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.
Contents
- 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 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 1 (Transformation Rules)
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 Numbers)
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, we have from which .
~twotothetenthis1024
Solution 3 (Vector Dot Product)
Using the same method as in Solution 1, we can obtain that the point before the reflection is . If we let the original point be , then we can use that the starting point is to obtain two vectors and . We know that two vectors are perpendicular if their dot product is equal to , and that both points are the same distance () from .
Therefore, we can write two equations using these vectors: (from distance and pythagorean theorem) and (from dot product)
Solving, we simplify the second equation to , and plug it into the first equation. We obtain . We can simplify this to the quadratic . When we factor out , we find that or . However, cannot equal . Therefore, , and plugging this into the second equation gives us that . Since the point is , we compute .
~saturnrocket
Solution 4 (Vector Dot Product scuffed version)
Using the same method as in Solution 1 reflecting about the line gives us
Let the original point be From point we form the vectors and that extend out from the initial point. If they are perpendicular, we know that their dot product has to equal zero. Therefore, Now, we have to do some guess and check from the multiple choices. Let where is one of the answer choices. Then, By intuition and logical reasoning we deduce that must be so that brings our potential answers down to and If from then which we can quickly rule out since we know thar rotated counterclockwise not clockwise. Hence, is the answer.
~peelybonehead
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
Video Solution (Just 3 min!)
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.