How Thick Is the Sierpi\'nski Triangle?

TL;DR

The paper proves that the Sierpiński triangle has a precise non-flat thickness of aaa, showing it contains equilateral triangles of proportional size at every point and scale.

Contribution

It establishes the exact Feng--Wu thickness of the Sierpinski triangle as aaa, with an elementary, self-similarity-based proof.

Findings

The Feng--Wu thickness of the Sierpinski triangle is exactly aaa.

At every point and scale, the convex hull of nearby points contains an equilateral triangle of proportional size.

The constant aaa is proven to be optimal.

Abstract

Although the Sierpi\'nski 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 Sierpi\'nski 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 $1/2\le r\le 1$.