Vietnamese Mathematical Competitions

a. Suppose that two circles $(I)$ and $(K)$ tangent to each other at point $T$, then
$$IK - IA = IK - IT = TK$$
which is a constant.
Hence, point $I$ belongs to the hyperbola with two foci $A$ and $K$, which are fixed points.
Next, we have to prove that the angle $\angle MAN$ is constant.

We can see that $MA = MB$, $MA = MC$ so $M,N$ are on the radical axis of degenerate circle $(A)$ and the circle $(K)$.
Hence, $MN$ is the radical axis of these circles.
Note that $D,E$ belong to $MN$ so based on the properties of radical axis of two circles, we obtain $DA^2 = P_{D/(K)}$, $EA^2 = P_{E/(K)}$.
Assume that $xy$ is the inner common tangent line of two circles $(I),(K)$. Let $T'$ be the intersection of the line $DC$ and the circle $(K)$.
We have $P_{D/(K)} = DT' \cdot DC$ so $DA^2 = DT' \cdot DC \Rightarrow \triangle DAT' \sim \triangle DCA$ (c.g.c).
Hence $\angle DAT' = \angle DCA = \angle T'BC = \angle CT'y = \angle DT'x$, so $Tx$ is the tangent line of the circle $(O')$ or $T$ is the tangent point of two circles $(K),(O')$.
From this result, we have three points $D,T,C$ are collinear.
Similarly, we also have $E,T,B$ are collinear.
The quadrilateral $ADTC$ is inscribed so $\angle DAE = 180^\circ - \angle DTE = 180^\circ - \angle BTC$, which is a constant value (Q.E.D).

b. We will prove a generalization lemma for this part of the given problem.
Let $(O)$ be a fixed circle and two fixed distinct points $A,B$ on it, $d$ is a fixed line and not intersecting $(O)$. The point $C$ is moving on the circle $(O)$ and not coinciding with $A,B$. Two rays $AC,BC$ intersect $d$ at $D,E$, respectively. Prove that the circle with diameter $DE$ is always tangent to two certain fixed circles when $C$ moves on $(O)$.
Indeed, let $M$ be the projection of point $O$ on the line $d$ and $A',B'$ are the symmetric points of $A,B$ about $OM$. Assume that $AB',A'B$ cut $d$ at $F,G$, respectively. So $F,G$ are fixed.
Let $X,Y$ be two fixed points on the line $OM$ such that $MX^2 = MY^2 = FA \cdot GB$.
Based on the symmetrical properties, the point $M$ is the midpoint of the segment $FG$ and $GA' = FA$, $GB = FB'$, two lines $AA',BB'$ are parallel with the line $d$. $MF = MG = a$. We have $\angle GBE = \angle A'BC = \angle A'AC = \angle FDA$.

Similarly, $\angle GEB = \angle FAD$, then $\triangle GBE \sim \triangle FDA$ (g.g). Hence,
$$GE \cdot FD = GB \cdot FA = MX^2 = MY^2.$$
Let $D', E'$ be points on the line $d$ such that $MD' = FD$, $ME' = GE$ (two pairs of points $D, D'$ and $E, E'$ are on the same side to point $M$), then
$$MD' \cdot ME' = MX^2 = MY^2.$$
Based on the properties of power of point, we have the circle with diameter $D'E'$ passes through two points $X, Y$.
From $DD' = EE' = a$, we can see that two circles with diameter $DE, D'E'$ share a common center. Then the circle with diameter $DE$ is tangent to two circles with fixed centers $X, Y$ and radii $a$.
This proof also completes the solution for the given part.