How Thick Is the Sierpiński Triangle?
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arXiv
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| Format: | Preprint |
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2026
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| _version_ | 1866918478131232768 |
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| author | Kominers, Scott Duke |
| author_facet | Kominers, Scott Duke |
| contents | Although the Sierpiński triangle has planar area $0$, it is uniformly non-flat: at every point and every scale, its nearby points span a two-dimensional region of comparable size. We prove a sharp version of this statement, showing that the Feng--Wu thickness of $E$ is exactly $\sqrt{3}/6$, the inradius of a unit equilateral triangle. More precisely, if $E$ is the standard Sierpiński triangle of side length $1$ and $B(x,r)$ denotes the closed disk of radius $r$ centered at $x$, then for every $x\in E$ and every $0<r\le 1$, the convex hull of $E\cap B(x,r)$ contains an equilateral triangle of side length $r$. Consequently, $\operatorname{conv}(E\cap B(x,r))$ contains a closed disk of radius $(\sqrt{3}/6)r$; this constant is best possible. The proof is elementary -- boundary edges of all construction triangles survive in the limit set, and self-similarity reduces the problem to the normalized range $1/2\le r\le 1$. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2605_01476 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | How Thick Is the Sierpiński Triangle? Kominers, Scott Duke Metric Geometry Classical Analysis and ODEs Dynamical Systems Primary 28A80, Secondary 52A10, 28A75, 11P70 Although the Sierpiński triangle has planar area $0$, it is uniformly non-flat: at every point and every scale, its nearby points span a two-dimensional region of comparable size. We prove a sharp version of this statement, showing that the Feng--Wu thickness of $E$ is exactly $\sqrt{3}/6$, the inradius of a unit equilateral triangle. More precisely, if $E$ is the standard Sierpiński triangle of side length $1$ and $B(x,r)$ denotes the closed disk of radius $r$ centered at $x$, then for every $x\in E$ and every $0<r\le 1$, the convex hull of $E\cap B(x,r)$ contains an equilateral triangle of side length $r$. Consequently, $\operatorname{conv}(E\cap B(x,r))$ contains a closed disk of radius $(\sqrt{3}/6)r$; this constant is best possible. The proof is elementary -- boundary edges of all construction triangles survive in the limit set, and self-similarity reduces the problem to the normalized range $1/2\le r\le 1$. |
| title | How Thick Is the Sierpiński Triangle? |
| topic | Metric Geometry Classical Analysis and ODEs Dynamical Systems Primary 28A80, Secondary 52A10, 28A75, 11P70 |
| url | https://arxiv.org/abs/2605.01476 |