2023 AMC 10A Problems/Problem 19
Contents
Problem
The line segment formed by and is rotated to the line segment formed by and about the point . What is ?
Solution 1
Due to rotations preserving an equal distance, we can bash the answer with the distance formula. , and . Thus we will square our equations to yield: , and . Canceling from the second equation makes it clear that equals .
Substituting will yield
.
Now .
-Antifreeze5420
Solution 2
Due to rotations preserving distance, we have that , as well as . From here, we can see that P must be on the perpendicular bisector of due to the property of perpendicular bisectors keeping the distance to two points constant.
From here, we proceed to find the perpendicular bisector of . We can see that this is just a horizontal line segment with midpoint at . This means that the equation of the perpendicular bisector is .
Similarly, we find the perpendicular bisector of . We find the slope to be , so our new slope will be . The midpoint of and is , which we can use with our slope to get another equation of .
Now, point P has to lie on both of these perpendicular bisectors, meaning that it has to satisfy both equations. Plugging in the value of we found earlier, we find that . This means that .
-DEVSAXENA
Solution 3 (Coordinate Geometry)
To find the center of rotation, we find the intersection point of the perpendicular bisectors of and .
We can find that the equation of the line is , and that the equation of the line is .
When we solve for the perpendicular bisector of , we determine that it has a slope of 2, and it runs through . Plugging in , we get than . Therefore our perpendicular bisector is . Next, we solve for the perpendicular of . We know that it has an undefined slope, and it runs through . We can determine that our second perpendicular bisector is .
Setting the equations equal to each other, we get . Solving for x, we get that . Therefore, .
~aydenlee & wuwang2002
Solution 4
We use the complex numbers approach to solve this problem. Denote by the angle that rotates about in the counterclockwise direction.
Thus, and .
Taking ratio of these two equations, we get
By solving this equation, we get . Therefore, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Power Solve (easy to understand!)
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=fIzCR4x4x-M
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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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