Three Points on Sphere in Same Hemisphere
Theorem
Let $S$ be a sphere.
Let $A$, $B$ and $C$ be points on $S$ which do not all lie on the same great circle.
Then it is possible to divide $S$ into two hemispheres such that $A$, $B$ and $C$ all lie on the same hemisphere.
Proof
Because $A$, $B$ and $C$ do not lie on the same great circle, no two of these points are the endpoints of the same diameter of $S$.
Otherwise it would be possible to construct a great circle passing through all $3$ points $A$, $B$ and $C$.
Let a great circle $E$ be constructed through $A$ and $B$.
Then as $C$ is not on $E$, it is on either one side or the other.
Hence there is a finite spherical angle $\phi$ between $E$ and $C$.
Let a diameter $PQ$ of $S$ be constructed whose endpoints are on $E$ such that:
- neither $P$ nor $Q$ coincides with $A$ or $B$
- both $A$ and $B$ are on the same semicircle of $E$ into which $PQ$ divides $E$.
Let two great circles $F$ and $G$ be constructed through $PQ$ which are at a spherical angle $\dfrac \phi 2$ from $E$, one in one direction and one in the other.
Then $F$ and $G$ both divide $S$ into two hemispheres:
- one such division has a hemisphere which contains $A$ and $B$ but does not contain $E$
- the other such division has a hemisphere which contains $A$, $B$ and $E$.
Hence the result.
$\blacksquare$
Sources
- 1976: W.M. Smart: Textbook on Spherical Astronomy (6th ed.) ... (previous) ... (next): Chapter $\text I$. Spherical Trigonometry: $2$. The spherical triangle.