Difference between revisions of "2015 AMC 10A Problems/Problem 17"
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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 | + | ==Solution 3== |
Let the intersection point between the line <math>y = 1 + \frac{\sqrt{3}}{3}x</math> and the line that crosses the origin be <math>P</math>. | Let the intersection point between the line <math>y = 1 + \frac{\sqrt{3}}{3}x</math> and the line that crosses the origin be <math>P</math>. | ||
Latest revision as of 21:51, 22 September 2024
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 
(which is must be, given 60 degree angle of the triangle, relative to vertical) 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
Let the intersection point between the line 
and the line that crosses the origin be 
.
We drop an altitude from onto the line . Since the overall triangle is an equilateral triangle, we are splitting the base (on ) in half. As the y-axis is parallel to the line , the altitude from P will also split the y-axis from 
to 
in half. From this, we can get that the y-value of P is 
.
Plugging this into the equation , we get that 
, and thus our height for the equilateral triangle is 
. Using that, we can calculate the perimeter to be .
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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