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 cirle. 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]

Solution

Let O be the center of the circle with ABCDE on it. Let x=ED=DC=CB=BA and y=EF=FG=GH=HI=IA. $\angle ECA$ is therefore 5y by way of circle C and 180-2x by way of circle O. $\angle ABD$ is 180-3x/2 by way of circle O, and $\angle AHG$ is 180-3y/2 by way of circle C. This means that 180-3x/2=180-3y/2+12, which when simplified yields 3x/2+12=3y/2, or x+8=y. Since 5y=180-2x, 5x+40=180-2x 7x=140 x=20 y=28. $\angle BAG$ is equal to $\angle BAE$ + $\angle EAG$, which equates to 3x/2+y. Plugging in yields 30+28, or 058

See Also

2015 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions

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