Difference between revisions of "2015 AMC 10A Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | A line that passes through the origin | + | A line that passes through the origin intersects both the line <math> x = 1</math> and the line <math>y=1+ \frac{\sqrt{3}}{3} x</math>. The three lines create an equilateral triangle. What is the perimeter of the triangle? |
<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 1== | ||
− | ==Solution== | + | 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 <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. |
+ | |||
+ | <math>y = -\frac{\sqrt{3}}{3}</math> | ||
+ | |||
+ | <math>y = 1 + \frac{\sqrt{3}}{3}</math> | ||
+ | |||
+ | We now have the coordinates of two vertices, <math>\left(1, -\frac{\sqrt{3}}{3}\right)</math> and <math>\left(1, 1 + \frac{\sqrt{3}}{3}\right)</math>. The length of one side is the distance between the y-coordinates, or <math>1 + \frac{2\sqrt{3}}{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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Draw a line from the y-intercept of the equation <math>y=1+ \frac{\sqrt{3}}{3} x</math> perpendicular to the line <math>x=1</math>. 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 <math>2\left(\frac{1}{\sqrt{3}}\right) + 1</math>. After multiplying the side length by 3 and rationalizing, you get <math>\boxed{\textbf{(D) }3 + 2\sqrt{3}}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/-l1Kawq_hds | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | Video Solution: | ||
+ | |||
+ | https://www.youtube.com/watch?v=2kvSRL8KMac | ||
+ | |||
+ | |||
+ | {{AMC10 box|year=2015|ab=A|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}}a |
Latest revision as of 15:44, 31 January 2021
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 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 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
Draw a line from the y-intercept of the equation perpendicular to the line . 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 . After multiplying the side length by 3 and rationalizing, you get .
Video Solution
~savannahsolver
See Also
Video Solution:
https://www.youtube.com/watch?v=2kvSRL8KMac
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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