Difference between revisions of "2015 AMC 10A Problems/Problem 17"
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We now have the coordinates of two vertices. <math>(1, -\frac{\sqrt{3}}{3})</math> and <math>(1, 1 + \frac{\sqrt{3}}{3})</math>. Apply the distance formula, <math>\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}</math>. | We now have the coordinates of two vertices. <math>(1, -\frac{\sqrt{3}}{3})</math> and <math>(1, 1 + \frac{\sqrt{3}}{3})</math>. Apply the distance formula, <math>\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}</math>. | ||
− | <math>\sqrt{(1-1)^2 + (-\frac{\sqrt{3}}{3} - (1 + \frac{\sqrt{3}}{3}))^2}</math> | + | <math>\sqrt{(1-1)^2 + \left(-\frac{\sqrt{3}}{3} - \left(1 + \frac{\sqrt{3}}{3}\right)\right)^2}</math> |
− | <math>\sqrt{(-\frac{\sqrt{3}}{3} - 1 - \frac{\sqrt{3}}{3})^2}</math> | + | <math>\sqrt{\left(-\frac{\sqrt{3}}{3} - 1 - \frac{\sqrt{3}}{3}\right)^2}</math> |
The length of one side is <math>1 + \frac{2\sqrt{3}}{3}</math> | The length of one side is <math>1 + \frac{2\sqrt{3}}{3}</math> | ||
The perimeter of the triangle is <math>3 * (1 + \frac{2\sqrt{3}}{3})</math>, so the answer is <math>\boxed{\textbf{(D) }3 + 2\sqrt{3}}</math> | The perimeter of the triangle is <math>3 * (1 + \frac{2\sqrt{3}}{3})</math>, so the answer is <math>\boxed{\textbf{(D) }3 + 2\sqrt{3}}</math> |
Revision as of 18:24, 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 . Apply the distance formula, .
The length of one side is
The perimeter of the triangle is , so the answer is