AMC 10 · 2015 · #20
Grade 7 geometry-3dPick an answer.
AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
A blind hunt through all corner-to-corner routes is hopeless, so the plan first **shrinks the problem**, then counts. Draw the cube (Tool #1) and color its corners like a 3D checkerboard (Tool #17): every edge flips dark↔light, so a 7-edge (odd) trip must finish on the opposite color. The 'can't return home' clue (Tool #3, Eliminate Possibilities) then forces the finish to be the single far corner. With both ends pinned, counting is easy by **working backwards** from that far corner (Tool #11): only the first couple of backward steps are real choices, the rest are forced. A plain systematic list (Tool #2) cross-checks the total. We avoid algebra entirely — the whole thing is coloring plus a tiny multiplication.
Draw the cube and mark start S. Every corner touches exactly 3 edges, so visiting all 8 corners along 7 edges is a single route.
Drawing the cube's corner-and-edge skeleton makes the hidden structure of the crawl visible, so the routes become countable.
1.G.A.2Draw A DiagramColor the corners like a 3D checkerboard so every edge flips the color. From dark S, an even step count ends on dark, an odd one on light.
Each step changes the color exactly once, so where you end up depends only on whether the number of steps is odd or even.
2.OA.C.3Visualize Spatial Relationships7 is odd, so Erin ends on a light corner: S's 3 neighbors or the far one. 'No edge home' rules out the neighbors, leaving the far corner F.
An odd-length trip must land on the opposite color, and 'no edge back home' leaves only the single far corner of that color.
2.OA.C.3Eliminate PossibilitiesCount S-to-F routes through all corners backwards: the last edge into F has 3 choices, the next has 2 choices, and the rest is forced.
Tracing the count backwards as a short tree — 3 branches, then 2, then nothing — shows almost the entire route is forced once both ends are fixed.
7.SP.C.8Work BackwardsMultiply the independent choices: 3 × 2 = 6 complete routes end at the far corner — a forward check from S confirms it, matching choice (A).
Two independent stages with 3 and 2 options combine by multiplying, giving 6 routes in all.
3.OA.A.1Make A Systematic ListColor the cube like a checkerboard: an odd 7-step trip has to end on the far corner, and from there only 3× 2 = 6 all-corner routes exist.