2015 AMC 10A Problems/Problem 17


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 triangle must have a horizontal line of symmetry, and therefore the other two sides will have opposite slopes. The slope of the other given line is $\frac{\sqrt{3}}{3}$ (can be find out easily by using 60 degree angle of the triangle) 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 problem 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 $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$.

Solution 3 (Intuitive)

We have $y = mx.$

With the first condition, we have that $y=m.$

Then, we have $1 + \frac{\sqrt{3}}{3} x = mx$

Dividing both sides by $x$ on the second and putting over a common denominator gets us $\frac{3 + x \sqrt{3}}{3x} = m.$ The only answer in the answer choices that satisfies this is (D)

Solution 4

Let the intersection point between the line $y = 1 + \frac{\sqrt{3}}{3}x$ and the line that crosses the origin be $P$.

We drop an altitude from $P$ onto the line $x = 1$. Since the overall triangle is an equilateral triangle, we are splitting the base (on $x=1$) in half. As the y-axis is parallel to the line $x=1$, the altitude from P will also split the y-axis from $y=0$ to $y=1$ in half. From this, we can get that the y-value of P is $\frac{1}{2}$.

Plugging this into the equation $y = 1 + \frac{\sqrt{3}}{3}x$, we get that $x = -\frac{\sqrt{3}}{2}$, and thus our height for the equilateral triangle is $1 + \frac{\sqrt{3}}{2}$. Using that, we can calculate the perimeter to be $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$.

Video Solution



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2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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