Triangles with given inradius, A057721

[Joseph S. Myers]

On Sat, 10 Jun 2006, all at abouthugo.de wrote:

> and checking the minimum and maximum lengths of the longest triangle side c for a given inradius n one finds
> 
> n cmin  cmax
> 1   5     5
> 2  10    29
> 3  12   109
> 4  15   305
> 5  25   701
> 
> http://www.research.att.com/~njas/sequences/A057721
> n^4 + 3*n^2 + 1. Author: njas
> seems to be the upper bound for the longest side. Can someone try to explain?

If the sides are a, b, c, semiperimeter s=(a+b+c)/2, write u=s-a, v=s-b, 
w=s-c (all positive, either all integers or all halves of odd integers), 
so for inradius r we have the area = rs = sqrt(suvw) so uvw = sr^2 = 
(u+v+w)r^2 (so u, v, w are all integers for integer inradius).

Suppose 1 <= u <= v <= w and we wish to maximise v+w (the longest side).  
We can achieve the given r^4+3r^2+1 using (1, r^2+1, r^4+2r^2); so suppose 
v+w >= r^4 + 3r^2 + 2.  Now u+v+w <= 3w so uv <= 3r^2 so v <= 3r^2 so w >= 
r^4 + 2 so u + v <= 3r^2 + 1 < w(3/r^2 + 1/w) so uv < r^2 + 3 + r^2/w so 
uv <= r^2 + 3.

So v <= r^2 + 3 and w >= r^4 + 2r^2 - 1, so u + v <= r^2 + 4 < w(1/r^2 + 
3/w) so uv < r^2 + 1 + 3r^2/w.  If r = 1 we consider the finitely many 
cases for u and v ((1,1), (1,2), (1,3): w = (u+v)r^2/(uv-r^2) so (1,2,3) 
is the only case with u <= v <= w and this gives the required v+w=5), 
otherwise suppose r >= 2 so uv <= r^2 + 1 so uv = r^2 + 1 (since clearly 
uv > r^2).

Now v <= r^2 + 1 so w >= r^4 + 2r^2 + 1, so u + v <= r^2 + 2 < w/r^2 so uv 
< r^2 + 1 and we have a contradiction.

-- 
Joseph S. Myers
jsm at polyomino.org.uk