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Difference between revisions of "2014 AMC 8 Problems/Problem 15"

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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>
 
<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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==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/3QHH9xV-QDw
 +
 +
~Education, the Study of Everything
 +
 +
  
 
==Video Solution==
 
==Video Solution==
−
https://www.youtube.com/watch?v=qseG63LK4AU
+
https://www.youtube.com/watch?v=qseG63LK4AU   ~David
  
 
https://youtu.be/aZhjhb3mMfg ~savannahsolver
 
https://youtu.be/aZhjhb3mMfg ~savannahsolver
 +
 +
==Video Solution ==
 +
https://youtu.be/abSgjn4Qs34?t=3242
  
 
==Solution==
 
==Solution==
−
For this problem, it is useful to know that 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>\left( \frac{360}{12} \right) ^{\circ}=30^{\circ}</math>. Then, we know that the central angle of x = 60, so inscribed angle = 30. Also, central angle of y = 120, so inscirbed angle = 60. Summing both inscribed angles gives 30 + 60 = 90.
+
For this problem, it is useful to know that 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>\left( \frac{360}{12} \right) ^{\circ}=30^{\circ}</math>. Then, we know that the central angle of x = <math>60</math>, so inscribed angle = <math>30</math>. Also, central angle of y = <math>120</math>, so inscribed angle = <math>60</math>. Summing both inscribed angles gives <math>30 + 60 = \boxed{(C)\ 90}.</math>
  
 
==See Also==
 
==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 18:06, 17 June 2024

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]

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

Video Solution (CREATIVE THINKING)

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

~Education, the Study of Everything


Video Solution

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

https://youtu.be/aZhjhb3mMfg ~savannahsolver

Video Solution

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

Solution

For this problem, it is useful to know that 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 $\left( \frac{360}{12} \right) ^{\circ}=30^{\circ}$. Then, we know that the central angle of x = $60$, so inscribed angle = $30$. Also, central angle of y = $120$, so inscribed angle = $60$. Summing both inscribed angles gives $30 + 60 = \boxed{(C)\ 90}.$

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

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