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 AMC 12B Problems/Problem 5

The following problem is from both the 2021 AMC 10B #9 and 2021 AMC 12B #5, so both problems redirect to this page.

Problem

The point $P(a,b)$ in the $xy$-plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$. The image of $P$ after these two transformations is at $(-6,3)$. What is $b - a ?$

$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$

Solution 1 (Transformation Rules)

The final image of $P$ is $(-6,3)$. We know the reflection rule for reflecting over $y=-x$ is $(x,y) \rightarrow (-y, -x)$. So before the reflection and after rotation the point is $(-3,6)$.

By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$. The first slope is $\frac{5-6}{1-(-3)} = \frac{-1}{4}$. This means the slope of $P$ and $(1,5)$ is $4$.

Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from $(-3,6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$.

Therefore point $P$ is located at $(1+1, 5+4) = (2,9)$. The answer is $9-2 = 7 = \boxed{\textbf{(D)} ~7}$.

-abhinavg0627

Solution 2 (Complex Numbers)

Let us reconstruct that coordinate plane as the complex plane. Then, the point $P(a, b)$ becomes $a+b\cdot{i}$. A $90^\circ$ rotation around the point $(1, 5)$ can be done by translating the point $(1, 5)$ to the origin, rotating around the origin by $90^\circ$, and then translating the origin back to the point $(1, 5)$. \[a+b\cdot{i} \implies (a-1)+(b-5)\cdot{i} \implies ((a-1)+(b-5)\cdot{i})\cdot{i} = 5-b+(a-1)i \implies 5+1-b+(a-1+5)i = 6-b+(a+4)i.\] By basis reflection rules, the reflection of $(-6, 3)$ about the line $y = -x$ is $(-3, 6)$. Hence, we have \[6-b+(a+4)i = -3+6i \implies b=9, a=2,\] from which $b-a = 9-2 = \boxed{\textbf{(D)} ~7}$.

~twotothetenthis1024

Solution 3 (Reverso)

The problem gives a series of transformations and proceeds to give the resultant point, being $(-6,3)=P$. Therefore, all we must do is reverse it. First, we reverse the last transformation by computing the distance from the point $(-6,3)$ to $y=-x$ by using the formula $d=\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$. Where $Ax+By+C$ is the standard form of a line. Computing, we obtain that $d=\frac{3\sqrt{2}}{2}$. We also know that this is magnitude is at an angle of $\frac{\pi}{4}$. Therefore, to do the transformation, we double that vector and add it to the point. We get:

$P^{'}=P+2\overrightarrow{v}$ $\implies P'=(-6,3)+\left\langle 3\sqrt{2}\cdot \frac{\sqrt{2}}{2},3\sqrt{2}\cdot \frac{\sqrt{2}}{2} \right \rangle$ $\implies P'=(-3,6)$.

Now, we must reverse the second transformation. To do so, realize that $P'-(1,5)\Leftrightarrow \overrightarrow{v_{2}}=\left\langle -4,1 \right\rangle$. Simply make this vector perpendicular by switching the $x$ and $y$ components and switching the sign of the initial $y$ component. Therefore, we get $\overrightarrow{v_{2\bot }}=\left\langle 1,4 \right\rangle$. Therefore, adding this vector to $(1,5)$ yields $(2,9)$, which leads us to $9-2 = \boxed{\textbf{(D)} ~7}$.

~justgiveup JoshKosh

Solution 4 (Vector Dot Product)

Using the same method as in Solution 1, we can obtain that the point before the reflection is $(-3,6)$. If we let the original point be $(x, y)$, then we can use that the starting point is $(1,5)$ to obtain two vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$. We know that two vectors are perpendicular if their dot product is equal to $0$, and that both points are the same distance ($\sqrt {17}$) from $(1,5)$.

Therefore, we can write two equations using these vectors: $(x-1)^2 + (y-5)^2 = 17$ (from distance and pythagorean theorem) and $-4x+y-1 = 0$ (from dot product)

Solving, we simplify the second equation to $y=4x+1$, and plug it into the first equation. We obtain $(x-1)^2 + (4x-4)^2 = 17$. We can simplify this to the quadratic $17x^2-34x=0$. When we factor out $17x$, we find that $x = 2$ or $x = 0$. However, $x$ cannot equal $0$. Therefore, $x = 2$, and plugging this into the second equation gives us that $y = 9$. Since the point is $(9, 2)$, we compute $9-2 = \boxed{\textbf{(D)} ~7}$.

~saturnrocket


Solution 5 (Vector Dot Product scuffed version)

Using the same method as in Solution 1 reflecting $(-6,3)$ about the line $y = -x$ gives us $(-3,6).$

Let the original point be $\langle x,y \rangle.$ From point $(1,5),$ we form the vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ that extend out from the initial point. If they are perpendicular, we know that their dot product has to equal zero. Therefore, \[\langle -4,1 \rangle \cdot \langle x-1, y-5 \rangle = 0 \implies -4x+y-1= 0.\]Now, we have to do some guess and check from the multiple choices. Let $y - x = A$ where $A$ is one of the answer choices. Then, $A -3x = 1.$ By intuition and logical reasoning we deduce that $A$ must be $1 \pmod 3$ so that brings our potential answers down to $\text{\textbf{(A)}}$ and $\text{\textbf{(C)}}.$ If $A = 1$ from $\text{\textbf{(A)}},$ then $x = 0,$ which we can quickly rule out since we know thar $P$ rotated counterclockwise not clockwise. Hence, $\boxed{\textbf{(D)} ~7}$ is the answer.

~peelybonehead

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=qpvS2PVkI8A&t=335s

Video Solution by OmegaLearn (Rotation & Reflection tricks)

https://youtu.be/VyRWjgGIsRQ

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=VzwxbsuSQ80

Video Solution by TheBeautyofMath

https://youtu.be/GYpAm8v1h-U?t=860 (for AMC 10B)

https://youtu.be/EMzdnr1nZcE?t=814 (for AMC 12B)

~IceMatrix

Video Solution by Interstigation

https://youtu.be/DvpN56Ob6Zw?t=776

~Interstigation

Video Solution (Just 3 min!)

https://youtu.be/j39KCUC2Qz8

~Education, the Study of Everything

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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
2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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_AMC_12B_Problems/Problem_5&oldid=232920"