Difference between revisions of "2022 AMC 10B Problems/Problem 20"
Mathfan2020 (talk | contribs) (Added Solutions 3 and 4) |
Pi is 3.14 (talk | contribs) |
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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+ | == Video Solution by OmegaLearn Using Clever Similar Triangles and Angle Chasing == | ||
+ | https://youtu.be/lEmCprb20n4 | ||
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+ | ~ pi_is_3.14 | ||
Revision as of 14:43, 19 November 2022
Contents
[hide]Problem
Let be a rhombus with . Let be the midpoint of , and let be the point on such that is perpendicular to . What is the degree measure of ?
Solution (Law of Sines and Law of Cosines)
Without loss of generality, we assume the length of each side of is 2. Because is the midpoint of , .
Because is a rhombus, .
In , following from the law of sines,
We have .
Hence,
By solving this equation, we get .
Because ,
In , following from the law of sines,
Because , the equation above can be converted as
Therefore,
Therefore, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
Extend segments and until they meet at point .
Because , we have and , so by AA.
Because is a rhombus, , so , meaning that is a midpoint of segment .
Now, , so is right and median .
So now, because is a rhombus, . This means that there exists a circle from with radius that passes through , , and .
AG is a diameter of this circle because . This means that , so , which means that
~popop614
Solution 3
Let meet at , then is cyclic and . Also, , so , thus by SAS, and , then , and
~mathfan2020
Solution 4
Observe that all answer choices are close to . A quick solve shows that having yields , meaning that increases with . Substituting,
~mathfan2020
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Clever Similar Triangles and Angle Chasing
~ pi_is_3.14
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.