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

2021 Fall AMC 10B Problems/Problem 17

Revision as of 11:03, 14 December 2021 by Hurdler (talk | contribs) (→‎Solution 2)

Problem

Distinct lines $\ell$ and $m$ lie in the $xy$-plane. They intersect at the origin. Point $P(-1, 4)$ is reflected about line $\ell$ to point $P'$, and then $P'$ is reflected about line $m$ to point $P''$. The equation of line $\ell$ is $5x - y = 0$, and the coordinates of $P''$ are $(4,1)$. What is the equation of line $m?$

$(\textbf{A})\: 5x+2y=0\qquad(\textbf{B}) \: 3x+2y=0\qquad(\textbf{C}) \: x-3y=0\qquad(\textbf{D}) \: 2x-3y=0\qquad(\textbf{E}) \: 5x-3y=0$


Solution 1

It is well known that the composition of 2 reflections , one after another, about two lines $l$ and $m$, respectively, that meet at an angle $\theta$ is a rotation by $2\theta$ around the intersection of $l$ and $m$.

Now, we note that $(4,1)$ is a 90 degree rotation clockwise of $(-1,4)$ about the origin, which is also where $l$ and $m$ intersect. So $m$ is a 45 degree rotation of $l$ about the origin clockwise.

To rotate $l$ 90 degrees clockwise, we build a square with adjacent vertices $(0,0)$ and $(1,5)$. The other two vertices are at $(5,-1)$ and $(6,4)$. The center of the square is at $(3,2)$, which is the midpoint of $(1,5)$ and $(5,-1)$. The line $m$ passes through the origin and the center of the square we built, namely at $(0,0)$ and $(3,2)$. Thus the line is $y = \frac{2}{3} x$. The answer is (D) $\boxed{3y - 2x = 0}$.

~hurdler

Solution 2

We know that the equation of line $\ell$ is $y = 5x$. This means that $P'$ is $(-1,4)$ reflected over the line $y = 5x$. This means that the line with $P$ and $P'$ is perpendicular to $\ell$, so it has slope $-\frac{1}{5}$. Then the equation of this perpendicular line is $y = -\frac{1}{5}x + c$, and plugging in $(-1,4)$ for $x$ and $y$ yields $c = \frac{19}{5}$.

The midpoint of $P'$ and $P$ lies at the intersection of $y = 5x$ and $y = -\frac{1}{5}x + \frac{19}{5}$. Solving, we get the x-value of the intersection is $\frac{19}{26}$ and the y-value is $\frac{95}{26}$. Let the x-value of $P'$ be $x'$ - then by the midpoint formula, $\frac{x' - 1}{2} = \frac{19}{26} \implies x' = \frac{32}{13}$. We can find the y-value of $P'$ the same way, so $P' = (\frac{32}{13},\frac{43}{13})$.

Now we have to reflect $P'$ over $m$ to get to $(4,1)$. The midpoint of $P'$ and $P''$ will lie on $m$, and this midpoint is, by the midpoint formula, $(\frac{42}{13},\frac{28}{13})$. $y = mx$ must satisfy this point, so $m = \frac{\frac{28}{13}}{\frac{42}{13}} = \frac{28}{42} = \frac{2}{3}$.

Now the equation of line $m$ is $y = \frac{2}{3}x \implies 2x-3y = 0 = \boxed{D}$


~KingRavi

Video Solution

Solution 2021 Fall 10B #17

~hurdler

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_17&oldid=167680"