Peaceful Giants

In the image above shown are the front views of the positions of two giants, standing at points $G_1,G_2$ on the sphere, with center at $O$ . Each of their horizons touch the ground at an angle $\theta=\cos^{-1}\left(\frac{R}{R+h}\right)=\cos^{-1}\left(\frac{6000}{6928.203}\right)=\frac{\pi}{6}$ . So, the arc joining the two points $G_1,G_2$ makes an angle $\alpha$ at the center, with $\alpha\ge \frac{\pi}{3}$ , that is the angle between the two vectors $\vec{OG_1},\ \vec{OG_2}$ , with the same magnitudes $R$ , is larger than $\frac{\pi}{3}$ . Then the problem becomes finding the maximum number of such vectors of same magnitudes such that the angle between any two of them is larger than $\frac{\pi}{3}$ . This is known as the kissing number problem , which is a special case of the more general problem, called the spherical codes problem . The general statement of the problem (in Euclidean space) is the following:

Let $S=\{\mathbf{x}_i, 1\le i\le M,\ \mathbf{x}_i\in \mathbb{R}^n: \|\mathbf{x}_i\|_2=1,\ i=1,2,\cdots,\ M,\ \ \langle\mathbf{x}_i,\mathbf{x}_j\rangle\le t,\ \forall i\ne j\}$ be a finite set of unit vectors such that their inner products are smaller than some number $t,\ t\in [-1,1]$ . Then find the maximum cardinality of such a set $S$ for fixed $n$ .

This problem is in general an open problem, and for a few cases of small $n$ , the answers are known. For $n=3,\ t=1/2$ , the answer is $12$ (though the proof is quite non-trivial. An elementary proof can be found in this paper ). So the answer to our problem is also $\boxed{12}$ .

It's not a complete solution. I'm just sharing the last step of it. Think of the sphere as the Earth (of course a completely spherical one). Now, you can position 6 giants along the equator. And, the north and the south pole can each contain 3 giant's roaming ground. So, in total you can put 6+3+3=12 giants in the spherical Earth.

At the end, I should explain why each pole can contain 3 giants. Consider one of the giants in the equator. How can we put another giant closest to this one? Of course their roaming ground should just touch (Let's call the touching point A). Now, the second giant can walk around in a restricted way. The opposite point of A in the roaming ground of the second giant now creates a locus bisecting the Earth. Now, consider another giant in the equator next to the first one. Similarly, his closest giant should make another such trajectory. You would see the two trajectories intersect in such a way that they enclose 120 degrees in the longitudinal measure. So, there must be 3 enclosed areas such as the above one in each pole.

[I've tried hard to explain it simply. But, I'm not sure about the clarity of the explanation. So, I'd appreciate your efforts if you help me find out the ambiguities here. Thank you.]