AMC 8 · 2012 · #21
Grade 6 geometry-3dPick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Every face of the cube is treated the same way, so by symmetry the 300 sq ft of green paint splits evenly into 6 equal portions — one per face. That is Tool #11 (Use Symmetry): instead of tracking the whole cube, we only have to think about one face. Then Tool #7 (Identify Subproblems) handles that one face: its 100 sq ft area is made of two pieces, white square + green border, so the white area is just 100 minus the green portion. No square roots or Pythagoras needed — the radical choices are distractors.
Each face of the cube is a 10-by-10 square, so one face has area 100 sq ft.
Area of a rectangle as side × side is the Grade 3 area standard.
3.MD.C.7Work BackwardsBy symmetry the 300 sq ft of green paint splits evenly over 6 faces, so each face gets 50 sq ft of green.
Splitting the painted surface evenly across 6 identical faces is exactly the "surface area as the sum of face areas" idea from Grade 6 nets.
6.G.A.4Work BackwardsOn one face the centered white square and green border fill it with no gaps, so their areas sum to 100 sq ft.
Treating a face as the sum of its non-overlapping regions is the Grade 4 "area as additive" principle.
4.MD.A.3Identify SubproblemsSubtract the green from the whole face: white area = 100 - 50 = 50 sq ft.
Subtracting the green piece from the whole face leaves the white piece — just additive area.
4.MD.A.3Identify SubproblemsThis AMC 8 problem only needs the Grade 6 idea that a cube has 6 equal faces — then it's just 100 - 50 = 50 on one face.