AMC 10 · 2016 · #18

Grade 7 geometry-3d
Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role
spatial-visualizationcombinations-basic caseworksymmetry-argument ↑ Prerequisites: spatial-visualization
📏 Long solution 💡 4 insights
Problem
Label the eight corners of a cube with the integers 1 through 8, each used once, so that the four corners of every face add to the same total. Two labelings that can be turned into each other by rotating the cube count as one. Count how many different labelings are possible.

Pick an answer.

(A)
1
(B)
3
(C)
6
(D)
12
(E)
24

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

The question asks "how many," so tool #2 (Make a Systematic List) drives the count once the structure is pinned down. First tool #17 (Visualize Spatial Relationships) fixes the cube facts — each corner on three faces, each edge on two — which forces the common face total and reveals where the extreme numbers 1 and 8 must sit. Tool #3 (Eliminate Possibilities) then rules out every placement except 1 and 8 sharing an edge, collapsing a messy search into three clean cases. Tool #1 (Draw a Diagram) keeps the two faces meeting at that edge, and the edge across the cube, straight in mind while the pairs are placed.

1STEP 1

Every face totals 18

All eight labels sum to 36; adding all six faces counts every corner three times, giving 6S=3×36=108, so each face totals 18.

1+2+…+8=36, 6S=3× 36=108, S=18
2STEP 2

1 and 8 must share an edge

Summing the three faces at 8's corner forces 8's neighbors small and 1's large; these can't both hold unless 1 and 8 share an edge.

N=o+2 ≤ 9, N'=o'+16 ≥ 18, {2,3,4}∩{5,6,7}=∅
3STEP 3

Fill each face with a pair summing to 9

Since 1+8=9, each face needs two more numbers summing to 9; only {2,7}, {3,6}, {4,5} work, and which one takes the far edge gives 3 cases.

{2,7}, {3,6}, {4,5}; far-edge pair: 3 choices
4STEP 4

Each case gives a shape and its mirror

Once all three pairs sit on their edges, the only freedom is a mirror flip, not a rotation, so each case gives 2 labelings — total 3×2=6.

3 cases× 2 mirrors=6 → (C)
Answer
6
The face total 18 is forced, and the count 6 sits in the middle of the answer list, fitting that rotations (but not reflections) are merged. Each of the three pairs {2,7},{3,6},{4,5} takes one turn on the far edge, and the two mirror copies in each case are genuinely different labelings, so 3× 2=6 with nothing double-counted and nothing missed.
💡Key takeaway

Every face adds to 18, that forces 1 and 8 onto one edge, and then you just pick which pair sits on the far edge (3 ways) and a mirror flip (2 ways) — six arrangements in all.

  • Every face totals 18
  • 1 and 8 must share an edge
  • Fill each face with a pair summing to 9
  • Each case gives a shape and its mirror

More like this

Same archetype — closest grade level first.