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## Main Question or Discussion Point

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

- Thread starter Loren Booda
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- #1

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

- #2

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I don't know. How many?

- #3

Mark44

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Since the volume of such a sphere is ~ 9.42 cm

- #4

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Try again.

Since the volume of such a sphere is ~ 9.42 cm^{3}, no more than 9 unbroken cubes could possibly fit into this sphere

- #5

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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

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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

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Nice question :)

- #8

DaveC426913

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I say 4, arranged tetrahedrally.

[EDIT] Doh! I came up with that answer before seeing Diffy's.

[EDIT] Doh! I came up with that answer before seeing Diffy's.

- #9

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I would say 6.

- #10

DaveC426913

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Arranged how?I would say 6.

- #11

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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:Arranged how?

[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

DaveC426913

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

P.S. Must. Try. So Hard. To Take. You Seriously.

Username... Seriousness fading...

- #13

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Imagine a Cartesian coordinate system. It cuts the sphere at six points. You place the cubes "near" these points.Uh. OK. So arranged how? You said six cubes. How do you see them being arranged?

- #14

DaveC426913

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Still having trouble.Imagine a Cartesian coordinate system. It cuts the sphere at six points. You place the cubes "near" these points.

1] Seems to me it would cut the sphere via

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

- #15

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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.Still having trouble.

1] Seems to me it would cut the sphere viaplanes, 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 mayor may nottouch each other.

Ahh:

They will block each other. In this way, I will only be able to put 2.

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DaveC426913

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Right.Ahh:

They will block each other. In this way, I will only be able to put 2.

- #17

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My guess is that the tetrahedron, if it fits, would be asymmetric. Please describe your array.I say 4, arranged tetrahedrally.

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Since the volume of such a sphere is ~ 9.42 cm^{3}, 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.

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whether we allow the vertices of the cubes to touch the surface of the sphere or not, I say just 2 cubes fit inside.How many cubes of side 1 centimeter can fit into a sphere of diameter 3 centimeters?

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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... :(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.

- #22

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[PLAIN]http://img231.imageshack.us/img231/3039/cubes.png [Broken]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... :(

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- #23

DaveC426913

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No. The tetrahedron is the same as the 1x2x2 square, just with one pair rotated 90 degrees.My guess is that the tetrahedron, if it fits, would be asymmetric. Please describe your array.

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.

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