Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 7: Line 7:
 
== Solution ==
 
== Solution ==
  
−
The slope of the line is <math>\frac{3}{5}</math>. We must transform it by <math>45^{\circ}</math>. <math>45^{\circ}</math> creates an isosceles right triangle since the sum of the angles of the triangle must be <math>180^{\circ}</math> and one angle is <math>90^{\circ}</math> which means the last leg angle must also be <math>45^{\circ}</math>. In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of <math>\frac{3}{5}</math> slope on graph paper. That line with <math>\frac{3}{5}</math> slope starts at <math>(0,0)</math> and will go to <math>(5,3)</math>, the vector <math><5,3></math>. Construct another line from <math>(0,0)</math> to <math>(3,-5)</math>, the vector <math><3,-5></math>. This is <math>\perp</math> and equal to the original line segment. The difference between the two vectors is <math><2,8></math>, which is the slope <math>4</math>, and that is the slope of line <math>k</math>. 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
+
The slope of the line is <math>\frac{3}{5}</math>. We must transform it by <math>45^{\circ}</math>.  
 +
<math>45^{\circ}</math> creates an isosceles right triangle since the sum of the angles of the triangle must be <math>180^{\circ}</math> and one angle is <math>90^{\circ}</math> which means the last leg angle must also be <math>45^{\circ}</math>.  
 +
In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of <math>\frac{3}{5}</math> slope on graph paper. That line with <math>\frac{3}{5}</math> slope starts at <math>(0,0)</math> and will go to <math>(5,3)</math>, the vector <math><5,3></math>. Construct another line from <math>(0,0)</math> to <math>(3,-5)</math>, the vector <math><3,-5></math>. This is <math>\perp</math> and equal to the original line segment. The difference between the two vectors is <math><2,8></math>, which is the slope <math>4</math>, and that is the slope of line <math>k</math>.  
 +
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

Revision as of 14:12, 1 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_12&oldid=116301"