In $\triangle A B C, A B=2, B C=\sqrt{3}$ and $\angle A B C=150^{\circ}$. $P$ is a point on the plane such that $\angle A P B=45^{\circ}$ and $\angle B P C=120^{\circ}$. Find $B P$.
(2 marks)
In $\triangle A B C, A B=2, B C=\sqrt{3}$ and $\angle A B C=150^{\circ}$. $P$ is a point on the plane such that $\angle A P B=45^{\circ}$ and $\angle B P C=120^{\circ}$. Find $B P$.
12. $\frac{2}{\sqrt{5}}$
12. As shown in the figure, let $O_{1}$ and $O_{2}$ be the circumcenters of $\triangle A B P$ and $\triangle B C P$ respectively, $M$ be the midpoint of $B P$ and $N$ be the midpoint of $A B$. Since the angle at the center is twice the angle at the circumference, we get $\angle A O_{1} B=90^{\circ}$ and the reflex $\angle B O_{2} C=240^{\circ}$. Hence
$$
\angle C B O_{2}=\frac{180^{\circ}-\left(360^{\circ}-240^{\circ}\right)}{2}=30^{\circ}
$$
and so $A, B, O_{2}$ are collinear.
Note that $O_{1} N \perp A B$ and $O_{1}, M, O_{2}$ are collinear. Furthermore, we have $B P \perp O_{1} O_{2}$ and simple computation gives $O_{1} N=N B=B O_{2}=1$. Note also that $\Delta B M O_{2} \sim \Delta O_{1} N O_{2}$, so $\frac{B M}{O_{1} N}=\frac{B O_{2}}{O_{1} O_{2}}$, or $\frac{B M}{1}=\frac{1}{\sqrt{1^{2}+2^{2}}}$, so that $B P=2 B M=\frac{2}{\sqrt{5}}$.