2022 AMC 10B Problems/Problem 20
Contents
- 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 Solution 6 (Simplification/Reduction)
- 9 Video Solution (⚡️Just 1 min!⚡️)
- 10 Video Solution
- 11 Video Solution
- 12 Video Solution by OmegaLearn Using Clever Similar Triangles and Angle Chasing
- 13 Video Solution, best solution (not family friendly, no circles drawn)
- 14 Video Solution, by Challenge 25
- 15 Video Solution by Interstigation
- 16 Video Solution (Cool Solution)
- 17 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).
Solution 6 (Simplification/Reduction)
If angle was a right angle, it would be much easier. Thus, first pretend that is a right angle. is now a square. WLOG, let each of the side lengths be 1. We can use the Pythagorean Theorem to find the length of line , which is . We want the measure of angle , so to work closer to it, we should try finding the length of line . Angle and angle are complementary. Angle and angle are also complementary. Thus, . . Since ,and , . It follows now that .
Now, zoom in on triangle . To use the Law of Cosines on triangle , we need the length of . Use the Law of Cosines on triangle . Cos . Thus, after using the Law of Cosines, .
Since we now have SSS on , we can get use the Law of Cosines. . is 45, but if the cosine is negative that means that the angle is the supplement of the positive cosine value. . Angle is .
Realize that, around point F, there will always be 3 right angles, regardless of what angle is. There are only two angles that change when changes. Break up angle into angle , which is always 90 degrees, and angle , which we have discovered to to be half of . Thus, when angle is 46 degrees, then will be 23. . Angle is degrees.
Video Solution (⚡️Just 1 min!⚡️)
~Education, the Study of Everything
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 (not family friendly, no circles drawn)
https://www.youtube.com/watch?v=vwI3I7dxw0Q
Video Solution, by Challenge 25
https://youtu.be/W1jbMaO8BIQ (cyclic quads)
Video Solution by Interstigation
~Interstigation
Video Solution (Cool Solution)
https://www.youtube.com/watch?v=cZcaeU9P25s&ab_channel=Chillin
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 |
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