# 2014 AMC 8 Problems/Problem 15

## 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 $y$? $[asy] size(230); defaultpen(linewidth(0.65)); pair O=origin; pair[] circum = new pair; string[] let = {"A","B","C","D","E","F","G","H","I","J","K","L"}; draw(unitcircle); for(int i=0;i<=11;i=i+1) { circum[i]=dir(120-30*i); dot(circum[i],linewidth(2.5)); label(let[i],circum[i],2*dir(circum[i])); } draw(O--circum--circum--circum--circum--cycle); label("x",circum,2.75*(dir(circum--circum)+dir(circum--circum))); label("y",circum,1.75*(dir(circum--circum)+dir(circum--circum))); label("O",O,dir(60)); [/asy]$ $\textbf{(A) }75\qquad\textbf{(B) }80\qquad\textbf{(C) }90\qquad\textbf{(D) }120\qquad\textbf{(E) }150$

## Solution

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 inscribed arc of $\angle x=60^{\circ}$ so $m\angle x=30^{\circ}$; and the inscribed arc of $\angle y=120^{\circ}$ so $m\angle y=60^{\circ}$. $m\angle x+m\angle y=30+60=\framebox{(C) 90}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 