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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 6"

(Created page with "==Problem== As shown in the figure below, point <math>E</math> lies on the opposite half-plane determined by line <math>CD</math> from point <math>A</math> so that <math>\ang...")
 
(→‎Solution 2)
 
(33 intermediate revisions by 10 users not shown)
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{{duplicate|[[2021 Fall AMC 10A Problems/Problem 7|2021 Fall AMC 10A #7]] and [[2021 Fall AMC 12A Problems/Problem 6|2021 Fall AMC 12A #6]]}}
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==Problem==
 
==Problem==
  
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<asy>
 
<asy>
−
usepackage("mathptmx");
 
 
size(6cm);
 
size(6cm);
 
pair A = (0,10);
 
pair A = (0,10);
Line 27: Line 28:
 
<math>\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174</math>
 
<math>\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174</math>
  
−
==Solution==
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==Solution 1==
 
By angle subtraction, we have <math>\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.</math> Note that <math>\triangle DEF</math> is isosceles, so <math>\angle DFE = \frac{180^\circ - \angle ADE}{2}=10^\circ.</math> Finally, we get <math>\angle AFE = 180^\circ - \angle DFE = \boxed{\textbf{(D) }170}</math> degrees.
 
By angle subtraction, we have <math>\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.</math> Note that <math>\triangle DEF</math> is isosceles, so <math>\angle DFE = \frac{180^\circ - \angle ADE}{2}=10^\circ.</math> Finally, we get <math>\angle AFE = 180^\circ - \angle DFE = \boxed{\textbf{(D) }170}</math> degrees.
  
 
~MRENTHUSIASM ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
 
~MRENTHUSIASM ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
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 +
==Solution 2 (Extension)==
 +
We can extend <math>\overline{AD}</math> to <math>G</math>, making <math>\angle CDG</math> a right angle. It follows that <math>\angle GDE</math> is <math>110^\circ - 90^\circ = 20^\circ</math>, as shown below.
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<asy>
 +
size(6cm);
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pair A = (0,10);
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label("$A$", A, N);
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pair B = (0,0);
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label("$B$", B, S);
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pair C = (10,0);
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label("$C$", C, S);
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pair D = (10,10);
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label("$D$", D, SW);
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pair EE = (15,11.8);
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label("$E$", EE, N);
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pair F = (3,10);
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label("$F$", F, N);
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pair G = (15,10);
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label("$G$", G, E);
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filldraw(D--arc(D,2.5,270,380)--cycle,lightgray);
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dot(A^^B^^C^^D^^EE^^F^^G);
 +
draw(A--B--C--D--G--cycle);
 +
draw(D--EE--F--cycle);
 +
</asy>
 +
Since <math>\angle DFE = \angle DEF</math>, we see that <math>\angle DFE = \angle DEF = \frac{20}{2} = 10^\circ</math>. Thus, <math>\angle AFE = 180^\circ - 10^\circ = \boxed{\textbf{(D)} ~170}</math> degrees.
 +
 +
~MrThinker
 +
 +
==Video Solution (Simple and Quick)==
 +
https://youtu.be/cBLyn2HZ5YY
 +
 +
~Education, the Study of Everything
 +
 +
 +
==Video Solution by A+ Whiz==
 +
 +
https://www.youtube.com/watch?v=x4cF3o3Fzj8&t=260s
 +
 +
==Video Solution by TheBeautyofMath==
 +
for AMC 10: https://youtu.be/ycRZHCOKTVk?t=232
 +
 +
for AMC 12: https://youtu.be/wlDlByKI7A8
 +
 +
~IceMatrix
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/9nUZhyhi9_o
 +
 +
~savannahsolver
 +
 +
==Video Solution by HS Competition Academy==
 +
https://youtu.be/l3nnd-eWOI0
 +
 +
~Charles3829
 +
 +
==Video Solution==
 +
https://youtu.be/T4NhPER6SrM
 +
 +
~Lucas
  
 
==See Also==
 
==See Also==

Latest revision as of 21:21, 12 July 2023

The following problem is from both the 2021 Fall AMC 10A #7 and 2021 Fall AMC 12A #6, so both problems redirect to this page.

Problem

As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$. Point $F$ lies on $\overline{AD}$ so that $DE=DF$, and $ABCD$ is a square. What is the degree measure of $\angle AFE$?

[asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy]

$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$

Solution 1

By angle subtraction, we have $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.$ Note that $\triangle DEF$ is isosceles, so $\angle DFE = \frac{180^\circ - \angle ADE}{2}=10^\circ.$ Finally, we get $\angle AFE = 180^\circ - \angle DFE = \boxed{\textbf{(D) }170}$ degrees.

~MRENTHUSIASM ~Aops-g5-gethsemanea2

Solution 2 (Extension)

We can extend $\overline{AD}$ to $G$, making $\angle CDG$ a right angle. It follows that $\angle GDE$ is $110^\circ - 90^\circ = 20^\circ$, as shown below. [asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); pair G = (15,10); label("$G$", G, E); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F^^G); draw(A--B--C--D--G--cycle); draw(D--EE--F--cycle); [/asy] Since $\angle DFE = \angle DEF$, we see that $\angle DFE = \angle DEF = \frac{20}{2} = 10^\circ$. Thus, $\angle AFE = 180^\circ - 10^\circ = \boxed{\textbf{(D)} ~170}$ degrees.

~MrThinker

Video Solution (Simple and Quick)

https://youtu.be/cBLyn2HZ5YY

~Education, the Study of Everything


Video Solution by A+ Whiz

https://www.youtube.com/watch?v=x4cF3o3Fzj8&t=260s

Video Solution by TheBeautyofMath

for AMC 10: https://youtu.be/ycRZHCOKTVk?t=232

for AMC 12: https://youtu.be/wlDlByKI7A8

~IceMatrix

Video Solution by WhyMath

https://youtu.be/9nUZhyhi9_o

~savannahsolver

Video Solution by HS Competition Academy

https://youtu.be/l3nnd-eWOI0

~Charles3829

Video Solution

https://youtu.be/T4NhPER6SrM

~Lucas

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 12 Problems and Solutions
2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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. AMC logo.png

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