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| − | ==Problem==
| + | #REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_6]] |
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| − | 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>?
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| − | <asy>
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| − | usepackage("mathptmx");
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| − | size(6cm);
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| − | pair A = (0,10);
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| − | label("$A$", A, N);
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| − | pair B = (0,0);
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| − | label("$B$", B, S);
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| − | pair C = (10,0);
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| − | label("$C$", C, S);
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| − | pair D = (10,10);
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| − | label("$D$", D, SW);
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| − | pair EE = (15,11.8);
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| − | label("$E$", EE, N);
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| − | pair F = (3,10);
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| − | label("$F$", F, N);
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| − | filldraw(D--arc(D,2.5,270,380)--cycle,lightgray);
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| − | dot(A^^B^^C^^D^^EE^^F);
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| − | draw(A--B--C--D--cycle);
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| − | draw(D--EE--F--cycle);
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| − | label("$110^\circ$", (15,9), SW);
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| − | </asy>
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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>
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| − | ==Solution==
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| − | By angle subtraction, we have <math>\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.</math>
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| − | Note that <math>\triangle DEF</math> is an isosceles triangle, 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.
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| − | ~MRENTHUSIASM
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| − | ==See Also==
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| − | {{AMC10 box|year=2021 Fall|ab=A|num-b=6|num-a=8}}
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| − | {{MAA Notice}}
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