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'Incircles' printed from https://nrich.maths.org/
Ruoyi Sun, Sarah and Elizabeth from the North London Collegiate
School Puzzle Club sent this neat solution.
"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 $$3 - r + 4 -r = 5$$ hence
$r = 1.$ For the 5-12-13 triangle the equivalent formula is $$5 - r
+ 12 - r = 13$$ and hence $r = 2.$"
Sue Liu of Madras College went further to find a formula for other
Pythagorean triples for right angled triangles with incircles of
radius $k$ for any integer $k$ and this is Sue's method.
"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 $${1\over 2}ar +{1\over
2}br +{1\over 2}cr ={1\over 2}ab.$$ $$r = {ab\over a + b + c}.$$
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 $$a = 2mn, \ b = m^2 - n^2, \ c = m^2 +
n^2$$ 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 $2mn, \ m^2 - n^2, \ m^2 + n^2$ Again by equating areas as
before, $${1\over 2} (2mnr + (m^2 - n^2)r + (m^2 + n^2)r) = {1\over
2}(m^2 - n^2)2mn$$ Hence $$r = {2mn(m^2 - n^2) \over 2m(m + n)} =
n(m -n).$$ 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 $c = 41$."
Sue's generalisation of this problem to isosceles triangles is
given as a Further Inspiration.