AMC 8 · 2009 · #9
Grade 4 geometry-2dProblem
Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?
Pick an answer.
AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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?
Givens: Polygons used, in order: triangle ($3$), square ($4$), pentagon ($5$), hexagon ($6$), heptagon ($7$), octagon ($8$); Each new polygon is glued onto exactly one side of the previous polygon; "Non-adjacent" just means the new polygon sticks out into open space, not back through the chain; Answer choices: (A) $21$, (B) $23$, (C) $25$, (D) $27$, (E) $29$
Unknowns: The number of sides on the outline of the final composite shape
Understand
Restated: 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?
Givens: Polygons used, in order: triangle ($3$), square ($4$), pentagon ($5$), hexagon ($6$), heptagon ($7$), octagon ($8$); Each new polygon is glued onto exactly one side of the previous polygon; "Non-adjacent" just means the new polygon sticks out into open space, not back through the chain; Answer choices: (A) $21$, (B) $23$, (C) $25$, (D) $27$, (E) $29$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
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.
Execute — Answer: B
2.G.A.1 Step 1 - Sketch the chain.
- Draw the triangle, then the square sharing one side with it, then the pentagon sharing one side with the square, and so on through the octagon.
- The picture shows six polygons in a row, each connected to its neighbor by exactly one shared side.
💡 Recognizing each shape by its number of sides ($3, 4, 5, 6, 7, 8$) is Grade 2 "name shapes by their attributes" thinking.
4.G.A.2 Step 2 - Split the count into one subproblem per polygon.
- For each polygon, count its total sides, then subtract the sides that are glued (interior) — $1$ for the two end polygons, $2$ for the four middle polygons.
💡 Knowing a regular $n$-gon has $n$ equal sides is Grade 4 "classify $2$-D figures" knowledge.
3.OA.D.8 Step 3 - Triangle (end of chain): $3$ sides total, $1$ glued to the square.
- It contributes $3 - 1 = 2$ outline sides.
💡 A single subtraction word-problem — Grade 3 two-step operations.
3.OA.D.8 Step 4 - Square (middle): $4$ sides total, $2$ glued (one to triangle, one to pentagon).
- It contributes $4 - 2 = 2$ outline sides.
💡 Same subtraction logic, but now subtracting $2$ because the square is sandwiched between two neighbors.
3.OA.D.8 Step 5 - Pentagon (middle): $5 - 2 = 3$.
- Hexagon (middle): $6 - 2 = 4$.
- Heptagon (middle): $7 - 2 = 5$.
- Each middle polygon loses $2$ sides to its two neighbors.
💡 The middle polygons follow a clean pattern: contribution $= n - 2$ for an $n$-gon.
3.OA.D.8 Step 6 - Octagon (end of chain): $8$ sides total, $1$ glued to the heptagon.
- It contributes $8 - 1 = 7$ outline sides.
💡 Like the triangle, the octagon only has one neighbor, so it loses just one side.
4.OA.A.3 Step 7 Add all six contributions to get the total number of outline sides.
💡 Summing six small numbers to finish a multi-step word problem is Grade 4 "solve multistep problems".
2.G.A.1 Sketch the chain. Draw the triangle, then the square sharing one side with it, t 4.G.A.2 Split the count into one subproblem per polygon. For each polygon, count its tot 3.OA.D.8 Triangle (end of chain): $3$ sides total, $1$ glued to the square. It contribute 3.OA.D.8 Square (middle): $4$ sides total, $2$ glued (one to triangle, one to pentagon). 3.OA.D.8 Pentagon (middle): $5 - 2 = 3$. Hexagon (middle): $6 - 2 = 4$. Heptagon (middle) 3.OA.D.8 Octagon (end of chain): $8$ sides total, $1$ glued to the heptagon. It contribut 4.OA.A.3 Add all six contributions to get the total number of outline sides. Review
Reasonableness: 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 $5 \times 2 = 10$ sides disappear from the outline. $33 - 10 = 23$. Matches answer (B).
Alternative: Tool #5 (Look for a Pattern): each middle $n$-gon contributes $n - 2$ outline sides, and the two end polygons contribute (sides $- 1$). So the total is $(3 - 1) + (4 - 2) + (5 - 2) + (6 - 2) + (7 - 2) + (8 - 1) = 2 + 2 + 3 + 4 + 5 + 7 = 23$. Same answer (B).
CCSS standards used (min grade 4)
2.G.A.1Recognize and draw shapes having specified attributes, such as a given number of angles (Naming the six polygons in the chain by their number of sides ($3, 4, 5, 6, 7, 8$).)3.OA.D.8Solve two-step word problems using the four operations (Subtracting shared sides from each polygon's total side count (e.g., $3 - 1 = 2$, $4 - 2 = 2$).)4.G.A.2Classify two-dimensional figures based on properties of their sides and angles (Using the fact that a regular $n$-gon has $n$ equal sides to set up each polygon's contribution.)4.OA.A.3Solve multistep word problems using the four operations (Combining the six subtraction results into the final sum $2 + 2 + 3 + 4 + 5 + 7 = 23$.)
⭐ This AMC 8 problem just needs Grade 4 thinking: name the shapes, subtract the glued sides, then add it all up.
⭐ 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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