10.4 In triangle $A B C$, the median $A M$ is drawn. Can the radius of the circle inscribed in triangle $A B M$ be exactly twice the radius of the circle inscribed in triangle $A C M$?
See the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_fac6f5a99db6b6e1e5c4g-2.jpg?height=411&width=797&top_left_y=1074&top_left_x=481)

\[
\begin{aligned}
& AB=c, \\
& BC=2a, \\
& CA=b, \\
& AM=m, \\
& S_{ABM}=S_{ACM}=S
\end{aligned}
\]

We will use the known formula for the radius \( r \) of the inscribed circle in a triangle with sides \( a, b, c \):

\[
r = \frac{2S}{a+b+c}, \text{ where } S \text{ is the area of the triangle.}
\]

Assume that \( r_{ABM} = 2r_{ACM} \).

We have: \(\frac{2S}{c+a+m} = 2 \cdot \frac{2S}{a+b+m}\).

Then \( a+b+m = 2(c+a+m) \),

from which \( m = b - a - 2c \).

Since \( b > a \) (triangle inequality).