Competition problem — Combinatorics

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Combinatorics Difficulty 4.6 AIME Prove it China

Let $A = \{a_1, a_2, a_3, a_4\}$ . Suppose the set of sums of all the elements in every ternary subset of $A$ is $B = \{-1, 3, 5, 8\}$ . Then $A = \underline{\hspace{2cm}}$ .

Solution

Obviously, every element of $A$ appears three times in all the ternary subsets. Then we have
$3(a_1 + a_2 + a_3 + a_4) = (-1) + 3 + 5 + 8 = 15,$
or $a_1 + a_2 + a_3 + a_4 = 5$ . Therefore, the four elements of $A$ are $5 - (-1) = 6$ , $5 - 3 = 2$ , $5 - 5 = 0$ , $5 - 8 = -3$ , respectively.
The answer is $A = \{-3, 0, 2, 6\}$ . $\square$

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