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?
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...
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.
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.)
$(c_1+c_2)^2=a^2+b^2 \Rightarrow (a-r+b-r)^2=a^2+b^2 \Rightarrow (a+b-2r)^2=a^2+b^2$
$(a+b)^2-4r(a+b)+4r^2=a^2+b^2 \Rightarrow 4r^2-4r(a+b)+2ab=0$
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 $r=5$
Your question has been answered but there is a way to find primitive Pythagorean triples for any given inradius if that is desired. We begin with a variation of Euclid's formula shown here as: \begin{align*} &A=(2n-1)^2+ 2(2n-1)k\\ &B=\phantom{(2n-1)^2+{}} 2(2n-1)k+2k^2\\ &C=(2n-1)^2+ 2(2n-1)k+2k^2 \end{align*} We let $\text{inradius}(r)=\dfrac{\text{area}(a)}{\text{semiperimeter}(s)}$ and we let $j=(2n-1)$
$a=\dfrac{AB}{2} =\dfrac{(j^2+2jk)(2jk+2k^2)}{2} =jk(j^2+3jk+2k^2)$
$s=\dfrac{A+B+C}{2} =\dfrac{(j^2+2jk)+(2jk+2k^2)+(j^2+2jk+2k^2)}{2} =j^2+3jk+2k^2$
$r=\dfrac{a}{s} =\dfrac{jk(j^2+3jk+2k^2)}{(j^2+3jk+2k^2)}=jk =(2n-1)k$
Now, given $\,r=(2n-1)k,\,$ the factors of $\,r\,$ indicate which values of $\,(n,k)\,$ will produce a triple with the given inradius.
If $\,r=5,\,$ we have $\,\,(2n-1)=5\implies n=3,\,k=1\,\,$ or $\,\,(2n-1)=1\implies n=1,\,k=5\,\,$ and the formula generates $\,\,F(3,1)= (35,12,37)\,\,$ or $\,\,F(1,5)= (11,60,61).$
