How to find the number of right-angled triangles with integer sides and inradius $2009$?

Question

Problem: How to find the number of right-angled triangles with integer sides and inradius $2009$.

Please help, as I have no clue how to proceed with this problem. I do know that the inradius of a right-angle triangle with sides $a$, $b$, and $c$ is given by

$$r = \frac{ab}{a+b+c}.$$

Answer 1

We also know that $a=k(u^2-v^2)$, $b=2kuv$, $c=k(u^2+v^2)$ for some integers $k,u,v$, so $$ 2k^2(u^2-v^2)uv=2009k\cdot(u^2-v^2+2uv+u^2+v^2)$$ i.e. $$ k(u-v)v = 2009.$$ Now match the factors on the left with factors of $2009$.

Answer 2

I assume you are asking "how many primitive Pythagorean triples have inradius $r$ equal to 2009?" The answer depends on how many way we can write the factors $r=q'q$ where $q'$ is odd and where $q',q$ are coprime. Obviously, we can write $q'=1, q= 2009$ or as $q'=2009, q=1$. Two ways.

Note also that $r=2009$ has prime factors $7^2$ and $41$. Therefore, there are two more ways to write the inradius as the product of the 2 factors $r=q'q$ as defined above. Either $q'=41$ or $q'=7^2=49$.

The generalized Fibonacci sequence $[q',q,p,p']$ can be used to determine the integer sides of these four right triangles $(a,b,c)$ with inradius $r= q'q$.

One example is

$(q',q,p,p')=(49,41,90,131)$. This corresponds to $(a,b,c)$ where

$r=q'q= (49*41)=2009$

$a=q'p'=6419$

$b=2qp=7380$

$c=pp'-qq'= 9781$.

Using the equations in the example above we can also determine the other 3 triples using the generalized Fibonacci sequences

$(q',q,p,p')=(41,49,90,139)$ yields $(a,b,c)=(5699, 8820,10501)$

$(q',q,p,p')=(1,2009,2010,4019)$ yields $(a,b,c)=(4019, 8076180, 8076181)$

$(q',q,p,p')=(2009,1,2010,2011)$ yields $(a,b,c)=(4040099, 4020,4040101)$

Answer 3

The following formula generates the subset of Pythagorean triples where $GCD(A,B,C)$ is an odd square (which includes all primitives).

\begin{alignat*}{3} &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{alignat*}

Using this formula, the inradius is $\space r=(2n-1)k.$

Proof: We let $\text{inradius}(r)=\dfrac{\text{area}(a)}{\text{semiperimeter}(s)}$ and we let $j=(2n-1)$

\begin{align*}a&=\dfrac{AB}{2} =\dfrac{(j^2+2jk)(2jk+2k^2)}{2}=jk(j^2+3jk+2k^2) \end{align*}

\begin{align*} \qquad s &=\dfrac{A+B+C}{2}=\dfrac{(j^2+2jk)+(2jk+2k^2)+(j^2+2jk+2k^2)}{2}\\ &=j^2+3jk+2k^2 \end{align*}

\begin{align*} r&=\dfrac{a}{s}=\dfrac{jk(j^2+3jk+2k^2)}{(j^2+3jk+2k^2)}=jk=(2n-1)k \end{align*}

The desired triples may be found as a function of these cofactors. The factors of $2009$ are $1\,|\,7\,| 41\,|\,49\,|\,287\,|\,2009$ which implies $ n\in{1,4,21,25,144,1005}\space\land\space k\in\{1,7,41,49,287,2009).\quad$

The triples with an inradius of $\space 2009\space$ are, therefore \begin{align*} F(1,2009)&= (4019,8076180,8076181)\\ F(4,287)&= (4067,168756,168805)\\ F(21,49)&= (5699,8820,10501)\\ F(25,41)&= (6419,7380,9781)\\ F(144,7)&= (86387,4116,86485)\\ F(1005,1)&= (4040099,4020,4040101)\\ \end{align*}