Competition · AMC preparation · step 4 of 4
AMC 8 · 2009 · #9
Grade 4 geometry-2d
Pick an answer.
AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem is fundamentally visual: shapes are being glued edge-to-edge in a chain, and every shared edge disappears from the outline. Tool #1 (Draw a Diagram) — even just a quick sketch of the chain — makes it obvious that the two end polygons (triangle and octagon) each lose 1 side to gluing, while the four middle polygons each lose 2 sides. Tool #7 (Identify Subproblems) then turns the big count into six tiny counts — "how many outline sides does each polygon contribute?" — that we add at the end. No algebra needed.
Sketch the polygon chain
Sketch the chain: triangle, square, pentagon, hexagon, heptagon, octagon — six polygons in a row, each sharing a side with its neighbor.
Recognizing each shape by its number of sides (3, 4, 5, 6, 7, 8) is Grade 2 "name shapes by their attributes" thinking.
2.G.A.1Draw A DiagramCount sides polygon by polygon
Handle one polygon at a time: outline sides = total minus glued sides — subtract 1 at each end, 2 in the middle.
Knowing a regular n-gon has n equal sides is Grade 4 "classify 2-D figures" knowledge.
For each polygon in the chain, the number of its sides that stay on the final outline equals its total number of sides minus the number of sides it shares with a neighbor.
▸ Why?
A side that is glued to a neighboring polygon sits between the two shapes, so it is inside the combined figure and never appears on the outline.
▸ Why?
When two polygons are joined edge to edge with no gap and no overlap, the shared edge becomes an inner seam with a shape on both sides, so only the unglued edges form the outer boundary.
▸ Why?
So each polygon puts all of its sides on the outline except the glued ones, and the outline count is its total sides with the glued sides taken away.
▸ Why?
A polygon's glued sides and its outline sides together are all of its sides, so removing the glued sides from the total leaves exactly the outline sides.
Count the triangle's outer sides
Triangle (chain end): 3 sides, 1 glued to the square, so it adds 2 sides to the outline.
A single subtraction word-problem — Grade 3 two-step operations.
3.OA.D.8Identify SubproblemsCount the square's outer sides
Square (middle): 4 sides, 2 glued (triangle and pentagon), so it adds 2 sides.
Same subtraction logic, but now subtracting 2 because the square is sandwiched between two neighbors.
3.OA.D.8Identify SubproblemsCount the middle polygons
Pentagon 5-2=3, hexagon 6-2=4, heptagon 7-2=5 — each middle n-gon adds n-2 sides.
The middle polygons follow a clean pattern: contribution = n - 2 for an n-gon.
3.OA.D.8Identify SubproblemsCount the octagon's outer sides
Octagon (chain end): 8 sides, 1 glued to the heptagon, so it adds 7 sides.
Like the triangle, the octagon only has one neighbor, so it loses just one side.
3.OA.D.8Identify SubproblemsAdd the six contributions
Add the six contributions: 2+2+3+4+5+7 = 23 outline sides — choice (B).
Summing six small numbers to finish a multi-step word problem is Grade 4 "solve multistep problems".
4.OA.A.3Identify SubproblemsThis AMC 8 problem just needs Grade 4 thinking: name the shapes, subtract the glued sides, then add it all up.
- Sketch the polygon chain
- Count sides polygon by polygon
- Count the triangle's outer sides
- Count the square's outer sides
- Count the middle polygons
- Count the octagon's outer sides
- Add the six contributions
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