Estonia — Open Contests

Take a point $X$ on the common tangent to the circles $c_1$ and $c$ which lies on the other side of the line $AB$ from the point $C$. Then $\angle ALM = \angle XAM = \angle XAB = \angle ACB$ (Fig. 4). Consequently $ML \parallel BC$. Similarly $KM \parallel CA$ and $LK \parallel AB$. If $\frac{|AM|}{|AB|} = \lambda$, then
$$\frac{|BK|}{|BC|} = \frac{|BM|}{|BA|} = 1 - \lambda \quad \text{and} \quad \frac{|CL|}{|CA|} = \frac{|CK|}{|CB|} = 1 - (1 - \lambda) = \lambda, \quad \text{whence} \quad \lambda = \frac{|AM|}{|AB|} = \frac{|AL|}{|AC|} = 1 - \lambda. \quad \text{Hence} \quad \lambda = \frac{1}{2}, \text{ therefore the triangles } AML, MBK \text{ and } LKC \text{ are all similar}$$
to $ABC$ with the factor $\frac{1}{2}$. Thus the radii of their circumcircles $c_1$, $c_2$ and $c_3$ are equal to half of the radius of the circumcircle $c$ of the triangle $ABC$. Since the circles $c$ and $c_1$ are tangent, the diameter of $c_1$ and the radius of $c$, both drawn from the tangent point $A$, coincide. Hence the circle $c_1$ goes through the center of the circle $c$; similarly the circles $c_2$ and $c_3$ go through the center of the circle $c$.

Figure 4