Today’s puzzle is from The Riddler, and has to do with spherical geometry.

A guardian constantly patrols a spherical planet, protecting it from alien invaders that threaten its very existence. One fateful day, the sirens blare: A pair of hostile aliens have landed at two random locations on the surface of the planet. Each has one piece of a weapon that, if combined with the other piece, will destroy the planet instantly. The two aliens race to meet each other at their midpoint on the surface to assemble the weapon. The guardian, who begins at another random location on the surface, detects the landing positions of both intruders. If she reaches them before they meet, she can stop the attack.

The aliens move at the same speed as one another. What is the probability that the guardian saves the planet if her linear speed is 20 times that of the aliens’?

Here is my solution for the case of interest, where the guardian is faster than the intruders.

[Show Solution]

### Fast guardian, flat planet

The fact that everything takes place on a sphere makes matters complicated, so let’s start with a simpler problem: suppose the world is flat, and we would like to know the set of possible starting points for the guardian such that the planet is saved. Let the alien starting locations be $A$ and $B$, and denote their rendezvous point $M$ (the midpoint of $AB$). Let $X$ be the point along $AB$ where the guardian intercepts one of the intruders. If the guardian is $q$ times faster than the intruders, then by the time an intruder has traveled to $X$, the guardian may have traveled a distance $q$ times greater. Therefore, the set of locations the guardian could have come from is a circle centered at $X$ and of radius $r = \min(AX,BX)$. See below for an illustration.

Since the guardian is *faster* than the intruders ($q \ge 1$), then the guardian might as well intercept at $M$. In other words, the possible guardian starting points for $M$ contains the starting points for all other intercept locations! The following animation shows the case $q=1.5$. The red circle shows the set of starting points for the guardian that leads to saving the planet (it’s a circle of radius $\tfrac{1}{2} q \cdot AB$ centered at $M$).

### Fast guardian, spherical planet

Things get interesting when the planet is spherical. Here, the solution is the same in principle, but wrapped around a sphere. Again, let $A$ and $B$ be the intruders’ positions. The shortest path connecting $A$ and $B$ is a great circle of angle $0 \le \theta \le \pi$. If $q \ge 1$, the guardian’s positions that guarantee safety will be a spherical cap defined by an angle $\tfrac{1}{2}q \theta$.

If $\tfrac{1}{2}q \theta \ge \pi$, then the guardian will always be able to intercept the intruders, leading to certain safety. If the angle is less than $\pi$, the ratio of unsafe area to total area is

\[

\frac{\text{unsafe area}}{\text{total area}} = \frac{2\pi R^2 \left(1 – \cos(\pi – \tfrac{1}{2}q \theta)\right)}{4\pi R^2} = \frac{1}{2}\left(1 + \cos(\tfrac{1}{2}q \theta)\right)

\]

Therefore, the probability that the planet is destroyed as a function of $\theta$ is:

\[

f(\theta) = \begin{cases}

\frac{1}{2}\left(1 + \cos(\tfrac{1}{2}q \theta)\right) & \theta \le \frac{2\pi}{q} \\

0 & \theta > \frac{2\pi}{q}

\end{cases}

\]

Note that if $1 \le q < 2$, then $\frac{2\pi}{q} > \pi$ and so the second case never occurs; planetary destruction is always a possibility if the guardian is less than twice as fast as the intruders! In order to calculate the total probability, we need to average over all $\theta$, since the distribution of the intruders’ starting locations is random. However, it’s not as simple as averaging over $\theta$, as doing so would bias the distribution towards small $\theta$. Since we want the distribution to be uniform over the sphere, the angle $\theta$ will have probability distribution $p(\theta) = \tfrac{1}{2}\sin\theta$. For more information about where this comes from see this article. The total probability of destruction is therefore

\begin{align}

\int_0^\pi f(\theta) p(\theta) \, d\theta

&= \int_0^{\min(\pi,\,2\pi/q)} \frac{1}{2}\left(1 + \cos\bigl(\tfrac{1}{2}q \theta\bigr)\right) \frac{1}{2}\sin\theta\, d\theta \\

&= \begin{cases}

\frac{6-q^2+2 \cos\bigl(\tfrac{\pi q}{2}\bigr)}{2 (4-q^2)} & 1 \le q < 2 \\
\frac{q^2-8-q^2 \cos\bigl(\tfrac{2 \pi }{q}\bigr)}{4 (q^2-4)} & q \ge 2
\end{cases}
\end{align}
To find the probability of safety, we compute $1$ minus the probability of destruction, and we obtain:

$\displaystyle

\mathbb{P}(\text{planet is saved}) = \begin{cases}

\frac{2-q^2-2 \cos\bigl(\tfrac{\pi q}{2}\bigr)-6}{2 (4-q^2)} & 1 \le q < 2 \\
\frac{3q^2-8-q^2 \cos\bigl(\tfrac{2 \pi }{q}\bigr)}{4 (q^2-4)} & q \ge 2
\end{cases}
$

Here is a plot of the probability of safety as a function of the guardian’s relative speed.

To answer the specific question asked, if the guardian’s speed is $20$ times the speed of the intruders, the probability that the planet will be saved is

$\displaystyle

\mathbb{P}(\text{planet is saved}) =\frac{149+50\cos(\pi/10)}{198} \approx 99.27\%

$

So with a guardian this fast, there is a very good chance the planet will be safe from the aliens!

Here is a partial solution for the more complicated case where the guardian is slower than the intruders.

[Show Solution]

If the guardian is *slower* than the intruders ($0 < q < 1$), then the solution is no longer a circle, because it might happen that the guardian cannot intercept at $M$ but *can* intercept at some other point along $AB$. The following animation illustrates the case $q=0.5$. The red shape shows the set of starting points for the guardian that leads to saving the planet (it’s a circle of radius $\tfrac{1}{2} q \cdot AB$ centered at $M$ together with the region enclosed by the tangent segments drawn to $A$ and $B$).

Unfortunately, things get very ugly from here. If we map the situation onto a sphere, then the segments on either side are no longer straight lines, of course, but they aren’t great circles either! To find the boundary of interest, let’s say the intruder that starts at $B$ is intercepted at $T$, and the guardian leaves from $D$. If we assume the guardian just barely makes it to $T$, then the arc $\widehat{DT}$ should be $q$ times the arc $\widehat{BT}$. Let’s suppose $\phi$ is fixed, and we’ll seek the maximum possible $\psi$ such that there exists some $\tau$ such that $D$ reaches $T$ in time.

After some ugly math, we can solve for the set of $(\phi,\psi)$ as follows: fix $\phi$, then solve for $\tau$ such that $\tan(\tau-\phi) = q \tan(q\tau)$. If the associated $\tau$ is greater than $\tfrac{1}{2}\theta$, then simply set $\tau = \tfrac{1}{2}\theta$. Then, find $\psi$ such that $\cos(\tau-\phi)\cos\psi = \cos (q\tau)$.

So for each fixed $\phi$, we can use the above procedure to find $\tau$ and $\psi$. By symmetry, we can repeat on the other side to get the part of the curve where we intercept the intruder leaving $A$ instead. Unfortunately, these equations do not admit an analytic solution, so they must be solved numerically. We can visualize what the region will look like, but there doesn’t appear to be a nice way to solve for its boundary or compute its area. Nevertheless, here is an illustration of the “safe zone” where the guardian will be able to intercept the intruders, for the case $q=0.5$.