# 2015 AIME I Problems/Problem 6

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## Problem

Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$. Find the degree measure of $\angle BAG$. $[asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label("A",A,dir(0)); label("B",B,dir(75)); label("C",C,dir(90)); label("D",D,dir(105)); label("E",E,dir(180)); label("F",F,dir(225)); label("G",G,dir(260)); label("H",H,dir(280)); label("I",I,dir(315)); [/asy]$

## The Only Solution

Let $O$ be the center of the circle with $ABCDE$ on it.

Let $x$ be the degree measurement of $\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ in circle $O$ and $y$ be the degree measurement of $\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$ in circle $C$. $\angle ECA$ is, therefore, $5y$ by way of circle $C$ and $$\frac{360-4x}{2}=180-2x$$ by way of circle $O$. $\angle ABD$ is $180 - \frac{3x}{2}$ by way of circle $O$, and $$\angle AHG = 180 - \frac{3y}{2}$$ by way of circle $C$.

This means that: $$180-\frac{3x}{2}=180-\frac{3y}{2}+12$$

which when simplified yields $$\frac{3x}{2}+12=\frac{3y}{2}$$ or $$x+8=y$$ Since: $$5y=180-2x$$ and $$5x+40=180-2x$$ So: $$7x=140\Longleftrightarrow x=20$$ $$y=28$$ $\angle BAG$ is equal to $\angle BAE$ + $\angle EAG$, which equates to $\frac{3x}{2} + y$. Plugging in yields $30+28$, or $\boxed{058}$.

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