Difference between revisions of "2021 Fall AMC 10A Problems/Problem 7"

(Tag: New redirect)
 
(10 intermediate revisions by 2 users not shown)
Line 1: Line 1:
==Problem==
+
#REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_6]]
 
 
As shown in the figure below, point <math>E</math> lies on the opposite half-plane determined by line <math>CD</math> from point <math>A</math> so that <math>\angle CDE = 110^\circ</math>. Point <math>F</math> lies on <math>\overline{AD}</math> so that <math>DE=DF</math>, and <math>ABCD</math> is a square. What is the degree measure of <math>\angle AFE</math>?
 
 
 
<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>
 
 
 
<math>\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174</math>
 
 
 
==Solution==
 
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 EFD = \frac{180^\circ - \angle ADE}{2}=10^\circ.</math> Finally, we get <math>\angle AFE = 180^\circ - \angle EFD = \boxed{\textbf{(D) }170}</math> degrees.
 
 
 
~MRENTHUSIASM
 
 
 
==Solution 2 (same as Solution 1 but by another user)==
 
Since <math>\angle CDE=110^\circ</math> and <math>\angle ADC=90^\circ,</math> we know that <math>\angle FDE=360^\circ-110^\circ-90^\circ=160^\circ.</math> From the given, <math>DE=DF,</math> so that means <math>\triangle DEF</math> is isosceles. So, <math>\angle EFD=\left(\frac{180-160}{2}\right)^\circ=10^\circ.</math> Hence, <math>\angle AFE=180^\circ-10^\circ=170^\circ</math> since it is supplementary to <math>\angle EFD.</math> This gives us answer choice <math>\boxed{\textbf{(D)}}.</math>
 
 
 
[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 18:56, 22 November 2021 (EST)
 
This solution conflicted with the above solution due to two people editing at the same time.
 
 
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=6|num-a=8}}
 
{{MAA Notice}}
 

Latest revision as of 19:03, 23 November 2021