2015 AMC 10A Problems/Problem 17

Problem

A line that passes through the origin in tersects 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 one of the other lines is $\frac{\sqrt{3}}{3}$ so the other must be $-\frac{\sqrt{3}}{3}$ . Since this other 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 $(1, 1 + \frac{\sqrt{3}}{3})$ . Apply the distance formula, $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ .

$\sqrt{(1-1)^2 + (-\frac{\sqrt{3}}{3} - (1 + \frac{\sqrt{3}}{3}))^2}$

$\sqrt{(-\frac{\sqrt{3}}{3} - 1 - \frac{\sqrt{3}}{3}))^2}$

$1 + 2 \frac{\sqrt{3}}{3}$

Page

Toolbox

Search

2015 AMC 10A Problems/Problem 17

Problem

Solution