Intersecting arcs on a sphere

Question

I'm working through this paper and I'm hung up on Proposition $3.1$. To strip away the context of the problem and present it in another light: suppose there are two intersecting arcs $ab$ and $cd$ on the unit sphere of equal length $d$, then one of the arcs $ac$, $ad$, $bc$, $bd$ has length less than $d$. I'm not very keen on spherical geometry so this is a bit inaccessible to me.

I tried doing a (spherical) law of cosines argument but couldn't get anywhere. I had four equations but they didn't seem easy to manipulate. I also considered the following proposition (which I don't know to be true or not): suppose $d < ac$, then the angle opposite $ac$ must be greater than $\frac{\pi}{2}$. If this is the case, then the inner angles would all add up to something greater than $2\pi$ which would be a contradiction. This would say that one of them has to be $\frac{\pi}{2}$ or smaller, meaning that $ac \le d$ which (I assume) would nearly prove the result.

Thanks for any assistance!

Answer 1

If you can prove the analogous statement in the Euclidean plane, you are done...the four segments produced are strictly shorter, on the sphere, than the four matching segments in the plane. You can make this explicit with the appropriate law of cosines. The general area is Toponogov's theorem.

So, is this thing true in the plane?

Put another way, the way to maximize the shortest of the four segments is to make them all equal, by putting your two original segments crossing at their midpoints and orthogonally. Then you have the appropriate Pythagorean Theorem in either setting.

Answer 2

With Will's help, I've put together a solution. It is not very brief but here goes. All arcs here are made by great circles.

By the spherical law of cosines, if $abc$ makes a triangle then

$$\begin{alignat}{3} & \cos(a) &&= \cos(b)\cos(c)+\sin(b)\sin(c)\cos(A)\\ \cos(b)\cos(c)-\sin(b)\sin(c)\le & &&\le \cos(b)\cos(c)+\sin(b)\sin(c) \\ \cos(b+c) = & &&= \cos(b-c)\end{alignat}$$

Hence $b-c\le a \le b+c$. Consider the diagram below.

Suppose arcs $\stackrel{\frown}{ab}$ and $\stackrel{\frown}{cd}$ have length $D$. Let the arc $\stackrel{\frown}{Od}$ have length $\eta$ and $\stackrel{\frown}{Ob}$ have length $\xi$. Then by the triangle inequality,

$$\begin{align}\lvert\stackrel{\frown}{bd}\rvert&\le \xi+\eta \\ \lvert\stackrel{\frown}{ad}\rvert&\le D-\xi+\eta \\ \lvert\stackrel{\frown}{ac}\rvert&\le D-\xi+D-\eta \\ \lvert\stackrel{\frown}{bc}\rvert&\le \xi+D-\eta\end{align}$$

Adding the first and third inequalities gives $\lvert\stackrel{\frown}{bd}\rvert+\lvert\stackrel{\frown}{ac}\rvert\le 2D$. So either one of $\lvert\stackrel{\frown}{bd}\rvert$ or $\lvert\stackrel{\frown}{ac}\rvert < D$ or $\lvert\stackrel{\frown}{bd}\rvert = D = \lvert\stackrel{\frown}{ac}\rvert$.

By symmetry, the same inequalities hold for the other two arcs. Suppose by contradiction that all of the outside arcs have length $D$, then by the triangle condition above we have that $\lvert\stackrel{\frown}{bc}\rvert-\lvert\stackrel{\frown}{Oc}\rvert\le \xi$ but since $\lvert\stackrel{\frown}{bc}\rvert = D$ and $\lvert\stackrel{\frown}{Oc}\rvert = D-\eta$, we see that $\eta\le\xi$. By symmetry, we also have $\xi\le\eta$ and so $\xi=\eta$.

By the first inequality above, $D-\eta\le\xi$. But since $\eta=\xi$, we see that

$$D-\xi\le\xi\le2D-\xi$$

and hence $\dfrac{D}{2}\le\xi$ and so by symmetry (with $D-\xi$), we see that $\xi=\dfrac{D}{2}$. Since all sides of the triangles in the drawing are equal, the angles $\angle dOa=\angle doB = \angle bOc = \angle cOa = \dfrac{\pi}{2}$ and so the triangles are right triangles. By the spherical Pythagorean theorem (which follows directly from the spherical law of cosines), we have that

$$\cos(\lvert\stackrel{\frown}{ad}\rvert) = \cos(\lvert\stackrel{\frown}{Od}\rvert)\cos(\lvert\stackrel{\frown}{Oa}\rvert).$$

However since $\lvert\stackrel{\frown}{ad}\rvert = D$ and $\lvert\stackrel{\frown}{Od}\rvert = \dfrac{D}{2} = \lvert\stackrel{\frown}{Oa}\rvert$, we have that

$$\cos\left(2\frac{D}{2}\right) = \cos^2\left(\frac{D}{2}\right).$$

Since $\cos(2x) = 2\cos^2x-1$, we see that $\cos^2\left(\dfrac{D}{2}\right) = 2\cos^2\left(\dfrac{D}{2}\right)-1$. This implies that $\frac{D}{2} = 0,2\pi$ but neither of these are admissible since we required that it be a triangle. Hence we conclude that at least one of the outside arcs has length less than $D$.