[ [product of the lengths of the segments of chords and the lengths of the segments of secants ] [ Radical axis $]$

Two circles intersect at points $A$ and $B$. Chord $C D$ of the first circle has a common point $M$ with chord $E F$ of the second circle. It is known that $B M=2, A B=3 C M=9 E M, M D=2 C M, M F=6 C M$. What values can the length of the segment $A M$ take?
Prove that points $C, D, E$, and $F$ lie on the same circle.

## Solution

Let points $C$ and $D$ lie on circle $\Omega_{1}$, and points $E$ and $F$ lie on circle $\Omega_{2}$. Denote $C M=3 a$. Then $A B = 9 a, M D = 6 a, M F = 18 a, E M = a$. Since $D M \cdot M C = 18 a^{2} = E M \cdot F M$, the points $D, E, C$, and $F$ lie on the same circle. Denote it as $\Omega_{3}$. We will prove that the chord $A B$ passes through point $M$.

Point $M$ is obviously the radical center of circles $\Omega_{1}, \Omega_{2}$, and $\Omega_{3}$. Therefore, the chords $A B$ and $C D$ of circle $\Omega_{1}$ intersect at point $M$, and $(9 a - 2)^{2} = A M \cdot M B = D M \cdot M C = 18 a^{2}$, from which $a = 2 / 3$ or $1 / 3$. Consequently, $A M = 4$ or 1.

## Answer

4 or 1.