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.

1 Problem
2 Solution 1 (Transformation Rules)
3 Solution 2 (Complex Numbers)
4 Solution 3 (Vector Dot Product)
5 Solution 4 (Vector Dot Product scuffed version)
6 Video Solution by Punxsutawney Phil
7 Video Solution by OmegaLearn (Rotation & Reflection tricks)
8 Video Solution by Hawk Math
9 Video Solution by TheBeautyofMath
10 Video Solution by Interstigation
11 Video Solution (Just 3 min!)
12 See Also

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 (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 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.

~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

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

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