We found that by drawing the angle bisectors to find the centre of the incircle, and then drawing in 3 radii, we had created three pairs of congruent triangles. Therefore we found that part of the hypotenuse of the 3-4-5 triangle must have length $4-r$ and the other part $3-r$. We formed an equation

${\displaystyle 3-r+4-r=5}$

hence $r=1.$ For the 5-12-13 triangle the equivalent formula is

${\displaystyle 5-r+12-r=13}$

and hence $r=2.$

Clearly the largest circle that fits into a triangle is the incircle where the circle touches the three sides of the triangle. For a right angled triangle we can draw radii of length $r$ from the centre of the incircle perpendicular to each of the three sides $a$, $b$ and $c$. By equating areas we get

${\displaystyle \frac{1}{2}\mathrm{ar}+\frac{1}{2}\mathrm{br}+\frac{1}{2}\mathrm{cr}=\frac{1}{2}\mathrm{ab}.}$

${\displaystyle r=\frac{\mathrm{ab}}{a+b+c}.}$

For the 3-4-5 triangle $r=12/(3+4+5)=1$ so the incircle has radius 1. For the 5-12-13 triangle $r=60/(5+12+13)=2$ and the inradius is 2. The next part of the question asks us to find right angled triangles with incircle radius 3 and sides which are a primitive Pythagorean triples. Pythagorean triples $a,b,c$ are given parametrically by

${\displaystyle a=2\mathrm{mn},\hspace{0.5em}b=m^2-n^2,\hspace{0.5em}c=m^2+n^2}$

where the integers $m$ and $n$ are coprime, one even and the other odd, and $m>n.$ We can consider a triangle with side lengths $2\mathrm{mn},\hspace{0.5em}{m}^2-n^2,\hspace{0.5em}{m}^2+n^2$ Again by equating areas as before,

${\displaystyle \frac{1}{2}(2\mathrm{mnr}+(m^2-n^2)r+(m^2+n^2)r)=\frac{1}{2}(m^2-n^2)2\mathrm{mn}}$

Hence

${\displaystyle r=\frac{2\mathrm{mn}(m^2-n^2)}{2m(m+n)}=n(m-n).}$

By taking $n=1$ and $m=k+1$ or alternatively $n=k$ and $m=k+1$ we get $r=k$ for any integer $k$ (and of course the triangle has inradius $k$ even when $k$ is not an integer). For $r=3$ we have $n=1$ and $m=4$ giving the triangle with sides 8, 15 and 17 or alternatively $n=3$, $m=4$ in which case $a=24$, $b=7$ and $c=25$. For $r=4$ we can take $n=4$, $m=5$ which gives the Pythagorean triple $a=40$, $b=9$ and $c=41$.