# 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?

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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...

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Actually, $a+b-c$ is an even positive integer, making the in-radius a positive integer. That just leaves $5$ as the only candidate. – Srivatsan Oct 3 '11 at 15:46
Thanks for pointing that out :-) – Listing Oct 3 '11 at 15:51

Listing gives a much better answer. Here I explain how I would approach the problem. The following two facts spring to my mind:

1. 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).

2. 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:

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

2. $n=5$ 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.)

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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.

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"Thus in a very cheap way we can get all positive integers as inradius" - I get that the in-radius is a positive integer, but I don't see the cheap way to get all positive integers. Can you elaborate a bit more? – Srivatsan Oct 3 '11 at 16:00
@Sri: $\frac{(3p)(4p)}{3p+4p+5p}=p$ – J. M. Oct 3 '11 at 16:01
@J.M. Oh, yes. Thanks. I missed the scaling bit in the answer. (Well, this just makes my answer seem silly. :-/) – Srivatsan Oct 3 '11 at 16:03
@Srivatsan Narayanan: Not silly at all, it addresses the more interesting question about inradii of primitive triples. Might want to fix the assertion that all Pythagorean triples are given by $(2mn, m^2-n^2,m^2+n^2)$. – André Nicolas Oct 3 '11 at 16:14
@André Done. :) – Srivatsan Oct 3 '11 at 17:42

$(c_1+c_2)^2=a^2+b^2 \Rightarrow (a-r+b-r)^2=a^2+b^2 \Rightarrow (a+b-2r)^2=a^2+b^2$

$(a+b)^2-4r(a+b)+4r^2=a^2+b^2 \Rightarrow 4r^2-4r(a+b)+2ab=0$

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 $r=5$

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