Iberoamerican Mathematical Olympiad

Solution:

Let $C$, $C'$ have centers $O$, $O'$ respectively and radii $r$, $r'$ respectively. Let a circle center $P$ bisect $C$. Suppose it meets $C$ at $A$ and $B$. Then $AB$ is perpendicular to $OP$ and is a diameter of $C$. Hence $PA^{2} = OP^{2} + r^{2}$. Conversely, the circle center $P$, radius $\sqrt{OP^{2} + r^{2}}$ bisects $C$. So $P$ will bisect $C$ and $C'$ iff $OP^{2} + r^{2} = OP'^{2} + r'^{2}$.

It is well-known that the locus of points $P'$ with equal tangents to $C$ and $C'$ is the radical axis. Call the radical axis $R$. For a point $P'$ on the radical axis we have $P'O^{2} - r^{2} = P'O'^{2} - r'^{2}$. If we reflect $P'$ in the perpendicular bisector of $OO'$ to get $P$, then $PO = P'O'$, $PO' = P'O$, so $PO'^{2} - r^{2} = PO^{2} - r^{2}$ and hence $PO^{2} + r^{2}$. Call the reflection of $R$ in the perpendicular bisector of $OO'$ the line $R'$. We have established that points on $R'$ form part of the locus. Conversely, if $P'$ is such that there is a circle center $P'$ bisecting both circles, then $OP'^{2} + r^{2} = O'P'^{2} + r'^{2}$, so if $P$ is the reflection of $P'$, then $OP^{2} - r^{2} = O'P^{2} - r'^{2}$ and hence $P$ lies on the radical axis $R$. Hence $P'$ must lie on $R'$.

We have $PT^{2} = PO^{2} - r^{2} = PX^{2} + OX^{2} - r^{2}$, and similarly $PT'^{2} = PX^{2} + O'X^{2} - r'^{2}$. So $PT = PT'$ iff $OX^{2} - r^{2} = O'X^{2} - r'^{2}$. There is evidently a unique point $X$ for which that is true, so the locus of such $P$ is the line through $X$ perpendicular to $OO'$.

If the circles intersect, then the point $X$ evidently lies on the line joining the two common points, because $OX^{2} - r^{2} = -XY^{2} = O'X^{2} - r'^{2}$. In any case the midpoint of each common tangent evidently lies on the line, so that provides a way of constructing it.