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Difference between revisions of "2014 AMC 8 Problems/Problem 15"
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== Problem ==
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The circumference of the circle with center <math>O</math> is divided into <math>12</math> equal arcs, marked the letters <math>A</math> through <math>L</math> as seen below. What is the number of degrees in the sum of the angles <math>x</math> and <math>y</math>?
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<asy>
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size(230);
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defaultpen(linewidth(0.65));
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pair O=origin;
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pair[] circum = new pair[12];
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string[] let = {"$A$","$B$","$C$","$D$","$E$","$F$","$G$","$H$","$I$","$J$","$K$","$L$"};
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draw(unitcircle);
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for(int i=0;i<=11;i=i+1)
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{
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circum[i]=dir(120-30*i);
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dot(circum[i],linewidth(2.5));
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label(let[i],circum[i],2*dir(circum[i]));
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}
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draw(O--circum[4]--circum[0]--circum[6]--circum[8]--cycle);
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label("$x$",circum[0],2.75*(dir(circum[0]--circum[4])+dir(circum[0]--circum[6])));
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label("$y$",circum[6],1.75*(dir(circum[6]--circum[0])+dir(circum[6]--circum[8])));
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label("$O$",O,dir(60));
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</asy>
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<math> \textbf{(A) }75\qquad\textbf{(B) }80\qquad\textbf{(C) }90\qquad\textbf{(D) }120\qquad\textbf{(E) }150 </math>
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== Solution 1 ==
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The measure of an inscribed angle is half the measure of its corresponding central angle. Since each unit arc is <math>\frac{1}{12}</math> of the circle's circumference, each unit central angle measures <math>\frac{360}{12}^{\circ}=30^{\circ}</math>. From this, <math>\angle EOG = 60^{\circ}</math>, so <math>x = 30^{\circ}</math>. Also, <math>\angle AOI = 120^{\circ}</math>, so <math>y = 60^{\circ}</math>. The number of degrees in the sum of both angles is <math>30 + 60 = \boxed{(C)\ 90}.</math>
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== Solution 2 ==
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Since <math>\triangle AOE</math> is isosceles and <math>\angle AOE = \frac{4}{12} \cdot 360^{\circ} = 120^{\circ}</math>, <math>x = 30^{\circ}</math>. Since <math>\triangle GOI</math> is isosceles and <math>\angle GOI = \frac{2}{12} \cdot 360^{\circ} = 60^{\circ}</math>, <math>x = 60^{\circ}</math>. The number of degrees in the sum of both angles is <math>30+60 = \boxed{(C)\ 90}</math>.
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== Video Solution 1 ==
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https://youtu.be/3QHH9xV-QDw
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~Education, the Study of Everything
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== Video Solution 2 ==
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https://www.youtube.com/watch?v=qseG63LK4AU
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~David
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== Video Solution 3 ==
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https://youtu.be/aZhjhb3mMfg
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~savannahsolver
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== Video Solution 4 ==
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https://youtu.be/abSgjn4Qs34?t=3242
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== See Also ==
  
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==See Also==
 
 
{{AMC8 box|year=2014|num-b=14|num-a=16}}
 
{{AMC8 box|year=2014|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:40, 30 January 2025

Problem

The circumference of the circle with center $O$ is divided into $12$ equal arcs, marked the letters $A$ through $L$ as seen below. What is the number of degrees in the sum of the angles $x$ and $y$?

[asy] size(230); defaultpen(linewidth(0.65)); pair O=origin; pair[] circum = new pair[12]; string[] let = {"$A$","$B$","$C$","$D$","$E$","$F$","$G$","$H$","$I$","$J$","$K$","$L$"}; draw(unitcircle); for(int i=0;i<=11;i=i+1) { circum[i]=dir(120-30*i); dot(circum[i],linewidth(2.5)); label(let[i],circum[i],2*dir(circum[i])); } draw(O--circum[4]--circum[0]--circum[6]--circum[8]--cycle); label("$x$",circum[0],2.75*(dir(circum[0]--circum[4])+dir(circum[0]--circum[6]))); label("$y$",circum[6],1.75*(dir(circum[6]--circum[0])+dir(circum[6]--circum[8]))); label("$O$",O,dir(60)); [/asy]

$\textbf{(A) }75\qquad\textbf{(B) }80\qquad\textbf{(C) }90\qquad\textbf{(D) }120\qquad\textbf{(E) }150$

Solution 1

The measure of an inscribed angle is half the measure of its corresponding central angle. Since each unit arc is $\frac{1}{12}$ of the circle's circumference, each unit central angle measures $\frac{360}{12}^{\circ}=30^{\circ}$. From this, $\angle EOG = 60^{\circ}$, so $x = 30^{\circ}$. Also, $\angle AOI = 120^{\circ}$, so $y = 60^{\circ}$. The number of degrees in the sum of both angles is $30 + 60 = \boxed{(C)\ 90}.$

Solution 2

Since $\triangle AOE$ is isosceles and $\angle AOE = \frac{4}{12} \cdot 360^{\circ} = 120^{\circ}$, $x = 30^{\circ}$. Since $\triangle GOI$ is isosceles and $\angle GOI = \frac{2}{12} \cdot 360^{\circ} = 60^{\circ}$, $x = 60^{\circ}$. The number of degrees in the sum of both angles is $30+60 = \boxed{(C)\ 90}$.

Video Solution 1

https://youtu.be/3QHH9xV-QDw

~Education, the Study of Everything

Video Solution 2

https://www.youtube.com/watch?v=qseG63LK4AU ~David

Video Solution 3

https://youtu.be/aZhjhb3mMfg ~savannahsolver

Video Solution 4

https://youtu.be/abSgjn4Qs34?t=3242

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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