Competition · AMC preparation · step 4 of 4
AMC 8 · 2003 · #15
Grade 5 geometry-3d
Pick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem hands us two 2D views and asks for a 3D figure — exactly the trigger for Tool #10 (Create a Physical Representation): place real or imagined unit cubes on a grid until both shadows match. Tool #17 (Visualize Spatial Relationships) then helps us share cubes between the two views by lining them up along a single axis so one cube counts in both projections. Tool #3 (Eliminate Possibilities) is the multiple-choice safety net: the smallest choice is 3, and the front view alone has 3 squares — checking whether 3 cubes can satisfy both views and the connectivity rule lets us rule it out and lock in 4.
Set up the axes
Set up axes so the views become projections: the front view collapses y and shows (x,z); the side view collapses x and shows (y,z).
Reading a 3D location as two 2D pictures is the Grade 5 coordinate-axes idea, just extended to a third axis.
5.G.A.1Visualize Spatial RelationshipsDecode the two views
Decode the views into grid squares: the front L lights (0,0),(0,1),(1,0); the side L lights (0,0),(1,0),(1,1).
Listing the lit squares of each view as ordered pairs turns the picture into a checklist we can match.
5.G.A.1Create A Physical RepresentationLine cubes up to save cubes
Line cubes along a viewing axis: 4 cubes at (0,0,0),(0,0,1),(1,0,0),(1,1,0) project to the front and side L, and every cube shares a face.
Sharing one z=1 cube between the two views (it projects to both top squares) is the minimum-cube trick: stack along the axis the views collapse.
5.G.A.1Create A Physical RepresentationRule out 3 cubes
Try 3 cubes: the shared top cube forces the two bases to different x and y, so one floats with no face-neighbor — connectivity fails.
Even when 3 cubes could in principle cast the right shadows, the "every cube must touch another" rule forces an extra cube — so 4 is the true minimum.
Three unit cubes cannot produce both the front and the side view while every cube still shares a full face with another cube.
▸ Why?
Each view shows exactly three lit squares, and one cube casts exactly one square in each view, so three cubes must cover the three squares one for one — no cube is spare to patch a gap or double up.
▸ Why?
That forced one-for-one covering pins the three cubes into a bent shape: both L's carry a raised top square, so one cube must sit on the upper level, and because the two views face opposite ways, its floor partner lies directly under it while the third cube sits off in the other ground direction — so the two floor cubes end up offset in both ground directions.
▸ Why?
Two unit cubes offset in both ground directions can never be slid so that a whole face of one lays exactly onto a whole face of the other, so they meet at most along an edge and share no full face; the single raised cube rests fully on just one floor cube, leaving the other floor cube with no full-face neighbor, so the touching rule is broken.
Conclude the minimum
The 4-cube figure works and 3 cubes cannot, so the minimum is 4.
From the answer choices, 4 is the only one consistent with both the construction and the elimination.
5.G.A.1Eliminate PossibilitiesTwo flat views, one 3D answer: stack cubes along the line the views collapse, then add one more cube so nothing floats — Grade 5 coordinate thinking pins this AMC 8 problem to 4 cubes.
- Set up the axes
- Decode the two views
- Line cubes up to save cubes
- Rule out 3 cubes
- Conclude the minimum
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