Difference between revisions of "2020 AIME II Problems/Problem 4"
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==Problem== | ==Problem== | ||
Triangles <math>\triangle ABC</math> and <math>\triangle A'B'C'</math> lie in the coordinate plane with vertices <math>A(0,0)</math>, <math>B(0,12)</math>, <math>C(16,0)</math>, <math>A'(24,18)</math>, <math>B'(36,18)</math>, <math>C'(24,2)</math>. A rotation of <math>m</math> degrees clockwise around the point <math>(x,y)</math> where <math>0<m<180</math>, will transform <math>\triangle ABC</math> to <math>\triangle A'B'C'</math>. Find <math>m+x+y</math>. | Triangles <math>\triangle ABC</math> and <math>\triangle A'B'C'</math> lie in the coordinate plane with vertices <math>A(0,0)</math>, <math>B(0,12)</math>, <math>C(16,0)</math>, <math>A'(24,18)</math>, <math>B'(36,18)</math>, <math>C'(24,2)</math>. A rotation of <math>m</math> degrees clockwise around the point <math>(x,y)</math> where <math>0<m<180</math>, will transform <math>\triangle ABC</math> to <math>\triangle A'B'C'</math>. Find <math>m+x+y</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | After sketching, it is clear a <math>90^{\circ}</math> rotation is done about <math>(x,y)</math>. Looking between <math>C</math> and <math>C'</math>, <math>x+y-16=2</math> and <math>x-y=24</math>. Solving gives <math>(x,y)\implies(21,-3)</math>. Thus <math>90+21-3=\boxed{108}</math>. | ||
+ | ~mn28407 | ||
==Video Solution== | ==Video Solution== |
Revision as of 02:14, 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
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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