Assume $\angle{BAC}=90^\circ$ and denote $BC=a, AC=b, AB=c; r,R,p$ - the radius of the inscribed circle, the radius of the circumscribed circle, respective the semi-perimeter of the $\triangle{ABC}$. $p=\dfrac{a+b+c}{2}; S=\dfrac{bc}{2}$; $r=\dfrac{S}{p}=\dfrac{bc}{a+b+c}; S_1=\pi r^2=\dfrac{\pi b^2c^2}{(a+b+c)^2}$; $R=\dfrac{a}{2}; S_2=\pi R^2=\dfrac{\pi a^2}{4}$. $a^2=b^2+c^2\ge2bc\Longrightarrow k=\dfrac{bc}{a^2}\le\dfrac{1}{2}$. $2a^2=2b^2+2c^2\ge (b+c)^2\Longrightarrow b+c\le a\sqrt2$. Results: $\pi\dfrac{S-S_1}{S_2}=\pi\dfrac{\dfrac{bc}{2}-\dfrac{\pi b^2c^2}{(a+b+c)^2}}{\dfrac{\pi a^2}{4}}=\dfrac{2bc}{a^2}-\dfrac{4\pi b^2c^2}{a^2(a+b+c)^2}=$ $=2k-\dfrac{4\pi k^2}{\left(1+\dfrac{b+c}{a}\right)^2}\le 2k-\dfrac{4\pi k^2}{(1+\sqrt2)^2}$. The maximum of the second degree polynomial $P(k)=2k-\dfrac{4\pi k^2}{(1+\sqrt2)^2}$ occurs for $k_0=\dfrac{(1+\sqrt2)^2}{4\pi}<\dfrac{1}{2}$: $P(k)=2k-\dfrac{4\pi k^2}{(1+\sqrt2)^2}\le P(k_0)=\dfrac{(1+\sqrt2)^2}{4\pi}$. $\dfrac{(1+\sqrt2)^2}{4\pi}<\dfrac{1}{\pi-1}\Longleftrightarrow [(1+\sqrt2)^2-4)]\pi<(1+\sqrt2)^2$. Using $1.41<\sqrt2<1.415, \pi<3.15$ results: $[(1+\sqrt2)^2-4]\pi<[(1+1.415)^2-4]\cdot 3.15=5.77150875<(1.41+1)^2=5.8081<(1+\sqrt2)^2$. Hence: $\pi\dfrac{S-S_1}{S_2}<\dfrac{1}{\pi-1}$.

WallyWalrus wrote: Assume $\angle{BAC}=90^\circ$ and denote $BC=a, AC=b, AB=c; r,R,p$ - the radius of the inscribed circle, the radius of the circumscribed circle, respective the semi-perimeter of the $\triangle{ABC}$. $p=\dfrac{a+b+c}{2}; S=\dfrac{bc}{2}$; $r=\dfrac{S}{p}=\dfrac{bc}{a+b+c}; S_1=\pi r^2=\dfrac{\pi b^2c^2}{(a+b+c)^2}$; $R=\dfrac{a}{2}; S_2=\pi R^2=\dfrac{\pi a^2}{4}$. $a^2=b^2+c^2\ge2bc\Longrightarrow k=\dfrac{bc}{a^2}\le\dfrac{1}{2}$. $2a^2=2b^2+2c^2\ge (b+c)^2\Longrightarrow b+c\le a\sqrt2$. Results: $\pi\dfrac{S-S_1}{S_2}=\pi\dfrac{\dfrac{bc}{2}-\dfrac{\pi b^2c^2}{(a+b+c)^2}}{\dfrac{\pi a^2}{4}}=\dfrac{2bc}{a^2}-\dfrac{4\pi b^2c^2}{a^2(a+b+c)^2}=$ $=2k-\dfrac{4\pi k^2}{\left(1+\dfrac{b+c}{a}\right)^2}\le 2k-\dfrac{4\pi k^2}{(1+\sqrt2)^2}$. The maximum of the second degree polynomial $P(k)=2k-\dfrac{4\pi k^2}{(1+\sqrt2)^2}$ occurs for $k_0=\dfrac{(1+\sqrt2)^2}{4\pi}<\dfrac{1}{2}$: $P(k)=2k-\dfrac{4\pi k^2}{(1+\sqrt2)^2}\le P(k_0)=\dfrac{(1+\sqrt2)^2}{4\pi}$. $\dfrac{(1+\sqrt2)^2}{4\pi}<\dfrac{1}{\pi-1}\Longleftrightarrow [(1+\sqrt2)^2-4)]\pi<(1+\sqrt2)^2$. Using $1.41<\sqrt2<1.415, \pi<3.15$ results: $[(1+\sqrt2)^2-4]\pi<[(1+1.415)^2-4]\cdot 3.15=5.77150875<(1.41+1)^2=5.8081<(1+\sqrt2)^2$. Hence: $\pi\dfrac{S-S_1}{S_2}<\dfrac{1}{\pi-1}$. beautiful solution!