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

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The 2021 AMC 12B will be held on February 10th, 2021. The problems will not be made public until 24 hours after that.
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{{duplicate|[[2021 AMC 10B Problems#Problem 9|2021 AMC 10B #9]] and [[2021 AMC 12B Problems#Problem 5|2021 AMC 12B #5]]}}
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==Problem==
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The point <math>P(a,b)</math> in the <math>xy</math>-plane is first rotated counterclockwise by <math>90^\circ</math> around the point <math>(1,5)</math> and then reflected about the line <math>y = -x</math>. The image of <math>P</math> after these two transformations is at <math>(-6,3)</math>. What is <math>b - a ?</math>
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<math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9</math>
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==Solution==
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The final image of <math>P</math> is <math>(-6,3)</math>. We know the reflection rule for reflecting over <math>y=-x</math> is <math>(x,y) --> (-y, -x)</math>. So before the reflection and after rotation the point is <math>(-3,6)</math>.
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By definition of rotation, the slope between <math>(-3,6)</math> and <math>(1,5)</math> must be perpendicular to the slope between <math>(a,b)</math> and <math>(1,5)</math>. The first slope is <math>\frac{5-6}{1-(-3)} = \frac{-1}{4}</math>. This means the slope of <math>P</math> and <math>(1,5)</math> is <math>4</math>.
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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>.
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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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--abhinavg0627
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==Solution 2 (complex)==
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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>.
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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
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<math>90^\circ</math>, and then translating the origin back to the point <math>(1, 5)</math>.
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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>.
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Hence, <math>6-b+(a+4)i = -3+6i \implies b=9, a=2.</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 by Punxsutawney Phil==
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https://youtube.com/watch?v=qpvS2PVkI8A&t=335s
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== Video Solution by OmegaLearn (Rotation & Reflection tricks) ==
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https://youtu.be/VyRWjgGIsRQ
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~ pi_is_3.14
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==Video Solution by Hawk Math==
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https://www.youtube.com/watch?v=VzwxbsuSQ80
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==Video Solution by TheBeautyofMath==
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https://youtu.be/GYpAm8v1h-U?t=860 (for AMC 10B)
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https://youtu.be/EMzdnr1nZcE?t=814 (for AMC 12B)
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~IceMatrix
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==Video Solution by Interstigation==
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https://youtu.be/DvpN56Ob6Zw?t=776
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~Interstigation
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==See Also==
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{{AMC12 box|year=2021|ab=B|num-b=4|num-a=6}}
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{{AMC10 box|year=2021|ab=B|num-b=8|num-a=10}}
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{{MAA Notice}}

Revision as of 17:26, 6 May 2021

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

The final image of $P$ is $(-6,3)$. We know the reflection rule for reflecting over $y=-x$ is $(x,y) --> (-y, -x)$. 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 $9-2 = 7 = \boxed{\textbf{(D)}}$.

--abhinavg0627

Solution 2 (complex)

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, $6-b+(a+4)i = -3+6i \implies b=9, a=2.$ $b-a = 9-2 =7 = \boxed{\textbf{(D)}}$. ~ 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

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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