Solution:

Let the entry and exit points be A and B. Then all posible paths of length
101 lie in an ellipsoid such that the sum of the distance from A and that from
B is less than 101 units. But the centre of the sphere cannot be inside this
ellipsoid as the sum of the distance from A to the centre and from B to the
centre is 102.

Now the plane through the centre normal to the plane containing the points
A, B and the centre does not intersect the ellipsoid, and so divides the sphere
(ignoring the hole) into two equal hemisphere.

RonL

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Truth does not change because it is, or is not, believed by a majority of the people.

Giordano Bruno