It is known that two large spherical surfaces with a radius of 1 are tangent to the inner surface of a cylinder with a radius of 1, and the other small spherical surface is circumscribed by the two large spherical surfaces and inscribed in the cylinder. The plane passing through the center of the small ball and the center of the big ball intersects with the cylindrical surface to form an ellipse, and the maximum value of e is found.

Solution: Let the center of a big ball be A, the tangent point of two big balls be B, and the center of a small ball be C. After A and C, many planes can be made, and the intersection lines of these planes and cylinders are ellipses; But there is only one ellipse that maximizes eccentricity e, and the short semi-axis of this ellipse b= cylinder radius1; The long semi-axis of the ellipse a=AD (as shown in the figure). Let the radius of the ball be r; Then in △ABC, AC= 1+r, AB =1; BC= 1-r, so there is an equation in RT△ABC: (1+r)? =( 1-r)？ +1, that is, 1+2r+r? = 1-2r+r？ +1, so r= 1/4.

Sin ∠ CAD = BC/AC = (1-kloc-0//4)/(1+1/4) = 3/5, so a = ad =1/sin ∠. - 1? )=4/3

The maximum eccentricity of an ellipse e=c/a=(4/3)/(5/3)=4/5.