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

(Video Solution)
(Video Solution)
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==Video Solution==
 
==Video Solution==
 
https://www.youtube.com/watch?v=qseG63LK4AU
 
https://www.youtube.com/watch?v=qseG63LK4AU
 
https://www.youtube.com/watch?v=CfCdHd4l3n8 - BadBoyHalo
 
  
 
==Solution==
 
==Solution==

Revision as of 14:01, 5 January 2022

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

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

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 inscirbed angle = 60. Summing both inscribed angles gives 30 + 60 = 90.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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