A lady and a monster

Solution 1:

Since you seem to know the answer, I will give it here.

Suppose that $v_l = v_m / k $ and the radius of the lake is $r$. Then the lady can reach a distance $\frac{r}{k}$ from the centre and keep the monster directly behind her, a distance $r\left(1 + \frac{1}{k}\right)$ away. One way would be to swim in a spiral gradually edging outwards as the monster runs trying to close the distance; another would be to swim in a semi-circle of radius $\frac{r}{2k}$ away from the monster once it starts to run. And the lady can sustain this distance by going round in a circle as the monster tries in vain to close the distance.

The next stage is for the lady to try to swim direct to shore at some point away from the direction the monster is running. If the monster starts at the point $(-r,0)$ running anti-clockwise and the lady starts at the point $\left(\frac{r}{k},0\right)$ her best strategy is to head off in a straight line initially at right angles to the line between her and the monster: a less steep angle and the monster has proportionately less far to run than the lady has to swim, but a steeper angle and it is worth the monster changing direction. (If the monster changes direction in this right-angle case, the lady changes too but now starts closer to shore.) As they are both trying to get to the point $\left(\frac{r}{k},r \sqrt{1-\frac{1}{k^2}}\right)$ then they will arrive at the same time if $ \pi + \cos^{-1}(1/k) = k \sqrt{1 -1/k^2}$ which by numerical methods gives $k \approx 4.6033$.

So if the monster is less than 4.6033 times as fast as the lady, the lady can escape; if not then she stays in the lake and the monster stays on the edge and they live unhappily ever after.

Solution 2:

edit: I corrected a silly mistake, now I get the same answer as Henry.

Let $k=v_m/v_l$. We can suppose $v_l=1$, hence $v_m=k$, and that the radius of the lake is 1. Let's reformulate the problem in this way: lady swims as before, but monster stands still and turns the lake with speed $\leq k$. The (vector) speed of the lady is a point in a disc of radius $1$, and the monster can control the center of the disc - he can move it from $0$ in the tangent direction by at most $kr$, where $r$ is the distance of the lady from the center.

If $r<1/k$ then $0$ is inside the disc, so the monster has no control over the direction of lady's movement. She can therefore get to $r=1/k$ to the point away from monster.

When $r>1/k$ then $0$ is no longer in the disc and the monster can force (by turning at full speed) the constraint $|dr/d\phi|\leq r/\sqrt{k^2r^2-1}$ (where $\phi$ is the angle of the position of the lady). The question is whether she can get from $r=1/k$, $\phi=0$, to $r=1$, $\phi<\pi$ ($r=1$, $\phi=\pi$ is the position of monster). This is possible iff $\int_{1/k}^1 \sqrt{k^2-r^{-2}}\, dr <\pi$.

edit: Here is why $|dr/d\phi|\leq r/\sqrt{k^2r^2-1}$ (it's a bit difficult to explain without a picture, but I'll try). The possible speeds of the lady form a disc with the center at $(0,kr)$ and with the radius $1$. The speed with the largest slope is the point of tangency from $0$ to the circle. Its slope can be seen from the right-angled triangle, with hypotenuse $kr$ and two other sides $1$ and $\sqrt{(kr)^2-1}$ - so the slope is $1/\sqrt{(kr)^2-1}$.

Solution 3:

I don't see that there is any game theory to be done here.

For the lady to have a winning strategy means that there exists a legal lady path $\gamma_l$, which reaches the shore, such that for all legal monster paths $\gamma_m$, a monster following path $\gamma_m$ does not catch the lady.

For any choice of $v_l$ and $v_m$, either the lady has a winning strategy or she doesn't, and it sounds like you know how to find out which is which. If she doesn't, then inverting the quantifiers, for every legal lady path which reaches the shore there exists a monster path which catches it. So in this case she cannot reach the shore without being caught.

However, the lady can always force a draw by not reaching the shore.

Solution 4:

The monster can know if he can win or not at any time, likwise the lady.
They can't agree to enter a phase where the probability that either wins is something other than 0% or 100%.

If the lady can't win, so the monster won't win if she stays in the lake, they can toss a coin to determine her faith.

Solution 5:

This is my first answer on this site. I don't know how to format it beautiful yet. Sorry for the inconvenience. The solution provided by Henry seems plausible.

1. Why the first statement is true? "the lady can reach a distance $\frac{r}{k}$ from the centre and keep the monster directly behind her, a distance $r(1+\frac{1}{k})$ away. "
   I know that the result is given by the equation, $$\frac{\frac{r}{k}*\theta}{1} = \frac{r*\theta}{k}$$ which means the angle they go are the same. But still, it is a little bit ambiguous to me.

2. If we take the first argument as granted, the second choice is how the lady swim to the shore. The answer provided by Henry has two implicit assumptions, 1. the monster cannot change direction 2. right-angle , therefore, he comes up with the equation $$ \pi + \cos^{-1}(1/k) = k\sqrt{1-1/k^2}$$ The thing is, if the lady wants to fool the monster, and once monster starts to run, the lady changes the direction and monster still runs the same direction, the equation above is correct. But if the monster is a rational animal, it will change direction too (all happens in a very short time, otherwise the equation cannot hold).
   Also, why right angle? As assumed, they meet at the shore, the meeting point to the center is angle $\alpha$, I come up with the equation (use Law of cosines) below $$\frac{\sqrt{(r/k)^2+r^2-2r(r/k)\cos{\alpha}}}{1}=\frac{(\pi-\alpha)r}{k}$$ If I choose $\alpha = \cos^{-1}(1/k)$ and change $\pi-\alpha$ to $\pi + \alpha$, I will have the same equation as $$ \pi + \cos^{-1}(1/k) = k\sqrt{1-1/k^2}$$ So it is a function between $\alpha$ and k. It is not any trajectory problem, so we can expect when lady choose go straight, the distance is shortest, and it is the fastest routine. Rearrange the terms, we solve the quadratic equation to get the expression of $k$ as $$k=\cos{\alpha}+\sqrt{\cos^2{\alpha}-1+(\pi-\alpha)^2}$$ It is easy to prove and use python to visualize that it is a decreasing function of $\alpha$, so the max $k$ is achieved when $\alpha=0$, and $k_{max}=1+\pi\approx 4.142$