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

(Recently while mocking this AMC10, I found a solution that seemed to be quicker than all of the given ones on this page, so I felt like I should put it here.)
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==Solution 1==
 
==Solution 1==
  
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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 <math>\frac{\sqrt{3}}{3}</math> (can be find out easily by using 60 degree angle of the triangle) so the third must be <math>-\frac{\sqrt{3}}{3}</math>. Since this third 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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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 <math>\frac{\sqrt{3}}{3}</math> (which is must be, given 60 degree angle of the triangle, relative to vertical) so the third must be <math>-\frac{\sqrt{3}}{3}</math>. Since this third 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.  
  
 
<math>y = -\frac{\sqrt{3}}{3}</math>
 
<math>y = -\frac{\sqrt{3}}{3}</math>
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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>.
  
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==Solution 3 (Intuitive)==
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==Solution 3==
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We have <math>y = mx.</math>
 
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With the first condition, we have that <math>y=m.</math>
 
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Then, we have <math>1 + \frac{\sqrt{3}}{3} x = mx</math>
 
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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)
 
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==Solution 4==
 
 
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 $x = 1$ and the line $y=1+ \frac{\sqrt{3}}{3} x$. The three lines create an equilateral triangle. What is the perimeter of the triangle?

$\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)} \ 2+2\sqrt{3} \qquad\textbf{(C)} \ 6 \qquad\textbf{(D)} \ 3 + 2\sqrt{3} \qquad\textbf{(E)} \ 6 + \frac{\sqrt{3}}{3}$

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 $\frac{\sqrt{3}}{3}$ (which is must be, given 60 degree angle of the triangle, relative to vertical) so the third must be $-\frac{\sqrt{3}}{3}$. Since this third line passes through the origin, its equation is simply $y = -\frac{\sqrt{3}}{3}x$. To find two vertices of the triangle, plug in $x=1$ to both the other equations.

$y = -\frac{\sqrt{3}}{3}$

$y = 1 + \frac{\sqrt{3}}{3}$

We now have the coordinates of two vertices, $\left(1, -\frac{\sqrt{3}}{3}\right)$ and $\left(1, 1 + \frac{\sqrt{3}}{3}\right)$. The length of one side is the distance between the y-coordinates, or $1 + \frac{2\sqrt{3}}{3}$.

The perimeter of the triangle is thus $3\left(1 + \frac{2\sqrt{3}}{3}\right)$, so the answer is $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$

Solution 2

Draw a line from the y-intercept of the equation $y=1+ \frac{\sqrt{3}}{3} x$ perpendicular to the line $x=1$. 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 $2\left(\frac{1}{\sqrt{3}}\right) + 1$. After multiplying the side length by 3 and rationalizing, you get $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$.

Solution 3

Let the intersection point between the line $y = 1 + \frac{\sqrt{3}}{3}x$ and the line that crosses the origin be $P$.

We drop an altitude from $P$ onto the line $x = 1$. Since the overall triangle is an equilateral triangle, we are splitting the base (on $x=1$) in half. As the y-axis is parallel to the line $x=1$, the altitude from P will also split the y-axis from $y=0$ to $y=1$ in half. From this, we can get that the y-value of P is $\frac{1}{2}$.

Plugging this into the equation $y = 1 + \frac{\sqrt{3}}{3}x$, we get that $x = -\frac{\sqrt{3}}{2}$, and thus our height for the equilateral triangle is $1 + \frac{\sqrt{3}}{2}$. Using that, we can calculate the perimeter to be $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$.

Video Solution

https://youtu.be/-l1Kawq_hds

~savannahsolver

See Also

Video Solution:

https://www.youtube.com/watch?v=2kvSRL8KMac


2015 AMC 10A (Problems • Answer Key • Resources)
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Problem 16 		Followed by
Problem 18
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All AMC 10 Problems and Solutions

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