Three circles with centres $A, B$ and $C$ have radii $\sqrt{3}-1,3-\sqrt{3}$ and $1+\sqrt{3}$ respectively. Each circle is externally tangent to the other two as shown. The area of the shaded region is of the form $a \sqrt{3}+b \pi+c \pi \sqrt{3}$ for some rational numbers $a, b$ and $c$. What is the value of $a+b+c$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_03a8986aaed00af3ee97g-2.jpg?height=569&width=721&top_left_y=1559&top_left_x=756)
We will use the fact that if two circles are externally tangent, then the line connecting their centres passes through the point of tangency.

The side lengths of $\triangle A B C$ are $A B=(\sqrt{3}-1)+(3-\sqrt{3})=2, A C=(\sqrt{3}-1)+(1+\sqrt{3})=2 \sqrt{3}$, and $B C=(3-\sqrt{3})+(1+\sqrt{3})=4$.

Notice that

$$
\begin{aligned}
A B^{2}+A C^{2} & =2^{2}+(2 \sqrt{3})^{2} \\
& =4+12 \\
& =16 \\
& =4^{2} \\
& =B C^{2}
\end{aligned}
$$

and so the sides of $\triangle A B C$ satisfy $B C^{2}=A B^{2}+A C^{2}$. By the converse of the Pythagorean theorem, $\triangle A B C$ has $\angle C A B=90^{\circ}$.

Therefore, the area of $\triangle A B C$ is $\frac{1}{2} \times A B \times A C=\frac{1}{2} \times 2 \times 2 \sqrt{3}=2 \sqrt{3}$.

Since $\angle C A B=90^{\circ}$, the unshaded sector that lies inside the triangle has area $\frac{90^{\circ}}{360^{\circ}}=\frac{1}{4}$ of the area of the circle centred at $A$.

The area of the circle centred at $A$ is $\pi(\sqrt{3}-1)^{2}=\pi(4-2 \sqrt{3})$.

The area of the shaded region is therefore

$$
2 \sqrt{3}-\frac{\pi(4-2 \sqrt{3})}{4}=2 \sqrt{3}-\pi+\frac{1}{2} \pi \sqrt{3}
$$

which means $a=2, b=-1$, and $c=\frac{1}{2}$, so $a+b+c=\frac{3}{2}$.

ANSWER: $\frac{3}{2}$