A line that passes through the origin intersects both the line $y=5$ and the line $y=\sqrt3x+1$. The three lines create an equilateral triangle. What is the perimeter of the triangle?

1. The line $y=5$ is horizontal (direction $0^\circ$) and the line $y=\sqrt3x+1$ has slope $\sqrt3=\tan60^\circ$ (direction $60^\circ$). Three lines form an equilateral triangle when their directions are $0^\circ$, $60^\circ$ and $120^\circ$, so the third line has slope $-\sqrt3$; through the origin it is $y=-\sqrt3x$.
2. The vertices on $y=5$ are found from $\sqrt3x+1=5$, $x=\frac4{\sqrt3}$, and from $-\sqrt3x=5$, $x=-\frac5{\sqrt3}$.
3. The side on $y=5$ has length $\frac4{\sqrt3}+\frac5{\sqrt3}=\frac9{\sqrt3}=3\sqrt3$, so the perimeter is $3\cdot3\sqrt3=\boxed{9\sqrt3}$.