Main Question or Discussion Point
How many cubes of side 1 centimeter can fit into a sphere of diameter 3 centimeters?
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?
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.
Still having trouble.Imagine a Cartesian coordinate system. It cuts the sphere at six points. You place the cubes "near" these points.
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 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.
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
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.