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

m (Solution)
(Solution)
 
(3 intermediate revisions by the same user not shown)
Line 28: Line 28:
 
</asy>
 
</asy>
  
==Solution==
+
==The Only Solution==
  
 
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.  
  
 
Let <math>x</math> be the degree measurement of <math>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}</math> in circle <math>O</math> and <math>y</math> be the degree measurement of <math>\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}</math> in circle <math>C</math>.
 
Let <math>x</math> be the degree measurement of <math>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}</math> in circle <math>O</math> and <math>y</math> be the degree measurement of <math>\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}</math> in circle <math>C</math>.
<math>\angle ECA</math> is therefore <math>5y</math> by way of circle <math>C</math> and <math>\frac{360-4x}{2}=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 <cmath>\frac{360-4x}{2}=180-2x</cmath> 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 <cmath>\angle AHG = 180 - \frac{3y}{2}</cmath> by way of circle <math>C</math>.
  
 
This means that:
 
This means that:
  
<math>180-\frac{3x}{2}=180-\frac{3y}{2}+12</math>,
+
<cmath>180-\frac{3x}{2}=180-\frac{3y}{2}+12</cmath>
  
which when simplified yields <math>3x/2+12=3y/2</math>, or <math>x+8=y</math>.
+
which when simplified yields <cmath>\frac{3x}{2}+12=\frac{3y}{2}</cmath> or <cmath>x+8=y</cmath>
 
Since:
 
Since:
<math>5y=180-2x</math><math>5x+40=180-2x</math>
+
<cmath>5y=180-2x</cmath> and <cmath>5x+40=180-2x</cmath>
 
So:
 
So:
<math>7x=140, x=20</math>  
+
<cmath>7x=140\Longleftrightarrow x=20</cmath>  
<math>y=28.</math>
+
<cmath>y=28</cmath>
 
<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>.

Latest revision as of 22:46, 1 September 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]

The Only Solution

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 $\boxed{058}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS