A line that passes through the origin intersects both the line $x=1$ and the line $y=1+\frac{\sqrt{3}}{3}x$. The three lines create an equilateral triangle. What is the perimeter of the triangle? $\textbf{(A) }2\sqrt{6}\qquad$ $\textbf{(B) }2+2\sqrt{3}\qquad$ $\textbf{(C) }6\qquad$ $\textbf{(D) }3+2\sqrt{3}\qquad$ $\textbf{(E) }6+\frac{\sqrt{3}}{3}$
Solution
Let $P=(1,1+\frac1{\sqrt3})$ be the intersection of $x=1$ and the slanted line $y=1+\frac1{\sqrt3}x$, which has inclination $30^\circ$ (slope $\tan30^\circ=\frac1{\sqrt3}$). The vertical line makes an angle of $90^\circ-30^\circ=60^\circ$ with it, so the angle of the triangle at $P$ is $60^\circ$.
Let the third line $\ell$ through the origin meet $x=1$ at $Q$ and the slanted line at $R$. The triangle $PQR$ is equilateral iff the angle at $Q$ is also $60^\circ$ (then the angle at $R$ is $60^\circ$ too).
So $\ell$ must make a $60^\circ$ angle with the vertical line $x=1$, i.e.\ its inclination is $30^\circ$ or $150^\circ$. Inclination $30^\circ$ is parallel to the slanted line (no triangle), so $\ell$ has inclination $150^\circ$ and slope $-\frac1{\sqrt3}$, i.e.\ $\ell:y=-\frac1{\sqrt3}x$.
Then $Q=(1,-\frac1{\sqrt3})$ and the side is $PQ=\left(1+\frac1{\sqrt3}\right)-\left(-\frac1{\sqrt3}\right)=1+\frac2{\sqrt3}$.
The perimeter is $3PQ=3+\frac6{\sqrt3}=3+2\sqrt3$, answer (D).
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?
