Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following puzzle I'm trying to solve:

"A man lost on the Nullarbor Plain in Australia hears a train whistle due west of him. He cannot see the train but he knows that it runs on a very long, very straight track. His only chance to avoid perishing from thirst is to reach the track before the train has passed. Assuming that he and the train both travel at constant speeds, in which direction should he walk?"

I'm not quite sure what to do here. Given what the actual train tracks look like I'll assume that they are a straight line from east to west (from left to right). But where is the man? Above or below this line? Also, the train might be out of reach, or am I missing anything? Thanks for helps.

share|cite|improve this question
He has no idea how far away the train is, so his goal should be to minimize his expected time to reach the track. Since it might be north of him and might be south, walking due north forever doesn't do this: he might never reach the track! Instead he should walk north for a distance $d_1$, then south for some $d_2>d_1$, then back north again for $d_3>d_2$, and so on. This is the one-dimensional "lost at sea" problem. The solution depends in part on the assumed distribution of distances to the track. See for one approach. – mjqxxxx Sep 29 '12 at 18:17
up vote 1 down vote accepted

(This is not an answer, but it is too big for a comment, so I am posting it CW.)

Rotate the problem so that the track runs west to east and the man is south of it by distance $s$, and he hears the whistle in a northwesterly direction. Presumably the man is slower than the train; say the man's speed is 1 and the train's is $V$ with $V>1$.

An obvious strategy is to walk due north to point $N$.

The only other direction that is not an obvious loser is northeasterly to some point $d$ units east of $N$. The man has traveled a total of $s'=\sqrt{d^2+s^2}$, which takes longer, but only by $s' - s$. The train has traveled an extra $d$ distance. $d<s'$, but we might have $d > s'-s$, and if ${d\over s'-s} > V$, then the man gets a benefit from going northeast. So it's at least conceivable that the man could get a win by doing this.

share|cite|improve this answer
Ok. You assume that the tracks are north of him. I assume it's ok to do that since the other case follows by symmetry. – Rudy the Reindeer Sep 29 '12 at 17:51
Perhaps one could assume that the train and the man move at the same or at comparable speed to make the question less boring. – Rudy the Reindeer Sep 29 '12 at 17:52
I'm not assuming that the tracks are north of the man. I'm observing that we can draw a picture of the situation and then rotate the paper so that the tracks are north of him, without changing anything significant in the problem. This doesn't change the problem, but it makes it easier to discuss. As stated, the tracks can't be due north, because he hears the train whistle due west. But the exact direction of the whistle isn't important to the solution, since we don't know which direction the tracks go. – MJD Sep 29 '12 at 17:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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