Searcharxiv⌕ Search
Home/Archive/arXiv

arXiv · math/0605120

The GGU-model and Generation of the Developmental Paradigms

Robert A. Herrmann
View original source ↗Read source document ↗Export metadata ↓

Abstract

In this brief note, certain illustrated methods for generating developmental paradigms (event sequences) are formalized and compared to previous methods.

Explore related subjects

math.GM

Keep this discovery

BibTeXRIS

Robert A. Herrmann. 2014-02-17. The GGU-model and Generation of the Developmental Paradigms. https://arxiv.org/abs/math/0605120

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↗