IMO 2011 (Problem 1)

**FernandoRevilla** · July 18th 2011, 09:55 AM

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

**waqarhaider** · July 18th 2011, 10:48 AM

are i and j must be distinct

**TheodorMunteanu** · July 19th 2011, 02:40 AM

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

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