Puzzle: How Can Carter’s Killer Rabbit Escape?

Oops, I should mention that the rabbit is swimming not running

And sorry Alex, but I’m taken.

The rabbit should don his carmen miranda outfit and sashay past the agent in a seductive manner.

bahaha love the monty python reference

John, I don’t think the agent would be able to. If the rabbit just swam straight from the center to the edge, the agent would get there edge first, because the agent would have to cover pi times the rabbit’s distance, and he can travel faster than pi times the rabbit’s speed.

I think, though, that if while the rabbit swam it kept turning so that it was always swimming directly away from the agent, the rabbit would pretty quickly find itself in a spot that is only .78 (or pi/4) of the distance from the point that was at that moment directly opposite the agent. From there, the rabbit makes a break for it and escapes.

I could be visualizing this improperly, though. Somebody should set up a computer model to figure this out.

MadMolecule is correct. This problem appeared in one of the books of mathematical puzzles written by Martin Gardner that were published several decades ago when he was writing the “Mathematical Games” column for Scientific American.

It seems like MadMolecule answer should work. I tried a quick calculation only to realize you probably need a bit more then straigh GEO to get the job done.

But on second thought all the rabbit has to do is be able to get out to a radius away from the center that the rabbit can cover in the same time it takes the agent to run around the pond. This will take some time for the rabbit since it need to ensure the agent is always directly behind it. Using MadMolecule’s idea. Once it reaches that radius (distance) from the center of the pond it can make a mad dash at the shore line.

MATH::
Pond has Radius = R
Agents Speed = 4x
Rabbit Speed = x

If the rabbit moves outward to a distance of (1/4)R from the center always keeping the agent behind the rabbit can swim around in circles and the agent will always remain at the opposite shore assuming the agent never stop pursuit.

This is because the rabbit here will be covering a distance of 2(1/4)(pi)R or (1/2)(pi)R at speed x taking ((1/2)(pi)R)/x or (1/2x)(pi)R and the agent will traveling 2(pi)R at 4x which takes (2(pi)R)/4x = (1/2x)(pi)R

Now the rabbit is R – (1/4)R away from the edge or (3/4)R and can cover that in ((3/4)R)/x or (3R)/(4x). The agent is half way around the pond and will need to cover (1/2)(2(pi)R) or (pi)R at a rate of 4x. That will take him ((pi)R)/(4x). Since (pi) is about 3.14 the agent will take longer to reach the other side of the pond then the rabbit in this scenario.

The agent will have to run slower than((pi)+1) times faster than the rabbit. Hopefully I will write back the process later, its time to go home.

I think if the rabbit swims underwater, it should be fine.

If the rabbit is smart he’ll go back and take Jimmy hostage.then he can negotiate his escape.

So. Who do I get to smooch?

Obviously I’m not first, but my answer was “No, pi.”

Forget the puzzle; I’d like to know how many here even remember the incident with the mad water-borne rabbit.

Deadskin’s right. The rabbit can maintain a position opposite the agent while he stays 1/4 of the way to shore. From there it’s a race – the rabbit goes (3/4)R at speed x and the agent goes (pi)R at speed 4x. Time for the rabbit is distance/speed, or (3/4)(R/x). Time for the agent is (pi/4)(R/x). Pi/4 is about 0.79, which is longer than than the rabbit, which is 0.75. Go, bunny, go!

For the maximum agent speed, set (3/4)(R/x) = (pi/max)(R/x). Solve for max = (4/3)(pi)

Steohawk, I think your math is wrong:

“The rabbit would be able to swim from the center to the south edge in 2 seconds, while the agent would only be able to run from the north end to the south end in 3.14 seconds.”

Actually the agent would be able to get from the north end to the south end in 1/2(pi) seconds; with a 2-meter radius, the pond’s circumference is 4(pi), so half that is 2(pi). At 4 meters/second, 2(pi) meters would take about 1.57 seconds, and he would get there 0.43 seconds before the rabbit does.

I’ve now checked your work, and I can say that in your last step, you’ve made a mistake. It should be ‘you’re’ not ‘your’.

Ha!

In a straight race from center to shore, Secret Service guy can run the max half-circumference faster than the rabbit can swim the radius.

So to beat the man, the rabbit needs to do more work than that. Kinda thinking a spiral pattern might work but I’m too lazy to do that math.

Let’s try something simpler. If the rabbit can get himself sufficiently far in one direction, while forcing Secret Service man to be 180 degrees opposite, from there he clearly can win the race. If the pond has radius r, and the rabbit is x away from the center, then a straight swim to the edge is distance (r-x), in which time the man can run 4(r-x) and has to cover pi*r distance. So the rabbit has to be within pi*r/4 of the shore, or almost 1/4 of the way to the shore or closer, to win. The remaining question is can the rabbit get to that point, and force Secret Agent Man to the diametrically opposite side?

Well, rabbit swims 1/4 as fast as Man can run. So if rabbit is anywhere closer to the center than r/4, he can begin to swim in a circle and cover the angle faster than the man can run, and therefore swim in that circle until he forces man to be 180 degrees opposite. Then he can bolt for the shore, and to Rabbit Freedom.

So rabbit can swim to some point in between r/4 from the center and (1 – pi/4)r from the center. Regardless where the man is, Rabbi can then begin swimming his loop until the man is 180 opposite, then dash.