Difference between revisions of "2015 AMC 10A Problems/Problem 17"
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<math> \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} </math> | <math> \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} </math> | ||
| − | ==Solution== | + | ==Solution 1== |
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 <math>\frac{\sqrt{3}}{3}</math> so the third must be <math>-\frac{\sqrt{3}}{3}</math>. Since this third line passes through the origin, its equation is simply <math>y = -\frac{\sqrt{3}}{3}x</math>. To find two vertices of the triangle, plug in <math>x=1</math> to both the other equations. | 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 <math>\frac{\sqrt{3}}{3}</math> so the third must be <math>-\frac{\sqrt{3}}{3}</math>. Since this third line passes through the origin, its equation is simply <math>y = -\frac{\sqrt{3}}{3}x</math>. To find two vertices of the triangle, plug in <math>x=1</math> to both the other equations. | ||
Revision as of 23:21, 2 March 2015
Contents
Problem
A line that passes through the origin intersects both the line 
and the line 
. The three lines create an equilateral triangle. What is the perimeter of the triangle?
Solution 1
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 
so the third must be 
. Since this third line passes through the origin, its equation is simply 
. To find two vertices of the triangle, plug in 
to both the other equations.
We now have the coordinates of two vertices, 
and 
. The length of one side is the distance between the y-coordinates, or 
.
The perimeter of the triangle is thus 
, so the answer is 
Solution 2
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(
) + 1. After multiplying the side length by 3 and rationalizing, you get .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
