5. In $\triangle A B C, B C=4$ and $\angle B A C=60^{\circ}$. Let $I$ be the incentre of $\triangle A B C$. The circle passing through $B, I$ and $C$ meets the perpendicular bisector of $B C$ at a point $X$ inside $\triangle A B C$. Find $A X$.在 $\triangle A B C$ 中, $B C=4$ 及 $\angle B A C=60^{\circ}$ 。設 $I$ 為 $\triangle A B C$ 的內心。穿過 $B 、 I 、 C$ 的圓與 $B C$ 的垂直平分線相交於 $\triangle A B C$ 內的一點 $X \circ$ 求 $A X \circ$
(1 mark)
5. $\frac{4 \sqrt{3}}{3}$
5. We claim that $X$ is the circumcentre of $\triangle A B C$. Assuming the claim, we know from the sine law that $\frac{4}{\sin 60^{\circ}}=2 A X$, so that $A X=\frac{4 \sqrt{3}}{3}$.
It remains to prove the claim. Since $X$ is on the perpendicular bisector of $B C$, we have $X B=X C$. It therefore suffices to show that $\angle B X C=2 \angle B A C=120^{\circ}$. But this is clear, because
$$
\begin{aligned}
\angle B X C & =\angle B I C \\
& =180^{\circ}-\angle I B C-\angle I C B \\
& =180^{\circ}-\frac{\angle A B C}{2}-\frac{\angle A C B}{2} \\
& =180^{\circ}-\frac{180^{\circ}-\angle B A C}{2} \\
& =120^{\circ}
\end{aligned}
$$