Sphere itself

Main Question or Discussion Point

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

Answers and Replies

I don't know. How many?

Mark44
Mentor
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'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.

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

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.

Nice question :)

DaveC426913
Gold Member
I say 4, arranged tetrahedrally.

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

I would say 6.

DaveC426913
Gold Member
I would say 6.
Arranged how?

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:

$$h = a + R - \sqrt{R^{2} - \frac{a^{2}}{2}}$$

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:

$$b = 2 R - 2 h = \sqrt{(2 R)^{2} - 2 a^{2}} - 2 a = \sqrt{7} - 2 = 0.646 < 1 = a$$

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

DaveC426913
Gold Member
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...

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.

DaveC426913
Gold Member
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.

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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DaveC426913
Gold Member
Ahh:

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

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

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.

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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.

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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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.

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... :(

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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DaveC426913
Gold Member
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.