Triangles with given inradius, A057721

[David Wilson]

T(r) = set of integer triangles with inradius r, so that { |T(n)| } = 
A120062.
P(r) = set of primitive integer triangles with inradius r.

Let sequences t = { |T(n)| } = A120062, p = { |P(n)| }

You suggest that T(r) = UNION(d | r, r/d x P(d)), by which I mean that T(r) 
consists of the primitive triangles in P(d) magnified by a factor of r/d, 
for each d dividing r.  The union is disjoint, which would imply

[1]  t(r) = SUM(d | r, p(d))

Unfortunately, this is not quite true.  By [1], we would expect

[2]  t(3) = p(1) + p(3).

Now, there is only one primitive triangle with inradius 1, namely (3,4,5), 
so p(1) = 1, and there are ten primitive triangles of inradius 3,

(7,24,25) (7,65,68) (8,15,17) (11,13,20) (12,55,65)
(13,40,51) (15,28,41) (16,25,39) (19,20,37) (11,100,109)

which would suggest

t(10) = 1 + 10 = 11.

However, the computer gives t(3) = 13.  The two extra triangles in T(3) are

(10,10,12) (8,26,30).

These triangles are 2x magnifications of the primitive triangles

(5,5,6) (4,13,15)

which have inradius 3/2.  So really

t(3) = p(1) + p(3/2) + p(3) = 1 + 2 + 10 = 13.

So the proper generalization of [1] seems to be

[3]  t(r) = SUM(d in X and r/d in Z, p(d))

where X is the set of rational inradii of integer triangles.  X seems 
complicated at first glance, for example, all integer k > 1 are in X, but 
k/2 seems at a glance to be in X only if k >= 1 is not a product of primes 
== 1 (mod 4) (complement of A004613 with respect to Z+).  X contains 
rationals with arbitrarily large denominator, for example, primitive 
triangle ((k^2+1)/2, (k^2+1)/2, k^2-1) has inradius (k^2-1)/(2k).

This of course blows away any hope of using a Moebius transform to count 
primitive triangles.

However, I have written a program which can count all triangles or only 
primitive triangles of radius 1 <= n <= 1000.  Using it, I have already 
submitted b120062.txt, and can generate a similar list of primitive triangle 
counts once the sequence is submitted.

----- Original Message ----- 
From: "Graeme McRae" <g_m at mcraefamily.com>
To: "seqfan" <seqfan at ext.jussieu.fr>
Sent: Monday, June 12, 2006 6:22 PM
Subject: Re: Triangles with given inradius, A057721


> Mathworld has an article on Heronian Triangles in which a parameterized 
> formula is given that gives all rational-sided triangles with rational 
> area. Using integer values for the parameters, and scaling down all 
> triangles that are not "primitive" (in this sense, primitive means no 
> common divisor among the sides or inradius), I was able to quickly 
> generate all the primitive triangles with inradii 1, 2, 3, ... up to 12.
>
> The number of primitive integer-sided triangles for each radius is 1, 4, 
> 12, 13, 14, 28, 23, 27, 38, 32, 25, 81.  Can someone check that?
>
> Then, assuming that's right, the total number of triangles is the sum of 
> the primitive triangles for the given radius plus those of all the radii 
> that are factors of the given radius.  So, for example, when the inradius 
> is 12, the number of triangles is 81+1+4+12+13+28=139 -- that is, the 81 
> primitive triangles plus the primitive triangles for each of 12's factors 
> 1, 2, 3, 4, and 6.
>
> Using this method, the sequence of integer-sided triangles with inradius n 
> is 1, 5, 13, 18, 15, 45, 24, 45, 51, 51, 26, 139.  Can someone check that, 
> too?
>
> Then, after someone has verified these numbers, I will go ahead and extend 
> A120062, and add the reference to mathworld, along with the method of 
> calculating the sequence.  I will also do a superseeker, and then submit 
> another sequence (and/or comment an existing sequence) that contains just 
> the primitive triangles, and cross-reference them to each other.  Thanks 
> very much!
>
>
>
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