Competition · AMC preparation · step 4 of 4
AMC 8 · 2002 · #22
Grade 6 geometry-3d
Pick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Counting visible faces one by one on a lumpy stack is error-prone. Tool #16 (Count the Complement) flips the job: start with all the faces the 6 cubes would have if they were separate (6 × 6 = 36), then subtract only the faces that got hidden by being pressed against another cube. Tool #1 (Draw a Diagram) makes the hidden faces easy to find — each place where two cubes touch is one contact, and one contact hides two unit faces. Tool #7 (Identify Subproblems) splits the count into two clean pieces: (a) total faces of separate cubes, (b) hidden faces from contacts.
Add all six cubes' faces
Pretend the 6 cubes are separate: each has 6 faces of area 1, so 6 × 6 = 36 unit faces in all.
Grade 6 surface-area-from-nets: a cube's net is 6 unit squares, so 6 cubes contribute 36 unit squares before any are glued.
6.G.A.4Identify SubproblemsCount the touching pairs
Trace the figure cube by cube: the pairs that share a whole face give 5 contacts.
Drawing or tracing the figure cube-by-cube makes each contact visible exactly once, so nothing is double-counted.
6.G.A.4Draw A DiagramTurn contacts into hidden faces
Each contact hides 2 faces (one per cube), so 5 × 2 = 10 unit faces disappear.
Each glued seam takes two unit squares off the outside — one from each cube — exactly like closing the flaps on a cardboard net.
6.G.A.4Change Focus Count The ComplementSubtract the hidden area
Subtract hidden from the separated total: 36 - 10 = 26 in² → (C).
"All faces minus the hidden ones" is the complement move applied directly to surface area.
The glued solid's surface area equals the six separate cubes' total face area minus the area of the faces buried where cubes are joined.
▸ Why?
Every unit face of the six cubes ends up in exactly one of two groups — still showing on the outside, or pressed flat against a neighbor and buried — with no face in both and none left out, so the outside area is the full total minus the buried part.
▸ Why?
That full total is 36 unit faces, because the six cubes form six equal groups of the same six faces each.
▸ Why?
The buried part comes from the 5 places where two cubes are fastened, and each such place buries exactly 2 faces — one given up by each of the two cubes that meet there.
▸ Why?
At a join, a single face of one cube lies exactly against a single face of the other, pairing the two faces one-for-one, so each join accounts for two buried faces and no more.
▸ Why?
Five joins that each bury two faces make 10 buried faces in all, because that is five equal groups of two.
Imagine the cubes apart (36 faces), then erase 2 faces for every spot where two cubes touch (5 spots = 10 faces). What's left is the surface area: 36 - 10 = 26.
- Add all six cubes' faces
- Count the touching pairs
- Turn contacts into hidden faces
- Subtract the hidden area
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