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Difference between revisions of "2015 AMC 10A Problems/Problem 17"

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We now have the coordinates of two vertices, <math>(1, -\frac{\sqrt{3}}{3})</math> and <math>(1, 1 + \frac{\sqrt{3}}{3})</math>. The length of one side is the distance between the y-coordinates, or <math>1 +  \frac{2\sqrt{3}}{3}</math>.
 
We now have the coordinates of two vertices, <math>(1, -\frac{\sqrt{3}}{3})</math> and <math>(1, 1 + \frac{\sqrt{3}}{3})</math>. The length of one side is the distance between the y-coordinates, or <math>1 +  \frac{2\sqrt{3}}{3}</math>.
  
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The perimeter of the triangle is thus <math>3 * (1 +  \frac{2\sqrt{3}}{3})</math>, so the answer is <math>\boxed{\textbf{(D) }3 + 2\sqrt{3}}</math>
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The perimeter of the triangle is thus <math>3(1 +  \frac{2\sqrt{3}}{3})</math>, so the answer is <math>\boxed{\textbf{(D) }3 + 2\sqrt{3}}</math>

Revision as of 18:32, 4 February 2015

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

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

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