# 2015 AMC 10A Problems/Problem 17

## Problem

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

Since the triangle is equilateral and one of the sides is a vertical line, the other two sides will have opposite slopes. The slope of the other given line is $\frac{\sqrt{3}}{3}$ so the third must be $-\frac{\sqrt{3}}{3}$. Since this third line passes through the origin, its equation is simply $y = -\frac{\sqrt{3}}{3}x$. To find two vertices of the triangle, plug in $x=1$ to both the other equations. $y = -\frac{\sqrt{3}}{3}$ $y = 1 + \frac{\sqrt{3}}{3}$

We now have the coordinates of two vertices, $(1, -\frac{\sqrt{3}}{3})$ and $(1, 1 + \frac{\sqrt{3}}{3})$. The length of one side is the distance between the y-coordinates, or $1 + \frac{2\sqrt{3}}{3}$.

The perimeter of the triangle is thus $3(1 + \frac{2\sqrt{3}}{3})$, so the answer is $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$