2015 AMC 10A Problem 17

Turn the slopes into angles. An equilateral triangle needs every angle to be $60^\circ$.

1. The line $y = 1 + \frac{\sqrt{3}}{3}x$ has slope $\frac{\sqrt{3}}{3}$. In a $30$-$60$-$90$ triangle the side across from $30^\circ$ is $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ times the other leg, so this line rises at $30^\circ$ from horizontal.
2. The line $x = 1$ is vertical. A line tilted $30^\circ$ from horizontal is tilted $90^\circ - 30^\circ = 60^\circ$ from vertical. So these two lines already meet at $60^\circ$.
3. The third line goes through the origin and must also meet $x = 1$ at $60^\circ$, so it is also tilted $30^\circ$ from horizontal. It cannot rise the same way (it would be parallel to the given line and never meet it), so it falls at $30^\circ$. Its slope is $-\frac{\sqrt{3}}{3}$:
$$y = -\frac{\sqrt{3}}{3}x.$$
The two slanted lines, one up $30^\circ$ and one down $30^\circ$, meet each other at $60^\circ$ too, so the triangle is equilateral.
4. The easiest side to measure is the one on $x = 1$. Put $x = 1$ into each slanted line:
$$\text{top: } y = 1 + \frac{\sqrt{3}}{3}, \qquad \text{bottom: } y = -\frac{\sqrt{3}}{3}.$$
5. The side length is the gap between them:
$$1 + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = 1 + \frac{2\sqrt{3}}{3}.$$
6. All three sides are equal, so the perimeter is
$$3\left(1 + \frac{2\sqrt{3}}{3}\right) = 3 + 2\sqrt{3}.$$

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

A slope is "rise over run", which is the same as the tangent of the line's angle. Slopes $\frac{\sqrt{3}}{3}$ and $\sqrt{3}$ mean $30^\circ$ and $60^\circ$; learn to spot them. Once you know the triangle is equilateral, you only need one side, so choose the side on the vertical line, where the lengths are simple subtractions.

Find the third corner and measure a slanted side as a check. Setting $-\frac{\sqrt{3}}{3}x = 1 + \frac{\sqrt{3}}{3}x$ gives $x = -\frac{\sqrt{3}}{2}$ and $y = \frac{1}{2}$. The distance from this point to the bottom corner $(1, -\frac{\sqrt{3}}{3})$ works out to $1 + \frac{2\sqrt{3}}{3}$ again, the same as the vertical side.