$2021$ points on the plane in the convex position, no three collinear and no four concyclic, are given. Prove that there exist two of them such that every circle passing through these two points contains at least $673$ of the other points in its interior. (A finite set of points on the plane are in convex position if the points are the vertices of a convex polygon.)
Problem
Source: IGO 2021 Advanced P4
Tags: geometry, IGO
08.01.2022 15:58
Proposed by Morteza Saghafian - Iran
24.01.2022 11:49
Finally, this was kinda tricky call a triangle with its vertices between our points, "nice" if all other points lie inside its circumcircle. First, we prove there exists a triangulation of this polygon such that all triangles are nice. Choose a random edge $AB$ on the convex hull. between all other points $P$, there exists a unique one such that $\widehat{APB}$ is minimal (because no four are collinear). it is clear now that $\Delta APB$ is nice. To finish this part, we prove a lemma: if $X,Y$ are points such that $XY$ such that $XY$ is not an edge of the convex hull and there exists a $P$ such that $\Delta XPY$ is nice, then there is another point $Q$ on the other side of $XY$ such that $\Delta XQY$ is also nice. proof: Using $XY$, divide the polygon $S$ into two parts $S_1,S_2$ such that $P\in S_1$ and call the circumcircle of $XYP$, $\Gamma$. it is clear that except $X,Y$ the rest of $S_2$ lies strictly inside $\Gamma$. now continuously change $\Gamma$ such that the side not containing $P$ shrinks and the side containing $P$ grows. we know it still contains $S_1$. do this until $\Gamma$ touches $S_2$ at a third point. if we choose this point as $Q$ it is clear that $\Delta XYQ$ is nice.(and of course $Q$ is vertex and not in the middle of an edge) so the lemma is proven. so now using this lemma starting from $\Delta APB$ as long as the polygon isn't fully triangulated, we can add another triangle to it. note that because we are adding triangles in the parts of the polygon that were previously untouched, this actually does create a triangulation. so we now know that there exists a triangulation consisting solely of nice triangles. we finish the problem using this famous fact: in any triangulation of a convex polygon with $n+2$ vertices, there exists an edge in the triangulation such that the endpoints are at least $\frac n3$ points apart on the perimeter both clockwise and counter-clockwise. for the proof one can set the vertices of the $n+2$-gon on the vertices of a regular $n+2$-gon (such that the order of the points on the convex hull remains the same) so its enough to prove it for a regular polygon. which can be done by taking the longest edge in the triangulation. if it's ends are less than $n/3$ apart, then one of the two triangles on each side of it has a longer edge, a contradiction (look on the perimeter of its circumcircle) . taking two such points if we divide the polygon using this line, we find that each circle passing through both contains at least one part of the polygon and we knew that each part has at least $\frac{2021-2}{3}=673$ points! so we are done.
24.01.2022 17:27
$\frac{2021-2}{3}=667$? But still astonishing
25.01.2022 21:50
Q.C.Ignorant wrote: $\frac{2021-2}{3}=667$? But still astonishing Huh? $\frac{2021-2}{3}=673$
28.04.2022 00:41
01.10.2023 04:41
Good question!