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 $△QBP\sim △RPC$.

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 $△QDP\sim △REP$ since $D$, $Q$, $R$ and $E$ are concyclic, and thus $QP:RP=DP:EP=1:3$.

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