Prove that if nine points are placed arbitrarily on a sphere S,then two of these points are no more than $r\sqrt{2}$ apart,where r is the radius of S.

Question

Prove that if nine points are placed arbitrarily on a sphere $S$, then two of these points are no more than $r\sqrt{2}$ apart, where $r$ is the radius of $S$.

I divided my sphere into 8 symmetric regions (great circle), then by the Pigeonhole principle, at least two of the nine points will be on the same region. From there, I don't know how to "maximise" the distance of two points.

Answer 1

Consider one octant of the sphere. Think of each point in the sphere as a vector in $\mathbb{R}^3$ with magnitude $r$. Then the distance between points $u$ and $v$ is $\lvert u-v\rvert =\sqrt{u\cdot u -2u\cdot v + v\cdot v}$

We know that $\lvert u\rvert=\lvert v\rvert = r$, so $u\cdot u=v\cdot v=r^2$. Also, $u\cdot v=\lvert u\rvert \lvert v\rvert \cos \theta$ where $\theta$ is the angle between the $u$ and $v$. Since $u$ and $v$ are in the same octant, $0\le\theta\le \frac{\pi}{2}$ and so $\cos \theta\ge 0$ Hence, $$\lvert u-v\rvert =\sqrt{2r^2-2r^2\cos\theta}\le \sqrt{2r^2}=r\sqrt 2$$

Answer 2

The two points that are in the same octant will uniquely determine a great circle. The two points being in the same octant will be within $\frac {\pi}2r$ distance along the circle from each other.

Orient your coordinate system for this great circle and the plane it occupies so that one of the points is $(0,r)$ and the other is $(r\cos \theta, r\sin \theta)$ where $\theta$ is the angle between these two points ($0 < \theta \le \frac{\pi}2$).

The distance between them is ... well, now it's just work... I hate work.... $\sqrt{(r\cos\theta - 0)^2 + (r\sin \theta - r)^2} = r\sqrt{\cos^2 \theta + (1 - \sin\theta)^2} = r\sqrt{2-2\sin \theta} \le \sqrt 2 r$.