The snowflake with an infinite edge
Tonight's idea
Start with an equilateral triangle. Call its side 1 and its area 1 (so areas are measured in triangles). Apply one rule to every side: cut out the middle third and replace it with two sides of a smaller equilateral triangle, pointing outward. Repeat forever. Helge von Koch drew it in 1904.
Perimeter. Each step turns every side into four sides of one third the length, so the perimeter is multiplied by \(\tfrac43\): \(3,\; 4,\; \tfrac{16}{3},\; \tfrac{64}{9},\; \ldots\) Repeated, that passes any number you name. The edge is infinitely long.
Area. Stage \(n\) adds one small triangle to each of the \(3 \times 4^{n-1}\) sides of stage \(n-1\). Each new triangle has side \(\tfrac{1}{3^n}\), so area \(\tfrac{1}{9^n}\) (lengths shrink by 3, area by 9). The area added at stage \(n\) is
\(\dfrac{3 \times 4^{n-1}}{9^n} = \dfrac13 \left(\dfrac49\right)^{n-1}\): each stage adds \(\tfrac49\) of what the stage before added.
Write \(S = 1 + \tfrac49 + \left(\tfrac49\right)^2 + \cdots\). Then \(\tfrac49 S = S - 1\) (the move that turns 0.272727… into a fraction), so \(S = \tfrac95\) and the total area is \(1 + \tfrac13 \times \tfrac95 = \tfrac85\). An edge longer than any road, around a region smaller than two triangles.
Watch one: stage 2 by hand
Stage 2 has \(3 \times 4 \times 4 = 48\) sides, each of length \(\tfrac19\). Perimeter: \(\tfrac{48}{9} = \tfrac{16}{3}\), which is \(3 \times \tfrac43 \times \tfrac43\), as it should be.
Area: stage 1 added 3 triangles of side \(\tfrac13\), area \(\tfrac19\) each, total \(\tfrac13\). Stage 2 added 12 triangles of side \(\tfrac19\), area \(\tfrac1{81}\) each, total \(\tfrac{12}{81} = \tfrac{4}{27}\). So \(\text{area}_2 = 1 + \tfrac13 + \tfrac{4}{27} = \tfrac{40}{27} \approx 1.48\). The bumps are already getting small while the perimeter keeps climbing.
Step through the stages and watch the two numbers at mathdetours.com/demos/koch.html.
Your turn
- Stage 3: how many sides, how long is each, and what is the perimeter? Give the perimeter as a fraction and as a decimal.
- Each stage multiplies the perimeter by \(\tfrac43\). After how many stages does the perimeter first exceed 30, ten times the original? Show the powers you computed.
- How much area does stage 3 add? Give the total area after stage 3 as a single fraction and as a decimal.
- (a) Explain, by counting triangles and comparing their sizes, why the area added at any stage is exactly \(\tfrac49\) of the area added at the stage before. (b) The gap between the final area \(\tfrac85\) and the area after stage 2 is \(\tfrac85 - \tfrac{40}{27}\). Compute it, and check it equals \(\tfrac35 \times \left(\tfrac49\right)^2\). Why does the gap shrink by a factor of \(\tfrac49\) every stage?
Stretch: the anti-snowflake
Apply the same rule but point every new triangle inward. The perimeter at each stage is the same as before. What happens to the area, stage by stage, and what does it settle down to? Give the final area as a fraction of the original triangle.
Answers
- \(3 \times 4^3 = 192\) sides, each \(\tfrac{1}{27}\) long. Perimeter \(\tfrac{192}{27} = \tfrac{64}{9} \approx 7.11\).
- You need \(\left(\tfrac43\right)^n > 10\). The powers run \(1.33, 1.78, 2.37, 3.16, 4.21, 5.62, 7.49, 9.99, 13.3\). Eight stages give \(9.99\), just short; nine stages give \(13.3\). So nine stages.
- Stage 3 adds \(3 \times 16 = 48\) triangles of side \(\tfrac{1}{27}\), area \(\tfrac{1}{729}\) each: \(\tfrac{48}{729} = \tfrac{16}{243}\). Total \(\tfrac{40}{27} + \tfrac{16}{243} = \tfrac{376}{243} \approx 1.547\).
- (a) Each stage has four times as many sides as the one before, so four times as many new triangles; each new triangle has one third the side and therefore one ninth the area. Four times as many, each a ninth the size: \(\tfrac49\). (b) \(\tfrac85 - \tfrac{40}{27} = \tfrac{216 - 200}{135} = \tfrac{16}{135}\), and \(\tfrac35 \times \tfrac{16}{81} = \tfrac{48}{405} = \tfrac{16}{135}\). The gap is the sum of all the additions still to come. Every one of those is \(\tfrac49\) of the corresponding addition one stage earlier, so the whole remaining sum is \(\tfrac49\) of the previous gap.
Stretch. The same triangles are removed instead of added, so the area is \(1 - \tfrac13 - \tfrac{4}{27} - \cdots\). The subtractions total \(\tfrac13 \times \tfrac95 = \tfrac35\), so the area settles at \(1 - \tfrac35 = \tfrac25\) of the original triangle. The perimeter is still infinite.