# NY Times math problem

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DId anyone here see the problem presented in the Science section of NY Times last week? Quite startling, to see something so sophisticated in a 'general readership' publication.

Is it solvable without a calculus of variations approach?

-- Mark

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Didn't see it.

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I imagine this is what you're referring to:

puzzle/

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It conjures memories of rip tides, swimmers and lifeguards.

Sue...

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Yes.

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Since the agent can run four times as fast as the rabbit can swim, no matter what the agent does, the rabbit can swim away from the centre, and keep the agent on the far side as long as the rabbit is no more than

1/4 of the radius from the centre. So the rabbit can reach a point 3/4 of a radius from the edge while the agent is still on the opposite side.

From that point, if the rabbit swims directly to the edge, he has to swim 3/4 of a radius. In that time, the agent can run at most 4 times as far, or 3 * radius. But the distance to the rabbit's exit point is Pi * radius, or a bit further, so the rabbit is free.

For the second part, instead of 4, write m, and let R be the radius.

The rabbit can reach a point R/m from the centre while keeping the agent opposite. From that point, the rabbit needs to swim R * (m - 1) / m to reach the edge. In the available time, the agent can run R * (m - 1). So the break even point is where R * (m - 1) = R * Pi, or where m = Pi + 1.

Sylvia.

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No, the rabbit can do better than swimming in a straight line toward the closest point on shore.

For example, suppose the shore were straight instead of circular, with the agent starting distance pi from the closest point, while the rabbit is at distance pi/m. If the rabbit swims in a straight line, they will reach that point at exactly the same time.

If the rabbit swims at angle A, then the agent must run extra distance (pi/m) tan A, while the rabbit must swim an extra (pi/m) (sec A - 1). The rabbit is better off if tan A > m (sec A - 1), which has solutions in A for any m. So for a straight shore, the rabbit is always better off swimming at some nonzero angle.

The circular shore does change the distances, but the same pattern holds: the rabbit is always better off swimming at an angle, even if a small one.

- Tim

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I solved this when it appeared in rec.puzzles in 1992 (using a duck and a fox).

See

The bottom line is that there is a strategy by which the rabbit can escape as long as the agent can run no more than v =3D

4.6033388487517003525565820291030165130674... times as fast as the rabbit can swim. The number v is the reciprocal of the solution of the transcendental equation sqrt(1-r*r) =3D r*(pi+arccos(r)), which can be derived using trigonometry.

Dave

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The rabbit is in the middle of a circular pond. The agent is on the edge of the pond. The agent can run 4-times as fast as the rabbit can swim. Can the rabbit get away ?

Okay the radius of the pond is 1 so the rabbit must swim 1 . The pond circumference is Pi*radius*2 but the agent must run only half the circumference for a distance of Pi . Then rabbit can swim at a rate of 1 so the agent can run at a rate of 4.

Rate * Time = Distance & Time = Distance / Rate

So the rabbit can swim to the shore in

Time = 1 / 1 or 1 .

And the agent can run to the opposite shore in

Time = Pi / 4 or 0.785 .

And so I seem to have the wrong answer. But the second part of the problem based on my answer would be

Pi / Rate = 1 Rate = 3.14

If the agent runs only 3.14 times as fast as the rabbit then they time...and a tie goes to the base runner.

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So the rabbit swims an arc. The agent is running a counterclockwise circular curve while the rabbit is swimming a counterclockwise...outward spiral.

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See...in the extreme...at each instantaneous moment of the agents movement the rabbit makes a new line from the agent to the rabbit to the shore.

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The rabbit initially moves to stay on the same diameter as the agent while maximizing the distance between them.

If the agent acts optimally, the rabbit can get no more than 5/8 of that diameter ( or 1 and 1/4 radii) away from the agent along that common diameter.

From there the can rabbit race for the nearest point on shore.

The rabbit's distance from that point is (3/4)*r and the agent's is pi*r.

Since (3/4)*r is slightly less than 1/4 of pi*r, ...

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Where I live we shoot rabbits and outgrabe mome raths.

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No. The rabbit wants to swim in a straight line. It is the shortest distance from where he is to a point on the shore. If he swims in an arc, it will take longer than on a line, meaning that the agent has more time to reach him.

Dave

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By swimming so as to keep on the same diameter as the agent while moving as fast as possible away from the agent, the rabbit can arrive at a point that is only 3/4 of a radius from a point on the edge while the agent is still half a circumference's running distance from that point.

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"> > The rabbit is in the middle of a circular pond. The agent is on the edge

And the solution hereabouts says that the rabbit keeps turning away from the agent until a point where he can break for shore. But how does the rabbit know when he can break for shore ?

I'm saying that with each movement of the agent...the rabbit turns away from the agent. So the rabbit swims an outward spiral that does prove that the rabbit can get away. See...the rabbit stays in the water until the point when he touches shore with the agent some distance away...

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Seems clear to me. If the rabbit travels in a straight line, regardless of which direction is taken, that line is always a radius. Therefore, the point on the shore closest to the agent (without reaching the agent) is just four radii of circumference (that is, 4 radians angle) from where he was standing.

However, the agent can reach the other side of the pond sooner than

3.14 radii, that is, the optimal point exactly opposite the agent. So a straight-line solution is very easy: the ratio of speeds is pi, and the rabbit cannot escape.

But that would be too easy, wouldn't it? If the rabbit is allowed a curved path, then I would believe some of these complicated solutions (a trancendental equation, eh?).

Tim

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If the rabbit swims in a straight line from the centre, the agent can beat him to the shore. So he must get closer to the shore while remaining diametrically opposite the agent.

So long as the rabbit is less than r/4 from the centre, he can maintain diametric opposition. If the agent runs clockwise, the rabbit swims anti-clockwise and vice-versa. The rabbit moves at 1/4 of the speed, but is at less than 1/4 the radius, so his maximum angular speed is higher than the agent's. He only needs to match the angular speed, so whatever speed is left over can be used for outward motion.

Once he reaches r/4 from the center, the rabbit can maintain diametric opposition, but not while moving outward. But at this point, he is only

3r/4 from the closest point on the shore, while the agent is pi*r from that point. So he can head in a straight line to the shore and get there before the agent does.
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Only if the agent stands still until the last possible moment. I think the agent will move continuously, and as soon as the rabbit starts moving in any direction, the agent will move toward the point that is the closest to the rabbit at that moment. Let's call the agent 'Xeno'.

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No - until the rabbit is 1/4 of the radius away from the centre, he can swim fast enough, in a suitable direction, to keep the agent exactly opposite. So it doesn't matter if the agent moves.

Sylvia.

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Which shows, I think (to return to the very original question about whether this needs horrible maths to solve) that there is a simple proof that pretty-well anyone can follow that shows the rabbit can escape in the specific case (it goes to where it can just swim faster than the agent, swims round in a circle until it gets to 180 degrees away from the agent, and can then make it to the shore faster than the agent can run round), but that the optimum strategy, and hence the answer to the unasked question about which size of ponds or relative speeds allow the rabbit to escape do require the horrid maths.

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