1. A regular hexagon is inscribed in another regular hexagon such that each vertex of the inscribed hexagon divides a side of the original hexagon into two parts in the ratio $2: 1$. Find the ratio of the area of the inscribed hexagon to the area of the larger hexagon.
Answer: $\frac{7}{9}$
Solution: Without loss of generality, we may assume that the original hexagon has side length 3 . Let $s$ then be the side length of the inscribed hexagon. Notice that there are six triangles with sides 1 , 2 , and $s$, and the angle between the sides of lengths 1 and 2 is $120^{\circ}$, as shown below:

By the cosine law, we have
$$
s^{2}=1^{2}+2^{2}-2 \cdot 1 \cdot 2 \cdot-\frac{1}{2}=7
$$
and so $s=\sqrt{7}$. Since the ratio of the areas of two similar shapes is equal to the square of the ratio of their sides, the desired ratio is thus $\left(\frac{\sqrt{7}}{3}\right)^{2}=\frac{7}{9}$.