Find the minimum positive integer $n\ge 3$, such that there exist $n$ points $A_1,A_2,\cdots, A_n$ satisfying no three points are collinear and for any $1\le i\le n$, there exist $1\le j \le n (j\neq i)$, segment $A_jA_{j+1}$ pass through the midpoint of segment $A_iA_{i+1}$, where $A_{n+1}=A_1$
Problem
Source: China MO 2023 P4
Tags: combinatorics
a22886
30.12.2022 09:40
It's fairly easy to show $n=3,4$ don't work. Now let $n=5$.
There are no parallelograms forms by any four of $A_i$.
WLOG $A_3A_4$ bisects $A_1A_2$, then
$A_5A_1$ bisects $A_3A_4,A_2A_3$ bisects $A_5A_1,A_4A_5$ bisects $A_2A_3,A_1A_2$ bisects $A_4A_5$.
The rest can be solved by bashing $\Big(A_3(0,0),A_2(2,0)\Big)$. There are no such five points.
WLOG $A_1A_4A_2A_3$ is a parallelogram. Then $A_2A_3$ has to be bisected by one of $A_4A_5$ and $A_1A_5$, WLOG $A_1A_5$.
Consider the point $A_5':=2A_2-A_4$, the reflection of $A_4$ wrt $A_2$. By the condition that no three points are collinear, $A_5\neq A_5'$.
But then $A_4A_5$ has to be bisected by one of $A_1A_2$ or $A_2A_3$ which means one of them has to be parallel to $A_5A_5'$, contradiction.
youthdoo
30.12.2022 13:17
[asy][asy]
pair a1=dir(60);pair a2=-a1;pair a3=dir(120);pair a6=-a3;pair a4=E;pair a5=-a4;
draw(a1--a2--a3--a4--a5--a6--cycle);label("$A_1$",a1,NE);label("$A_2$",a2,SW);label("$A_3$",a3,NW);label("$A_4$",a4,E);label("$A_5$",a5,W);label("$A_6$",a6,SE);
[/asy][/asy]
IAmTheHazard
26.01.2023 22:25
The answer is $6$, which can be achieved by selecting the points $(0,0),(1,1),(2,0),(0,3),(1,-2),(2,3)$, which works. To show that this is minimal we can easily see that $n=3,4$ fail because we can't have every edge intersected by another. For $n=5$ casework bash.