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Thread: IMO 2011 (Problem 1)

  1. #1
    MHF Contributor FernandoRevilla's Avatar
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    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 $\displaystyle n_A$ .
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    Re: IMO 2011 (Problem 1)

    are i and j must be distinct
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    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.
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