Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
(→‎Solution 2: Moved this solution to Discussion Page for now, as it is very similar to Solution 1. I wish to keep the page clean. Let me know if you disagree ...)
Line 35: Line 35:
  
 
~MRENTHUSIASM ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
 
~MRENTHUSIASM ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
−
 
−
==Solution 2==
 
−
Note that <math>\angle ADC = 90</math>, meaning that the reflex of <math>\angle ADE = 90+110=200^\circ</math>, so <math>\angle ADE = 360-200=160^\circ</math>. It is given that <math>\triangle DEF</math> has two sides of equal length, so it is isosceles, thus having two congruent angles.
 
−
 
−
The sum of these two angles is <math>180-160=20^\circ</math>, so the measure of both <math>\angle DFE</math> and angle <math>\angle FED</math> is <math>10^\circ</math>. Since <math>\angle AFE</math> is the supplement to <math>\angle DFE</math>, and <math>\angle DFE = 10^\circ</math>, <math>\angle AFE = 180-10 = \boxed{\textbf{(D)}170}</math> degrees.
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=6|num-a=8}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:44, 22 November 2021

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] 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]

$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$

Solution

By angle subtraction, we have $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.$

Note that $\triangle DEF$ is isosceles, so $\angle EFD = \frac{180^\circ - \angle ADE}{2}=10^\circ.$

Finally, we get $\angle AFE = 180^\circ - \angle EFD = \boxed{\textbf{(D) }170}$ degrees.

~MRENTHUSIASM ~Aops-g5-gethsemanea2

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_7&oldid=165238"