Problem 12 · AMC 8 Stretch
Core
Geometry & Measurement
Algebra & PatternsArithmetic & Operations
pattern-recognitionorganizing-datareduce-and-expand
In the same Sierpinski pattern, give the stage-0 triangle an area of 1 and a perimeter of 1. Each stage replaces every triangle with 3 half-size copies. What is the total shaded area at stage 5, as a fraction?
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Answer: 243/1024 (about 0.24); perimeter is (3/2)^5 = 243/32
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Hint 1 of 4
Each kept triangle is half as wide and half as tall as its parent. What does cutting the side length in half do to ONE triangle's area? to its perimeter?
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Hint 2 of 4
Area: half the side length makes \(\tfrac14\) the area. There are 3 kept triangles, so total area multiplies by \(3 \times \tfrac14 = \tfrac34\) each stage.
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Hint 3 of 4
Perimeter: half the side length makes half the perimeter. With 3 triangles, total perimeter multiplies by \(3 \times \tfrac12 = \tfrac32\) each stage.
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Approach: Track how area and perimeter scale each stage
- Halving the side length makes each triangle's area \(\left(\tfrac12\right)^2 = \tfrac14\) as big, and there are 3 copies, so total area multiplies by \(3 \times \tfrac14 = \tfrac34\) every stage: \(\text{Area}(n) = \left(\tfrac34\right)^n\).
- Halving the side length makes each perimeter \(\tfrac12\) as big, and there are 3 copies, so total perimeter multiplies by \(3 \times \tfrac12 = \tfrac32\) every stage: \(\text{Perimeter}(n) = \left(\tfrac32\right)^n\).
- At stage 5: area \(= \left(\tfrac34\right)^5 = \tfrac{243}{1024} \approx 0.24\); perimeter \(= \left(\tfrac32\right)^5 = \tfrac{243}{32} \approx 7.59\).
- So the stage-5 area is \(\tfrac{243}{1024}\). Forever: area shrinks toward 0 while perimeter grows without limit — almost no area but an endlessly long edge.
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