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=108</math>. | ||
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+ | ==Solution 4== | ||
+ | For the above reasons, the transformation is simply a <math>90^\circ</math> degree 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 yields the desired point to be <math>(21, -3) \rightarrow 90+21-3=\boxed{108}</math> | ||
==Video Solution== | ==Video Solution== |
Revision as of 12:42, 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 degree 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 yields the desired point to be
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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