It's answer

. I'll add in an actual solution later.

Let the perimeter be n, side of the triangle are a,b,c. Then area =$\frac{abc}{4R}=rs =>\frac{4abc}{65}=n$ Thus 13 divides exactly one of a,b,c and same for 5. From here, hopefully we can solve? (Sorry for the incomplete solution)

adik7 wrote:

It's answer

. I'll add in an actual solution later. I’m sure you will!

You also need to prove that $\lceil \min(p)\rceil=\lfloor \max(p)\rfloor$

$a,b,c<2R<17$ $abc=4SR=2PRr=65P$ Let $13|a$ obvious, that $a=13$ $bc=5(13+b+c)$ Let $b=5d,d<4$ $cd=13+5d+c$ $(c-5)(d-1)=18$ If $d=2,c=23$ but $c<17$ If $d=3,b=15,c=14$ Answer$(13,14,15)$

unless there is some typo we have in the statement that the perimeter is an integer but it s not mentioned that $a,b,c$ are natural numbers