Auswahlwettbewerb Zur IMO

Solution:

We denote the centers of the circles $k_{1}$ and $k_{2}$ by $O_{1}$ and $O_{2}$, respectively. The solution now assumes that $\angle C D B$ is obtuse (see Fig.); otherwise only some angles need to be exchanged modulo $180^{\circ}$.
Let $P$ be the circumcenter of triangle $BDC$; then, by the inscribed angle theorem (central angle theorem),
$$\angle B P C = 2 \cdot (180^{\circ} - \angle C D B) = 360^{\circ} - 2 \cdot \angle C D B$$

$P$ lies on the circumcircle of triangle $A B C$ if and only if $A C P B$ is a cyclic quadrilateral; this holds if and only if $\angle B P C = 180^{\circ} - \angle C A B$.
To prove this, we compute $\angle C D B$. Since $k_{1}$ and $k_{2}$ are tangent to each other at $D$, $D$ lies on the segment $O_{1} O_{2}$ and we have
$$\angle C D B = 180^{\circ} + \angle O_{1} D B - \angle O_{2} D C$$
(2). Since $k_{1}$ is tangent to the line $A B$, we have $\angle A B O_{1} = 90^{\circ}$, hence $\angle A B D + \angle D B O_{1} = 90^{\circ}$. Since $B$ and $D$ lie on $k_{1}$, the triangle $B D O_{1}$ is isosceles with $\angle O_{1} D B = \angle D B O_{1}$. Thus
$$\angle O_{1} D B = 90^{\circ} - \angle A B D$$
(3). Since $k_{2}$ is tangent to the line $A C$, we have $\angle A C O_{2} = 90^{\circ}$, hence $\angle A C D + \angle D C O_{2} = 90^{\circ}$. Since $C$ and $D$ lie on $k_{2}$, the triangle $C D O_{2}$ is isosceles with $\angle O_{2} D C = \angle D C O_{2}$. Thus
$$\angle O_{2} D C = 90^{\circ} - \angle A C D$$
(4). Substituting (3) and (4) into (2) gives
$$\angle C D B = 180^{\circ} + (90^{\circ} - \angle A B D) - (90^{\circ} - \angle A C D) = 180^{\circ} - \angle A B D + \angle A C D$$
(5).
Now we consider the intersection point $R$ of $A C$ and $B D$ and conclude
$$\angle A B D = \angle A B R = 180^{\circ} - \angle R A B - \angle B R A = 180^{\circ} - \angle C A B - \angle B R A$$
as well as
$$\angle A C D = \angle R C D = 180^{\circ} - \angle C D R - \angle D R C = 180^{\circ} - \angle C D B - \angle B R A$$
From (5) it follows that
$$\angle C D B = 180^{\circ} + \angle C A B - \angle C D B$$
hence
$$2 \cdot \angle C D B = 180^{\circ} + \angle C A B$$
Substituting into (1) gives
$$\angle B P C = 360^{\circ} - (180^{\circ} + \angle C A B) = 180^{\circ} - \angle C A B$$
Thus everything is shown.