B3. In the diagram below, right triangle $A B C$ has side lengths $A C=3$ units, $A B=4$ units, and $B C=5$ units. Circles centred around the corners of the triangle all have the same radius, and the circle with centre $O$ has area 4 times that of the circle with centre $P$. The shaded area is $k \pi$ square units. What is $k$ ?
Solution: Since $A_{\circ}=\pi r^{2}$, the ratio of the areas $4=\frac{A_{\odot}}{A_{\odot}}=\frac{\pi r_{\odot}^{2}}{\pi r_{©}^{2}}$ translates into a ratio $\frac{r_{\odot}}{r_{\circledR}}=\sqrt{4}=2$, so $r_{\odot}=2 r_{\circledR}$. Let $x$ be $r_{\circledR}$ and $y$ be the radius of the corner circles. Then $r_{\odot}=2 x$ and $y+2 x+y=3$, $y+4 x+y=4$. Subtracting $2 x+2 y=3$ from $4 x+2 y=4$ we have $x=1 / 2$. Then $y=1$. Thus, $r_{\odot}=1$, and $r_{®}=1 / 2$.
The sum of internal angles in a triangle is $\pi$, so the shaded region is
$$
A=\frac{\pi y^{2}}{2}+\frac{\pi r_{©}^{2}}{2}+\frac{\pi r_{©}^{2}}{2}=\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi\left(\frac{1}{2}\right)^{2}}{2}=\frac{9}{8} \pi \Longrightarrow k=\frac{9}{8} \text {. }
$$

Answer: $k=\frac{9}{8}$.