2015 AMC 10A Problem 17

The line $y = 1 + \frac{\sqrt{3}}{3}x$ has slope $\tan 30^\circ$, so it makes a $30^\circ$ angle with the horizontal and therefore a $60^\circ$ angle with the vertical line $x = 1$. Good: that is one angle of the equilateral triangle.

The third line passes through the origin and must also make $60^\circ$ with $x = 1$, meaning $30^\circ$ with the horizontal. It cannot be parallel to the given line, so its slope is $-\tan 30^\circ$: the line is $y = -\frac{\sqrt{3}}{3}x$. (It then makes $60^\circ$ with the given line as well, so all three angles are $60^\circ$.)

The vertical side lies on $x = 1$. Its endpoints are where the two slanted lines cross $x = 1$:
$$y = 1 + \frac{\sqrt{3}}{3} \qquad\text{and}\qquad y = -\frac{\sqrt{3}}{3}.$$
So the side length is $1 + \frac{2\sqrt{3}}{3}$, and the perimeter is
$$3\left(1 + \frac{2\sqrt{3}}{3}\right) = 3 + 2\sqrt{3}.$$

The answer is $\boxed{\textbf{(D)}\ 3+2\sqrt{3}}$.

Slopes are tangents of angles, so $\pm\frac{\sqrt{3}}{3}$ immediately signals $\pm 30^\circ$ lines, which are symmetric about the horizontal and meet a vertical line at $60^\circ$ each. Once the equilateral triangle is identified, computing one side (the vertical one, easiest) gives the perimeter without finding the third vertex.

Find the third vertex: $-\frac{\sqrt{3}}{3}x = 1 + \frac{\sqrt{3}}{3}x$ gives $x = -\frac{\sqrt{3}}{2}$, $y = \frac{1}{2}$. Its distance to $(1, -\frac{\sqrt{3}}{3})$ is $\sqrt{(1 + \frac{\sqrt{3}}{2})^2 + (\frac{1}{2} + \frac{\sqrt{3}}{3})^2}$, which simplifies to $1 + \frac{2\sqrt{3}}{3}$, confirming the side length.

• Assuming the line through the origin must be perpendicular to the given line, or measuring the side as the distance from the origin to $x = 1$ (which is $1$).
• Forgetting the $y$-intercept $1$ of the given line, which makes the vertical side $\frac{2\sqrt{3}}{3}$ and the perimeter $2\sqrt{3}$.