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2016 AMC 10B Problem 17

Problem 17 of 25IntermediateAlgebraGeometry

All the numbers 2,2,3, 3, 4,4, 5,5,6, 6,7 7 are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

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Solution

Pair opposite faces as (a1,a2)(a_1,a_2), (b1,b2)(b_1,b_2), and (c1,c2)(c_1,c_2). Each vertex product uses one number from each pair, so the sum of all eight vertex products is (a1+a2)(b1+b2)(c1+c2).(a_1+a_2)(b_1+b_2)(c_1+c_2). The six face labels sum to 2+3+4+5+6+7=272+3+4+5+6+7=27, so the three opposite-pair sums have total 2727. Their product is maximized when the sums are as equal as possible, namely 9,9,99,9,9, giving at most 93=7299^3=729. This maximum is attainable by pairing 22 with 77, 33 with 66, and 44 with 55. Hence the greatest possible sum is 729729. Thus, the correct answer is D.

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Concepts: cube geometry · optimization · AM-GM Inequality

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.