11. Two concentric circles have radii 2006 and 2007. $A B C$ is an equilateral triangle inscribed in the smaller circle and $P$ is a point on the circumference of the larger circle. Given that a triangle with side lengths equal to $P A, P B$ and $P C$ has area $\frac{a \sqrt{b}}{c}$, where $a, b, c$ are positive integers such that $a$ and $c$ are relatively prime and $b$ is not divisible by the square of any prime, find $a+b+c$.
(2 marks)
兩個同心圓的半徑分別是 2006 和 $2007 \circ A B C$ 是個內接於小圓的一個等邊三角形，而 $P$ 則是大圓的圓周上的一點。已知一個邊長分別等於 $P A 、 P B$ 和 $P C$ 的三角形的面積為 $\frac{a \sqrt{b}}{c}$, 其中 $a$ 、 $c$ 皆是正整數， $a$ 與 $c$ 互質，而且 $b$ 不可被任何質數的平方整除。求 $a+b+c$ 。
11. 4020
11. Let $O$ be the common centre of the two circles, and let $P^{\prime}$ be a point such that $\triangle A P P^{\prime}$ is equilateral and that $B$ and $P^{\prime}$ are on different sides of $A P$. Observe that $P A=P P^{\prime}$ and $P B=C P^{\prime}$ (as $\triangle B A P \cong$ $\triangle C A P^{\prime}$ ), so $\triangle P C P^{\prime}$ is the desired triangle.
Let $R=2007, r=2006, \angle P O B=x$ and $\angle P O C=y$. Then we have $x+y=120^{\circ}$ and $\frac{x-y}{2}=60^{\circ}-y$. Now
$$
\begin{aligned}
{\left[P C P^{\prime}\right] } & =\left[A P P^{\prime}\right]-[A P C]-\left[A P^{\prime} C\right] \\
& =\left[A P P^{\prime}\right]-[A P C]-[A P B] \\
& =\left[A P P^{\prime}\right]-[A B P C] \\
& =\left[A P P^{\prime}\right]-[O A B]-[O A C]-[O B P]-[O C P] \\
& =\frac{\sqrt{3}}{4}\left[R^{2}+r^{2}-2 R r \cos \left(120^{\circ}+y\right)\right]-\frac{\sqrt{3}}{4}\left(r^{2}+r^{2}\right)-\frac{1}{2} R r(\sin x+\sin y) \\
& =\frac{\sqrt{3}}{4}\left(R^{2}-r^{2}\right)+\frac{\sqrt{3}}{2} R r \cos \left(60^{\circ}-y\right)-\frac{1}{2} R r \cdot 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2} \\
& =\frac{\sqrt{3}}{4}\left(R^{2}-r^{2}\right)+\frac{\sqrt{3}}{2} R r \cos \left(60^{\circ}-y\right)-\frac{\sqrt{3}}{2} R r \cos \left(60^{\circ}-y\right) \\
& =\frac{\sqrt{3}}{4}\left(R^{2}-r^{2}\right) \\
& =\frac{4013 \sqrt{3}}{4}
\end{aligned}
$$

Hence $a+b+c=4013+3+4=4020$.
Remark. It would be easy to guess the answer if one considers the special case where $A, O, P$ are collinear.