Difference between revisions of "2022 AMC 10B Problems/Problem 20"
MRENTHUSIASM (talk | contribs) m (→Solution 1 (Law of Sines and Law of Cosines)) |
MRENTHUSIASM (talk | contribs) (→Solution 5 (Similarity & Circle Geometry): Made the solution more concise, and I deleted Point H (unnecessary to mention).) |
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~mathfan2020 | ~mathfan2020 | ||
− | ==Solution 5 (Similarity | + | ==Solution 5 (Similarity and Circle Geometry)== |
Let's make a diagram, but extend <math>AD</math> and <math>BE</math> to point <math>G</math>. | Let's make a diagram, but extend <math>AD</math> and <math>BE</math> to point <math>G</math>. | ||
<asy> | <asy> | ||
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label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); | label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); | ||
</asy> | </asy> | ||
− | We know that <math>AB= | + | We know that <math>AB=AD=2</math> and <math>CE=DE=1</math>. |
− | By | + | By AA Similarity, <math>\triangle ABG \sim \triangle DEG</math> with a ratio of <math>2:1</math>. This implies that <math>2AD=AG</math> and <math>AD \cong DG</math>, so <math>AG=2AD=2\cdot2=4</math>. That is, <math>D</math> is the midpoint of <math>AG</math>. |
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− | This | ||
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− | <math>AG=2AD=2 | ||
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Now, let's redraw our previous diagram, but construct 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. | Now, let's redraw our previous diagram, but construct 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. | ||
Line 165: | Line 159: | ||
*/ | */ | ||
size(300); | size(300); | ||
− | pair A, B, C, D, E, F, G | + | pair A, B, C, D, E, F, G; |
D = origin; | D = origin; | ||
A = 6*dir(46); | A = 6*dir(46); | ||
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F = foot(A,B,E); | F = foot(A,B,E); | ||
G = 6*dir(226); | G = 6*dir(226); | ||
− | |||
dot("$A$",A,1.5*NE,linewidth(5)); | dot("$A$",A,1.5*NE,linewidth(5)); | ||
dot("$B$",B,1.5*NE,linewidth(5)); | dot("$B$",B,1.5*NE,linewidth(5)); | ||
Line 181: | Line 174: | ||
dot("$F$",F,1.5*dir(-20),linewidth(5)); | dot("$F$",F,1.5*dir(-20),linewidth(5)); | ||
dot("$G$",G,1.5*SW,linewidth(5)); | dot("$G$",G,1.5*SW,linewidth(5)); | ||
− | |||
markscalefactor=0.04; | markscalefactor=0.04; | ||
draw(rightanglemark(A,F,B),red); | draw(rightanglemark(A,F,B),red); | ||
− | draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G | + | draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); |
label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); | label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); | ||
draw(Circle(D,6),dashed); | draw(Circle(D,6),dashed); | ||
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Notice how <math>F</math> and <math>C</math> are on the circle and that <math>\angle CFE</math> intercepts with <math>\overset{\Large\frown} {CG}</math>. | Notice how <math>F</math> and <math>C</math> are on the circle and that <math>\angle CFE</math> intercepts with <math>\overset{\Large\frown} {CG}</math>. | ||
− | + | Let's call <math>\angle CFE = \theta</math>. | |
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− | <math>\angle | + | Note that <math>\angle CDG</math> also intercepts <math>\overset{\Large\frown} {CG}</math>, So <math>\angle CDG = 2\angle CFE</math>. |
− | <math>\angle BFC= | + | Let <math>\angle CDG = 2\theta</math>. Notice how <math>\angle CDG</math> and <math>\angle ADC</math> are supplementary to each other. We conclude that <cmath>\begin{align*} |
+ | 2\theta &= 180-\angle ADC \ | ||
+ | 2\theta &= 180-46 \ | ||
+ | 2\theta &= 134 \ | ||
+ | \theta &= 67. | ||
+ | \end{align*}</cmath> | ||
+ | Since <math>\angle BFC=180-\theta</math>, we have <math>\angle BFC=180-67=\boxed{\textbf{(D)} \ 113}</math>. | ||
~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits). | ~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits). |
Revision as of 23:18, 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 . 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 and Circle Geometry)
Let's make a diagram, but extend and to point . 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. 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
~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.