Difference between revisions of "2022 AMC 10B Problems/Problem 20"
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~mathfan2020 | ~mathfan2020 | ||
+ | =Solution 5 (Similarity & Circle Geometry)= | ||
+ | Let's make a diagram, but extend <math>AD</math> and <math>BE</math> to point <math>G</math>. | ||
+ | |||
+ | <asy> | ||
+ | pair A = (0,0); | ||
+ | label("$D$", A, NW); | ||
+ | pair B = (2.25,3); | ||
+ | label("$A$", B, NW); | ||
+ | pair C = (6,3); | ||
+ | label("$B$", C, NE); | ||
+ | pair D = (3.75,0); | ||
+ | label("$C$", D, SE); | ||
+ | pair E = (1.875,0); | ||
+ | label("$E$", E, S); | ||
+ | draw(C--E); | ||
+ | draw(A--B--C--D--E--cycle); | ||
+ | label("$F$", (3.3,1), S); | ||
+ | draw(B--(3.5,1.2)); | ||
+ | draw(rightanglemark(B,(3.5,1.2),E)); | ||
+ | draw((3.5,1.2)--D); | ||
+ | label("$G$", (-2.25, -3), SW); | ||
+ | draw(A--(-2.25, -3)); | ||
+ | draw(E--(-2.25, -3)); | ||
+ | </asy> | ||
+ | |||
+ | We know that <math>AB=2, AD=2, DE=1</math>, and <math>CE=1</math>. | ||
+ | |||
+ | By SAS Similarity, <math>\triangle ABG \sim \triangle DEG</math> with a ratio of <math>2:1</math>. | ||
+ | |||
+ | This means that, <math>2AD=AG</math> and <math>AD \cong DG</math>. | ||
+ | |||
+ | <math>AG=2AD=2(2)=4</math>. | ||
+ | |||
+ | This also can prove that <math>D</math> is the midpoint of <math>AG</math>. | ||
+ | |||
+ | Now, let's redraw our previous diagram, but add a circle with radius <math>AD</math> or <math>2</math> centered at <math>D</math> and by extending <math>CD</math> to point <math>H</math>, which is on the circle. | ||
+ | |||
+ | <asy> | ||
+ | pair A = (0,0); | ||
+ | label("$D$", A, NW); | ||
+ | pair B = (2.25,3); | ||
+ | label("$A$", B, NW); | ||
+ | pair C = (6,3); | ||
+ | label("$B$", C, NE); | ||
+ | pair D = (3.75,0); | ||
+ | label("$C$", D, SE); | ||
+ | pair E = (1.875,0); | ||
+ | label("$E$", E, S); | ||
+ | draw(C--E); | ||
+ | draw(A--B--C--D--E--cycle); | ||
+ | label("$F$", (3.3,1), S); | ||
+ | draw(B--(3.5,1.2)); | ||
+ | draw(rightanglemark(B,(3.5,1.2),E)); | ||
+ | draw((3.5,1.2)--D); | ||
+ | label("$G$", (-2.25, -3), SW); | ||
+ | draw(A--(-2.25, -3)); | ||
+ | draw(E--(-2.25, -3)); | ||
+ | pair O1 = (0,0); | ||
+ | draw(circle(O1,3.75)); | ||
+ | label("$H$", (-3.75,0), SW); | ||
+ | draw(D--(-3.75,0)); | ||
+ | </asy> | ||
+ | |||
+ | Notice how <math>F</math> and <math>C</math> are on the circle and that <math>\angle CFE</math> intercepts with arch <math>CG</math>. | ||
+ | |||
+ | Lets call <math>\angle CFE = \theta</math>. | ||
+ | |||
+ | <math>\angle FDG</math> also intercepts arch <math>CG</math>, but it's vertical angle (<math>\angle ADH</math>), also intercepts an arch. So <math>\angle FDG = 2\angle CFE</math>. | ||
+ | |||
+ | <math>\angle FDG = 2\theta</math>. | ||
+ | |||
+ | Notice how <math>FDG</math> and <math>ADC</math> are supplementary to each other. | ||
+ | |||
+ | This concludes that, <math>2\theta=180-\angle ADC</math>. | ||
+ | |||
+ | <math>2\theta=180-46</math> | ||
+ | |||
+ | <math>2\theta=134</math> | ||
+ | |||
+ | <math>\theta=67°</math>. | ||
+ | |||
+ | Realize how <math>\angle BFC=180-\theta</math> | ||
+ | |||
+ | <math>\angle BFC=180-67</math> | ||
+ | |||
+ | <math>\angle BFC= 113.</math> Which means the answer is <math>\boxed{\textbf{(D)} \ 113}</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 00:52, 20 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
Solution 5 (Similarity & Circle Geometry)
Let's make a diagram, but extend and to point .
We know that , and .
By SAS Similarity, with a ratio of .
This means that, and .
.
This also can prove that is the midpoint of .
Now, let's redraw our previous diagram, but add a circle with radius or centered at and by extending to point , which is on the circle.
Notice how and are on the circle and that intercepts with arch .
Lets call .
also intercepts arch , but it's vertical angle (), also intercepts an arch. So .
.
Notice how and are supplementary to each other.
This concludes that, .
.
Realize how
Which means the answer is .
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.