Bulgarian Mathematical Competitions

Solution:

Let $K = CO \cap AB$ and $\text{AKC =}$. Denote by $M$ and $N$ the intersection points of $QO$ with $CB$ and $CA$, respectively. The Sine theorem for $\triangle APQ$ gives
$$\frac{AQ}{PQ} = \frac{\sin \left(90^\circ + \beta\right)}{\sin \frac{\gamma}{2}} = \frac{\cos \beta}{\sin \frac{\gamma}{2}}$$
From the right $\triangle KPQ$ we find
$$\frac{KQ}{PQ} = \frac{1}{\sin \varphi}$$
Hence
$$\frac{AQ}{QK} = \frac{\cos \beta \sin \varphi}{\sin \frac{\gamma}{2}}$$

It follows from $\triangle AKC$ ($AO$ is the bisector of $\text{KAC}$) that
$$\frac{KO}{OC} = \frac{AK}{AC} = \frac{\sin \frac{\gamma}{2}}{\sin \varphi}$$
On the other hand applying the Menelaus theorem for $\triangle AKC$ and the line $OQ$ we get
$$\frac{AQ}{QK} \cdot \frac{KO}{OC} \cdot \frac{CN}{NA} = 1$$
Now plugging (1) and (2) in (3) we obtain $\frac{CN}{NA} = \frac{1}{\cos \beta}$. Hence it follows that $\frac{CN}{CA} = \frac{1}{1 + \cos \beta}$, i.e. $CN = \frac{2R \sin \beta}{2 \cos^2 \frac{\beta}{2}} = 2R \tan \frac{\beta}{2}$.

Similarly, $CM = 2R \tan \frac{\alpha}{2}$. Therefore
$$\frac{CN}{CM} = \frac{\tan \frac{\beta}{2}}{\tan \frac{\alpha}{2}}$$
It is well known that $CB_1 = p - a = r \cot \frac{\alpha}{2}$ and $CA_1 = p - b = r \cot \frac{\beta}{2}$. Hence using (4) we get
$$\frac{CB_1}{CA_1} = \frac{\cot \frac{\alpha}{2}}{\cot \frac{\beta}{2}} = \frac{\tan \frac{\beta}{2}}{\tan \frac{\alpha}{2}} = \frac{CN}{CM}$$
Therefore $A_1B_1 \parallel MN$.