Difference between revisions of "2020 AIME II Problems/Problem 4"
Lopkiloinm (talk | contribs) (→Solution 2 (Official MAA)) |
Lopkiloinm (talk | contribs) (→Solution 3) |
||
Line 8: | Line 8: | ||
==Solution 2 (Official MAA)== | ==Solution 2 (Official MAA)== | ||
Because the rotation sends the vertical segment <math>\overline{AB}</math> to the horizontal segment <math>\overline{A'B'}</math>, the angle of rotation is <math>90^\circ</math> degrees clockwise. For any point <math>(x,y)</math> not at the origin, the line segments from <math>(0,0)</math> to <math>(x,y)</math> and from <math>(x,y)</math> to <math>(x-y,y+x)</math> are perpendicular and are the same length. Thus a <math>90^\circ</math> clockwise rotation around the point <math>(x,y)</math> sends the point <math>A(0,0)</math> to the point <math>(x-y,y+x) = A'(24,18)</math>. This has the solution <math>(x,y) = (21,-3)</math>. The requested sum is <math>90+21-3=108</math>. | Because the rotation sends the vertical segment <math>\overline{AB}</math> to the horizontal segment <math>\overline{A'B'}</math>, the angle of rotation is <math>90^\circ</math> degrees clockwise. For any point <math>(x,y)</math> not at the origin, the line segments from <math>(0,0)</math> to <math>(x,y)</math> and from <math>(x,y)</math> to <math>(x-y,y+x)</math> are perpendicular and are the same length. Thus a <math>90^\circ</math> clockwise rotation around the point <math>(x,y)</math> sends the point <math>A(0,0)</math> to the point <math>(x-y,y+x) = A'(24,18)</math>. This has the solution <math>(x,y) = (21,-3)</math>. The requested sum is <math>90+21-3=108</math>. | ||
− | |||
− | |||
− | |||
==Solution 3== | ==Solution 3== |
Revision as of 13:24, 8 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 or (21, -3). The sum 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.