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Question: Caught in the Hemisphere Two points on a sphere of radius 1 are joined by an arc of length less than 2, lying inside the sphere. Prove that the arc must lie in some hemisphere of the given sphere. Answer: Let A = (a_x, a_y, a_z) and B = (b_x, b_y, b_z) be vectors in R^3 to denote
the two points on the unit sphere and l the arc through A and B: l = A +
t(B-A) for t in [0, 1].
Let U be the plane through zero (0, 0, 0), perpendicular to vector M = (A +
B) / 2. So U: (a_x+b_x)/2 * x + (a_y+b_y)/2 * y + (a_z+b_z)/2 * z = 0. Plane
U divides the unit sphere into two hemispheres.
By construction, the arc l is parallel to the plane U. This leaves us with
two possibilities: all points of arc l lie on U or no points at all.
Suppose all points of arc l lie on U.
If zero is on the arc l, then M must be zero and thus, the length of arc l
is 2. This is a contradiction.
If zero is not on the arc l, then M is non-zero. But then U is not
perpendicular to M. This is also a contradiction.
Therefore, arc l shares no points with plane U at all. So the arc l lies in
one of the two hemispheres that U creates.
QED.
-Ron L.J. van den Burg
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