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 of the y-coordiantes, 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 $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$

Page

Toolbox

Search

2015 AMC 10A Problems/Problem 17

Problem

Solution