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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.
Incircles
Age 16 to 18
Challenge Level 
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.
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