$\left[\begin{array}{l}{[\text { Ratio of areas of triangles with a common angle }} \\ {\left[\begin{array}{l}\text { Law of Cosines }\end{array}\right]}\end{array}\right.$

In triangle $A B C$, angle $A$ is $60^{\circ} ; A B: A C=3: 2$. Points $M$ and $N$ are located on sides $A B$ and $A C$ respectively, such that $B M=M N=N C$. Find the ratio of the area of triangle $A M N$ to the area of triangle $A B C$.
Let $AB = 3t, AC = 2t, BM = MN = NC = x$. By the Law of Cosines,

$$
MN^2 = AM^2 + AN^2 - 2 \cdot AM \cdot AN \cdot \cos 60^{\circ}, \quad x^2 = (3t - x)^2 + (2t - x)^2 - (3t - x)(2t - x)
$$

from which $x = \frac{7}{5} t$. Then

$$
AM = 3t - x = 3t - \frac{7}{5} t = \frac{8}{5} t, \quad AN = 2t - x = 2t - \frac{7}{5} t = \frac{3}{5} t,
$$

therefore,

$$
\frac{S_{\triangle AMN}}{S_{\triangle ABC}} = \frac{AM}{AB} \cdot \frac{AN}{AC} = \frac{8/5 t}{3t} \cdot \frac{3/5 t}{2t} = \frac{4}{25}
$$

## Answer

$\frac{4}{25}$