63rd Czech and Slovak Mathematical Olympiad

In the first part of our solution, we will assume that $X$ is any point with the required property. It is clear that $X$ lies in the exteriors of the circles $k_1$ and $k_2$ and that the points $S_1$, $S_2$ and $X$ are vertices of a triangle whose sides $S_{1}X$, $S_{2}X$ are intersected successively by the circles $k_1$ and $k_2$ in such points $Y_1$, $Y_2$ which lie on the same line parallel to $S_1{S}_2$ (Fig. 1). Since the triangles $X{Y}_1{Y}_2$ and $X{S}_1{S}_2$ are similar (by theorem AA), it holds $$\frac{|X{Y}_1|}{|X{S}_1|}=\frac{|X{Y}_2|}{|X{S}_2|},(1)$$ which can be rewritten, because of the equalities $$|X{Y}_1|=|X{S}_1|-r_1\text{a}|X{Y}_2|=|X{S}_2|-r_2,(2)$$ as an equation for lengths of the segments $X{S}_1$ and $X{S}_2$: $$\begin{gathered} \frac{|X{S}_1|-r_1}{|X{S}_1|}=\frac{|X{S}_2|-r_2}{|X{S}_2|}, \\ \frac{|X{S}_1|}{|X{S}_2|}=\frac{r_1}{r_2}.(3) \end{gathered}$$ Since the points $S_1$ and $S_2$ are fixed as well as the ratio $r_1/r_2$, the locus of points $X$

Fig. 1

satisfying equation (3) is a circle of Apollonius (which evidently becomes a straight line if the ratio $r_1/r_2$ equals 1). As known for the case when $r_1/r_2\ne 1$, there are exactly two solutions $X=H_1$ and $X=H_2$ of the equation (3) that lie on the line $S_1{S}_2$ and form a diameter $H_1{H}_2$ of the resulting circle of Apollonius. For our situation, let us add the fact that the points $H_1$ and $H_2$ are centres of both homotheties of the initially given circles $k_1$ and $k_2$.

In the second part of our solution, we will conversely assume that $X$ is a point of the circle of Apollonius given by the equation (3) and that $X$ lies outside of the line $S_1{S}_2$, i.e. $X\ne H_1$ and $X\ne H_2$. In view of the condition that $|S_1{S}_2|>r_1+r_2$, the whole circle of Apollonius (with diameter $H_1{H}_2$) lies in the exteriors of the circles $k_1$ and $k_2$. Indeed, the last fact follows from a known position of the homothety centres $H_1$ and $H_2$ on the line $S_1{S}_2$: If for example $r_1<r_2$, then the diameter $H_1{H}_2$ contains the diameter of $k_1$, while the diameter of $k_2$ and the diameter $H_1{H}_2$ are disjoint. (See also Remark below.)

The proven property implies that $X{S}_1{S}_2$ is a triangle with $|X{S}_1|>r_1$ and $|X{S}_2|>r_2$. Thus there are points $Y_1\in k_1$ and $Y_2\in k_2$ lying on the segments $S_{1}X$ and $S_{2}X$, respectively. Since the equalities (2) are valid again, it is possible to transform the equation (3) to equation (1). Consequently, the triangles $X{S}_1{S}_2$ and $X{Y}_1{Y}_2$ are similar (now by SAS theorem) and hence $S_1{S}_2\parallel Y_1{Y}_2$. Therefore the distances of $Y_1$ and $Y_2$ to the line $S_1{S}_2$ are equal which proves the required property of the point $X$.

Answer. If $r_1\ne r_2$, the locus of points $X$ is the circle of Apollonius which is given by the above equation (3), excepting the two points on the line $S_1{S}_2$. If $r_1=r_2$, the locus is the perpendicular bisector of the segment $S_1{S}_2$, with exception of the midpoint of $S_1{S}_2$.

Fig. 1