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 $$\frac{|BK|}{|BC|}=\frac{|BM|}{|BA|}=1-\lambda \text{and}\frac{|CL|}{|CA|}=\frac{|CK|}{|CB|}=1-(1-\lambda )=\lambda ,\text{whence}\lambda =\frac{|AM|}{|AB|}=\frac{|AL|}{|AC|}=1-\lambda .\text{Hence}\lambda =\frac{1}{2},\text{ therefore the triangles }AML,MBK\text{ and }LKC\text{ are all similar}$$ 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