Difference between revisions of "2022 AMC 10B Problems/Problem 20"
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Let <math>ABCD</math> be a rhombus with <math>\angle ADC = 46^\circ</math>. Let <math>E</math> be the midpoint of <math>\overline{CD}</math>, and let <math>F</math> be the point | Let <math>ABCD</math> be a rhombus with <math>\angle ADC = 46^\circ</math>. Let <math>E</math> be the midpoint of <math>\overline{CD}</math>, and let <math>F</math> be the point | ||
on <math>\overline{BE}</math> such that <math>\overline{AF}</math> is perpendicular to <math>\overline{BE}</math>. What is the degree measure of <math>\angle BFC</math>? | on <math>\overline{BE}</math> such that <math>\overline{AF}</math> is perpendicular to <math>\overline{BE}</math>. What is the degree measure of <math>\angle BFC</math>? | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(250); | ||
+ | pair A, B, C, D, E, F; | ||
+ | D = origin; | ||
+ | A = 6*dir(46); | ||
+ | C = (6,0); | ||
+ | B = C + (A-D); | ||
+ | E = midpoint(C--D); | ||
+ | F = foot(A,B,E); | ||
+ | dot("$A$",A,1.5*NW,linewidth(5)); | ||
+ | dot("$B$",B,1.5*NE,linewidth(5)); | ||
+ | dot("$C$",C,1.5*SE,linewidth(5)); | ||
+ | dot("$D$",D,1.5*SW,linewidth(5)); | ||
+ | dot("$E$",E,1.5*S,linewidth(5)); | ||
+ | dot("$F$",F,1.5*dir(-20),linewidth(5)); | ||
+ | markscalefactor=0.04; | ||
+ | draw(rightanglemark(A,F,B),red); | ||
+ | draw(A--B--C--D--cycle^^A--F--C^^B--E); | ||
+ | label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); | ||
+ | </asy> | ||
+ | ~MRENTHUSIASM | ||
==Solution 1 (Law of Sines and Law of Cosines)== | ==Solution 1 (Law of Sines and Law of Cosines)== |
Revision as of 08:21, 28 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 ?
Diagram
~MRENTHUSIASM
Solution 1 (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 construct 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 .
Lets call .
also intercepts , but it's vertical angle (), also intercepts an arch congruent to . So .
.
Notice how and are supplementary to each other.
This concludes that, .
.
Realize how
Which means the answer is .
~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits).
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.