The angle \(A\) of triangle \(ABC\) is a right angle and the sides \(BC\), \(CA\) and \(AB\) are of lengths \(a\), \(b\) and \(c\) respectively. Each side of the triangle is tangent to the circle \(S_1\) which is of radius \(r\). Show that \(2r=b+c-a\).

We need a good diagram…

Can we find any lengths that are equal?

Each vertex of the triangle lies on the circle \(S_2\). The ratio of the area of the region between \(S_1\) and the triangle to the area of \(S_2\) is denoted by \(R\). Show that \[\pi R=-(\pi-1)q^2+2\pi q-(\pi+1),\] where \(q=\dfrac{b+c}{a}\).

Can we find an expression for \(R\) in terms of \(a\), \(b\) and \(c\)? How can we then eliminate \(a, b\) and \(c\) leaving just \(q\)?

Deduce that \[R\leq\dfrac{1}{\pi(\pi-1)}.\]

What is the radius of the circle \(S_2\)?

Can we find the areas of \(S_1\), \(S_2\) and the triangle \(ABC\)?