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

Revision as of 14:47, 9 August 2022

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

If we extend $\overline{AD}$ to a new point G, so that $\angle CDG = 90^\circ$ . We find $\angle GDE = 110-90= 20^\circ$ .

Since $\triangle DEF$ is isosceles, $\angle DEF = \angle DFE = 20 \div 2 = 10^\circ$ . Hence, $\angle DFE = 180-10= \boxed{\textbf{(D) }170}$ degrees.

~MrThinker

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

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.

@@ Line 36: / Line 36: @@
 If we extend <math>\overline{AD}</math> to a new point G, so that <math>\angle CDG = 90^\circ</math>. We find <math>\angle GDE = 110-90= 20^\circ</math>.
-Since <math>\triangle FDE</math> is isosceles, <math>\angle DEF = \angle DFE = 20 \div 2 = 10^\circ</math>. Hence, <math>\angle DFE = 180-10= \boxed{\textbf{(D) }170}</math> degrees.
+Since <math>\triangle DEF</math> is isosceles, <math>\angle DEF = \angle DFE = 20 \div 2 = 10^\circ</math>. Hence, <math>\angle DFE = 180-10= \boxed{\textbf{(D) }170}</math> degrees.
 ~MrThinker

Page

Toolbox

Search