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 (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. | ||
| + | <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 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== | ==Video Solution by TheBeautyofMath== | ||
Latest revision as of 20:21, 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.
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 
?
![[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]](http://latex.artofproblemsolving.com/8/d/9/8d9ddc29373bd395df63dedaac4b8f1eb11201cd.png)
Solution 1
By angle subtraction, we have 
Note that 
is isosceles, so 
Finally, we get 
degrees.
~MRENTHUSIASM ~Aops-g5-gethsemanea2
Solution 2 (Extension)
We can extend to 
, making 
a right angle. It follows that 
is 
, 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]](http://latex.artofproblemsolving.com/9/1/a/91ab5d77b49f589e0ebd38313d10ed34ae5489a0.png)
Since 
, we see that 
. Thus, 
degrees.
~MrThinker
Video Solution (Simple and Quick)
~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
~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. 
