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

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The perimeter of the triangle is thus <math>3\left(1 +  \frac{2\sqrt{3}}{3}\right)</math>, so the answer is <math>\boxed{\textbf{(D) }3 + 2\sqrt{3}}</math>
 
The perimeter of the triangle is thus <math>3\left(1 +  \frac{2\sqrt{3}}{3}\right)</math>, so the answer is <math>\boxed{\textbf{(D) }3 + 2\sqrt{3}}</math>
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==Solution 1==
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Once you draw the diagram, draw a line from the y-intercept of the equation : 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 a equilateral triangle with a square of side length 1 inscribed in it. The side length is 2(<math>\frac{1}{\sqrt{3}}</math>) + 1. After multiplying the side length by 3 and rationalizing, you get <math>\boxed{\textbf{(D) }3 + 2\sqrt{3}}</math>.
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2015|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:20, 2 March 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, $\left(1, -\frac{\sqrt{3}}{3}\right)$ 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 1

Once you draw the diagram, draw a line from the y-intercept of the equation : 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 a equilateral triangle with a square of side length 1 inscribed in it. The side length is 2($\frac{1}{\sqrt{3}}$) + 1. After multiplying the side length by 3 and rationalizing, you get $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$.

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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