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

m (→‎Solution)
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==Problem==
 
==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>?
+
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  
−
 
+
<math>K</math>","<math>L</math>"};
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<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);
 
draw(unitcircle);
 
for(int i=0;i<=11;i=i+1)
 
for(int i=0;i<=11;i=i+1)
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}
 
}
 
draw(O--circum[4]--circum[0]--circum[6]--circum[8]--cycle);
 
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])));
+
label("<math>x</math>",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])));
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label("<math>y</math>",circum[6],1.75*(dir(circum[6]--circum[0])+dir(circum[6]--circum[8])));
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label("$O$",O,dir(60));
+
label("<math>O</math>",O,dir(60));
 
</asy>
 
</asy>
  

Revision as of 11:39, 19 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 $K$","$L$"}; 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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