Difference between revisions of "2020 AIME II Problems/Problem 4"
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==Solution 3== | ==Solution 3== | ||
− | A <math> | + | <asy> |
+ | /* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
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+ | |||
+ | We first draw a diagram with the correct Cartesian coordinates and a center of rotation <math>P</math>. Note that <math>PC=PC'</math> because <math>P</math> lies on the perpendicular bisector of <math>CC'</math> (it must be equidistant from <math>C</math> and <math>C'</math> by properties of a rotation). | ||
+ | |||
+ | Since <math>AB</math> is vertical while <math>A'B'</math> is horizontal, we have that the angle of rotation must be <math>90^{\circ}</math>, and therefore <math>\angle P = 90^{\circ}</math>. Therefore, <math>CPC'</math> is a 45-45-90 right triangle, and <math>CD=DP</math>. | ||
+ | |||
+ | We calculate <math>D</math> to be <math>(20,1)</math>. Since we translate <math>4</math> right and <math>1</math> up to get from point <math>C</math> to point <math>D</math>, we must translate <math>1</math> right and <math>4</math> down to get to point <math>P</math>. This gives us <math>P(21,-3)</math>. Our answer is then <math>90+21-3=\boxed{108}</math>. ~Lopkiloinm & samrocksnature | ||
==Solution 4== | ==Solution 4== |
Latest revision as of 23:48, 10 April 2021
Contents
Problem
Triangles and lie in the coordinate plane with vertices , , , , , . A rotation of degrees clockwise around the point where , will transform to . Find .
Solution
After sketching, it is clear a rotation is done about . Looking between and , and . Solving gives . Thus . ~mn28407
Solution 2 (Official MAA)
Because the rotation sends the vertical segment to the horizontal segment , the angle of rotation is degrees clockwise. For any point not at the origin, the line segments from to and from to are perpendicular and are the same length. Thus a clockwise rotation around the point sends the point to the point . This has the solution . The requested sum is .
Solution 3
We first draw a diagram with the correct Cartesian coordinates and a center of rotation . Note that because lies on the perpendicular bisector of (it must be equidistant from and by properties of a rotation).
Since is vertical while is horizontal, we have that the angle of rotation must be , and therefore . Therefore, is a 45-45-90 right triangle, and .
We calculate to be . Since we translate right and up to get from point to point , we must translate right and down to get to point . This gives us . Our answer is then . ~Lopkiloinm & samrocksnature
Solution 4
For the above reasons, the transformation is simply a rotation. Proceed with complex numbers on the points and . Let be the origin. Thus, and . The transformation from to is a multiplication of , which yields . Equating the real and complex terms results in the equations and . Solving,
~beastgert
Solution 5
We know that the rotation point has to be equidistant from both and so it has to lie on the line that is on the midpoint of the segment and also the line has to be perpendicular to . Solving, we get the line is . Doing the same for and , we get that . Since the point of rotation must lie on both of these lines, we set them equal, solve and get: ,. We can also easily see that the degree of rotation is since is initially vertical, and now it is horizontal. Also, we can just sketch this on a coordinate plane and easily realize the same. Hence, the answer is
Video Solution
https://www.youtube.com/watch?v=iJkNkSAmqhg
~North America Math Contest Go Go Go
Video Solution
~IceMatrix
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.