The quick brown fox and the lazy duck

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.

The answer is not $\pi$ or $<\pi$ !

A first thought of many (including me) is to say the answer is $\pi$ . This corresponds to the prey simply running for the point opposite to the predator on the other side of the pond. This turns out to be wrong.

Here is a better strategy. Decompose the motion of the prey (and predator) 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. However the predator has always a fixed angular speed (he could choose to go clockwise or counterclockwise, but everything less is useless).

The prey starts to 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 $\tfrac{1}{g}$ 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 having the predator on the opposite side of the circle.

After a very long time, being as close as desired to the circle of radius $\tfrac{1}{g}$ the prey can try to dash for the point opposite the predator.

This means that the prey has $1-\tfrac{1}{g}$ time units to the boundary while the predator would need $\tfrac{\pi}{g}$ to reach that point. $\begin{array}{rrl} & 1-\tfrac{1}{g} & < \tfrac{\pi}{g} \\ \iff & g - 1 < \pi \\ \iff & g < \pi +1 \end{array}$ This shows that as long as $g < \pi +1$ the prey can win. So the answer is $\fbox{It's \$>\pi\$}$ .

Now obviously, you can still improve upon that later strategy (the prey can re-adjust the point on the edge of the pond at which it is aiming, see below). The actual maximal speed is [according to the reference] about $\approx 4.6$ . See this site for details.

Another strategy.

Another strategy is that the duck aims [at first] for the point opposite the fox on the edge. The fox then starts to run either clockwise or counter-clockwise. When the duck has overcome half the radius, he can start to swim in a different direction (his path gets longer, but the path of the fox gets even longer). It turns out that if he then makes an angle (w.r.t. his current direction) of $\approx 37°$ , he can beat any fox with a speed $g < 3.3$ . It's far from being the best strategy, but it's still sufficient to see that the answer is not $\leq \pi$ .

One can try to optimise the moment at which the duck should go in a different direction and the angle of that direction. (e.g. if you go for a third of the radius and an angle of $\approx 26°$ you can beat speeds $g < 3.5$ ).

This solution may not show the actual maximal value of $s$ , but is sufficient to show that for the following strategy, the maximal value of $s$ is greater than $\pi$ :

The duck can start by swimming to the center $D$ of the pond, and then by swimming directly away from where the fox is sitting at point $F$ . The fox must pick a direction to chase the duck as soon as possible to catch it. Once it does, the duck should veer away from it by swimming along the arc of a circle with the same radius $r$ as the pond and a center $O$ such that $\angle ODF$ makes a right angle, to reach land at point $E$ .

Since $OD$ , $DE$ , and $OE$ are all equal to $r$ , $\triangle ODE$ is an equilateral triangle with $60°$ angles. Using this strategy, the total distance that the duck travels along arc $DE$ is $\frac{60°}{360°}2\pi r = \frac{\pi}{3}r$ , and the total distance that the fox travels along major arc FE is $\frac{360° - (90° + 60°)}{360°}2\pi r = \frac{7\pi}{6}r$ , which means the fox's speed must be at least $\frac{\frac{7\pi}{6}r}{\frac{\pi}{3}r} = \frac{7}{2}$ times faster than the duck in order to catch it.

Since the duck's speed is $1$ , the fox's speed must be at least $\frac{7}{2} \cdot 1 = \frac{7}{2}$ in order to catch the duck. If the fox's speed is less than $\frac{7}{2}$ , it will not catch the duck, so the maximal speed $s$ of the fox is $s < \frac{7}{2}$ , but since $\frac{7}{2} > \pi$ , the maximal speed s can also be $s > \pi$ .

If the duck is at the center of the lake and the fox is on the perimeter, the duck can head towards the shore of the lake while keeping opposite of the center from the fox. Regardless of whether the fox moves back and forth along perimeter or runs in one direct only (counterclockwise here), the distance the duck travels until it cannot get any closer to the shore in this way is the perimeter of the semicircle, its diameter being the maximum distance from the center while still staying opposite of the center from the fox. The distance is the same whether the duck goes from $D1$ to $D2$ , or from $D1$ to $D3$ , or from $D1$ to any other point on the blue circle depending how the fox attempts to head off the duck. See the following diagram: Once the duck has reached the maximum distance by this means, then the next thing the duck can do is to head straight for the shore, and hope to beat the fox that will have to run halfway around the circumference of the lake. See the following diagram: While the fox runs from $F1$ to $F2$ , the duck paddles from $D1$ to $D2$ , and then the fox has to run from $F2$ to $X3$ while the duck paddles from $D2$ to $X3$ . Let the speed of the duck be $v_{duck}=1$ and the radius of the blue semicircle be $r_{semicircle}=1$ also. Then we need to determine the speed of the fox $v_{fox}$ and the radius of the lake relative to the the semicircle $r_{lake}$ . We have the following two equations, for the two parts of the chase:

$\dfrac{2\pi r_{semicircle}}{2v_{duck}}=\dfrac{2\pi r_{lake}}{4v_{fox}}$
$\dfrac{r_{lake}-2r_{semicircle}}{v_{duck}}=\dfrac{2\pi r_{lake}}{2v_{fox}}$

The solution leads to $v_{fox}=\pi+1$ , which the fox has to exceed in order to reach the duck before it reaches the shore at $X3$ . But this is not the maximum fox speed.

Per Antoine, see this site for details of how the duck can instead go from $D2$ to $X3$ perpendicular the line the duck, the center, and the fox are all on, for the to have to exceed $4.60333884...$ to get to the duck before it gets to point $X3$ on the shore. This strategy is based on the fox's realizing that it cannot afford change directions once it has become clear to him what direction the duck is heading for the shore.

I'm not sure if this is even the maximum fox speed, though.

From the centre of the pond, duck can swim opposite from the fox, making a spiral, up until a radius of 1/s. At 1/s radius (red circle), duck can maintain position opposite from the fox indefinitely. There on duck can swim in straight line in one of the direction opposite from the direction the fox have chosen. Fox will start reducing the distance, but it will be too late if the speed ratio is approx. s<1.465 $\pi$ , the fox will not catch the duck.

In the picture, fox start on the right, at 0 mark. Blue circle is an optimal path for the duck, given a fox that prefers clockwise direction (upon equal choice of direction). After the red circle, the fox have to commit on the direction.

EDIT: It turns out that an optimal "spiral" path that duck should follows at the beginning of the escape is actually a half circle, with radius r=1/2s. It follows from simple geometric consideration (angles duck and fox traverse).

How I looked at it: The worst case scenario would be the Fox being at the edge of the pond on land and the Duck being at the edge of the pond on the same side as the Fox but in the water instead. If the duck travels to the opposite end of the pond it will give the fox the longest distance to travel. I just wanted to find out if anyone else considered this as a worst case scenario. I got the question right but my math didn't end up checking out