IMO Team Selection Contest

Let $L_1$ and $L_2$ be the points of tangency of the external common tangent of the circles, $N_1$ and $N_2$ be the points of tangency of an internal common tangent, and $O_1$ and $O_2$ be the centers of the two circles (see Fig. 18).

As all the lines $O_1{L}_1$, $AK$, and $O_2{L}_2$ are perpendicular to the line $L_1{L}_2$, they are parallel to each other and thus $$\frac{|L_{1}K|}{|L_{2}K|}=\frac{|O_{1}A|}{|O_{2}A|}.$$ The triangles $O_{1}A{N}_1$ and $O_{2}A{N}_2$ are similar because they are both right-angled and have the same vertical angles. Thus, $$\frac{|O_{1}A|}{|O_{2}A|}=\frac{|O_1{N}_1|}{|O_2{N}_2|}=\frac{|O_1{L}_1|}{|O_2{L}_2|}.$$ Therefore, the right-angled triangles $O_1{L}_{1}K$ and $O_2{L}_{2}K$ are similar due to proportionality of their legs. Hence, $\angle L_{1}K{O}_1=\angle L_{2}K{O}_2$.

As $\angle L_{1}K{M}_1=2\angle L_{1}K{O}_1$ and $\angle L_{2}K{M}_2=2\angle L_{2}K{O}_2$, we also get that $\angle L_{1}K{M}_1=\angle L_{2}K{M}_2$. Together with the equality $\angle L_{1}KA=\angle L_{2}KA=90^{\circ }$ this implies $\angle M_{1}KA=\angle M_{2}KA$.

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Alternative solution.

Both of the circles appear at the same angle, when viewed from the point $A$. To solve the problem, it is enough to show that both of the circles also appear at the same angle, when viewed from the point $K$.

Let the centers of the circles have the coordinates $O_1(a_1,b_1)$ and $O_2(a_2,b_2)$ and let $r_1$ and $r_2$ be the radii of the circles. The two circles appear at the same angle from the point $P(x,y)$ if and only if $$\frac{r_1}{|O_{1}P|}=\frac{r_2}{|O_{2}P|},$$ i.e., $$\frac{r_1}{\sqrt{(x-a_1{)}^2+(y-b_1{)}^2}}=\frac{r_2}{\sqrt{(x-a_2{)}^2+(y-b_2{)}^2}}.$$ Simple algebra shows that this equation is equivalent to $$(r_1^2-r_2^2)x^2+(r_1^2-r_2^2)y^2+c_{1}x+c_{2}y+c_3=0,$$ where $c_1,c_2$, and $c_3$ are some constants.

If $r_1=r_2$, then the statement clearly holds. If $r_1\ne r_2$, then the last equation is that of a circle. Point $A$ as well as the point $D$ of intersection of the external common tangents both lie on that circle, and from symmetry, the diameter of that circle is $AD$. As $AK$ is perpendicular to the external common tangent of the circles, the point $K$ also lies on that circle.

Fig. 18