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Difference between revisions of "2021 AMC 12B Problems/Problem 5"

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Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from <math>(3,-6)</math> to <math>(1,5)</math> it follows we shall only use the slope once to travel from <math>(1,5)</math> to <math>P</math>.
 
Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from <math>(3,-6)</math> to <math>(1,5)</math> it follows we shall only use the slope once to travel from <math>(1,5)</math> to <math>P</math>.
  
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Therefore point <math>P</math> is located at <math>(1+1, 5+4) = (2,9)</math>. The answer is <math>9-2 = 7 = \boxed{\textbf{(D)}}</math>.
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Therefore point <math>P</math> is located at <math>(1+1, 5+4) = (2,9)</math>. The answer is <math>9-2 = 7 = \boxed{\textbf{(D)} ~7}</math>.
  
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--abhinavg0627
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-abhinavg0627
  
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==Solution 2 (complex)==
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==Solution 2 (Complex Numbers)==
  
 
Let us reconstruct that coordinate plane as the complex plane. Then, the point <math>P(a, b)</math> becomes <math>a+b\cdot{i}</math>.  
 
Let us reconstruct that coordinate plane as the complex plane. Then, the point <math>P(a, b)</math> becomes <math>a+b\cdot{i}</math>.  
 
A <math>90^\circ</math> rotation around the point <math>(1, 5)</math> can be done by translating the point <math>(1, 5)</math> to the origin, rotating around the origin by  
 
A <math>90^\circ</math> rotation around the point <math>(1, 5)</math> can be done by translating the point <math>(1, 5)</math> to the origin, rotating around the origin by  
 
<math>90^\circ</math>, and then translating the origin back to the point <math>(1, 5)</math>.
 
<math>90^\circ</math>, and then translating the origin back to the point <math>(1, 5)</math>.
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<cmath>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.</cmath>  
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<math>a+b\cdot{i}  \implies (a-1)+(b-5)\cdot{i} \implies ((a-1)+(b-5)\cdot{i})\cdot{i} = </math>
 
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<math>5-b+(a-1)i \implies 5+1-b+(a-1+5)i = 6-b+(a+4)i</math>.
 
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By basis reflection rules, the reflection of <math>(-6, 3)</math> about the line <math>y = -x</math> is <math>(-3, 6)</math>.  
 
By basis reflection rules, the reflection of <math>(-6, 3)</math> about the line <math>y = -x</math> is <math>(-3, 6)</math>.  
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Hence, <math>6-b+(a+4)i = -3+6i \implies b=9, a=2.</math>  
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Hence, we have <cmath>6-b+(a+4)i = -3+6i \implies b=9, a=2,</cmath> from which <math>b-a = 9-2 = \boxed{\textbf{(D)} ~7}</math>.  
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<math>b-a = 9-2 =7 =
 
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\boxed{\textbf{(D)}}</math>.  
 
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~ twotothetenthis1024
 
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==Video Solution 1==
 
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https://youtu.be/j39KCUC2Qz8
 
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~Education, the Study of Everything
 
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~twotothetenthis1024
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==
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~Interstigation
 
~Interstigation
 +
 +
==Video Solution==
 +
https://youtu.be/j39KCUC2Qz8
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 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==

Revision as of 01:16, 19 August 2022

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

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

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

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