|
|
| (4 intermediate revisions by the same user 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 ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
| |
| − | | |
| − | ==See Also==
| |
| − | {{AMC10 box|year=2021 Fall|ab=A|num-b=6|num-a=8}}
| |
| − | {{MAA Notice}}
| |
