AMC 8 · 2008 · #16
Grade 6 geometry-3d
Pick an answer.
AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #7 (Break into Subproblems) splits the ratio into two clean counts: volume first, then surface area. Volume is just the cube count. For surface area, Tool #13 (Systematic Count) handles each of the 7 cubes in turn, asking how many of its 6 faces are exposed. Tool #15 (Visualize / Use Symmetry) makes the count almost trivial: the center cube is hidden by all 6 neighbors, and by symmetry every outer cube is in the same situation — exactly 1 glued face and 5 exposed faces.
Each of the 7 unit cubes has volume 1, so the total volume is 7 cubic units.
Grade 5 defines volume by counting unit cubes — exactly the situation here.
5.MD.C.3Identify SubproblemsApart, the 7 cubes show 7 × 6 = 42 faces; the 6 glued seams each hide 2 squares, leaving 30 exposed faces.
Grade 6 surface-area work: start with all faces, then subtract the ones that touch another cube.
6.G.A.4Convert To AlgebraBy symmetry the center cube shows 0 faces and each of the 6 outer cubes shows 5, giving 6 × 5 = 30 again.
Symmetry: the six outer cubes are interchangeable, so we only need to count one of them.
6.G.A.4Organize Information In More WaysVolume to surface area is 7 to 30, and gcd(7, 30) = 1, so is already in lowest terms → (D).
Grade 6 ratio language: pair the two totals and write them as a : b.
6.RP.A.1Identify SubproblemsVolume is just the cube count, and surface area is just face count — every glued seam hides exactly 2 unit squares. Spot that, and a 3D AMC 8 problem becomes a clean Grade 6 ratio.