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I was just wondering how rigorous a USA(J)MO proof has to be to receive full credit. If a problem can be explained by logically extending an argument, will it earn as many points as a proof by induction explaining essentially the same thing?

For example, an observation post is setup on each of 2nāˆ’1 planets. Each observation post observes another planet closest to it. Suppose the distances between any two planets are all distinct. Show that there is at least one planet that is not observed by any other posts.

Would saying something like, "The two planets with the closest distance between them must observe each other. Now, if any other planet observes one of these two planets, then at least one other planet is unobserved because there are 2n-1 planets and 2n-1 observations. Therefore, we must systematically "pair" each of the next closest planets until only one is left. This last planet is not observed by any other planet." be as good as an inductive proof?

dcxt
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  • Do you have any specific example(s) in mind? ā€“  Jul 24 '17 at 04:18
  • The other planet watching doesn't necessarily have to be within less than the minimal distance to watch one of the paired planets. Even if it is far away, it can still observe the planets if they are the closest relative to it. ā€“ dcxt Jul 24 '17 at 04:33
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    I can share some experience on the matter, although at a higher level. The one time I participated in the IMO, the person in my team that did best lost points on almost all problems because he, in the words of our coach, "didn't write enough trivial things". So I would personally err on the side of caution. If a judge is able to say "You just guessed here" about some part of your answer, it's not thorough enough. ā€“ Arthur Jul 24 '17 at 05:34

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Now, if any other planet observes one of these two planets, then at least one other planet is unobserved because there are 2n-1 planets and 2n-1 observations.

The problem I see with that is not the casual wording vs. formal induction, but rather that the "then ... because" part is not obvious enough to be just hand-waved.

A more convincing (still casual) argument could be: if a third planet observes one of these two planets then that third planet is either unobserved, or otherwise must itself be observed by a fourth planet, and so on until it fails by infinite descent. In the other case where no other planets observe either of the two closest planets, those two can be removed from the pool and the problem reduces to the case of $2n-3$ planets, where the previous argument can be reused recursively.

dxiv
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