Let $P$ be a set of $n \ge 2$ distinct points in the plane, which does not contain any triplet of aligned points. Let $S$ be the set of all segments whose endpoints are points of $P$. Given two segments $s_1, s_2 \in S$, we write $s_1 \otimes s_2$ if the intersection of $s_1$ with $s_2$ is a point other than the endpoints of $s_1$ and $s_2$. Prove that there exists a segment $s_0 \in S$ such that the set $\{s \in S | s_0 \otimes s\}$ has at least $\frac{1}{15}\dbinom{n-2}{2}$ elements
Problem
Source: 2009 Peru Iberoamerican TST problem 6
Tags: combinatorics, combinatorics unsolved
hendrata01
10.05.2023 14:28
What about this counterexample: Suppose $P_1 P_2 P_3$ form a triangle and $P_4$ is inside. Then the statement does not hold, except if we allow the segments to intersect outside the segments themselves. But if we allow segments to intersect outside the segments themselves, then the problem is sort of trivial, because every segment will intersect every other segment. There are $\binom{n}{2}$ segments. Pick any one of them to be $s_0$ and the set of segments that intersect $s_0$ not at the endpoints will be $\binom{n-2}{2}$
hectorleo123
11.05.2023 03:00
I could not tell you
hectorleo123
11.05.2023 19:39
if it is maximum integer does it comply?
ReaperGod
12.05.2023 00:02
hendrata01 wrote: What about this counterexample: Suppose $P_1 P_2 P_3$ form a triangle and $P_4$ is inside. Then the statement does not hold, except if we allow the segments to intersect outside the segments themselves. But if we allow segments to intersect outside the segments themselves, then the problem is sort of trivial, because every segment will intersect every other segment. There are $\binom{n}{2}$ segments. Pick any one of them to be $s_0$ and the set of segments that intersect $s_0$ not at the endpoints will be $\binom{n-2}{2}$ parallel lines exist hectorleo123 wrote: if it is maximum integer does it comply? yes but the bound would not make sense