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 1

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, $\left(1, -\frac{\sqrt{3}}{3}\right)$ and $\left(1, 1 + \frac{\sqrt{3}}{3}\right)$ . 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 $3\left(1 + \frac{2\sqrt{3}}{3}\right)$ , so the answer is $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$

Solution 2

Draw a line from the y-intercept of the equation $y=1+ \frac{\sqrt{3}}{3} x$ perpendicular to the line x=1. There is a square of side length 1 inscribed in the equilateral triangle. The problems becomes reduced to finding the perimeter of an equilateral triangle with a square of side length 1 inscribed in it. The side length is 2 $\left(\frac{1}{\sqrt{3}}\right)$ + 1. After multiplying the side length by 3 and rationalizing, you get $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$ .

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2015 AMC 10A Problems/Problem 17

Problem

Solution 1

Solution 2