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

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</asy>
 
</asy>
  
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==The Only Solution==
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==Solution 1==
  
 
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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<math>\angle BAG</math> is equal to <math>\angle BAE</math> + <math>\angle EAG</math>, which equates to <math>\frac{3x}{2} + y</math>.
 
<math>\angle BAG</math> is equal to <math>\angle BAE</math> + <math>\angle EAG</math>, which equates to <math>\frac{3x}{2} + y</math>.
 
Plugging in yields <math>30+28</math>, or <math>\boxed{058}</math>.
 
Plugging in yields <math>30+28</math>, or <math>\boxed{058}</math>.
 +
 +
==Solution 2==
 +
Let <math>m</math> be the degree measurement of <math>\angle GCH</math>. Since <math>G,H</math> lie on a circle with center <math>C</math>, <math>\angle GHC=\frac{180-m}{2}=90-\frac{m}{2}</math>.
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Since <math>\angle ACH=2 \angle GCH=2m</math>, <math>\angle AHC=\frac{180-2m}{2}=90-m</math>. Adding <math>\angle GHC</math> and <math>\angle AHC</math> gives <math>\angle AHG=180-\frac{3m}{2}</math>, and <math>\angle ABD=\angle AHG+12=192-\frac{3m}{2}</math>. Since <math>AE</math> is parallel to <math>BD</math>, <math>\angle DBA=180-\angle ABD=\frac{3m}{2}-12=\overarc{BE}</math>.
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We are given that <math>A,B,C,D,E</math> are evenly distributed on a circle. Hence,
 +
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<math>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}=\frac{\angle DBA}{3}=\frac{m}{2}-4</math>
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 +
Here comes the key: Draw a line through <math>C</math> parallel to <math>AE</math>, and select a point <math>X</math> to the right of point <math>C</math>. <math>\angle ACX=\overarc{AB}+\overarc{BC}=m-8</math>. Let the midpoint of <math>\overline{HG}</math> be <math>Y</math>, then <math>\angle YCX=\angle ACX+\angle ACY=(m-8)+\frac{5m}{2}=90</math>. Solving gives <math>m=28</math>
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 +
The rest of the solution proceeds as in solution 1. The correct answer is <math>\boxed{058}</math>
 +
 +
  
 
==See Also==
 
==See Also==

Revision as of 08:14, 18 December 2021

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. The correct answer is $\boxed{058}$


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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