Difference between revisions of "2020 AIME II Problems/Problem 4"
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− | For the above reasons, the transformation is simply a <math>90^\circ</math> rotation. Proceed with complex numbers on the points <math>C</math> and <math>C'</math>. Let <math>(x, y)</math> be the origin. Thus, <math>C \rightarrow (16-x)+(-y)i</math> and <math>C' \rightarrow (24-x)+(2-y)i</math>. The transformation from <math>C'</math> to <math>C</math> is a multiplication of <math>i</math>, which yields <math>(16-x)+(-y)i=(y-2)+(24-x)i</math>. Equating the real and complex terms results in the equations <math>16-x=y-2</math> and <math>-y=24-x</math>. Solving | + | For the above reasons, the transformation is simply a <math>90^\circ</math> rotation. Proceed with complex numbers on the points <math>C</math> and <math>C'</math>. Let <math>(x, y)</math> be the origin. Thus, <math>C \rightarrow (16-x)+(-y)i</math> and <math>C' \rightarrow (24-x)+(2-y)i</math>. The transformation from <math>C'</math> to <math>C</math> is a multiplication of <math>i</math>, which yields <math>(16-x)+(-y)i=(y-2)+(24-x)i</math>. Equating the real and complex terms results in the equations <math>16-x=y-2</math> and <math>-y=24-x</math>. Solving, <math>(x, y) : (21, -3) \rightarrow 90+21-3=\boxed{108}</math> |
~beastgert | ~beastgert |
Revision as of 14:08, 18 June 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 .
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
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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