Competition · AMC preparation · step 4 of 4
AMC 8 · 2007 · #23
Grade 6 geometry-2d
Pick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The pinwheel has slanted edges and a complicated outline, but the unshaded region is much friendlier — it's just four corner unit squares and four side triangles. That is the signature of Tool #16 (Count the Complement): measure what is easy and subtract from the whole. Tool #1 (Draw a Diagram) sets the bottom-left corner at (0,0) so the center sits at (2.5, 2.5) and every blade tip has clean coordinates. Tool #7 (Identify Subproblems) breaks the unshaded region into two simple pieces — four 1 × 1 squares and four congruent triangles — each of which falls to a basic Grade 6 area formula.
Find the whole grid's area
Place the grid's bottom-left corner at (0, 0), so the center is (2.5, 2.5) and the total area is 5 × 5 = 25.
Putting the figure on a coordinate grid is a Grade 5 graphing move — it gives every vertex a clean number.
5.G.A.1Draw A DiagramCount the corner squares
At each corner the pinwheel leaves an uncovered 1 × 1 square, so the four corner squares total 4.
Four identical unit squares — a Grade 3 area-of-a-square sum.
3.MD.C.7Identify SubproblemsFind one triangle's area
The bottom triangle has base 3 (from (1,0) to (4,0)) and height 2.5 (up to the center), so its area is 3.75.
When the base sits on a grid line, the height is just the vertical distance from the far vertex — a Grade 6 triangle-area setup.
6.G.A.1Identify SubproblemsUse symmetry for all four
By the pinwheel's four-fold symmetry the other three side triangles match the bottom one, so the four together total 15.
Rotating the bottom triangle a quarter turn at a time produces the other three, so all four have the same area.
6.G.A.1Identify SubproblemsSubtract the unshaded area
Subtract the unshaded total 4 + 15 = 19 from the grid's 25: 25 - 19 = 6, the shaded pinwheel's area.
Total minus the easy-to-measure complement is the standard "count what's left" move.
The shaded pinwheel's area equals the whole 5 × 5 grid's area minus the unshaded region that surrounds it.
▸ Why?
The pinwheel and the unshaded region tile the entire grid with no gaps and no overlaps, so their areas add up to the grid's total of 25; removing the unshaded part therefore leaves exactly the pinwheel.
▸ Why?
Subtracting is the smart move because the unshaded region is easy to total: it splits into four 1 × 1 corner squares and four congruent side triangles.
▸ Why?
The four side triangles are quarter-turn rotations of one another about the center, and a rotation lays a shape onto an exact copy, so all four have equal area and measuring one measures them all.
▸ Why?
One side triangle has base 3 along a grid line and height 2.5 straight up to the center, so its area is one-half base times height.
The pinwheel itself has slanted edges, but the empty space around it is plain: four unit squares and four equal triangles. Add those up, subtract from 25, and the pinwheel's area falls out as 6.
- Find the whole grid's area
- Count the corner squares
- Find one triangle's area
- Use symmetry for all four
- Subtract the unshaded area
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