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

m (→‎Solution)
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Let <math>O</math> be the center of the circle with <math>ABCDE</math> on it.  
 
Let <math>O</math> be the center of the circle with <math>ABCDE</math> on it.  
  
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Let <math>x=ED=DC=CB=BA</math> and <math>y=EF=FG=GH=HI=IA</math>.
+
Let <math>x=\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}</math> and <math>y=\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}</math>.
 
<math>\angle ECA</math> is therefore <math>5y</math> by way of circle <math>C</math> and <math>180-2x</math> by way of circle <math>O</math>.
 
<math>\angle ECA</math> is therefore <math>5y</math> by way of circle <math>C</math> and <math>180-2x</math> by way of circle <math>O</math>.
 
<math>\angle ABD</math> is <math>180 - \frac{3x}{2}</math> by way of circle <math>O</math>, and <math>\angle AHG</math> is <math>180 - \frac{3y}{2}</math> by way of circle <math>C</math>.
 
<math>\angle ABD</math> is <math>180 - \frac{3x}{2}</math> by way of circle <math>O</math>, and <math>\angle AHG</math> is <math>180 - \frac{3y}{2}</math> by way of circle <math>C</math>.

Revision as of 19:30, 18 August 2017

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

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

Let $x=\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ and $y=\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$. $\angle ECA$ 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 $\boxed{058}$.

See Also

2015 AIME I (Problems • Answer Key • Resources)
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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