Is the triple a P.P.T ?

$a = \dfrac{m}{n}(b + c)$ and $a + b + c = 1 \implies b + c = 1 - a \implies \dfrac{n}{m}a = 1 - a \implies$

$\implies na = m - ma \implies (n + m)a = m \implies \boxed{a = \dfrac{m}{n + m}} \implies$

$b + c = 1 - \dfrac{m}{n + m} = \dfrac{n}{n + m} \implies c = \dfrac{n}{n + m} - b = \dfrac{n - (n + m)b}{n + m} \implies$

$n^2 - 2n(n + m)b + (n + m)^2b^2 = m^2 + (n + m)^2b^2 \implies$

$n^2 - m^2 = 2n(n + m)b \implies \boxed{b = \dfrac{n - m}{2n}} \implies$

$c = \boxed{\dfrac{n^2 + m^2}{2n(n + m)}} \implies (a,b,c) =$ $(\dfrac{m}{n + m}, \dfrac{n - m}{2n}, \dfrac{n^2 + m^2}{2n(n + m)})$

Multiplying each side of right $\triangle{ABC}$ by $2n(n + m)$ we obtain

$(a,b,c) \sim (2nm, n^2 - m^2, n^2 + m^2) = (a^{*}, b^{*}, c^{*})$ and $gcf(n,m) = 1,$

$n$ is odd and $m$ is even and $n > m \implies (a^{*}, b^{*}, c^{*})$ is a primitive pythagorean triple

$\implies a^{*} + b^{*} + c^{*} = 2n^2 + 2nm = 2nm + 56n + 186 \implies$

$2(n^2 - 56n - 186) = 0 \implies 2(n + 3)(n - 31) = 0$ and $n > 0 \implies n = 31$

$\implies k = \lfloor{\dfrac{31}{2}\rfloor} = \boxed{15}$ .

Note: I used the following theorem below:

Let $n$ and $m$ be positive integers with $n > m$ and $a = n^2 - m^2, b = 2nm, c = n^2 + m^2$ .

The triple $(a,b,c)$ is a primitive pythagorean triple if and only if $gcf(m,n) = 1$ and $m$ and $n$ are not both odd.