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2020 AMC 12A Problem 7

Problem 7 of 25EasierAlgebraGeometry

Seven cubes, whose volumes are 1,1, 8,8, 27,27, 64,64, 125,125, 216,216, and 343343 cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?

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Solution

The side lengths are 1,2,,7.1, 2, \ldots, 7. The four side faces of cube kk contribute 4k2,4k^2, so the vertical faces total 4(12+22++72)4(1^2 + 2^2 + \cdots + 7^2) =4140=560.= 4 \cdot 140 = 560. Viewed from directly above, every upward-facing horizontal patch projects onto the 7×77 \times 7 base without overlap, giving 49.49. Viewed from below, the same is true, giving another 49.49. The total surface area is 560+49+49=658.560 + 49 + 49 = 658. Thus, B is the correct answer.

More practice

Concepts: surface area · sum of first n squares

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