Problem 11 · AMC 8 Stretch
Core
Geometry & Measurement
Number TheoryAlgebra & Patterns
pattern-recognitionorganizing-datareduce-and-expand
Start with one triangle (stage 0). Connect the midpoints of its sides to cut it into 4 small triangles, keep the 3 corner ones, and throw away the middle (that leaves a triangular HOLE). Do the same to every triangle you kept, again and again. At stage 5, how many shaded triangles are there, and how many holes?
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Answer: 243 triangles and 121 holes at stage 5
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Hint 1 of 4
Build a table for the first few stages. Each shaded triangle turns into how many shaded triangles next stage? And how many NEW holes appear?
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Hint 2 of 4
Triangles go 1, 3, 9, 27, 81, ... Each stage MULTIPLIES by 3. Write stage \(n\) as a power of 3.
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Hint 3 of 4
Holes: each stage you punch one new hole in every triangle that existed the stage before. New holes at stage \(n\) = number of triangles at stage \(n-1\) = \(3^{n-1}\). Add up all the holes made so far.
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Approach: Find the multiply-by-3 pattern and sum the new holes
- Each shaded triangle becomes 3 shaded triangles next stage, so the count triples: \(1, 3, 9, 27, 81, \dots\), giving triangles at stage \(n = 3^n\).
- Each stage you punch one new hole in every triangle that was there the stage before, so new holes at stage \(k\) equal \(3^{k-1}\). Holes pile up: holes at stage \(n = 1 + 3 + 9 + \cdots + 3^{n-1}\).
- At stage 5: triangles \(= 3^5 = 243\); holes \(= 1 + 3 + 9 + 27 + 81 = 121\).
- So stage 5 has 243 triangles and 121 holes. (In general the hole count equals \(\tfrac{3^n - 1}{2}\).)
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