Problem of the Month (December 2006)

This month we investigate triangulating polygons. If a regular polygon P with k sides can be cut into n triangles with sides no larger than 1, what is the largest value of the side length s of P? What are the best solutions for small values of k and n?

Gavin Theobald and Trevor Green filled in some gaps on a problem on which they have sent many solutions over the years.

The best known solutions are shown below.

Triangulations of a Triangle
 n=1 s = 1 n=4 s = 2 n=8 s = 1 + 2 / √3 = 2.154+(Gavin Theobald) n=9 s = 3 n=13 s = 2 + 4/√13 = 3.109+(Trevor Green) n=14 2 + 2/√3 = 3.154+ n=15 s = s = √3 + 3/2 = 3.232+ n=16 s = 4 n=20 s = (3√109 + 83) / 28 = 4.082+(Trevor Green) n=21 s = 4.102+(Trevor Green) n=22 s = 4.161+(Trevor Green) n=23 s = 4.205+ n=24 s = 9/2 = 4.500(Trevor Green) n=25 s = 5

Triangulations of a Square
 n=2 s = 1 / √2 = .707+ n=3 s = 2 / √5 = .894+ n=4 s = 1 n=7 s = 1 / √2 + √(3/8) = 1.319+ n=8 s = 8 / 5 = 1.600 n=9 s = 1.660+ n=10 s = 4 / √5 = 1.788+ n=11 s = (1+√7) / 2 = 1.822+(Gavin Theobald) n=12 s = 1 / √(2) + √(3/2) = 1.931+ n=13 s = 1.949+(Gavin Theobald) n=14 s = 2 n=15 s = 2.044+(Gavin Theobald) n=16 s = 2.247+(Gavin Theobald) n=17 s = 2.308+(Gavin Theobald) n=18 s = 6 - 2√3 = 2.535+(Gavin Theobald) n=19 s = 2.560+(Gavin Theobald) n=20 s = 2.604+(Gavin Theobald) n=21 s = 6 / √5 = 2.683+ n=22 s = 2.707+(Gavin Theobald) n=23 s = 2.740+(Gavin Theobald) n=24 s = 2.811+(Gavin Theobald) n=25 s = 2.864+(Gavin Theobald) n=26 s = 2.905+(Gavin Theobald) n=27 s = 3

Triangulations of a Pentagon
 n=3 s = (√5 - 1) / 2 = .618+ n=4 s = 2 / √(5 + 2√5) = .649+ n=5 s = 1 n=10 s = 1.326+(Gavin Theobald) n=11 s = 1.399+(Trevor Green) n=12 s = 1.411+(Trevor Green) n=13 s = 1.489+(Gavin Theobald) n=14 s = 1.595+(Gavin Theobald) n=15 s = 1.625+(Gavin Theobald) n=16 s = 1.723+(Gavin Theobald) n=17 s = 1.784+(Gavin Theobald) n=18 s = 1.922+(Gavin Theobald) n=19 s = 1.938+(Gavin Theobald) n=20 s = 2

Triangulations of a Hexagon
 n=4 s = 1 / √3 = .577+ n=6 s = 1 n=12 s = 2 / √3 = 1.154+ n=15 s = 1.251+(Gavin Theobald) n=16 s = 1.390+(Gavin Theobald) n=18 s = 3/2 = 1.500 n=19 s = 1.541+(Gavin Theobald) n=20 s = 1.549+(Gavin Theobald) n=21 s = 1 + 1 / √3 = 1.577+(Gavin Theobald) n=22 s = 1.709+(Gavin Theobald) n=24 s = 2

Triangulations of a Heptagon
 n=5 s = .445+(Trevor Green) n=6 s = .554(Trevor Green) n=7 s = .867+(Trevor Green) n=10 s = .890(Gavin Theobald) n=11 s = .985+(Gavin Theobald) n=12 s = 1(Trevor Green) n=17 s = 1.031+(Gavin Theobald) n=18 s = 1.124+(Gavin Theobald)

Triangulations of a Octagon
 n=6 s = √2 - 1 = .414+(Trevor Green) n=7 s = .484+(Gavin Theobald) n=8 s = .765+(Trevor Green) n=11 s = 2√2 - 2 = .828+(Gavin Theobald) n=12 s = .907+(Gavin Theobald) n=13 s = .927+(Gavin Theobald) n=14 s = .986+(Gavin Theobald) n=15 s = 1(Gavin Theobald) n=22 s = 1.097+(Trevor Green)

Trevor Green also sent an analysis of small triangulations of polygons with more than 6 sides. Among his results:

The optimal (n–2)-triangulation of a 3k-gon has s = 2 sin(π/n) / √3.
The optimal (n–2)-triangulation of a (3k+1)-gon has s = sin(π/n) / sin(π(n+2)/3n).
The optimal (n–2)-triangulation of a (3k+2)-gon has s = sin(π/n) / sin(π(n+1)/3n).

The optimal (n–1)-triangulation of a 5-gon or (3k+1)-gon can be improved slightly.

The optimal n-triangulation of an n-gon has s = 2 sin(π/n).

The optimal (n+3)-triangulation of an n-gon cannot be improved.

The optimal (n+4)-triangulation of a 3k-gon has s = 4 sin(π/n) / √3.
The optimal (n+4)-triangulation of a (3k+1)-gon has s = 2 sin(π/n) / sin(π(n+2)/3n).
The optimal (n+4)-triangulation of a (3k+2)-gon has s = 2 sin(π/n) / sin(π(n+1)/3n).

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 2/24/11.