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> | ||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2015|ab=A|num-b=16|num-a=18}} | ||
| + | {{MAA Notice}} | ||
Revision as of 21:14, 4 February 2015
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
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 
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. 
