Given positive integer $ n \ge 5 $ and a convex polygon $P$, namely $ A_1A_2...A_n $. No diagonals of $P$ are concurrent. Proof that it is possible to choose a point inside every quadrilateral $ A_iA_jA_kA_l (1\le i<j<k<l\le n) $ not on diagonals of $P$, such that the $ \tbinom{n}{4} $ points chosen are distinct, and any segment connecting these points intersect with some diagonal of P.
Problem
Source: 2021ChinaTST test3 day1 P1
Tags: combinatorial geometry, combinatorics, graph theory
22.04.2021 05:22
Are you sure this is right? For example, a regular 12-gons is divided into 444 regions by its diagonals,but $\tbinom{12}{4}=495$. So when we choose 495 points in the regular 12-gons,there must be some two points in the same region. Then the segment connecting these two points doesn't intersect with any diagonal of the polygon.
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22.04.2021 09:42
@above: You somehow think the intersection points of the diagonals are forbidden. But they are not, because every one of them lies in the interior of some quadrilateral $ A_iA_jA_kA_l$. So, if you add some of them to the chosen points, there would be no contradiction.
22.04.2021 10:28
dgrozev wrote: @above: You somehow think the intersection points of the diagonals are forbidden. But they are not, because every one of them lies in the interior of some quadrilateral $ A_iA_jA_kA_l$. So, if you add some of them to the chosen points, there would be no contradiction. Let the "region" mentioned above contains its boundary, then the intersection points doesn't make something different.
22.04.2021 19:03
Of course it makes difference. Any segment with one of its end points lying on a diagonal, will intersect/meet this diagonal, because its end point lies on it. The problem is trivial in case no 3 diagonals are concurrent. For any quadrilateral $ A_iA_jA_kA_l$ we choose the intersection point of its diagonals. The point is that the chosen points must be distinct. So , one cannot do it in case there are three diagonals that have a common point. Thus, the essential question is how to proceed when there are many diagonals with common points, i.e. when the polygon is too regular. EDIT. Ok, there is indeed something wrong with this problem, since with this interpretation, for any quadrilateral we can always choose a point on some of its diagonals that is distict from all the others and thus the problem would be trivial.
25.04.2021 11:00
According to one of the Chinese national team: The chosen points can not on the diagonals,and no three diagonals are concurrent.
25.04.2021 18:17
Just use Hall Theorem
28.04.2021 18:35
How to derive the estimate for Hall's condition? Could you elaborate?
07.05.2021 09:37
Bump, bump.
12.05.2021 10:20
14.05.2021 03:59
@xiejiesuo @dgrozev Sorry for the misleading due to the lost condition. I have added them according to Mr.Trick.