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AMC 8 · 2009 · #9

Grade 4 geometry-2d
Archetype Arithmetic Expression Decomposition
perimeterspatial-visualizationpattern-recognition identify-subproblemspattern-recognition ↑ Prerequisites: multi-digit-arithmetic
📏 Medium solution 💡 3 insights 📊 Diagram
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Problem

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.

(A)
21
(B)
23
(C)
25
(D)
27
(E)
29
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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

#1 Draw a Diagram 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.
$$\triangle - \square - \pentagon - \hexagon - \text{(7-gon)} - \text{(8-gon)}$$

💡 Recognizing each shape by its number of sides ($3, 4, 5, 6, 7, 8$) is Grade 2 "name shapes by their attributes" thinking.

#7 Identify Subproblems 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.
$$\text{contribution} = (\text{total sides}) - (\text{shared sides})$$

💡 Knowing a regular $n$-gon has $n$ equal sides is Grade 4 "classify $2$-D figures" knowledge.

#7 Identify Subproblems 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.
$$3 - 1 = 2$$

💡 A single subtraction word-problem — Grade 3 two-step operations.

#7 Identify Subproblems 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.
$$4 - 2 = 2$$

💡 Same subtraction logic, but now subtracting $2$ because the square is sandwiched between two neighbors.

#7 Identify Subproblems 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.
$$5 - 2 = 3,\;\; 6 - 2 = 4,\;\; 7 - 2 = 5$$

💡 The middle polygons follow a clean pattern: contribution $= n - 2$ for an $n$-gon.

#7 Identify Subproblems 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.
$$8 - 1 = 7$$

💡 Like the triangle, the octagon only has one neighbor, so it loses just one side.

#7 Identify Subproblems 4.OA.A.3 Step 7

Add all six contributions to get the total number of outline sides.

$$2 + 2 + 3 + 4 + 5 + 7 = 23 \;\Rightarrow\; \textbf{(B)}$$

💡 Summing six small numbers to finish a multi-step word problem is Grade 4 "solve multistep problems".

[1] #1 2.G.A.1 Sketch the chain. Draw the triangle, then the square sharing one side with it, t
[2] #7 4.G.A.2 Split the count into one subproblem per polygon. For each polygon, count its tot
[3] #7 3.OA.D.8 Triangle (end of chain): $3$ sides total, $1$ glued to the square. It contribute
[4] #7 3.OA.D.8 Square (middle): $4$ sides total, $2$ glued (one to triangle, one to pentagon).
[5] #7 3.OA.D.8 Pentagon (middle): $5 - 2 = 3$. Hexagon (middle): $6 - 2 = 4$. Heptagon (middle)
[6] #7 3.OA.D.8 Octagon (end of chain): $8$ sides total, $1$ glued to the heptagon. It contribut
[7] #7 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.1 Recognize 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.8 Solve 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.2 Classify 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.3 Solve 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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