Japan Mathematical Olympiad

$$\frac{\sqrt{33}}{11}$$
Let $Q$ be the intersection of line $EP$ and $AB$, and $R$ be the intersection of line $DP$ and $AC$. Since $\angle BPC + \angle CPE = \angle PEC + \angle CPE = 180^\circ - \angle ECP$, we have $\angle QPB = \angle ECP$. Similarly, we have $\angle RPC = \angle DBP$. Hence $\angle QPB = \angle RCP$ and $\angle QBP = \angle RPC$, which yields $\triangle QBP \sim \triangle RPC$.

Since $\angle QDR = \angle BDP = \angle PEC = \angle QER$, $D$, $Q$, $R$ and $E$ lie on the same circumference. Hence we have $\angle ABC = \angle DBC = 180^\circ - \angle CED = 180^\circ - \angle RED = \angle DQR = \angle AQR$ (note that $D$, $B$, $C$, $E$ are also concyclic). This means the line $BC$ is parallel to the line $QR$ and thus $BQ : CR = AB : AC = 9 : 11$. In addition, we have $\triangle QDP \sim \triangle REP$ since $D$, $Q$, $R$ and $E$ are concyclic, and thus $QP : RP = DP : EP = 1 : 3$.

The above argument yields $\frac{BP^2}{CP^2} = \frac{BQ \cdot QP}{PR \cdot RC} = \frac{BQ}{CR} \cdot \frac{QP}{RP} = \frac{9}{11} \cdot \frac{1}{3} = \frac{3}{11}$, hence $\frac{BP}{CP} = \frac{\sqrt{3}}{11} = \frac{\sqrt{33}}{11}$.