Difference between revisions of "2022 AMC 10B Problems/Problem 20"
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==Solution 4== | ==Solution 4== | ||
Observe that all answer choices are close to <math>112.5 = 90+\frac{45}{2}</math>. A quick solve shows that having <math>\angle D = 90^\circ</math> yields <math>\angle BFC = 135^\circ = 90 + \frac{90}{2}</math>, meaning that <math>\angle BFC</math> increases with <math>\angle D</math>. | Observe that all answer choices are close to <math>112.5 = 90+\frac{45}{2}</math>. A quick solve shows that having <math>\angle D = 90^\circ</math> yields <math>\angle BFC = 135^\circ = 90 + \frac{90}{2}</math>, meaning that <math>\angle BFC</math> increases with <math>\angle D</math>. | ||
− | Substituting, <math>\angle BFC = 90 + \frac{46}{2} = \boxed{\textbf{(D)} \ 113}</math> | + | Substituting, <math>\angle BFC = 90 + \frac{46}{2} = \boxed{\textbf{(D)} \ 113}</math>. |
~mathfan2020 | ~mathfan2020 |
Revision as of 22:53, 11 February 2023
Contents
[hide]- 1 Problem
- 2 Diagram
- 3 Solution 1 (Law of Sines and Law of Cosines)
- 4 Solution 2
- 5 Solution 3
- 6 Solution 4
- 7 Solution 5 (Similarity and Circle Geometry)
- 8 Video Solution
- 9 Video Solution
- 10 Video Solution by OmegaLearn Using Clever Similar Triangles and Angle Chasing
- 11 Video Solution, best solution (family friendly, no circles drawn)
- 12 Video Solution, by Challenge 25
- 13 See Also
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 ?
Diagram
~MRENTHUSIASM
Solution 1 (Law of Sines and Law of Cosines)
Without loss of generality, we assume the length of each side of is . 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
A little bit faster: is cyclic .
.
Therefore is cyclic.
Hence .
~asops
Solution 4
Observe that all answer choices are close to . A quick solve shows that having yields , meaning that increases with . Substituting, .
~mathfan2020
Solution 5 (Similarity and Circle Geometry)
This solution refers to the Diagram section.
We extend and to point , as shown below: We know that and .
By AA Similarity, with a ratio of . This implies that and , so . That is, is the midpoint of .
Now, let's redraw our previous diagram, but construct a circle with radius or centered at and by extending to point , which is on the circle, as shown below: Notice how and are on the circle and that intercepts with .
Let's call .
Note that also intercepts , So .
Let . Notice how and are supplementary to each other. We conclude that Since , we have .
~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits).
Video Solution
https://www.youtube.com/watch?v=HWJe96s_ugs&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=6
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
Video Solution, best solution (family friendly, no circles drawn)
https://www.youtube.com/watch?v=vwI3I7dxw0Q
Video Solution, by Challenge 25
https://youtu.be/W1jbMaO8BIQ (cyclic quads)
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.