2015 AIME I Problems/Problem 6

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]

Solution 1

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 $\boxed{058}$.

Solution 2

Let $m$ be the degree measurement of $\angle GCH$. Since $G,H$ lie on a circle with center $C$, $\angle GHC=\frac{180-m}{2}=90-\frac{m}{2}$.

Since $\angle ACH=2 \angle GCH=2m$, $\angle AHC=\frac{180-2m}{2}=90-m$. Adding $\angle GHC$ and $\angle AHC$ gives $\angle AHG=180-\frac{3m}{2}$, and $\angle ABD=\angle AHG+12=192-\frac{3m}{2}$. Since $AE$ is parallel to $BD$, $\angle DBA=180-\angle ABD=\frac{3m}{2}-12=$$\overarc{BE}$.

We are given that $A,B,C,D,E$ are evenly distributed on a circle. Hence,

$\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$$=\frac{\angle DBA}{3}=\frac{m}{2}-4$

Here comes the key: Draw a line through $C$ parallel to $AE$, and select a point $X$ to the right of point $C$.

$\angle ACX$ = $\overarc{AB}$ + $\overarc{BC}$ = $m-8$.

Let the midpoint of $\overline{HG}$ be $Y$, then $\angle YCX=\angle ACX+\angle ACY=(m-8)+\frac{5m}{2}=90$. Solving gives $m=28$

The rest of the solution proceeds as in solution 1, which gives $\boxed{058}$


Solution 3

2015 AIME I 6.png

Let $\angle GAH = \varphi \implies \overset{\Large\frown} {GH} = 2\varphi \implies$ \[\overset{\Large\frown} {EF} = \overset{\Large\frown} {FG} = \overset{\Large\frown} {HI} = \overset{\Large\frown} {IA} = 2\varphi \implies\] \[\angle AGH = 2\varphi, \angle ACE = 10 \varphi.\]

\[BD||GH \implies \angle AJB = \angle AGH = 2 \varphi.\] \[\triangle AHG: \hspace{10mm}  \angle AHG = \beta = 180^\circ – 3 \varphi.\] $\hspace{10mm} \triangle ABJ:  \hspace{10mm} \angle BAG + \angle ABD = \alpha + \gamma = 180^\circ + 2 \varphi.$

Let arc $\overset{\Large\frown} {AB} = 2\psi \implies$

$\angle ACE = \frac {360^\circ – 8 \psi}{2}= 180^\circ – 4 \psi,  \angle ABD = \gamma =\frac {360^\circ – 6 \psi}{2} =180^\circ – 3 \psi.$ $\gamma –  \beta = 3(\varphi – \psi) = 12^\circ \implies \psi = \varphi – 4^\circ \implies 10 \varphi = 180^\circ – 4(\varphi – 4^\circ) \implies 14 \varphi = 196^\circ \implies \varphi = 14^\circ.$

Therefore $\gamma = 180^\circ – 3 \cdot (14^\circ – 4^\circ) = 150^\circ \implies  \alpha = 180^\circ + 2 \cdot 14^\circ – 150^\circ = \boxed{\textbf{058}}.$

See Also

2015 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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