Japan Mathematical Olympiad

Let $P$, $Q$, $R$, $S$ be the points of tangency of the given circle to the sides $DA$, $AB$, $BC$, $CD$, respectively. Let, also $A'$, $B'$, $C'$, $D'$ be the midpoints of the line segments $PQ$, $QR$, $RS$, $SP$, respectively. Note that $A'$, $B'$, $C'$, $D'$ lie on the line segments $OA$, $OB$, $OC$, $OD$, respectively. Denote the length of the radius of the circle by $r$.

Since $\angle APO = \angle PA'O = 90^\circ$ and $\angle AOP = \angle POA'$, the triangles $APO$ and $PA'O$ are similar, and therefore, we have $AO : PO = PO : A'O$. From this we obtain $AO' = \frac{PO^2}{AO} = \frac{r^2}{AO}$. Similarly, we get $C'O = \frac{C'O}{r}$. Hence, we have $AO : CO = C'O : A'O$ and since $\angle AOC = \angle C'OA'$, we see that the triangles $AOC$ and $C'OA'$ are similar. This similarity carries the point $M$ to the midpoint $M'$ of the line segment $A'C'$, and therefore, we have $OM = OM' \cdot \frac{AO}{C'O} = OM' \cdot \frac{AO \cdot CO}{r^2}$.

In the same way, we conclude that if we denote by $N'$ the midpoint of the line segment $B'D'$ then $ON = ON' \cdot \frac{BO \cdot DO}{r^2}$. We see that
$$\overrightarrow{OM'} = \frac{1}{2}(\overrightarrow{OA'} + \overrightarrow{OC'}) = \frac{1}{4}(\overrightarrow{OP} + \overrightarrow{OQ} + \overrightarrow{OR} + \overrightarrow{OS}) = \frac{1}{2}(\overrightarrow{OB'} + \overrightarrow{OD'}) = \overrightarrow{ON'}$$

holds, and therefore, we conclude that $M' = N'$. Putting these facts together, we get
$$OM : ON = OM' \cdot \frac{AO \cdot CO}{r^2} : ON' \cdot \frac{BO \cdot DO}{r^2} = AO \cdot CO : BO \cdot DO = 35 : 48,$$
which gives the desired answer.