Given a finite nonempty set $X$ with real values, let $f(X) = \frac{1}{|X|} \displaystyle\sum\limits_{a\in X} a$, where $\left\lvert X \right\rvert$ denotes the cardinality of $X$. For ordered pairs of sets $(A,B)$ such that $A\cup B = \{1, 2, \dots , 100\}$ and $A\cap B = \emptyset$ where $1\leq |A| \leq 98$, select some $p\in B$, and let $A_{p} = A\cup \{p\}$ and $B_{p} = B - \{p\}.$ Over all such $(A,B)$ and $p\in B$ determine the maximum possible value of $(f(A_{p})-f(A))(f(B_{p})-f(B)).$
Problem
Source:
Tags: inequalities, algebra, China, CGMO
Complex2Liu
16.02.2016 11:46
Let $S=\left(f(A_p)-f(A)\right) \left(f(B_p)-f(B)\right)$, by the definition of $f$ we get \[\left(\mid A\mid+1\right)f(A_p)=\mid A\mid f(A)+p,\]\[\left(\mid B\mid-1\right)f(B_p)=\mid A\mid f(B)-p.\]Therefore \[S=\left(\frac{\mid A\mid f(A)+p}{\mid A\mid+1}-f(A)\right)\left(\frac{\mid B\mid f(B)-p}{\mid B\mid-1}-f(b)\right)=\frac{(p-f(A))(f(B)-p)}{(\mid A\mid+1)(\mid B\mid-1)},\qquad (1)\]and \[S=\frac{(p-f(A_p))(f(B_p)-p)}{\mid A\mid \cdot \mid B\mid}.\qquad (2)\] I am to claim $\mid f(A)-f(B)\mid\le 50$ and $\mid f(A_p)-f(B_p)\mid\le 50$. Note that $\mid A\mid+\mid B\mid=100$, so we get \[\begin{aligned} f(A)-f(B)&\le\frac{100+99+\cdots+(\mid B\mid+1))}{\mid A\mid}-\frac{1+2+\cdots+\mid B\mid}{\mid B\mid}\\ &=\frac{100+(\mid B\mid+1)}{2}-\frac{1+\mid B\mid}{2}=50,\\ f(A)-f(B)&\ge\frac{1+2+\cdots+\mid A\mid}{\mid A\mid}-\frac{100+99+\cdots+(\mid A\mid+1))}{\mid B\mid}\\ &=\frac{1+\mid A}{2}-\frac{100+(\mid A\mid +1)}{2}=-50. \end{aligned}\]So we conclude that $\mid f(A)-f(B)\mid\le 50$, and similarly $\mid f(A_p)-f(B_p)\mid \le 50$. Case 1. If $1\le \mid A\mid \le 97$, since \[(p-f(A))(f(B)-p)\le \left(\frac{(f(A)-f(B))}{2}\right)^2\le 625,\]and $(\mid A\mid +1)(\mid B\mid -1)\ge 2\times 98=196$, applying $(1)$ yields that $S\le\frac{625}{196}$. Case 2. If $\mid A\mid=98$, applying $(2)$ also yields that $S\le\frac{625}{196}$. At last, set $A=\{1\}, B=\{2,3,\cdots,100\}$ and $p=26$, by $(1)$ we get $S=\frac{625}{196}$. Hence the maximum is $\boxed{\frac{625}{196}.}$ This is the official solution .