0
$\begingroup$

Recalling the statement of the Happy Ending Problem, we see that

For any $k \in \mathbb{N}$ we may find a $n=n(k) \in \mathbb{N}$ such that every $n$ points in the plane, where no 3 are collinear, contains $k$ points in convex position

Is is clear to me the proof of it, however, I do not understand the reason for avoiding the problem of collinear points. My guess is:

Guess: Because, if you allow to have collinear points, you may have $n-k+1$ points in the same line, so, you won't have $k$ points in convex position. By the other hand, the probabilty of having 3 points in the same line is zero, so, why should we care about this?

Could someone give me a hand about this?

$\endgroup$
3
  • $\begingroup$ Your probability $0$ argument could then be extended to saying we should not care about equilateral triangles as the probability of having three points equidistant is zero. Your original question is not about probability and says the caveated statement is always true, not almost always true. $\endgroup$
    – Henry
    Commented Nov 15, 2022 at 17:21
  • $\begingroup$ @Henry yeah, I got the difference between true and almost always true, but, indeed, the point of avoiding collinear points is to avoid the problem that I pointed or is to avoid something else? $\endgroup$
    – Anyway142
    Commented Nov 15, 2022 at 17:42
  • $\begingroup$ Yes - perhaps even simpler: if all the points are collinear then at most two of them (or of a subset of them) will be in convex position. If $k>2$ then this would be a counter-example. $\endgroup$
    – Henry
    Commented Nov 15, 2022 at 18:04

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .