Difference between revisions of "2014 AMC 8 Problems/Problem 15"
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==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 | ||
Revision as of 18:10, 15 April 2023
Problem
The circumference of the circle with center 
is divided into 
equal arcs, marked the letters 
through 
as seen below. What is the number of degrees in the sum of the angles 
and 
?
![[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]](http://latex.artofproblemsolving.com/3/7/1/371f42726ae702ba57a0002d787f0e3c62767ec2.png)
Video Solution
https://www.youtube.com/watch?v=qseG63LK4AU ~David
https://youtu.be/aZhjhb3mMfg ~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=3242
~pi_is_3.14
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 
of the circle's circumference, each unit central angle measures 
. Then, we know that the central angle of x = 
, so inscribed angle = 
. Also, central angle of y = 
, so inscirbed angle = . Summing both inscribed angles gives 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
