Difference between revisions of "2015 AIME I Problems/Problem 6"

Revision as of 14:56, 19 May 2015

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 $180 - \frac{3x}{2}$ by way of circle $O$ , and $\angle AHG$ is $180 - \frac{3y}{2}$ 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: $5y=180-2x$ , $5x+40=180-2x$ So: $7x=140, x=20$ $y=28.$ $\angle BAG$ is equal to $\angle BAE$ + $\angle EAG$ , which equates to $\frac{3x}{2} + y$ . Plugging in yields $30+28$ , or $058$

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 {{AIME box|year=2015|n=I|num-b=5|num-a=7}}
 {{MAA Notice}}
+[[Category:Introductory Geometry Problems]]

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Difference between revisions of "2015 AIME I Problems/Problem 6"

Revision as of 14:56, 19 May 2015

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