In $\triangle ABC$, $AB=18$, $BC=30$, and $AC=33$. The bisector of $\angle B$ and the altitude from $C$ to line $AB$ intersect at point $P$. What is $BP$?

1. By the law of cosines, $\cos B=\dfrac{18^2+30^2-33^2}{2\cdot18\cdot30}=\dfrac{135}{1080}=\dfrac18$.
2. Let $H$ be the foot of the altitude from $C$ to line $AB$. In the right triangle $BHC$, $BH=BC\cos B=30\cdot\dfrac18=\dfrac{15}4$, which is less than $AB=18$, so $H$ is on segment $AB$.
3. In the right triangle $BHP$, $\angle PBH=\frac B2$, so $BP=\dfrac{BH}{\cos(B/2)}$.
4. $\cos^2\frac B2=\dfrac{1+\cos B}2=\dfrac{9}{16}$, so $\cos\frac B2=\dfrac34$ and $BP=\dfrac{15/4}{3/4}=\boxed{5}$.