Difference between revisions of "2015 AMC 10A Problems/Problem 17"
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==Solution 2== | ==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>. | 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>. | ||
+ | |||
+ | ==Solution 3 (Intuitive)== | ||
+ | We have <math>y = mx.</math> | ||
+ | |||
+ | With the first condition, we have that <math>y=m.</math> | ||
+ | |||
+ | Then, we have <math>1 + \frac{\sqrt{3}}{3} x = mx</math> | ||
+ | |||
+ | Dividing both sides by <math>x</math> on the second and putting over a common denominator gets us <math>\frac{3 + x \sqrt{3}}{3x} = m.</math> The only answer in the answer choices that satisfies this is (D) | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 01:15, 25 September 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 .
Solution 3 (Intuitive)
We have
With the first condition, we have that
Then, we have
Dividing both sides by on the second and putting over a common denominator gets us The only answer in the answer choices that satisfies this is (D)
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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All AMC 10 Problems and Solutions |
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