Japanese Mathematical Olympiad

$\boxed{\frac{\sqrt{37}}{10}}$

Since $C$ is the circumcenter of the triangle $PP_2P_3$, we have $P_2P_3 = 2PC \sin \angle P_2PP_3$ by the law of sines. Similarly, since $D$ is the circumcenter of the triangle $PP_4P_3$, we have

$P_4P_3 = 2PD \sin \angle P_4PP_3$. Since the lines $BC$ and $DE$ are parallel, $P_2$, $P$ and $P_4$ are collinear. Therefore, we have

$$P_2P_3 : P_4P_3 = 2PC \sin \angle P_2PP_3 : 2PD \sin \angle P_4PP_3 = PC : PD = 2 : 3.$$

Since $B$ is the circumcenter of the triangle $PP_1P_2$, and $P$ and $P_1$ are symmetric with respect to the line $AB$, we obtain $\angle PBA = \angle PP_2P_1$. Similarly, since $C$ is the circumcenter of the triangle $PP_3P_2$ and $P$ and $P_3$ are symmetric with respect to the line $CD$, we obtain $\angle PP_2P_3 = \angle PCD$. Therefore, we have

$$\angle P_3P_2P_1 = \angle PP_2P_1 - \angle PP_2P_3 = \angle PBA - \angle PCD = \angle PCA - \angle PCD = \angle DCA.$$

Similarly, we have $\angle P_3P_4P_5 = \angle CDA$. Since $\angle DCA = \angle CDA$, we have $\angle P_3P_2P_1 = \angle P_3P_4P_5$. From this and $P_2P_3 : P_4P_3 = 2 : 3 = P_2P_1 : P_4P_5$, the triangles $P_3P_2P_1$ and $P_3P_4P_5$ are similar with ratio $2 : 3$.

Since $B$ is the circumcenter of the triangle $PP_1P_2$, and $P$ and $P_2$ are symmetric with respect to the line $BC$, we have $\angle P_2P_1P = \angle CBP$. Similarly, since $A$ is the circumcenter of the triangle $PP_1P_5$, and $P$ and $P_5$ are symmetric with respect to the line $AE$, we have $\angle PP_1P_5 = \angle PAE$. Also the quadrilateral $BCDE$ is an isosceles trapezoid since $BC \parallel DE$. Therefore, we have

$$\angle P_2P_1P_5 = \angle P_2P_1P + \angle PP_1P_5 = \angle CBP + \angle PAE = \angle CBP + \angle PBE = \angle CBE = \angle BCD.$$

Similarly, we have $\angle P_4P_5P_1 = \angle EDC$. Hence, we have

$$\angle P_2P_1P_5 + \angle P_4P_5P_1 = \angle BCD + \angle EDC = 180^\circ,$$

which means that the lines $P_1P_2$ and $P_4P_5$ are parallel. Let $M_2$ and $M_4$ be the midpoints of the segments $PP_2$ and $PP_4$, respectively. Then $B$, $C$, and $M_2$ all lie on the perpendicular bisector of the segment $PP_2$, and $E$, $D$, and $M_4$ all lie on the perpendicular bisector of the segment $PP_4$. Therefore, we have

$$\sin \angle P_2P_1P_5 = \sin \angle BCD = \frac{M_2M_4}{CD} = \frac{P_2P_4}{2CD} = \frac{11}{8\sqrt{2}}.$$

Let $Q$ be the foot of the perpendicular from $P_5$ to the line $P_1P_2$, and let $R$ and $S$ be the feet of the perpendiculars from $P_3$ to the lines $P_1P_2$ and $P_4P_5$, respectively. Then the quadrilateral $QRSP_5$ is a rectangle. Since the triangles $P_3P_2P_1$ and $P_3P_4P_5$ are similar with ratio $2 : 3$, the triangles $P_3RP_1$ and $P_3SP_5$ are also similar with ratio $2 : 3$, and hence we have $P_1R : P_5S = P_3R : P_3S = 2 : 3$. Therefore, $Q$, $P_1$, and $R$ are collinear in this order, and we have $QP_1 : P_1R = 1 : 2$. Hence, we have

$$\frac{P_1R}{P_1P_5} = \frac{2QP_1}{P_1P_5} = 2|\cos \angle P_2P_1P_5| = 2\sqrt{1 - \left(\frac{11}{8\sqrt{2}}\right)^2} = \sqrt{\frac{7}{32}}.$$

We also have

$$\frac{RP_3}{P_1P_5} = \frac{2QP_5}{5P_1P_5} = \frac{2}{5} \sin \angle P_2P_1P_5 = \frac{11}{20\sqrt{2}}$$

Therefore, applying the Pythagorean theorem, we conclude

$$\frac{P_1P_3}{P_1P_5} = \sqrt{\left(\frac{P_1R}{P_1P_5}\right)^2 + \left(\frac{RP_3}{P_1P_5}\right)^2} = \sqrt{\frac{7}{32} + \left(\frac{11}{20\sqrt{2}}\right)^2} = \frac{\sqrt{37}}{10}$$