Puzzle: How Can Carter's Killer Rabbit Escape?

1. Aea
   April 14th, 2009 at 2:07 pm

   The rabbit can't escape, it's distance to the edge of the pond is r, if the rabbit decides to bolt in the opposite direction creating the longest distance the agent would have to run then the agent would need to run a distance of pi (since the circumference of a circle is 2*pi*d, we halve this since the rabbit only needs to run the radius).

   So if the rabbit needs to run a distance of r and the agent needs to run a distance pi*r, the agent needs to a run a distance that's greater by a factor of pi, since pi is 3.14, and the agent can run four times faster then the rabbit then the rabbit will always be caught.

   By the same logic any agent that runs slower then a ratio of pi:1 will not be able to catch the rabbit if it takes off in the opposite direction.

2. Aea
   April 14th, 2009 at 2:08 pm

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

3. MadMolecule
   April 14th, 2009 at 2:16 pm

   The rabbit should swim directly away from the Secret Service guy. The agent will begin to run around the pond, and the rabbit should keep turning as the agent moves around the pond, so that the rabbit is always swimming away from the agent. The rabbit will end up swimming in a spiral pattern, getting closer and closer to the edge, and eventually will be close enough to the edge opposite the agent that the bunny can just make a direct break for it.

4. MadMolecule
   April 14th, 2009 at 2:16 pm

   And sorry Alex, but I'm taken.

5. John
   April 14th, 2009 at 2:34 pm

   MadMolecule, wouldn't the Secret Service agent always be able to keep pace with the rabbit as he approaches the edge?

6. SarahW
   April 14th, 2009 at 2:40 pm

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

7. myleti
   April 14th, 2009 at 3:00 pm

   African or European rabbit?

   

8. Anon
   April 14th, 2009 at 3:02 pm

   bahaha love the monty python reference

9. Mike S.
   April 14th, 2009 at 3:02 pm

   I think John is right: No matter how many maneuvers, the agent will always outpace the rabbit by the time it reached the edge.

   Figure the radius of the lake (the direct path for the rabbit to reach the edge) is 10 meters. The agent must travel 10*? -- or half the lake's circumference -- in order to reach the point the rabbit reaches land: 31.41 meters.

   Considering, the agent runs 4 times as fast as the rabbit can swim -- 4 > 3.141 -- he'll still be able to catch it.

   The rabbit can only escape if the agent runs less than 3.141 times as fast as the rabbit.

   Of course, I could have all these calculations wrong.

10. MadMolecule
    April 14th, 2009 at 3:04 pm

    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.

11. jeffmshaw
    April 14th, 2009 at 3:05 pm

    The rabbit should dive underwater and swim to the edge of the pond. Since the agent can't tell where the rabbit will exit the pond, the rabbit should be able to escape.

    Kisses are great!

12. Sean
    April 14th, 2009 at 3:09 pm

    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.

13. Cathy
    April 14th, 2009 at 3:11 pm

    I remembered reading this in Scientific American, interesting article on Rabbit Escape problem. thanks

14. deadskin
    April 14th, 2009 at 3:19 pm

    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.

15. vonskippy
    April 14th, 2009 at 4:01 pm

    Can the rabbit out swim a bullet? If not, problem solved.

16. Chris P
    April 14th, 2009 at 4:24 pm

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

17. Steohawk
    April 14th, 2009 at 5:05 pm

    If the rabbit starts to swim north from the center, the agent will run to the north side of the pond. As soon as the agent gets in place, the rabbit could swim south.

    If the pond is 4 meters in diameter, it would be roughly 12.56 meters in circumference. If the rabbit can swim 1 meter per second, then the agent could run 4 meters per second. 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.

18. whitcwa
    April 14th, 2009 at 5:12 pm

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

19. Miss Cellania
    April 14th, 2009 at 5:22 pm

    I just came in here to see what Alex was going to say about that kiss remark. Maybe he hasn't noticed yet.

20. Alex
    April 14th, 2009 at 5:47 pm

    So. Who do I get to smooch?

    

21. DOJ
    April 14th, 2009 at 6:19 pm

    I think it depends on what proportion of the pond's radius is the agents reach with the net.

22. Paul Howard
    April 14th, 2009 at 6:22 pm

    Obviously I'm not first, but my answer was "No, pi."

23. ted
    April 14th, 2009 at 8:54 pm

    Instead of the Secret Service, they should have used a navy SEAL.

24. Dave
    April 15th, 2009 at 12:07 am

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

25. Scotchdrnkr
    April 15th, 2009 at 7:36 am

    I don't remember the Carter incident at all. But I didn't pay any attention to anything to do with Politicians back then.

26. idahogie
    April 15th, 2009 at 12:20 pm

    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)

27. deadskin
    April 15th, 2009 at 1:19 pm

    idahoqie , I think that you assume the rabbit can always swim out to 1/4 of the distance to the shore while being able to maintain an opposite position to the agent.

    This distance will vary on the speed of the agent , getting smaller as the agents speed increases. By casual observation we can see that that rabbit can swim out to the fraction 1/max to the shore based on a max speed for the agent. This would leave the rabbit a distance of (max-1)/max to swim at rate (x) while the agent was moving at rate (max)(x).

    So the time take should be equal to find the max rate of the agent.

    (((max-1)/max)R)/x = (R(pi))/((max)(x))

    If multiply both sides by x we get
    ((max-1)/max)R = (R(pi))/(max)

    We can divide both side by R and get
    (max-1)/max = (pi)/(max)

    We can multiple both sides by (max) and get
    max-1 = (pi)

    add one to both side and your done
    max=(pi)+1

28. MadMolecule
    April 15th, 2009 at 2:02 pm

    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.

29. idahogie
    April 15th, 2009 at 2:05 pm

    I haven't checked your math, deadskin, but I can see that I made the error that you describe. So I'm inclined to believe you on the extra credit question. Good work.

30. idahogie
    April 15th, 2009 at 2:07 pm

    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!

31. deadskin
    April 15th, 2009 at 2:09 pm

    As my wife can tell anyone , I am not a 'grammar/syntax/spelling' guy. Unless the semi-colon is missing at the end of a line of code , I wouldn't even know it was gone.

32. pjc
    April 15th, 2009 at 8:22 pm

    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.

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