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>.
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==Solution 3 (Intuitive)==
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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)
  
 
==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 $x = 1$ 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 $\frac{\sqrt{3}}{3}$ 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 (Intuitive)

We have $y = mx.$

With the first condition, we have that $y=m.$

Then, we have $1 + \frac{\sqrt{3}}{3} x = mx$

Dividing both sides by $x$ on the second and putting over a common denominator gets us $\frac{3 + x \sqrt{3}}{3x} = m.$ The only answer in the answer choices that satisfies this is (D)

Video Solution

https://youtu.be/-l1Kawq_hds

~savannahsolver

See Also

Video Solution:

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


2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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