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 == | ||
| + | If we extend <math>\overline{AD}</math> to a new point G, so that <math>\angle CDG</math> = 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. Hence, <math>\angle DFE = 180-10= \boxed{\textbf{(D) }170}</math> degrees. | ||
| + | |||
| + | ~MrThinker | ||
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
Revision as of 14:45, 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.
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
If we extend to a new point G, so that 
= 90^\circ
\angle GDE = 110-90= 20^\circ$.
Since$ (Error compiling LaTeX. Unknown error_msg)\triangle FDE
\angle DEF = \angle DFE = 20 \div 2 = 10^\circ. Hence, 
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
~savannahsolver
Video Solution by HS Competition Academy
~Charles3829
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. 
