A geometry problem by A Former Brilliant Member

I placed an equilateral triangle of base 1 on a 2-dimensional rectangular coordinate system, with one vertex at the origin and side along the positive x-axis. Note that this triangle has an area of $\sqrt{3}/4$ . Since the area of an equilateral triangle is $\frac{\sqrt{3}}{4}s^2$ , then the desired answer will be $s^2$ , where $s$ is the side length of the inner equilateral triangle. Next, I figured out the equations of the three lines that lie inside the triangle: $y=\frac{\sqrt{3}}{5}x$ , $y=-\frac{\sqrt{3}}{2}(x-1)$ , and $y=3\sqrt{3}(x-\frac{1}{3})$ . Then I found two intersection points of the three lines, namely, $(\frac{3}{7},\frac{2}{7}\sqrt{3})$ and $(\frac{5}{14},\frac{\sqrt{3}}{14})$ . So, $s^2=(\frac{3}{7}-\frac{5}{14})^2+(\frac{2}{7}\sqrt{3}-\frac{\sqrt{3}}{14})^2=\frac{1}{7}$ .