Searcharxiv⌕ Search
Home/Archive/arXiv

arXiv · 2503.11675

Dilute Hitomezashi on the Isometric Grid

Katherine A. Seaton
View original source ↗Read source document ↗Export metadata ↓

Abstract

This paper describes preliminary investigation of hitomezashi stitching designs created on the isometric grid. An imposed constraint is that only every second line of stitching in each of the possible three directions is present. Each vertex visited by the stitching has degree two. Motifs and wallpaper patterns with three- and six-fold rotational symmetry are illustrated, in particular designs featuring the von Koch snowflake iterates.

Explore related subjects

math.GM

Keep this discovery

BibTeXRIS

Katherine A. Seaton. 2025-02-28. Dilute Hitomezashi on the Isometric Grid. https://arxiv.org/abs/2503.11675

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

arXiv2026

Average Chord Lengths in a Triangle

Let $P$ be a point inside a triangle $T$. We consider the average length of the chords of $T$ through $P$, where the direction of the chord is chosen uniformly. An elementary formula is obtained in terms of the distances from $P$ to the sides and vertices of the triangle. Several classical triangle centers give especially simple specializations. For example, if $I$ is the incenter, then \[ M_T(I)=\frac{2r}{\pi} \log\left(\cot\frac A4\cot\frac B4\cot\frac C4\right). \] Our main result is the sharp inequality \[ M_T(P)\le \frac{p}{\pi\sqrt3}\log(2+\sqrt3), \] valid simultaneously for every triangle of perimeter $p$ and every interior point $P$. Thus, among all such pairs $(T,P)$, the largest possible average chord length occurs only when $T$ is equilateral and $P$ is its center. The proof is an elementary symmetrization argument. We close with brief remarks relating the problem to the radial center of a convex body, the electrostatic potential center of a triangle, and dual quermassintegrals.

math.GM↗
arXiv2026

A Proof of Liu's Conjecture on the Fundamental Triangle Inequality

Let $a,b,c$ be the side lengths of a triangle, and let $R$ and $r$ denote its circumradius and inradius, respectively. We prove a conjecture of Liu stating that \[\sum_{\mathrm{cyc}} \left(\frac{a(b+c-a)}{bc}\right)^k \geq 2+\left(\frac{2r}{R}\right)^k,~~k>1, \] with the reverse inequality for $0<k<1$. The proof reduces the problem to three positive variables with fixed sum and product. We also determine the equality cases.

math.GM↗