Chinese Mathematical Olympiad

Solution: First, we prove that $n=6$ satisfies the condition. Let $P_1,P_2,\dots ,P_6$ be the vertices of a regular hexagon in consecutive order. Let $A_1=P_1,A_2=P_3,A_3=P_5,A_4=P_2,A_5=P_6,A_6=P_4$. Then, these 6 points satisfy the conditions. See the figure below.

We now prove that $n\le 5$ does not satisfy the given conditions. Clearly, $n=3$ does not satisfy the condition because for any $1\le i\le n,1\le j\le n$ in the statement, we require $j-i\ne -1,0,1(\mathrm{mod}n)$.

When $n=4$, if $A_1,A_2,A_3,A_4$ satisfy the conditions, then $A_1{A}_2$ and $A_3{A}_4$ must bisect each other, and $A_2{A}_3$ and $A_4{A}_1$ must bisect each other, which is impossible.

Therefore, we only need to consider the case when $n=5$. Suppose $A_1,A_2,A_3,A_4,A_5$ satisfy the conditions. For any $1\le i\le 5$, we require $j\equiv i+2,i+3(\mathrm{mod}5)$ in the statement. We consider two cases.

Case 1: Two of the line segments $A_i{A}_{i+1}$, $i=1,2,3,4,5$ bisect each other. By symmetry, we assume that $A_1{A}_2$ and $A_3{A}_4$ bisect each other. Note that applying an affine transformation to the plane does not affect the conclusion, we can assume that $A_1{A}_2$ and $A_3{A}_4$ are perpendicular bisectors and have equal length. Let $O$ be their intersection point, and let us establish a Cartesian coordinate system with $O$ as the origin. Let $A_1(0,-2),A_2(0,2),A_3(2,0),A_4(-2,0)$. Then the midpoint $M$ of $A_2{A}_3$ must lie on either $A_4{A}_5$ or $A_1{A}_5$. Without loss of generality, we assume that $M$ lies on $A_1{A}_5$. Note that the slope of $A_{1}M$ is 3, so we can assume that the coordinates of $A_5$ are $(1+x,1+3x)$, where $x>0$. See the figure below.

Now consider the midpoint $N$ of $A_4{A}_5$. If it lies inside the segment $A_1{A}_2$, then $1+x+(-2)=0$ implies $x=1$. However, in this case $A_4,A_2,A_5$ are collinear, a contradiction. If $N$ lies inside the segment $A_2{A}_3$, then $$\frac{1+x+(-2)}{2}+\frac{1+3x}{2}=2,$$ also yields $x=1$, which is a contradiction.

Case 2: None of the segments $A_i{A}_{i+1}$, $i=1,2,3,4,5$ bisect each other. Without loss of generality, assume that the midpoint $B_1$ of $A_1{A}_2$ lies on $A_3{A}_4$. Then the midpoint $B_3$ of $A_3{A}_4$ lies on $A_5{A}_1$, the midpoint $B_5$ of $A_5{A}_1$ lies on $A_2{A}_3$, the midpoint $B_2$ of $A_2{A}_3$ lies on $A_4{A}_5$, and the midpoint $B_4$ of $A_4{A}_5$ lies on $A_1{A}_2$. This implies that connecting $A_1,A_2,A_3,A_4,A_5$ in sequence yields a five-pointed star.

By the Law of Sines, $$1=\prod _{i=1}^5\frac{\sin \angle A_i{B}_i{B}_{i+2}}{\sin \angle A_i{B}_{i+2}{B}_i}=\prod _{i=1}^5\frac{A_i{B}_{i+2}}{A_i{B}_i}=\prod _{i=1}^5\frac{A_{i+1}{B}_{i+3}}{A_i{B}_i}<\prod _{i=1}^5\frac{A_{i+1}{B}_i}{A_i{B}_i}=1,$$ where the subscripts are understood modulo 5. This is a contradiction.

Therefore, the minimum $n$ we seek is 6.