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Sphere itself

  1. May 26, 2010 #1
    How many cubes of side 1 centimeter can fit into a sphere of diameter 3 centimeters?
  2. jcsd
  3. May 26, 2010 #2
    I don't know. How many?
  4. May 26, 2010 #3


    Staff: 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
  5. May 26, 2010 #4
    Try again.
  6. May 26, 2010 #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
  7. May 26, 2010 #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.
  8. May 26, 2010 #7
    Nice question :)
  9. May 26, 2010 #8


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    I say 4, arranged tetrahedrally.

    [EDIT] Doh! I came up with that answer before seeing Diffy's.
  10. May 26, 2010 #9
    I would say 6.
  11. May 26, 2010 #10


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    Arranged how?
  12. May 26, 2010 #11
    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.
  13. May 26, 2010 #12


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    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...
  14. May 26, 2010 #13
    Imagine a Cartesian coordinate system. It cuts the sphere at six points. You place the cubes "near" these points.
  15. May 26, 2010 #14


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    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.
  16. May 26, 2010 #15
    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.


    They will block each other. In this way, I will only be able to put 2.
    Last edited: May 26, 2010
  17. May 26, 2010 #16


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  18. May 26, 2010 #17
    My guess is that the tetrahedron, if it fits, would be asymmetric. Please describe your array.
  19. May 27, 2010 #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.


    Attached Files:

  20. May 27, 2010 #19

    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.
  21. May 27, 2010 #20
    whether we allow the vertices of the cubes to touch the surface of the sphere or not, I say just 2 cubes fit inside.
    Last edited: May 27, 2010
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