Zorba the Hutt (zorbathut) wrote,
Zorba the Hutt
zorbathut

First off, we're going to ignore the situation where the duck is in the exact center of the pond. The duck can move minutely far to the side and suddenly it's not in the center, so we're just going to ignore the duck ever being in the center. It doesn't matter.

From here, it should be pretty obvious that the best position for the duck is when the duck is diametrically opposite from the fox. The fox will have the longest distance possible to run in order to get close to the duck. We consider this a Good Thing, and this is the situation we want. So: Assuming the duck can maintain this situation, how far can the duck swim before the fox catches up?

The fox has to travel pi*r units to get all the way around the pond, and does so at speed 4. Therefore, the duck can swim pi*r/4 units in that time. Therefore, if the duck can maintain a diametrically opposite position until it's pi*r/4 units or less from the edge of the pond (equal to r-pi*r/4 from the center) the duck can get there.

So the next question is, how long can the duck maintain that position? If the duck wants to maintain that position, it'll have to be able to outswim the fox. This is actually doable - imagine drawing a line from the center of the pond through the duck (this is why the duck is no longer allowed to go to the center of the pond) to the edge of the lake, and drawing a point there. Obviously, if the duck stays in the very center of the pond and swims around in tiny tiny circles, it can whip the point around far faster than the wolf can follow it. Thinking about it more mathematically, the only question is the radius of the circle that the duck can travel around as quickly as the fox can travel around his. pi*n = pi*r/4. (n being the radius of our inner circle, r being the radius of the pond, 4 being the speed of the fox.) Solving for n, we get n = r/4.

So: The duck can swim out to r/4 units from the center, and then, once he's r-pi*r/4 from the center, it can make a break for the shore and get there. For the sake of getting actual numbers, assume r=1 (it should be clear by now that it doesn't matter how large the pond actually is.) The duck can swim out to 0.25 units, and needs to get to about 0.215 units.

The duck wins.

I'm not about to prove it formally that this is the best solution, but if you think about it, these two numbers are the only important ones. As long as the duck can outpace the fox, there's no reason not to - as long as the duck can't, there's no point to even trying.

So the last question, which is what the cutoff is. If the fox travels at a speed of v, the duck can make a break for it once he's r-pi*r/v units from the center, and can maintain the best-situation until he's r/v units from the center. Therefore, solving r-pi*r/v = r/v for v should give us the cutoff point.

r - pi*r/v = r/v
r*v - pi*r = r
r*v = r + pi*r
v = 1 + pi

And there's our answer. If the fox travels faster than 1+pi, the duck loses - if the fox travels more slowly, the duck wins.

So here's another question: What happens if the fox travels precisely at a speed of 1+pi times the duck's speed?
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