation for a circle at the origin with diameter 3 is x^2 + y^2 = 2.25. So if you put the four corners of squares at the origin you can see they fit because the point (1,1) fits well within the circle since 1^2 + 1^2 < 2.25

However in 3 dimensions the equations change a bit. Here you have the sphere as x^2 + y^2 + z^2 = 2.25 and if you tried to fit 8 cubes all with corners meeting at the origin you would need the point (1 , 1, 1) to be inside the sphere. But alas, 1^2 + 1^2 + 1^2 > 2.25.

I think the answer here is again 4 cubes. And this depends on whether or not you allow the cubes to intersect the sphere in exactly one point.

For if the centered the four cubes at the origin their corners would hit the points (+/- 1, +/- 1, +/- .5) and here you have the corners intersecting with the sphere since 1^2 + 1^2 + .5^2 = 2.25
 
  • #6
Close enough.

How many cubes of side 1 centimeter can fit into a sphere of diameter 3 centimeters?

Four, stacked 1 x 2 x 2. By the Pythagorean theorem, 1^2+2^2+2^2=3^2.
 
  • #7
Nice question :)
 
  • #8
I say 4, arranged tetrahedrally.


[EDIT] Doh! I came up with that answer before seeing Diffy's.
 
  • #9
I would say 6.
 
  • #10
Dickfore said:
I would say 6.

Arranged how?
 
  • #11
DaveC426913 said:
Arranged how?

Imagine a cube with all vertices on one of its side touching the sphere. The height of the spherical cap cut by the opposite side of the cube is:

[tex]
h = a + R - \sqrt{R^{2} - \frac{a^{2}}{2}}
[/tex]

Inside the spherical cap, it's impossible to accommodate one more cube. Imagine doing this on six orthogonal directions. We cut 6 spherical caps and the central region is a cube with a side:

[tex]
b = 2 R - 2 h = \sqrt{(2 R)^{2} - 2 a^{2}} - 2 a = \sqrt{7} - 2 = 0.646 < 1 = a
[/tex]

Therefore, we cannot accommodate a single extra cube in this internal cube.
 
  • #12
Dickfore said:
Imagine a cube with all vertices on one of its side touching the sphere. The height of the spherical cap cut by the opposite side of the cube is:


Therefore, we cannot accommodate a single extra cube in this internal cube.

OK. So arranged how? You said six cubes. How do you see them being arranged?



P.S. Must. Try. So Hard. To Take. You Seriously.
Username... Seriousness fading...
 
  • #13
DaveC426913 said:
Uh. OK. So arranged how? You said six cubes. How do you see them being arranged?

Imagine a Cartesian coordinate system. It cuts the sphere at six points. You place the cubes "near" these points.
 
  • #14
Dickfore said:
Imagine a Cartesian coordinate system. It cuts the sphere at six points. You place the cubes "near" these points.

Still having trouble.
1] Seems to me it would cut the sphere via planes, not points.
2] Do these 6 cubes touch each other?

Oh, I think I see. We're all building a mass of cubes glued together in the middle. You're attaching the cubes to the inside surface of the sphere. So, they touch the sphere at 6 equidistant points. They may or may not touch each other.
 
  • #15
DaveC426913 said:
Still having trouble.
1] Seems to me it would cut the sphere via planes, not points.
2] Do these 6 cubes touch each other?

Oh, I think I see. We're all building a mass of cubes glued together in the middle. You're attaching the cubes to the inside surface of the sphere. So, they touch the sphere at 6 equidistant points. They may or may not touch each other.

No, each cube touches the sphere with 4 points, but there are 6 such cubes situated around the intersections of the coordinate axes with the sphere. They do touch each other. That is why we cannot insert a cube in the central cube. The plane in which the opposite sides of each cube lie form a cube inside the sphere.

Ahh:

They will block each other. In this way, I will only be able to put 2.
 
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  • #16
Dickfore said:
Ahh:

They will block each other. In this way, I will only be able to put 2.
Right.
 
  • #17
DaveC426913 said:
I say 4, arranged tetrahedrally.

My guess is that the tetrahedron, if it fits, would be asymmetric. Please describe your array.
 
  • #18
There can definitely be 4 cubes. Just order them to form a square prism with side 2 and height 1. Here is the relevant diagram.


attachment.php?attachmentid=26016&stc=1&d=1274937452.png

 

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  • #19
Mark44 said:
I'll say 8.

Since the volume of such a sphere is ~ 9.42 cm3, no more than 9 unbroken cubes could possibly fit into this sphere


I made it 14.137 cm3 which is reasonable as a 3x3x3 cube is 27cm3

sin 45 = 1.06 so you won't get a 2x2x2 cube in.

You will get a 2x2=4 in easy, then I think you can squeeze another in on top of that, but that's you lot. So I say 5.

Don't think you can get a sixth one in.
 
  • #20
Loren Booda said:
How many cubes of side 1 centimeter can fit into a sphere of diameter 3 centimeters?

whether we allow the vertices of the cubes to touch the surface of the sphere or not, I say just 2 cubes fit inside.
 
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  • #21
Dickfore said:
There can definitely be 4 cubes. Just order them to form a square prism with side 2 and height 1. Here is the relevant diagram.


attachment.php?attachmentid=26016&stc=1&d=1274937452.png


can you upload that image to imageshack or something? I'd like to see it and it's saying "pending attachment aproval" on my end... :(
 
  • #22
tauon said:
can you upload that image to imageshack or something? I'd like to see it and it's saying "pending attachment aproval" on my end... :(

[PLAIN]http://img231.imageshack.us/img231/3039/cubes.png [Broken]
 
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  • #23
Loren Booda said:
My guess is that the tetrahedron, if it fits, would be asymmetric. Please describe your array.

No. The tetrahedron is the same as the 1x2x2 square, just with one pair rotated 90 degrees.

So, if the 1x2x2 can fit, then any pair of the four can be rotated (and by any amount, not just 90 degrees).

This is why I realized I was overly-complicating things. Restricting it to a tetrahedron is assigning needless structure. It is simply 2 1x2s rotating at any angle (inclduing zero) through a common axis.
 

1. How do you define a rotation for a 1x2?

A rotation for a 1x2 refers to the movement of the two-dimensional shape in a circular motion around a fixed point, such as its center or a designated pivot point.

2. Is there a limit to the number of rotations a 1x2 can have?

There is no limit to the number of rotations a 1x2 can have. As long as the shape remains within the boundaries of the sphere and does not overlap with itself, it can rotate infinitely.

3. Can a 1x2 be rotated in any direction within the sphere?

Yes, a 1x2 can be rotated in any direction within the sphere as long as it does not intersect with itself or exceed the boundaries of the sphere.

4. How many unique ways can a 1x2 be rotated in a sphere?

There are an infinite number of unique ways a 1x2 can be rotated in a sphere. Each rotation may have a slightly different orientation or position within the sphere, making it unique.

5. Are there any limitations to how a 1x2 can be rotated in a sphere?

The only limitation to how a 1x2 can be rotated in a sphere is that it cannot intersect with itself or exceed the boundaries of the sphere. Other than that, it can rotate in any direction and orientation.

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