The answer is:
r = (sqrt(3/2)–1)*R = 0.2247R
A solution is:
If we connect the centers of all 4 blue spheres, you will see that this forms a regular tetrahedron, with edge length equal to 2R. The simplest method of approach is to realize that the center of the new red sphere must be at the centroid of the tetrahedron, due to symmetry. Once we find the centroid, we can find the distance D between the centers of the new red sphere and any of the blue spheres. The maximum radius of the new sphere occurs when R + r = D.
I could prove the following properties for you, but they have already been derived and are easily found either in geometry books or on the web (please note: a distance between a point and a plane will always be the perpendicular distance):
We will call one of the faces the Base of the tetrahedron, and we will call the distance from the Base to the center of the top blue sphere the Height, H. The centroid of the tetrahedron is located a distance H/4 up from the Base.
Notice that the centers of both the top blue sphere and the new red sphere are directly above the center of the Base. Well, if one is H away from the base, and the other is H/4 away from the base, this means that the two centers are a distance D = H- H/4 = 3/4H away from each other.
The Height of a tetrahedron is simply (EdgeLength)*sqrt(2/3) = 2R*sqrt(2/3) = H. This means:
D = 3/4*H = 3/4(2R*sqrt(2/3)) = 3/2*R*sqrt(2/3) = R*sqrt(3/2)
D = R + r so r = D-R
r = R*sqrt(3/2) – R = R[sqrt(3/2) – 1]
r = 0.2247R
There is also a very interesting approach by David Shin where he maps the spheres onto 4 vertices of a cube and solves from there. Check out his comment here. |
