So there is a question which was aked to me a long while ago (and I don't remember what was the answer) so here goes:
Captain Kirk is on a remote planet again and there is a monster trying to eat him. Fortunately, captain Kirk has a super-shielding device with him. The device casts a perfectly circular 10 meter around it. Captain Kirk uses it to protect himself from the monster. The problem is that while he is inside the protective range of the device, captain Kirk cannot be beamed away. As soon as he leaves the shield, he can (instantly) be beamed away.
What is the highest ratio $\dfrac{\text{speed of the monster}}{\text{speed of Kirk}}$ so that Kirk still has a chance to escape?
So in math terms:
the prey is inside a circle of radius $r$ (the actual value does not matter, so say it's 1).
the predator is outside the circle and cannot enter.
if the predator is at the point of exit of the prey, then the predator wins.
otherwise, the prey wins.
The answer should hold independently of the starting positions. (In other words, assume a worst-case scenario for the prey).
The answer is not $\pi$!
A first thought of many (including me) is to say the answer is $\pi$. This turns out to be wrong.
First, one can assume that the prey begins at the center of the circle and the monster on the boundary. This case encompasses the worst-case scenario: if the prey is not in the center, then it can go to the center; if the predator is not on the boundary of the circle, then it's only better for it to be on the boundary.
Next, consider the following strategy. The prey start to run for the point on the boundary opposite to the predator. The predator will then start to run along the circle either in clockwise or counterclockwise manner. Once the prey is at the middle of this radius, it makes an angle (about 37°) to go a point which is further away form the predator. Note that the angle is too small (so it's not worth for the predator to start to run in the opposite direction). Both the predator and the prey have now a longer way to go, but the predator more so. With the given data, the predator would need a speed of at least 3.3.
Even better
Here is a better strategy. Decompose the motion of the prey as radial and angular. Note that when the prey is closer to the center a small speed is sufficient to get a large angular speed.
Assume for simplicity that the speed of the prey is $1$ and that of the predator $v$. Then the prey can run to some point on the border (and the predator will start to run around the circle). As long as the prey is inside a smaller circle of $\tfrac{1}{v}$ the radius, it may have a higher angular speed than the predator. This means it can use parts of its speed to move radially until it gets as close as possible to this radius all the while haveing the predator on the opposite side of the circle.
After a very long time, being as close as desired to the circle of radius $\tfrac{1}{v}$ the prey can try to dash for the point opposite the predator.
This means that the prey has $1-\tfrac{1}{v}$ time units to the boundary while the predator would need $\tfrac{\pi}{v}$. This shows that as long as $v < \pi +1$ the prey can win.
Now obviously, you can still improve upon that later strategy (for example, by using the former strategy at some point). But in the end, I still don't know what is the actual supremum of the speed ratios.
