AMC 8 · 2009 · #9

Grade 4 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

perimeterspatial-visualizationpattern-recognition identify-subproblemspattern-recognition ↑ Prerequisites: multi-digit-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

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Problem

Start with an equilateral triangle (3 sides). Glue a square onto one of its sides. Glue a regular pentagon onto a non-adjacent side of the square. Keep gluing the next regular polygon (hexagon, heptagon, octagon) onto a non-adjacent side of the previous one, finishing with the octagon. The shapes form one big composite polygon. How many sides does the outline of that composite polygon have?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Sketch the chain: triangle, square, pentagon, hexagon, heptagon, octagon — six polygons in a row, each sharing a side with its neighbor.

△ - □ - pentagon - hexagon - (7-gon) - (8-gon)

2STEP 2

Handle one polygon at a time: outline sides = total minus glued sides — subtract 1 at each end, 2 in the middle.

contribution = (total sides) - (shared sides)

3STEP 3

Triangle (chain end): 3 sides, 1 glued to the square, so it adds 2 sides to the outline.

3 - 1 = 2

4STEP 4

Square (middle): 4 sides, 2 glued (triangle and pentagon), so it adds 2 sides.

4 - 2 = 2

5STEP 5

Pentagon 5-2=3, hexagon 6-2=4, heptagon 7-2=5 — each middle n-gon adds n-2 sides.

5 - 2 = 3, 6 - 2 = 4, 7 - 2 = 5

6STEP 6

Octagon (chain end): 8 sides, 1 glued to the heptagon, so it adds 7 sides.

8 - 1 = 7

7STEP 7

Add the six contributions: 2+2+3+4+5+7 = 23 outline sides — choice (B).

2 + 2 + 3 + 4 + 5 + 7 = 23 → (B)

Answer

Quick sanity check: the six polygons have 3+4+5+6+7+8 = 33 sides in total. The two end polygons share 1 side each (2 shared sides removed from the outline), and the four middle polygons share 2 sides each — but each glued seam is counted by both polygons that share it. There are 5 seams in the chain (triangle-square, square-pentagon, pentagon-hexagon, hexagon-heptagon, heptagon-octagon), so 5 × 2 = 10 sides disappear from the outline. 33 - 10 = 23. Matches answer (B).

💡Key takeaway

This AMC 8 problem just needs Grade 4 thinking: name the shapes, subtract the glued sides, then add it all up.

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