# Thread: IMO 2011 (Problem 1)

1. ## IMO 2011 (Problem 1)

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 $n_A$ .

2. ## Re: IMO 2011 (Problem 1)

are i and j must be distinct

3. ## Re: IMO 2011 (Problem 1)

Obviously $n_A \leq 6$.We may assume without loss of generality that $a_1 .
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.