Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
A lady and a monster

Here is another rather famous riddle - I've seen it several times, but only once in its full form that I quote here:

A duck is located in the center of a circular pond, near which a hungry fox is waiting. The fox cannot swim - it can only run around the pond. The duck can fly only if it gets out of the pond (don't ask, it's probably a genetically altered duck). What's the minimal ratio of the fox to duck speed such that the fox will always catch the duck? What's the duck's strategy to escape the fox if the fox is slower than that?

The lower bound of $\pi$ is immediate. Upon some further reflection one gets $\pi+1$, but I heard one can do better than that. Any ideas?

share|improve this question

marked as duplicate by Henning Makholm, Gerry Myerson, David Mitra, Henry, Austin Mohr Feb 3 '12 at 0:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2 Answers 2

up vote 1 down vote accepted

Here is a very brief solution recently posted in the Usenet group alt.math.recreational by Ray Koopman to essentially the same problem; I’ve converted the straight ASCII to something more readable. (Note that in this version the duck is you, and the fox is a bear.)

If the lake has center $(0,0)$ and radius $1$, the bear is at $(-1,0)$, the speed ratio is $k$, you are at $\left(\frac1k,0\right)$, and the bear starts moving counterclockwise, then you should head toward $\left(\frac1k, \sqrt{1 - \frac1{k^2}}\right)$ instead of $(1,0)$. You will always be on the same side of the lake as the bear, so he won’t reverse direction. You will win the race as long as $$\pi + \arccos \frac1k > \sqrt{k^2 - 1}\;,$$ which works out to $k < 4.6$ (approximately).

For more detail, see the first two answers to this earlier question.

share|improve this answer
You will be always on the same side - this is the key I kept missing! When you are out of 1/k distance from the center no matter what you do the angular distance will be decreasing (as long as you are going outbound; but going inbound makes no sense since you already start out in the best position and can't improve on it). Not going in a straight line also does not make sense since you are just wasting time. So the only question which direction between 0 and +90 degrees to choose (<0 - makes no sense, >90 - you are going inbound). And then it's the matter of simple calculus. –  malenkiy_scot Feb 4 '12 at 21:49

I found this page using google and it gives a number bigger than $\pi + 1$ but not much explanation.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.