Mongolian Mathematical Olympiad

Let us denote by $Q$, $R$ and $S$ the midpoints of $BC$, $AB$ and $AC$ respectively. Then $O_3R \perp AB$, $O_1R \perp AB$, $O_3S \perp AC$, $O_2S \perp AC$, $O_3Q \perp BC$ and $MQ \perp BC$. Assuming $\angle ABC = \beta$ and $\angle BCA = \gamma$, we get $\angle BO_3O_1 = \gamma$ and $\angle CO_3O_2 = \beta$, as $O_3$ is the circumcenter of $\triangle ABC$.

Since $BC$ is tangent to $\omega_1$, $\tilde{A}B = \beta$. So $\angle BO_1O_3 = \beta$. Similarly $\angle CO_3O_2 = \gamma$. Hence $\angle BO_3O_1 = \angle CO_2O_3 = \gamma$ and $\angle BO_1O_2 = \angle CO_3O_2 = \beta$. Thus, $\angle CO_2O_2 \sim \angle BO_1O_3$.

Let us denote the radii of $\omega_1$ by $r_1$, the radii of $\omega_2$ by $r_2$ and the radii of $\omega_3$ (the circumcircle of $\triangle ABC$) by $r_3$. Then $r_3 = CO_3 = BO_3 = AO_3 = AD$, $r_1 = O_1B = O_1A$ and $r_2 = O_2C = O_2A$. Also $\angle O_1BO_3 = \angle O_1AO_3$ and $\angle O_2CO_3 = \angle O_2AO_3$. It follows that $\angle O_1AO_3 = \angle O_2AO_3$, which means $\angle O_1AD = \angle O_2AD$. From $\angle CO_3O_2 \sim \angle BO_1O_3$, we have $\frac{O_1B}{O_3B} = \frac{CO_3}{CO_2}$, i.e $\frac{r_1}{r_3} = \frac{r_3}{r_2}$. Then $\frac{AO_1}{AD} = \frac{AD}{AO_2}$ holds. So $\angle O_1AD \sim \angle DAO_2$. Hence

$$\frac{DO_1}{DO_2} = \frac{AO_1}{AD} = \frac{r_1}{r_3} \quad (*)$$
$$\frac{r_1}{r_3} = \frac{AO_1}{AD} = \frac{\sin \angle ADO_1}{\sin \angle AO_1D} = \frac{\sin \angle ADO_1}{\sin \angle ADO_2} \quad (**)$$
Now we use the law of sines to $\triangle O_2MD$ and $\triangle DMO_1$.
$$\frac{O_2M}{\sin \angle O_2DM} = \frac{DO_2}{\sin \angle O_2MD}, \quad \frac{O_1M}{\sin \angle O_1DM} = \frac{DO_1}{\sin \angle O_1MD}$$
From thus, we'll get $\frac{DO_1}{DO_2} = \frac{\sin \angle O_2DM}{\sin \angle O_1DM}$. Then from () and (*),
$$\frac{DO_1}{DO_2} = \frac{r_1}{r_3} = \frac{\sin \angle ADO_1}{\sin \angle ADO_2}, \quad \frac{\sin \angle O_2DM}{\sin \angle O_1DM} = \frac{\sin \angle ADO_1}{\sin \angle ADO_2}$$
it is equivalent to
$$\frac{\sin(\angle O_1 DO_2 - \angle O_1 DM)}{\sin \angle O_1 DM} = \frac{\sin(\angle O_1 DO_2 - \angle ADO_2)}{\sin \angle ADO_2}$$
$$\frac{\sin \angle O_1 DO_2 \cos \angle O_1 DM - \cos \angle O_1 DO_2 \sin \angle O_1 DM}{\sin \angle O_1 DM} = \frac{\sin \angle O_1 DO_2 \cos \angle ADO_2 - \cos \angle O_1 DO_2 \sin \angle ADO_2}{\sin \angle ADO_2}$$
$$\sin \angle O_1 DO_2 \operatorname{ctg} \angle O_1 DM - \cos \angle O_1 DO_2 = \sin \angle O_1 DO_2 \operatorname{ctg} \angle ADO_2 - \cos \angle O_1 DO_2$$
Since $\sin \angle O_1 DO_2 \neq 0$, we obtain $\operatorname{ctg} \angle O_1 DM = \operatorname{ctg} \angle ADO_2$. Hence $\angle O_1 DM = \angle ADO_2$ ($0^\circ < \angle O_1 DM, \angle ADO_2 < 180^\circ$).