Competition · AMC preparation · step 4 of 4
AMC 8 · 2011 · #3
Grade 6 geometry-2d
Pick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The picture is the whole problem. Tool #1 (Draw a Diagram) lets us see the figure as nested squares — a 1 × 1 white center inside a 3 × 3 black ring inside a 5 × 5 white ring — and then mentally wrap a 7 × 7 black ring around it. Tool #7 (Identify Subproblems) splits the counting into two clean pieces: (a) how many black tiles are in the new outer ring (a difference of two square areas), and (b) what are the new totals once we add that ring to the old counts. The white count never changes, so the only real work is counting the new black tiles.
Picture the nested squares
See the figure as nested squares, so 3² - 1² = 8 black and 17 white — matching the given counts.
Counting tiles in a square ring by subtracting the inner square's area from the outer square's area is the Grade 3 "area as multiplication" idea.
3.MD.C.7Draw A DiagramFind the new side length
A one-tile border grows the side by 2, so the figure becomes a 7 × 7 square of 49 tiles.
Wrapping a square in a one-tile border always bumps the side length by 2 — drawing it once makes this stick.
3.MD.C.7Draw A DiagramCount the new ring of tiles
The new ring is the 7 × 7 minus the 5 × 5, so it holds 7² - 5² = 24 black tiles.
"Big square minus small square" is the same subproblem pattern we already used in Step 1 — it scales straight from 3 vs 1 to 7 vs 5.
The black tiles added around the square form a ring of 24 tiles, equal to the 7 × 7 square's tile count minus the 5 × 5 square's tile count.
▸ Why?
The border is everything in the new 7 × 7 square that is not part of the old 5 × 5 square, and the inner square and the border together fill the whole figure with no gap or overlap, so the border count is the whole minus the inner square.
▸ Why?
Cutting the 7 × 7 figure into the inner 5 × 5 block and the surrounding band leaves no tile uncounted and none counted twice, so the two pieces add back to the full 49 tiles.
▸ Why?
Since the inner block plus the border make the whole, taking the inner block away from the whole must leave exactly the border, because subtracting undoes that adding.
▸ Why?
The 7 × 7 square holds 7 × 7 = 49 tiles and the 5 × 5 square holds 5 × 5 = 25 tiles, because a square of unit tiles is that-many equal rows of that-many tiles.
▸ Why?
The new square is 7 tiles on a side because wrapping one tile band around the 5-tile side sets one tile against each end, and one plus 5 plus one makes 7.
Update the tile totals
Add old and new black: 8 + 24 = 32; white stays 17. Check: 32 + 17 = 49 = 7 × 7. ✓
Separating "old" from "new" turns the count into a small addition — and the total 49 checks our work.
3.OA.A.3Identify SubproblemsWrite the black-to-white ratio
Since 17 is prime and can't divide 32, black : white = 32 : 17 is already in lowest terms — choice (D).
A ratio just compares two counts; once we have 32 and 17, the answer is right there.
6.RP.A.1Identify SubproblemsThis AMC 8 problem only needs Grade 6 ratio language built on Grade 3 area thinking — "big square minus small square" — that you already know!
- Picture the nested squares
- Find the new side length
- Count the new ring of tiles
- Update the tile totals
- Write the black-to-white ratio
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