12. In $\triangle A B C, A B=3, B C=4$ and $C A=5$. A circle cuts $A B$ at $C_{1}$ and $C_{2}$, cuts $B C$ at $A_{1}$ and $A_{2}$, and cuts $C A$ at $B_{1}$ and $B_{2}$. If $A_{1} A_{2}=B_{1} B_{2}=C_{1} C_{2}=x$, and that the area of the hexagon formed by the vertices $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}$ and $C_{2}$ is 4 , find $x$.
(2 marks)
在 $\triangle A B C$ 中, $A B=3 、 B C=4 、 C A=5$ 。某圓與 $A B$ 交於 $C_{1}$ 和 $C_{2}$, 與 $B C$ 交於 $A_{1}$ 和 $A_{2}$, 與 $C A$ 交於 $B_{1}$ 和 $B_{2}$ 。若 $A_{1} A_{2}=B_{1} B_{2}=C_{1} C_{2}=x$, 且由 $A_{1}$ 、 $A_{2} 、 B_{1} 、 B_{2} 、 C_{1} 、 C_{2}$ 組成的六邊形的面積為 4 , 求 $x$ 。
12. $\frac{11-\sqrt{37}}{3}$
12. Since $A_{1} A_{2}=B_{1} B_{2}=C_{1} C_{2}$, the circle has the same centre as the inscribed circle of $\triangle A B C$. If we let $A_{0}$ be the midpoint of $A_{1} A_{2}$, then $A_{0}$ is the point where the inscribed circle of $\triangle A B C$ touches $B C$, and the same as true for $B_{0}$ and $C_{0}$. Since $A B_{0}=A C_{0}, B C_{0}=B A_{0}$ and $C A_{0}=C B_{0}$, it is easy to find that $A B_{0}=A C_{0}=2, B C_{0}=B A_{0}=1$ and $C A_{0}=C B_{0}=3$. Thus we have $A B_{2}=A C_{1}=2-\frac{x}{2}$, $B C_{2}=B A_{1}=1-\frac{x}{2}$ and $C A_{2}=C B_{1}=3-\frac{x}{2}$.

The area of $\triangle B C_{2} A_{1}$ is thus $\frac{1}{2}\left(1-\frac{x}{2}\right)^{2}$. As $\sin A=\frac{4}{5}$, the area of $\triangle A B_{2} C_{1}$ is $\frac{1}{2}\left(2-\frac{x}{2}\right)^{2}\left(\frac{4}{5}\right)$. Similarly, $\triangle C A_{2} B_{1}$ has area $\frac{1}{2}\left(3-\frac{x}{2}\right)^{2}\left(\frac{3}{5}\right)$. Finally, since the area of $\triangle A B C$ is 6 , we have
$$
\frac{1}{2}\left(1-\frac{x}{2}\right)^{2}+\frac{1}{2}\left(2-\frac{x}{2}\right)^{2}\left(\frac{4}{5}\right)+\frac{1}{2}\left(3-\frac{x}{2}\right)^{2}\left(\frac{3}{5}\right)=2
$$
and hence $x=\frac{11 \pm \sqrt{37}}{3}$. Obviously the negative square root should be taken as $x<3$.