Switzerland — Second Round, 2022

Solution:
First, observe that
(1) $\angle P A M = \angle P T M = 90^{\circ}$ since lines $P A$ and $P T$ are tangent to circle $k_{1}$.
(2) $P A$ is tangent to both circles $k_{1}$ and $k_{2}$ by the problem condition.
(3) $P A = P T$ since both lines $P A$ and $P T$ are tangent to circle $k_{1}$. Hence triangle $P A T$ is isosceles at $P$.

By (1), $P, A, M, T$ lie on a circle of diameter $P M$. By using (2) and (3), one gets
$$\angle A C P = \angle A C B \stackrel{(2)}{=} \angle P A B = \angle P A T \stackrel{(3)}{=} \angle P T A$$
so $P, A, C, T$ lie on a circle. Therefore, $P, A, C, M, T$ all lie on the circumcircle of $P A T$. In particular, $A T C M$ is a cyclic quadrilateral, as required.

Solution:
Let $N = P M \cap A T$. Since $P M$ is the perpendicular bisector of $A T$ (as $P T = P A$ and $M T = M A$), we get that $N$ is in fact the midpoint of segment $A T$. Now, note that triangles $P N A$ and $P A M$ are similar. Therefore, we get $\frac{P N}{P A} = \frac{P A}{P M}$. Hence, by power of a point and this identity, we get
$$P B \cdot P C = P A^{2} = P M \cdot P N$$
so $B, C, M, N$ lie on a circle. Therefore, since $\angle M N B = \angle M N A = 90^{\circ}$, $B, C, M, N$ lie on a circle of diameter $B M$. Thus, $\angle P C M = 90^{\circ}$. By (a), we conclude that $P, A, C, M, T$ lie on a circle of diameter $P M$.

Solution:
As before, we see that $P, A, M, T$ are a cyclic quadrilateral. Our aim is now to prove that $\angle A P T + \angle T C A = 180^{\circ}$. By the theorem of the radical center, we have that $(T C B)$ is tangent to $P T$, as the radical axes $P T, C B$ and $P A$ intersect in a point. Then by the tangent angle theorem, we have $\angle T C A = \angle T C B + \angle B C A = \angle T A P + \angle P T A$, allowing us to conclude.