Competition · AMC preparation · step 4 of 4
AMC 8 · 2002 · #23
Grade 4 geometry-2dcounting
Pick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The floor is infinite in the imagination but its tiling is periodic, so Tool #5 (Look for a Pattern) tells us the dark fraction is determined by one repeating block — we never have to count the whole floor. Tool #9 (Solve an Easier Related Problem) then shrinks the work even further: instead of working with the full corner shown, find the smallest square block that already captures the proportion. A small 3 × 3 block at the corner does the job. Tool #1 (Draw a Diagram) — really, reading the diagram already given — is what lets us mark each cell of that 3 × 3 block as dark or light and count.
Reduce to one block
Because the same pattern repeats across the floor, the dark fraction of one repeating block equals the dark fraction of the whole floor.
Grade 4 "generate and analyze a repeating pattern" — once the unit repeats, every copy looks the same, so any one copy answers the whole-floor question.
The fraction of tiles that are dark inside one repeating block is the same as the fraction of the whole floor that is dark.
▸ Why?
The whole floor is nothing but many identical copies of that one block, so its dark tiles and its total tiles are each just the block's count taken as many times as there are copies.
▸ Why?
Because the pattern repeats, every part of the floor is the same block slid into a new spot, and sliding the block lays it exactly onto its copy, so the dark tiles land on dark tiles and each copy keeps the very same dark count and the very same total.
▸ Why?
The copies fill the floor with no gaps and no overlaps, so the floor's dark tiles are exactly all the copies' dark tiles gathered back together, and with equal copies that gathering is the block's dark count taken once per copy.
▸ Why?
Fitting the copies edge to edge with nothing left uncovered and nothing counted twice means the floor's tiles are precisely the copies' tiles put back together into the whole.
▸ Why?
Adding the block's count once for each copy, over and over the same amount, is the same as taking that many equal groups of it, which is that count multiplied by the number of copies.
▸ Why?
So the floor's dark-over-total is the block's dark times the number of copies over the block's total times that same number of copies — that is the block's own fraction with its top and its bottom each scaled up by the same number of copies, and scaling a fraction's top and bottom by the same number leaves its value unchanged, so the floor's fraction is exactly the block's.
Pick the smallest block
Look at the 3 × 3 square of tiles in the very corner; since all four corners match, its dark fraction equals the whole floor's.
Solving an easier related problem: instead of counting hundreds of tiles, work with 9. The block size is small enough to count by eye.
4.OA.C.5Solve An Easier Related ProblemMark each cell dark or light
Read each cell of the 3 × 3 corner block from the diagram and mark it dark (D) or light (L), row by row from the top.
Grade 3 "partition a shape into equal parts" — the 3 × 3 grid is already partitioned into 9 unit squares; we just label each one.
3.G.A.2Draw A DiagramCount the dark cells
Row 1 has 1 dark, row 2 has 2, row 3 has 1, so the block holds 4 dark tiles.
Grade 3 multi-step addition: tally the dark tiles in each row, then add the row totals.
3.OA.D.8Look For A PatternForm the fraction
Divide dark tiles by the block's total tiles: 4 out of 9 is the whole floor's dark fraction — choice (B).
Grade 3 fractions: 4 shaded parts out of 9 equal parts is the fraction 4/9.
3.NF.A.1Look For A PatternWhen a pattern repeats, you don't need to count the whole floor — count one small block. A 3 × 3 corner has 4 dark tiles out of 9, so the answer is .
- Reduce to one block
- Pick the smallest block
- Mark each cell dark or light
- Count the dark cells
- Form the fraction
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