The 35th Japanese Mathematical Olympiad

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

Since $C$ is the circumcenter of the triangle $P{P}_2{P}_3$, we have $P_2{P}_3=2PC\sin \angle P_{2}P{P}_3$ by the law of sines. Similarly, since $D$ is the circumcenter of the triangle $P{P}_4{P}_3$, we have

$P_4{P}_3=2PD\sin \angle P_{4}P{P}_3$. Since the lines $BC$ and $DE$ are parallel, $P_2$, $P$ and $P_4$ are collinear. Therefore, we have

$$P_2{P}_3:P_4{P}_3=2PC\sin \angle P_{2}P{P}_3:2PD\sin \angle P_{4}P{P}_3=PC:PD=2:3.$$

Since $B$ is the circumcenter of the triangle $P{P}_1{P}_2$, and $P$ and $P_1$ are symmetric with respect to the line $AB$, we obtain $\angle PBA=\angle P{P}_2{P}_1$. Similarly, since $C$ is the circumcenter of the triangle $P{P}_3{P}_2$ and $P$ and $P_3$ are symmetric with respect to the line $CD$, we obtain $\angle P{P}_2{P}_3=\angle PCD$. Therefore, we have

$$\angle P_3{P}_2{P}_1=\angle P{P}_2{P}_1-\angle P{P}_2{P}_3=\angle PBA-\angle PCD=\angle PCA-\angle PCD=\angle DCA.$$

Similarly, we have $\angle P_3{P}_4{P}_5=\angle CDA$. Since $\angle DCA=\angle CDA$, we have $\angle P_3{P}_2{P}_1=\angle P_3{P}_4{P}_5$. From this and $P_2{P}_3:P_4{P}_3=2:3=P_2{P}_1:P_4{P}_5$, the triangles $P_3{P}_2{P}_1$ and $P_3{P}_4{P}_5$ are similar with ratio $2:3$.

Since $B$ is the circumcenter of the triangle $P{P}_1{P}_2$, and $P$ and $P_2$ are symmetric with respect to the line $BC$, we have $\angle P_2{P}_{1}P=\angle CBP$. Similarly, since $A$ is the circumcenter of the triangle $P{P}_1{P}_5$, and $P$ and $P_5$ are symmetric with respect to the line $AE$, we have $\angle P{P}_1{P}_5=\angle PAE$. Also the quadrilateral $BCDE$ is an isosceles trapezoid since $BC\parallel DE$. Therefore, we have

$$\angle P_2{P}_1{P}_5=\angle P_2{P}_{1}P+\angle P{P}_1{P}_5=\angle CBP+\angle PAE=\angle CBP+\angle PBE=\angle CBE=\angle BCD.$$

Similarly, we have $\angle P_4{P}_5{P}_1=\angle EDC$. Hence, we have

$$\angle P_2{P}_1{P}_5+\angle P_4{P}_5{P}_1=\angle BCD+\angle EDC=180^{\circ },$$

which means that the lines $P_1{P}_2$ and $P_4{P}_5$ are parallel. Let $M_2$ and $M_4$ be the midpoints of the segments $P{P}_2$ and $P{P}_4$, respectively. Then $B$, $C$, and $M_2$ all lie on the perpendicular bisector of the segment $P{P}_2$, and $E$, $D$, and $M_4$ all lie on the perpendicular bisector of the segment $P{P}_4$. Therefore, we have

$$\sin \angle P_2{P}_1{P}_5=\sin \angle BCD=\frac{M_2{M}_4}{CD}=\frac{P_2{P}_4}{2CD}=\frac{11}{8\sqrt{2}}.$$

Let $Q$ be the foot of the perpendicular from $P_5$ to the line $P_1{P}_2$, and let $R$ and $S$ be the feet of the perpendiculars from $P_3$ to the lines $P_1{P}_2$ and $P_4{P}_5$, respectively. Then the quadrilateral $QRS{P}_5$ is a rectangle. Since the triangles $P_3{P}_2{P}_1$ and $P_3{P}_4{P}_5$ are similar with ratio $2:3$, the triangles $P_{3}R{P}_1$ and $P_{3}S{P}_5$ are also similar with ratio $2:3$, and hence we have $P_{1}R:P_{5}S=P_{3}R:P_{3}S=2:3$. Therefore, $Q$, $P_1$, and $R$ are collinear in this order, and we have $Q{P}_1:P_{1}R=1:2$. Hence, we have

$$\frac{P_{1}R}{P_1{P}_5}=\frac{2Q{P}_1}{P_1{P}_5}=2|\cos \angle P_2{P}_1{P}_5|=2\sqrt{1-{\left(\frac{11}{8\sqrt{2}}\right)}^2}=\sqrt{\frac{7}{32}}.$$

We also have

$$\frac{R{P}_3}{P_1{P}_5}=\frac{2Q{P}_5}{5P_1{P}_5}=\frac{2}{5}\sin \angle P_2{P}_1{P}_5=\frac{11}{20\sqrt{2}}$$

Therefore, applying the Pythagorean theorem, we conclude

$$\frac{P_1{P}_3}{P_1{P}_5}=\sqrt{{\left(\frac{P_{1}R}{P_1{P}_5}\right)}^2+{\left(\frac{R{P}_3}{P_1{P}_5}\right)}^2}=\sqrt{\frac{7}{32}+{\left(\frac{11}{20\sqrt{2}}\right)}^2}=\frac{\sqrt{37}}{10}$$