Iberoamerican Mathematical Olympiad

Solution:

Let $O$ be the center of $C$. Let the line $AM$ meet $T$ at $N$. Let the perpendicular to $T$ at $N$ meet the line $OM$ at $O'$. Then $\angle O'NM = \angle MAB$ ($O'N$ parallel to $AB$, because both perpendicular to $T$) $= \angle OMA$ ($OM = OA$) $= \angle O'MN$. So $O'M = O'N$. Hence $O'$ is the center of $C'$.

Take $B$ to be the origin and $A$ to be the point $(2a, 0)$, so $O$ is $(a, 0)$ and $C$ has radius $a$. If $O'$ is $(x, y)$, then we require that $O'O = x + a$ or $(x - a)^2 + y^2 = (x + a)^2$, or $y^2 = 4a x$, which is a parabola with vertex $B$ and axis the $x$-axis.

Triangles $AMB$, $ABN$ are similar ($\angle AMB = \angle ABN = 90^\circ$), so $AM / AB = AB / AN$ and hence $AM \cdot AN = AB^2$. Now consider the circle center $A$ radius $AB$. It must meet the circle $C'$, because it contains the point $M$. Suppose it meets at $X$. Then $AX^2 = AB^2 = AM \cdot AN$, so $AX$ is tangent to $C'$ and hence the circles are orthogonal.