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Statement

Let \(K\) be the Koch curve, the attractor of the IFS of four similarities each of ratio \(r=1/3\) that replace a segment by a four-segment "bump". Then:

  1. The similarity (Hausdorff) dimension is $\(D=\frac{\log 4}{\log 3}\approx 1.2618595071429148.\)$
  2. The curve has infinite length: the length after \(n\) generations is \((4/3)^{n}L_0\to\infty\).
  3. The Koch snowflake (three Koch curves on the sides of an equilateral triangle of area \(A_0\)) encloses finite area $\(A_\infty=\frac{8}{5}A_0.\)$

Assumptions

  1. \(K\) is the attractor of \(N=4\) contracting similarities with equal ratio \(r=1/3\).
  2. Open set condition (OSC): the four maps send an open triangular/segment neighbourhood \(U\) into disjoint sub-copies, \(f_i(U)\cap f_j(U)=\varnothing\) for \(i\ne j\); this licenses similarity dimension \(=\) Hausdorff dimension.
  3. Each generation replaces every segment of length \(\ell\) by four segments of length \(\ell/3\); areas of added triangles scale as \((1/9)\) per generation.

Proof

  1. [asserted] By Moran/Hutchinson under the OSC, \(D\) solves the Moran equation \(\sum_{i=1}^{4}(1/3)^{D}=1\), i.e. \(4\cdot 3^{-D}=1\).
  2. [verified] Solving \(4\cdot 3^{-D}=1\) gives \(3^{D}=4\), so \(D=\log 4/\log 3\). SymPy solve(4*(1/3)**D - 1, D) returns 2*log(2)/log(3) \(=\log 4/\log 3\), numerically \(1.2618595071429148\).
  3. [verified] Infinite length. Generation \(n\) has \(4^{n}\) segments each of length \((1/3)^{n}L_0\), so total length \(L_n=(4/3)^{n}L_0\). SymPy limit((4/3)**n, n, oo) = oo; since \(4/3>1\) the length diverges, and \(K\) is not rectifiable.
  4. [verified] Finite snowflake area. Start from the equilateral triangle of area \(A_0\). At generation \(n\ge 1\) one adds \(3\cdot 4^{\,n-1}\) new triangles, each of area \(A_0\,(1/9)^{n}\) (each added bump triangle is scaled by \(1/3\) in length, hence \(1/9\) in area, relative to the previous generation). The total added area is the convergent geometric series $\(\sum_{n=1}^{\infty} 3\cdot 4^{\,n-1}\,A_0\,\Big(\tfrac{1}{9}\Big)^{n} =\frac{3A_0}{4}\sum_{n=1}^{\infty}\Big(\tfrac{4}{9}\Big)^{n} =\frac{3A_0}{4}\cdot\frac{4/9}{1-4/9}=\frac{3}{5}A_0,\)$ so the enclosed area is \(A_\infty=A_0+\tfrac{3}{5}A_0=\tfrac{8}{5}A_0\).
  5. [verified] SymPy A0 + summation(3*4**(n-1)*A0*(1/9)**n, (n,1,oo)) simplifies to 8*A0/5 exactly, confirming that a boundary of infinite length encloses a finite area.

Verification

  • Dimension (step 2): SymPy solve(4*3**(-D)-1, D) = 2*log(2)/log(3), float \(1.2618595071429148\); matches the Moran root for \(N=4,\ r=1/3\).
  • Perimeter (step 3): SymPy limit((4/3)**n, n, oo) = oo — length diverges geometrically with ratio \(4/3\).
  • Snowflake area (step 4–5): SymPy A0 + summation(3*4**(n-1)*A0*(1/9)**n, (n,1,oo)) simplifies to 8*A0/5 exactly (closed-form geometric series with ratio \(4/9\)): added area \(3A_0/5\), total \(8A_0/5\).
  • Asserted external result: Moran/Hutchinson OSC ⇒ Hausdorff-dimension identity (step 1).

Historical source

Helge von Koch defined the curve in 1904 as an elementary geometric example of a continuous curve lacking a tangent at every point (nowhere differentiable), predating and complementing Weierstrass's analytic 1872 example. The snowflake variant and its area/perimeter paradox are standard consequences.

Related concepts

Citations

  1. H. von Koch, "Sur une courbe continue sans tangente, obtenue par une construction géométrique élémentaire", Arkiv för Matematik 1 (1904), 681–704.
  2. J. E. Hutchinson, "Fractals and self-similarity", Indiana Univ. Math. J. 30 (1981), 713–747.
  3. K. Falconer, Fractal Geometry, 3rd ed., Wiley (2014), §2.2, §9.2.
  4. B. B. Mandelbrot, The Fractal Geometry of Nature, Freeman (1982), ch. 6.