# 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. And thanks to Gavin Theobald for rendering so many pictures for me.

The best known solutions are shown below.

Triangulations of a Triangle
 n=1s = 1 n=4s = 2 n=8s = 1 + 2 / √3 = 2.154+(Gavin Theobald) n=9s = 3 n=13s = 2 + 4/√13 = 3.109+(Trevor Green) n=142 + 2/√3 = 3.154+ n=15s = s = √3 + 3/2 = 3.232+ n=16s = 4 n=20s = (3√109 + 83) / 28 = 4.082+(Trevor Green) n=21s = 4.102+(Trevor Green) n=22s = 4.161+(Trevor Green) n=23s = 4.205+ n=24s = 9/2 = 4.500(Trevor Green) n=25s = 5

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

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

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

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

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