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Difference between revisions of "2020 AMC 12A Problems/Problem 12"

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Furthermore, the equation <math>3x-5y+40=0</math> passes straight through <math>(20,20)</math> since <math>3(20)-5(20)+40=60-100+40=0</math>, which means that any rotations about <math>(20,20)</math> would contain <math>(20,20)</math>. We can create a line of slope <math>4</math> through <math>(20,20)</math>. The <math>x</math>-intercept is therefore <math>20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}</math>~lopkiloinm
 
Furthermore, the equation <math>3x-5y+40=0</math> passes straight through <math>(20,20)</math> since <math>3(20)-5(20)+40=60-100+40=0</math>, which means that any rotations about <math>(20,20)</math> would contain <math>(20,20)</math>. We can create a line of slope <math>4</math> through <math>(20,20)</math>. The <math>x</math>-intercept is therefore <math>20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}</math>~lopkiloinm
  
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==Solution 2==
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Since the slope of the line is <math>\frac{3}{5}</math>, and the angle we are rotating around is x, then <math>\tan x = \frac{3}{5}</math>
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<math>\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4</math>
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Hence, the slope of the rotated line is <math>4</math>. Since we know the line intersects the point <math>(20,20)</math>, then we know the line is <math>y=4x-60</math>. Set <math>y=0</math> to find the x-intercept, and so <math>x=\boxed{15}</math>
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~Solution by IronicNinja
 
==See Also==
 
==See Also==
  
 
{{AMC12 box|year=2020|ab=A|num-b=11|num-a=13}}
 
{{AMC12 box|year=2020|ab=A|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:09, 2 February 2020

Problem

Line $l$ in the coordinate plane has equation $3x-5y+40=0$. This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$. What is the $x$-coordinate of the $x$-intercept of line $k?$

$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$

Solution

The slope of the line is $\frac{3}{5}$. We must transform it by $45^{\circ}$.

$45^{\circ}$ creates an isosceles right triangle since the sum of the angles of the triangle must be $180^{\circ}$ and one angle is $90^{\circ}$ which means the last leg angle must also be $45^{\circ}$.

In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of $\frac{3}{5}$ slope on graph paper. That line with $\frac{3}{5}$ slope starts at $(0,0)$ and will go to $(5,3)$, the vector $<5,3>$.

Construct another line from $(0,0)$ to $(3,-5)$, the vector $<3,-5>$. This is $\perp$ and equal to the original line segment. The difference between the two vectors is $<2,8>$, which is the slope $4$, and that is the slope of line $k$.

Furthermore, the equation $3x-5y+40=0$ passes straight through $(20,20)$ since $3(20)-5(20)+40=60-100+40=0$, which means that any rotations about $(20,20)$ would contain $(20,20)$. We can create a line of slope $4$ through $(20,20)$. The $x$-intercept is therefore $20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}$~lopkiloinm

Solution 2

Since the slope of the line is $\frac{3}{5}$, and the angle we are rotating around is x, then $\tan x = \frac{3}{5}$ $\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4$

Hence, the slope of the rotated line is $4$. Since we know the line intersects the point $(20,20)$, then we know the line is $y=4x-60$. Set $y=0$ to find the x-intercept, and so $x=\boxed{15}$

~Solution by IronicNinja

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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