That is to say, it can be understood as a positive space tetrahedron, whose vertex is the center of four spheres, and each side is 2R.

The center of the ball is the geometric center of this regular tetrahedron, and the distance from this geometric center to the four vertices is R+R, and this geometric center is also the center of the circumscribed circle of this regular tetrahedron.

The method for finding the radius of the circumscribed circle of a regular tetrahedron is:

When a regular tetrahedron is placed in a cube, the circumscribed sphere of the cube is the circumscribed sphere of the regular tetrahedron.

Let the side length of a regular tetrahedron be A, the side length of this cube be a sin45, and the radius of the circumscribed sphere of the cube is three times of the root sign of half its side length, then the radius of the circumscribed sphere of the public * * * is six a. R=√6a/4.

So the radius of the circumscribed circle of a regular tetrahedron is √6a/4, and if a=2R, it is √6R/2.

So the relationship of r+R=√6R/2 comes out ~ ~ ~

Can you understand this text? I'll draw a picture for you later, and I'll draw it for you with the drawing board.