Competition · AMC preparation · step 4 of 4
AMC 8 · 2014 · #19
Grade 6 geometry-3dPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
This is a 3D placement problem, exactly the trigger for Tool #10 — picture (or build with 27 blocks) a 3 × 3 × 3 cube and notice that not every slot exposes the same number of faces. Tool #7 (Identify Subproblems) splits the task into three clean pieces: (i) find the total surface area, (ii) classify the 27 slots by how many faces they expose, (iii) place the 6 white cubes greedily in the most-hidden slots. Tool #2 (Systematic List) handles step (ii): list the slot types in order from least exposed to most exposed and count each.
Find the total surface area
A cube has 6 square faces, each 3 × 3 = 9 sq in, so the total surface area is 54 square inches.
Splitting the cube's exterior into 6 square faces is the Grade 6 surface-area move — count the faces, find one face's area, multiply.
6.G.A.4Identify SubproblemsSort the slots by exposure
Sort the 27 slots by exposed faces: a corner shows 3, an edge 2, a face-center 1, and the dead center shows 0.
Physically handling (or imagining) the cube makes the four slot types obvious — corners stick out in 3 directions, edges in 2, face-centers in 1, the center in none.
6.G.A.4Create A Physical RepresentationCheck the slot counts
List the types by exposure and add the counts — they total 27, so no slot is missed.
A systematic list of the slot types — ordered by exposure 0,1,2,3 — guarantees no slot is double-counted or missed.
4.OA.A.3Make A Systematic ListHide the white cubes
Fill the most-hidden slots first: 1 white cube in the dead center (0 faces), the other 5 in face-centers (1 face each).
To minimize total exposed white, hide as many white cubes as possible and let the rest each show only 1 face — this is the greedy subproblem of minimizing a sum of exposures.
Hiding one white cube in the deep center slot that shows no faces and putting the other five in face-center slots that each show one face gives the smallest possible visible white area.
▸ Why?
The visible white area is the exposed white faces added up, so the smallest white area is the placement that makes that sum of exposures smallest.
▸ Why?
The white on the outside breaks into one face-patch per white cube with no gaps and no overlaps, so those patches add back to the whole white area.
▸ Why?
The deep center slot and the six face-center slots make seven slots that each show at most one face, which is more than the six white cubes, so all six whites can be placed without any of them landing in a two-face or three-face slot.
▸ Why?
Pairing each of the six white cubes with its own separate at-most-one-face slot still leaves a slot unused, so the six cubes fit inside the seven slots.
▸ Why?
No placement can beat giving every white cube an at-most-one-face slot: if a white cube sat in a more-exposed slot while some at-most-one-face slot held a red cube, swapping the two cubes would lower the white total.
▸ Why?
The swap trades one white cube's larger exposure for a smaller one and leaves every other white cube's exposure unchanged, and shrinking one part of a sum shrinks the whole sum.
Find the white surface area
The center cube shows 0, and each of the 5 face-center cubes shows a 1 × 1 square, so the white area is 5 sq in.
Adding up the exposed face count from each slot turns the placement into a number — exactly the Grade 6 surface-area calculation, just applied to the white portion only.
6.G.A.4Identify SubproblemsForm the fraction
Put white area over total to get , already in lowest terms since gcd(5, 54) = 1.
Comparing a part to the whole as a fraction is the Grade 3 fraction concept — 5 of the 54 unit squares on the surface are white.
3.NF.A.1Identify SubproblemsHide one white cube in the very middle, and the other five in face-centers so each shows just one square — that is the smallest white surface possible.
- Find the total surface area
- Sort the slots by exposure
- Check the slot counts
- Hide the white cubes
- Find the white surface area
- Form the fraction
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