IMC 2002 · Day 2 · P10

We can assume $OA = OB = OC = 1$. Intersect the unit sphere with center $O$ with the angle domains $AOB$, $BOC$ and $COA$; the intersections are “slices” and their areas are $\frac{1}{2}\gamma$, $\frac{1}{2}\alpha$ and $\frac{1}{2}\beta$, respectively.

Now project the slices $AOC$ and $COB$ to the plane $OAB$. Denote by $C'$ the projection of vertex $C$, and denote by $A'$ and $B'$ the reflections of vertices $A$ and $B$ with center $O$, respectively. By the projection, $OC' < 1$.

The projections of arcs $AC$ and $BC$ are segments of ellipses with long axes $AA'$ and $BB'$, respectively. (The ellipses can be degenerate if $\sigma$ or $\tau$ is right angle.) The two ellipses intersect each other in 4 points; both half ellipses connecting $A$ and $A'$ intersect both half ellipses connecting $B$ and $B'$. There exist no more intersection, because two different conics cannot have more than 4 common points.

The signed areas of the projections of slices $AOC$ and $COB$ are $\frac{1}{2} \alpha \cdot \cos\tau$ and $\frac{1}{2} \beta \cdot \cos\sigma$, respectively.

The statement says thet the sum of these signed areas is less than the area of slice $BOA$.

There are three significantly different cases with respect to the signs of $\cos\sigma$ and $\cos\tau$ (see Figure).

If both signs are positive (case (a)), then the projections of slices $OAC$ and $OBC$ are subsets of slice $OBC$ without common interior point, and they do not cover the whole slice $OBC$; this implies the statement. In cases (b) and (c) where at least one of the signs is negative, projections with positive sign are subsets of the slice $OBC$, so the statement is obvious again.