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Math Help - IMO 2011 (Problem 1)

  1. #1
    MHF Contributor FernandoRevilla's Avatar
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    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 .
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  2. #2
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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 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_3or n_A \leq3
    In the first case we have \left{\left\begin{matrix}<br />
\\ a_3+a_4=k_1(a_1+a_2)<br />
\\ a_1+a_4=a_2+a_3<br />
\\a_1+a_3=k_2(a_2+a_4)<br />
\end{matrix}\right\Leftrightarrow \left\{\begin{matrix}<br />
\\k_1a_1+k_1a_2-a_3-a_4=0 <br />
\\ a_1-a_2-a_3+a_4=0<br />
\\a_1-k_2a_2+a_3-k_2a_4=0<br />
\end{matrix}\right \Leftrightarrow<br />
\left\{\begin{matrix}<br />
\\k_1a_1+k_1a_2-a_3=a_4 <br />
\\ a_1-a_2-a_3=-a_4<br />
\\a_1-k_2a_2+a_3=k_2a_4<br />
\right\end{matrix}.
    We note a_4=\alpha and we have that \frac{\begin{vmatrix}<br />
\alpha &k_1  &-1 \\ <br />
 -\alpha&-1  &-1 \\ <br />
-k_2\alpha &-k_2  &1 <br />
\end{vmatrix}}\left{\begin{vmatrix}<br />
k_1 &k-1  &-1 \\ <br />
 1&-1  &-1 \\ <br />
 1&-k_2  &1 <br />
\right\end{vmatrix}}=a_1 and so we find a_2,a_3depending 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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