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.


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 $9-2 = 7 = \boxed{\textbf{(D)} ~7}$.


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 $b-a = 9-2 = \boxed{\textbf{(D)} ~7}$.


Solution 3 (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 $9-2 = \boxed{\textbf{(D)} ~7}$.


Solution 4 (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.


Video Solution by Punxsutawney Phil


Video Solution by OmegaLearn (Rotation & Reflection tricks)


~ pi_is_3.14

Video Solution by Hawk Math


Video Solution by TheBeautyofMath

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

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


Video Solution by Interstigation



Video Solution (Just 3 min!)


~Education, the Study of Everything

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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