Finding the number of possible right angled triangles

Question

If the inradius=$2013$ of a right angled triangle with integer sides. Find the no. of possible right angled triangles that can be formed using the above information. I have tried $r(a+b+c)=ab$ and $a^2+b^2=c^2$ , but couldn't reach further. Thanks in anticipation

Answer 1

One thing we can get is

$CF=CD=b-r$ and $BE=BD=c-r$. So, $BC=a=b+c-2r\quad (1)$.

By Euclide's formula we can look for a solution such that $\gcd(a,b,c)=1$ and $b=m^2-n^2$, $c=2mn$ and $a=m^2+n^2$ with, $\gcd(m,n)=1$ and not both odd.

We can then back to $(1)$ and get

$$a+2\cdot2013=b+c\\ m^2+n^2+2\cdot2013=m^2-n^2+2mn\\ n^2+2013=mn\to n(m-n)=2013$$

so, $n|2013$ and once $2013=3\cdot11\cdot61$ then we have the possibilities:

$$n\in\{3,11,61,3\cdot11,3\cdot61,11\cdot61,3\cdot11\cdot61\}$$

So you just have to try all possibilities for $n$ and find a suitable $m$.

For example, for $n=3$ you will find $m=674$ and then you have one solution which is $(a,b,c)=(674^2+3^2,674^2-3^2,2\cdot3\cdot674)$.

Now just find the others.

Answer 2

Answer to the question is easy: 8 primitive Pythagorean triples are there for the inradius 2013. There are no other triples (non-primitive Pythagorean triples).

Answer 3

My extended Answer to the question: 2013 has 3 prime factors such that 2013 = 3*11*61, thus, number of Pythagorean triples with inradius = 2^3 = 8. Therefore, there are 8 Pythagorean triples with the given inradius 2013. Ironically, all of them are primitive and no non-primitive Pythagorean triples present for the inradius of 2013.

Reference: Neville Robbins, On the number of primitive Pythagorean triangles with a given inradius, Fibonacci Quarterly 2006, 44(4), pp. 368–369.

For total Pythagorean triples, read: Tron Omland, How many Pythagorean triples with a given inradius?, Journal of Number Theory 2017, 170(1), 1–2.

Answer 4

We begin with a formula that generates the subset of Pythagorean triples where $\space GCD(A,B,C)=(2x-1)^2, x\in\mathbb{N}\space$ which includes all primitives where $GCD(A,B,C)=1.$

A variation of Euclid's formula $$A=(2n-1+k)^2-k^2\quad\ B=2(2n-1+k)k \quad C=(2n-1+k)^2+k^2$$ expands to become

\begin{align*} &A=(2n-1)^2+&&2(2n-1)k\\ &B= &&2(2n-1)k+2k^2\\ &C=(2n-1)^2+&&2(2n-1)k+2k^2 \end{align*} We let inradius(r)=area(a)/semiperimeter(s) and we let $j=(2n-1)$

$a=\dfrac{AB}{2} =\dfrac{(j^2+2jk)(2jk+2k^2)}{2}\\ \quad=jk(j^2+3jk+2k^2)$

$s=\dfrac{A+B+C}{2} =\dfrac{(j^2+2jk)+(2jk+2k^2)+(j^2+2jk+2k^2)}{2}\\ \quad =j^2+3jk+2k^2$

$r=\dfrac{a}{s} =\dfrac{jk(j^2+3jk+2k^2)}{(j^2+3jk+2k^2)}=jk =(2n-1)k$

The triples that have inradius $\space r\space$ will have cofactors $\space f_1=(2n-1)\space$ and $\space f_2=k\space$ where $\space f_1=(2n-1)\implies n=\dfrac{f_1+1}{2}.\space$ The $\space 2013\space$ factors $\space 1\,|\,3\,|\,11\,|\, 33\,|\,61\,|\,183\,|\,671\,|\,2013\space$ imply $\space n\in\{1, 2, 6, 17, 31, 92, 336, 1007\}.\quad$ The $\space 8\space$ triples with inradius $\space 2013 \space$ are

\begin{align*} F(1,2013)&= (4027,8108364,8108365)\\ F(2,671)&= (4035,904508,904517)\\ F(6,183)&= (4147,71004,71125)\\ F(17,61)&= (5115,11468,12557)\\ F(31,33)&= (7747,6204,9925)\\ F(92,11)&= (37515,4268,37757)\\ F(336,3)&= (454267,4044,454285)\\ F(1007,1)&= (4056195,4028,4056197)\\ \end{align*}