Japan Mathematical Olympiad Initial Round

$\boxed{\frac{60}{23}}$ Let $\ell$ be the length of a side of square $ABCD$, and $OA=OB=OC=OD=r$, $OP=a$, $OQ=b$, $OR=c$, $OS=d$. Also let $X$, $Y$, $Z$ be the point of intersection of lines $AB$ and $PQ$, lines $BC$ and $QR$, lines $CD$ and $RS$, respectively. Then, by Menelaus' theorem, we have $$\frac{OP}{AP}\cdot \frac{AX}{BX}\cdot \frac{BQ}{OQ}=\frac{a}{r-a}\cdot \frac{BX+\ell }{BX}\cdot \frac{r-b}{b}=1,$$ from which we obtain $BX=\frac{\ell a(r-b)}{r(b-a)}$. Similarly, we get $BY=\frac{\ell c(r-b)}{r(b-c)}$, $CZ=\frac{\ell d(r-c)}{r(c-d)}$. Also, we have $CY=BC+BY=\ell +\frac{\ell c(r-b)}{r(b-c)}=\frac{\ell b(r-c)}{r(b-c)}$. Since triangles $YBX$ and $YCZ$ are similar, we get $$\begin{array}{rl}& BY\cdot CZ=BX\cdot CY\\ ⟺& \frac{\ell c(r-b)}{r(b-c)}\cdot \frac{\ell d(r-c)}{r((c-d))}=\frac{\ell a(r-b)}{r(b-a)}\cdot \frac{\ell b(r-c)}{r(b-c)}\\ ⟺& cd(b-a)=ab(c-d).\end{array}$$

Solving for $d$ from the last equation above, we get $d=\frac{abc}{ab+bc-ca}$, and substituting the values $a=3$, $b=5$, $c=4$, we get the length of the line segment $OS$ to be equal to $d=\frac{3\cdot 4\cdot 5}{3\cdot 5+5\cdot 4-4\cdot 3}=\frac{60}{23}$.