2015 AMC 10A Problems/Problem 17
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 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 problems 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 .
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 | |
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.
a