China Mathematical Competition (Complementary Test)

Given $j$ ($1<j<n$), the number of ternary groups $(i,j,k)$ satisfying $1\le i<j<k\le n$ and $$\frac{a_j-a_i}{a_k-a_j}=r\stackrel{◯}{1}$$ is denoted as $g_j(r)$. For fixed $i,j$ with $i<j$, there is at most one $k$ satisfying ①; so there are $j-1$ ways to choose $i$, which means $g_j(r)\le j-1$. In a similar way, for fixed $j,k$ with $k>j$, there is at most one $i$ satisfying ①; so there are $n-j$ ways to choose $k$, which means $g_j(r)\le n-j$. Therefore, $$g_j(r)\le \min \{j-1,n-j\}.$$ Then, when $n$ is even (i.e., $n=2m$), we have $$\begin{array}{rl}{f}_n(r)& =\sum _{j=2}^{n-1}{g}_j(r)=\sum _{j=2}^{m-1}{g}_j(r)+\sum _{j=m}^{2m-1}{g}_j(r)\\ & \le \sum _{j=2}^m(j-1)+\sum _{j=m+1}^{2m-1}(2m-j)=\frac{m(m-1)}{2}+\frac{m(m-1)}{2}\\ & =m^2-m<m^2=\frac{n^2}{4}.\end{array}$$ When $n$ is odd (i.e. $n=2m+1$), we have $$\begin{array}{rl}{f}_n(r)& =\sum _{j=2}^{n-1}{g}_j(r)=\sum _{j=2}^m{g}_j(r)+\sum _{j=m+1}^{2m}{g}_j(r)\\ & \le \sum _{j=2}^m(j-1)+\sum _{j=m+1}^{2m}(2m+1-j)\\ & =m^2<\frac{n^2}{4}.\end{array}$$ The proof is completed. $\square$