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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 6"
(→‎Solution 2: Removed repetitive solution. Prof. Chen agreed to this through PM ...)
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~MRENTHUSIASM ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
 
~MRENTHUSIASM ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
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== Solution 2 ==
 
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Because <math>ABCD</math> is square, <math>\angle ADC = 90^\circ</math>.
 
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Hence, <math>\angle ADE = 360^\circ - \angle ADC - \angle EDC = 160^\circ</math>.
 
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Because <math>DF = DE</math>, <math>\angle EFD = \angle FED</math>.
 
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Hence, <math>\angle EFD = \frac{180^\circ - \angle ADE}{2} = 10^\circ</math>.
 
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Therefore, <math>\angle AFE = 180^\circ - \angle EFD = 170^\circ</math>.
 
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Therefore, the answer is <math>\boxed{\textbf{(D) }170^\circ}</math>.
 
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~Steven Chen (www.professorchenedu.com)
 
  
 
==See Also==
 
==See Also==

Revision as of 00:48, 26 November 2021

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] usepackage("mathptmx"); 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

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

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