NEW FEATURE: VOTE & EARN NEATOPOINTS!
Submit your own Neatorama post and vote for others' posts to earn NeatoPoints that you can redeem for T-shirts, hoodies and more over at the NeatoShop!


Puzzle: How Can Carter's Killer Rabbit Escape?

Do you remember when President Jimmy Carter was attacked by a rabbit while fishing in 1979? Carter became the butt of many jokes about this incident. Here's a puzzle from John Tierney based on the incident:

Suppose, the day after attacking President Carter, the rabbit finds itself alone in the middle of the pond, which is perfectly circular. Suppose there is a single Secret Service agent on the edge of the pond, armed with a small net to ensnare the swimming rabbit as it approaches the edge. This net is effective only if the rabbit is still in the water. If the rabbit reaches any point on the edge before the agent does, it can hop away to freedom; if the agent gets there first, the rabbit will be captured.

If the agent runs four times as fast as the rabbit swims, can the rabbit escape? If so, how?

For extra credit: What’s the fastest the agent can run (as a multiple of the rabbit’s speed) such that the rabbit can still escape?


What is your answer? The first correct answer wins a kiss from Alex.

Link via Instapundit

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.
Abusive comment hidden. (Show it anyway.)
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.
Abusive comment hidden. (Show it anyway.)
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.
Abusive comment hidden. (Show it anyway.)
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.
Abusive comment hidden. (Show it anyway.)
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!
Abusive comment hidden. (Show it anyway.)
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.
Abusive comment hidden. (Show it anyway.)
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.
Abusive comment hidden. (Show it anyway.)
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.
Abusive comment hidden. (Show it anyway.)
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)
Abusive comment hidden. (Show it anyway.)
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
Abusive comment hidden. (Show it anyway.)
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.
Abusive comment hidden. (Show it anyway.)
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.
Abusive comment hidden. (Show it anyway.)
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.
Abusive comment hidden. (Show it anyway.)
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.
Abusive comment hidden. (Show it anyway.)
Login to comment.
Click here to access all of this post's 32 comments




Email This Post to a Friend
"Puzzle: How Can Carter's Killer Rabbit Escape?"

Separate multiple emails with a comma. Limit 5.

 

Success! Your email has been sent!

close window

This website uses cookies.

This website uses cookies to improve user experience. By using this website you consent to all cookies in accordance with our Privacy Policy.

I agree
 
Learn More