# Maximum number of half great circles of length $\pi$ can be drawn on a sphere without any intersection

It is well-known that any two great circles intersects on a sphere. In fact, there are infinitely many half great circles can be drawn on a sphere with a common intersection. Intuitively, it seems to me that there is only finitely many half great circles can be drawn on a sphere if one do not allow intersection.

I try to investigate this by considering stereographic projection.

I came up with the above picture in which the yellow edge, blue edge and red edge on a plane correspond to the three geodesics of length $$\ge \pi$$ on the sphere. Since there is no intersection among the yellow edge, blue edge and red edge, the three geodesics do not intersect either.

In other words, this picture shows that $$3$$ geodesics of length $$\ge \pi$$ can be drawn on a sphere without any intersection. In particular, this implies that $$3$$ half great circles can be drawn on a sphere without any intersection.

If one draw the fourth circle on top of the three, it seems impossible to draw a fourth edge on the circle which corresponds to a geodesic of length $$\ge \pi$$ on the sphere.

Conjecture:

The maximum number of half great circles can be drawn on a sphere is $$3$$ if one do not allow intersection.

How to prove/disprove this conjecture rigorously?

You can have arbitrarily many disjoint great semicircles on a sphere. Picture cutting a small regular $$n$$ gon $$A_1A_2\dots A_{n}$$ centered at the north pole, and let $$B_1B_2\dots B_{n}$$ be the polygon around the south pole, where each pair $$A_i,B_i$$ is antipodal. Join $$A_1,B_1$$ by a great semicircle contained in the complement of both polygons. The other semicircles are obtained from the first one by rotating the sphere on the axis that joins the two poles. You can see that the intersection of the semicircles with any horizontal plane forms a regular polygon, and in particular the semicircles have no common intersection.

• I'm not convinced that this works, so some coordinates would help. Suppose that in cylindrical coordinates $(r,\theta,z)$, $A_i=(\sin \epsilon, 2i\pi/n, \cos \epsilon)$, $B_i = (\sin \epsilon, \pi+2i\pi/n, -\cos \epsilon)$. What are the coordinates for the midpoints of the intended great semicircles?
– user44143
Commented Dec 31, 2019 at 15:57
• @MattF. I believe that the $\theta$ coordinate of $B_i$ in your notation should be $\pi+2i\pi /n$. The coordinate for the midpoints are $(1, \pi/2+2i\pi/n, 0)$. See figure 3 here for a picture of $n=4$ heldermann-verlag.de/jgg/jgg13/j13h1pani.pdf ) Commented Dec 31, 2019 at 16:07
• Here's an animated gif of Gjergji's construction in the case $n=10$. <math.ubc.ca/~israel/problems/circles.gif> Commented Dec 31, 2019 at 16:14
• Yes, I agree now. It also helped me to do it in Mathematica: f[e_, i_, t_] := Cos[t] {Sin[e] Cos[i], Sin[e] Sin[i], Cos[e]} + Sin[t] {Sin[i], -Cos[i], 0}; ParametricPlot3D[Table[f[0.1, i, t], {i, 0, 2 Pi, 2 Pi/5}], {t, 0, Pi}]
– user44143
Commented Dec 31, 2019 at 16:33

Here is another way to interpret Gjergji's answer.

Imagine a satellite orbits a spherical Earth in low orbit along a circular path whose maximum north and south latitudes are some angle $$<90°$$. These maximum latitude points are antipodal to each other, and we can divide the orbit into two halves -- one running from the northern antipode to the southern and the other from the southern antipode to the northern as the satellite completes its loop.

Now take the north-to-south half and translate it eastwards by any amount. It becomes evident that the translated half-orbit does not intersect the original one, or any other translation. The south-to-north half-orbit has a similar property.

Thus a set of disjoint half-orbits is not only potentially infinite, but also potentially uncountably infinite, and such a set may be placed in one-to-one correspondence with all the half-orbits!

This construction depends on the half-orbits sharing a common small circle tangent at each endpoint. Counting possible disjoint half-orbits without such a pair of common tangents is less trivial.