Modern day Pythagorean Triples

$2018=2 \times 1009$ where $1009$ is prime number. Primitive triple has odd hypotenuse and odd one leg. If $2018$ were to be part of a primitive triple it would have to be of form $2 m n$ using Euclid formula for generating triples where $m>n>0$ and only one of $m,n$ values is odd. This is a contradiction. Therefore 2018 cannot be a part of any primitive triple. We will consider triples with 1009.

$1009 = 4 \times 252 + 1=m^2 + n^2$

Analysing all possibilities we get $m=28,n=15$ . This gives inradius $r=n(m-n)=15 \times 13 = 195$ . Now we consider triples with one leg equal to $1009$ . It would have to be in the form: $1009=m^2-n^2=(m-n)(m+n)$ $m-n=1 \Rightarrow 2 n + 1 = 1009 \Rightarrow m=505, n=504 \Rightarrow r=504$ The answer is $2(195+504)=\boxed{1398}$