(USAMO 1998) Let $\Gamma_{1}$ and $\Gamma_{2}$ be two concentric circles, with $\Gamma_{2}$ inside $\Gamma_{1}$. Let $A, B$ be points on $\Gamma_{1}$ and $\Gamma_{2}$ respectively such that (AB) is tangent to $\Gamma_{2}$.

Let $C$ be the second intersection point of $(A B)$ with $\Gamma_{1}$ and let $D$ be the midpoint of [AB]. A line passing through $A$ intersects $\Gamma_{2}$ at $E$ and $F$ such that the perpendicular bisectors of [DE] and $[C F]$ intersect at a point $M$ on the line (AB). Calculate the ratio $A M / M C$.
![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-084.jpg?height=797&width=810&top_left_y=1022&top_left_x=606)

By calculating the power of $A$ with respect to $\Gamma_{1}$, we obtain $A E \cdot A F = A B^{2} = A D \cdot A C$. Therefore, D, C, F, E belong to the same circle $\Omega$, so that $M$ is the center of $\Omega$. We deduce that $[\mathrm{DC}]$ is a diameter of $\Omega$ and that $M$ is the midpoint of $[\mathrm{DC}]$. Hence:

$$
\frac{M C}{A C} = \frac{D C}{2 A C} = \frac{3}{2 \cdot 4} = \frac{3}{8}
$$

Thus, $A M / M C = 3 / 5$.