Difference between revisions of "2020 AIME II Problems/Problem 4"
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==Solution 3== | ==Solution 3== | ||
− | A <math>90^\circ</math> degree rotation is obvious. Let's look at <math>C</math> and <math>C'</math>. They are very close to each other. Let's join <math>C</math> and <math>C'</math> with a line. Then construct a perpendicular bisector to <math>\overline{CC'}</math> with the midpoint being <math>M</math> which is at <math>(20, 1)</math>. We also draw a point <math>N</math> on the perpendicular bisector such that <math>\angle CNC'</math> is <math>90^\circ</math>. That point <math>N</math> is the same distance to <math>M</math> as <math>M</math> is to <math>C</math> but it is on a line perpendicular to <math>\overline{CM}</math> Therefore <math>N</math> is at <math>(20+1, 1-4)</math>. The sum is <math>90+20+1+1-4=108</math>. | + | A <math>90^\circ</math> degree rotation is obvious. Let's look at <math>C</math> and <math>C'</math>. They are very close to each other. Let's join <math>C</math> and <math>C'</math> with a line. Then construct a perpendicular bisector to <math>\overline{CC'}</math> with the midpoint being <math>M</math> which is at <math>(20, 1)</math>. We also draw a point <math>N</math> on the perpendicular bisector such that <math>\angle CNC'</math> is <math>90^\circ</math>. That point <math>N</math> is the same distance to <math>M</math> as <math>M</math> is to <math>C</math> but it is on a line perpendicular to <math>\overline{CM}</math> Therefore <math>N</math> is at <math>(20+1, 1-4)</math>. The sum is <math>90+20+1+1-4=\boxed{108}</math>. ~Lopkiloinm |
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==Solution 4== | ==Solution 4== |
Revision as of 13:16, 13 July 2020
Contents
[hide]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
A degree rotation is obvious. Let's look at and . They are very close to each other. Let's join and with a line. Then construct a perpendicular bisector to with the midpoint being which is at . We also draw a point on the perpendicular bisector such that is . That point is the same distance to as is to but it is on a line perpendicular to Therefore is at . The sum is . ~Lopkiloinm
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
~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 |
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