The 35th Japanese Mathematical Olympiad

$\boxed{\frac{15}{11}}$

Let the inscribed circle of quadrilateral $ABCD$ be tangent to sides $AB$, $BC$, $CD$, and $DA$ at points $S$, $T$, $U$, and $V$, respectively. Since $AS$ and $AV$ are tangent segments from $A$ to this incircle, we have $AS=AV$. Similarly, $BS=BT$, $CT=CU$ and $DU=DV$ hold. Thus we have $$AB+CD=AS+BS+CU+DU=AV+BT+CT+DV=AD+BC.(\ast )$$ Since quadrilateral $ABCD$ is concyclic, we have $\angle PCB=\angle PAD$. Therefore, triangles $PBC$ and $PDA$ are similar. Since the similarity ratio is same as the ratio of the radii of their inscribed circles, it follows that $BC:DA=5:6$. Similarly we have $CD:AB=3:6=1:2$. Now we can write $BC=5y$, $DA=6y$, $CD=x$ and $AB=2x$. From $(\ast )$ we have $2x+x=6y+5y$ and thus $x:y=11:3$. We conclude that $\frac{BC}{CD}=\frac{5y}{x}=\frac{5\cdot 3}{11}=\frac{15}{11}$.