1. IMO 2011 (Problem 1)

Given any set $\displaystyle A=\{a_1,a_2,a_3,a_4\}$ of four distinct positive integers, we denote the sum $\displaystyle a_1+a_2+a_3+a_4$ by $\displaystyle s_A$. Let $\displaystyle n_A$ denote the number of pairs $\displaystyle (i, j)$ with $\displaystyle 1 \leq i < j \leq 4$ for which $\displaystyle a_i + a_j$ divides $\displaystyle s_A$. Find all sets $\displaystyle A$ of four distinct positive integers which achieve the largest possible value of $\displaystyle n_A$ .

2. Re: IMO 2011 (Problem 1)

are i and j must be distinct

3. Re: IMO 2011 (Problem 1)

Obviously $\displaystyle n_A \leq 6$.We may assume without loss of generality that $\displaystyle a_1 <a_2<a_3<a_4$.
Well $\displaystyle (a_3+a_4)\nmid(a_1+a_2),(a_2+a_4)\nmid(a_1+a_3)$
So $\displaystyle n_A\leq 4$;
But either we have $\displaystyle a_1+a_4=a_2+a_3$or $\displaystyle n_A \leq3$
In the first case we have $\displaystyle \left{\left\begin{matrix} \\ a_3+a_4=k_1(a_1+a_2) \\ a_1+a_4=a_2+a_3 \\a_1+a_3=k_2(a_2+a_4) \end{matrix}\right\Leftrightarrow \left\{\begin{matrix} \\k_1a_1+k_1a_2-a_3-a_4=0 \\ a_1-a_2-a_3+a_4=0 \\a_1-k_2a_2+a_3-k_2a_4=0 \end{matrix}\right \Leftrightarrow \left\{\begin{matrix} \\k_1a_1+k_1a_2-a_3=a_4 \\ a_1-a_2-a_3=-a_4 \\a_1-k_2a_2+a_3=k_2a_4 \right\end{matrix}.$
We note $\displaystyle a_4=\alpha$ and we have that $\displaystyle \frac{\begin{vmatrix} \alpha &k_1 &-1 \\ -\alpha&-1 &-1 \\ -k_2\alpha &-k_2 &1 \end{vmatrix}}\left{\begin{vmatrix} k_1 &k-1 &-1 \\ 1&-1 &-1 \\ 1&-k_2 &1 \right\end{vmatrix}}=a_1$ and so we find $\displaystyle a_2,a_3$depending on$\displaystyle \alpha$
In this we had 4 pairs and all the solutions;
In the second case we would have 3 pairs so it doesn't matter.