# Difference between revisions of "2015 AMC 10A Problems/Problem 17"

(Created page with "==Problem== A line that passes through the origin in tersects both the line <math> x = 1</math> and the line <math>y=1+ \frac{\sqrt{3}}{3} x</math>. The three lines create an eq...") |
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+ | 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 one of the other lines is <math>\frac{\sqrt{3}}{3}</math> so the other must be <math>-\frac{\sqrt{3}}{3}</math>. Since this other 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. | ||

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+ | <math>y = -\frac{\sqrt{3}}{3}</math> | ||

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+ | <math>y = 1 + \frac{\sqrt{3}}{3}</math> | ||

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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>. | ||

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+ | <math>\sqrt{(1-1)^2 + (-\frac{\sqrt{3}}{3} - (1 + \frac{\sqrt{3}}{3}))^2}</math> | ||

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+ | <math>\sqrt{(-\frac{\sqrt{3}}{3} - 1 - \frac{\sqrt{3}}{3}))^2}</math> |

## Revision as of 17:24, 4 February 2015

## Problem

A line that passes through the origin in tersects 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 one of the other lines is so the other must be . Since this other 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, .