Difference between revisions of "2021 AMC 12B Problems/Problem 5"
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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]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
− | The point <math>P(a,b)</math> in the <math>xy</math>-plane is first rotated counterclockwise by <math>90\ | + | 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> |
<math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9</math> | <math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9</math> | ||
+ | |||
==Solution== | ==Solution== | ||
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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>. | 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>. | ||
+ | |||
+ | --abhinavg0627 | ||
+ | |||
+ | ==Solution 2 (complex)== | ||
+ | |||
+ | 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 | ||
+ | <math>90^\circ</math>, and then translating the origin back to the point <math>(1, 5)</math>. | ||
+ | |||
+ | <math>a+b\cdot{i} \implies (a-1)+(b-5)\cdot{i} \implies ((a-1)+(b-5)\cdot{i})\cdot{i} = </math> | ||
+ | |||
+ | <math>5-b+(a-1)i \implies 5+1-b+(a-1+5)i = 6-b+(a+4)i</math>. | ||
+ | |||
+ | By basis reflection rules, the reflection of <math>(-6, 3)</math> about the line <math>y = -x</math> is <math>(-3, 6)</math>. | ||
+ | Hence, <math>6-b+(a+4)i = -3+6i \implies b=9, a=2.</math> | ||
+ | <math>b-a = 9-2 =7 = | ||
+ | \boxed{\textbf{(D)}}</math>. | ||
+ | ~ twotothetenthis1024 | ||
+ | |||
+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://youtube.com/watch?v=qpvS2PVkI8A&t=335s | ||
== Video Solution by OmegaLearn (Rotation & Reflection tricks) == | == Video Solution by OmegaLearn (Rotation & Reflection tricks) == | ||
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~ pi_is_3.14 | ~ 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== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=4|num-a=6}} | {{AMC12 box|year=2021|ab=B|num-b=4|num-a=6}} | ||
+ | {{AMC10 box|year=2021|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest 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.
Contents
Problem
The point in the -plane is first rotated counterclockwise by around the point and then reflected about the line . The image of after these two transformations is at . What is
Solution
The final image of is . We know the reflection rule for reflecting over is . So before the reflection and after rotation the point is .
By definition of rotation, the slope between and must be perpendicular to the slope between and . The first slope is . This means the slope of and is .
Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from to it follows we shall only use the slope once to travel from to .
Therefore point is located at . The answer is .
--abhinavg0627
Solution 2 (complex)
Let us reconstruct that coordinate plane as the complex plane. Then, the point becomes . A rotation around the point can be done by translating the point to the origin, rotating around the origin by , and then translating the origin back to the point .
.
By basis reflection rules, the reflection of about the line is . Hence, . ~ twotothetenthis1024
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=qpvS2PVkI8A&t=335s
Video Solution by OmegaLearn (Rotation & Reflection tricks)
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.