Difference between revisions of "2021 Fall AMC 12A Problems/Problem 6"
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<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== | + | ==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]] | ||
+ | |||
+ | ==Solution 2== | ||
+ | We can extend <math>\overline{AD}</math> to <math>G</math>, making <math>\angle CDG</math> a right angle, and <math>\angle GDE</math> is <math>110^\circ - 90^\circ = 20^\circ</math>. | ||
+ | |||
+ | <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 <math>\angle EFD = \angle DEF</math>, we see that <math>\angle EFD = \angle DEF = \frac{20}{2} = 10^\circ</math>. Thus, <math>\angle AFE = 180^\circ - 10^\circ = \boxed{\textbf{(D)} ~170}</math> | ||
+ | |||
+ | ~MrThinker | ||
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== |
Revision as of 15:43, 25 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.
Contents
Problem
As shown in the figure below, point lies on the opposite half-plane determined by line
from point
so that
. Point
lies on
so that
, and
is a square. What is the degree measure of
?
Solution 1
By angle subtraction, we have Note that
is isosceles, so
Finally, we get
degrees.
~MRENTHUSIASM ~Aops-g5-gethsemanea2
Solution 2
We can extend to
, making
a right angle, and
is
.
Since , we see that
. Thus,
~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
~savannahsolver
Video Solution by HS Competition Academy
~Charles3829
Video Solution
~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.