Range of inradius of a right Triangle Today in my test i was asked a question regarding the values which inradius of a given right angled triangle with integer sides can take, options to whose answers were
a)2.25
b)5
c)3.5
i simply couldnt understand how to start, as in i tried with few basic triplets i knew like (3,4,5) and (5,12,13) etc but never the inradius was coming near to 2.25 leave alone other 2 options, so i think the question might be wrong.
any ideas?
 A: As given here a right triangle like this:

has an inradius of $r=\frac{1}{2}(a+b-c)$. You know that $(a+b-c)$ has to be a positive integer...
A: And, to continue the answer of @Listing, the Pythagorean triangle with sides 
$15$, $20$, $25$ has inradius $5$. Thus in a very cheap way we can get all positive integers as inradius, by suitably scaling the $(3,4,5)$. 
The primitive Pythagorean triangles with inradius $5$ are the $(12,35,37)$ and the $(11,60,61)$. That's all. 
Added: If you computed the inradius of the most familiar Pythagorean triangle, namely the $(3,4,5)$, and got the right answer, which is $1$, it is immediate that you can scale up by a factor of $d$, where $d$ is any positive integer, to get inradius $d$. Thus $5$ is certainly achievable. Since it is one of the provided answers, and (presumably) only one of the answers is right, you can tick (b) and go on to the next question.
A: Listing gives a much better answer. Here I explain how I would approach the problem. The following two facts spring to my mind:


*

*The in-radius of a triangle is 
$$
r = \frac{2\Delta}{P}
$$
where $\Delta$ and $P$ are the area and perimeter of the triangle. For a right triangle, it takes the simple form
$$
r = \frac{ab}{a+b+c},
$$
where $a,b,c$ are the sides ($c$ is the hypotenuse).

*The side lengths of an integer right triangle are given by the Pythagorean triples. I know that any Pythagorean triple1 can be written as $(2kmn, k(m^2-n^2), k(m^2+n^2))$, where $k$, $m$ and $n$ are positive integers. So the inradius becomes
$$
\frac{2kmnk(m^2-n^2)}{2kmn+(km^2-kn^2)+(km^2+kn^2)} = \frac{2mkn(m-n)(m+n)}{2m(m+n)} = kn(m-n),
$$
which is an integer. 
1If we want only primitive Pythagorean triples, then in the formula, we should restrict $k$, $m$ and $n$ so that $k=1$ (think of $k$ as the scaling factor multiplying a given primitive solution), $\gcd(m,n)=1$ and $m-n$ is an odd integer. (Exercise: What happens when $m$ and $n$ are both odd, so that $m-n$ is even?) 

That rules out 2 options, leaving only $5$ as a possibility. If exactly one of the options is guaranteed to be correct, then I would mark this option and move on ;). Otherwise...
We want to find if there are positive integral solutions to $n(m-n)=5$. This gives us two solutions:


*

*$n = 1$ and $m=6$. The triangle in this case is $(12, 35, 37)$. 

*$n=5$ and $m=6$. The triangle in this case is $(11, 60, 61)$ (rearranging the sides a bit). 

If we want a general solution, every positive integer $r$ can be the inradius of an integer right triangle whose sides are primitive integers. (André shows that the $(3r, 4r, 5r)$-triangle has in-radius $r$, which gives an easy solution to the problem if we remove the primitiveness restriction.) It is easy to see that the equation
$$
r = n(m-n)
$$
is satisfied by $n=r$ and $m=r+1$, which all the necessary conditions ($\gcd(m,n)=1$ and $m-n=1$, which is odd). This solution gives the triangle $(2r+1, 2r(r+1), 2r^2+2r+1)$. (Of course, this is not the only possible solution.)
A: $(c_1+c_2)^2=a^2+b^2 \Rightarrow (a-r+b-r)^2=a^2+b^2 \Rightarrow (a+b-2r)^2=a^2+b^2$
$(a+b)^2-4r(a+b)+4r^2=a^2+b^2 \Rightarrow 4r^2-4r(a+b)+2ab=0$ 
Now we have to plug in each solution for $r$ in last equation:
$a)$  we get 25+2ab=20(a+b)  , which is not correct since left hand side represent odd integer and right hand side even integer
$b)$ we get 20(a+b)=100+2ab , which may be correct since both left hand side and right hand side represents even integer..
$c)$ we get $14(a+b)=49+2ab$ , which is not correct since left hand side represents even integer and right hand side represents odd integer 
So, only possible solution is $r=5$
