Difference between revisions of "2022 AMC 10B Problems/Problem 20"
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Mathfan2020 (talk | contribs) (Added Solutions 3 and 4) |
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Therefore, <math>\angle BFC = | Therefore, <math>\angle BFC = | ||
− | \boxed{\textbf{(D) 113 | + | \boxed{\textbf{(D)} \ 113}</math>. |
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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So now, because <math>ABCD</math> is a rhombus, <math>FD = AD = CD</math>. This means that there exists a circle from <math>D</math> with radius <math>AD</math> that passes through <math>F</math>, <math>A</math>, and <math>C</math>. | So now, because <math>ABCD</math> is a rhombus, <math>FD = AD = CD</math>. This means that there exists a circle from <math>D</math> with radius <math>AD</math> that passes through <math>F</math>, <math>A</math>, and <math>C</math>. | ||
− | AG is a diameter of this circle because <math>\angle AFG=90^\circ</math>. This means that <math>\angle GFC = \angle GAC = \frac{1}{2} \angle GDC</math>, so <math>\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ</math>, which means that <math>\angle BFC = \boxed{\textbf{(D) 113 | + | AG is a diameter of this circle because <math>\angle AFG=90^\circ</math>. This means that <math>\angle GFC = \angle GAC = \frac{1}{2} \angle GDC</math>, so <math>\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ</math>, which means that <math>\angle BFC = \boxed{\textbf{(D)} \ 113}</math> |
~popop614 | ~popop614 | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>\overline{AC}</math> meet <math>\overline{BD}</math> at <math>O</math>, then <math>AOFB</math> is cyclic and <math>\angle FBO = \angle FAO</math>. Also, <math>AC \cdot BO = [ABCD] = 2 \cdot [ABE] = AF \cdot BE</math>, so <math>\frac{AF}{BO} = \frac{AC}{BE}</math>, thus <math>\triangle AFC \sim \triangle BOE</math> by SAS, and <math>\angle OEB = \angle ACF</math>, then <math>\angle CFE = \angle EOC = \angle DAC = 67^\circ</math>, and <math>\angle BFC = \boxed{\textbf{(D)} \ 113}</math> | ||
+ | |||
+ | ~mathfan2020 | ||
+ | |||
+ | ==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>. | ||
+ | Substituting, <math>\angle BFC = 90 + \frac{46}{2} = \boxed{\textbf{(D)} \ 113}</math> | ||
+ | |||
+ | ~mathfan2020 | ||
==Video Solution== | ==Video Solution== |
Revision as of 22:13, 18 November 2022
Contents
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)
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.