You may find the
Sphere Cube problem to be similar.
Four identical spheres (like the ones shown in blue in the above cube case) are arranged in a pyramid, such that each sphere is tangent to the other three. If the radius of the four spheres is R, what is the radius r of the largest sphere (such as the one shown in red on the cube picture) that could exist inside the pyramid without overlapping the other spheres?
Consider the tetrahedron with the centers of the four spheres of radius R as its vertices. (Let's start, for simplicity, by considering the radius of each sphere to be 1; we'll multiply by R at the end to scale it up.)
Also consider the center of this tetrahedron connected by straight lines to the four vertices, forming triangles with the edges.
On one of the spheres, the three faces of the tetrahedron that meet there form a spherical triangle on that sphere. Each the sides of that triangle have the measure of one of the angles on the face of the tetrahedron: 60 degrees. From this we can calculate one of the angles of this spherical triangle, which corresponds to a dihedral angle of the tetrahedron: cos 60 = (cos 60)^2 + (sin 60) ^ 2 * cos A, or cos A = (cos 60  (cos 60)^2) / (sin 60)^2 = 1/3.
The center of this spherical triangle is the intersection of the sphere with a line drawn from the vertex at the center of the sphere to the center of the tetrahedron. The arc from this point on the sphere to one of the vertices of the spherical triangle has the measure of one of the base angles of one of the isosceles triangles with a vertex at the center of the tetrahedron and an edge of the tetrahedron as a base. Two such arcs form an isosceles spherical triangle, three of which make up the equilateral triangle we had. As such, the base angles of this spherical isosceles triangle are half the angle A mentioned above, or half the angle that has a cosine = 1/3. The the halfangle formula each has a cosine equal to sqrt(2/3). We actually want the measure of the arc from the center to the vertex of the spherical triangle, and we can get it by solving this isosceles spherical triangle. It's apex angle is 360/3 = 120. Its base is 60 degrees. A base angle has cosine equal to sqrt(2/3) and so a sine of sqrt(1/3). Thus by the law of sines, sin 120 / sin 60 = sin x / sqrt(1/3) = 1, so x is also arcsin(sqrt(1/3)) so its cosine is sqrt(2/3).
Now this angle x is one base angle of the plane isosceles triangle with base at an edge of the tetrahedron and vertex at the center. Half the length of the edge is taken as 1 (the temporary assumed radius of each sphere). The distance from a vertex of the tetrahedron to the center is therefore 1/sqrt(2/3), or sqrt(3/2). Since the spheres take up 1 unit of this, sqrt(3/2)  1 is what is left for the radius of the small sphere at the center.
Multiplying by R, then, r = (sqrt(3/2)  1) * R, or approximately r = 0.224744871391589 * R.

Posted by Charlie
on 20050119 18:35:43 