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

Latest revision as of 15:04, 21 July 2023

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

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

Video Solution

https://youtu.be/IuwkX2Dv25s

~MathProblemSolvingSkills.com

@@ Line 28: / Line 28: @@
 </asy>
-==The Only Solution==
+==Solution 1==
 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 <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 <cmath>\angle AHG = 180 - \frac{3y}{2}</cmath> by way of circle <math>C</math>.
@@ Line 48: / Line 51: @@
 <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>.
+==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>.
+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=</math><math>\overarc{BE}</math>.
+We are given that <math>A,B,C,D,E</math> are evenly distributed on a circle. Hence,
+<math>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}</math><math>=\frac{\angle DBA}{3}=\frac{m}{2}-4</math>
+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</math> = <math>\overarc{AB}</math> + <math>\overarc{BC}</math> = <math>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>
+The rest of the solution proceeds as in solution 1, which gives <math>\boxed{058}</math>
+==Solution 3==
+[[File:2015 AIME I 6.png|500px|right]]
+Let <math>\angle GAH = \varphi \implies \overset{\Large\frown} {GH} = 2\varphi \implies</math>
+<cmath>\overset{\Large\frown} {EF} = \overset{\Large\frown} {FG} = \overset{\Large\frown} {HI} = \overset{\Large\frown} {IA} = 2\varphi \implies</cmath>
+<cmath>\angle AGH = 2\varphi, \angle ACE = 10 \varphi.</cmath>
+<cmath>BD||GH \implies \angle AJB = \angle AGH = 2 \varphi.</cmath>
+<cmath>\triangle AHG: \hspace{10mm}  \angle AHG = \beta = 180^\circ – 3 \varphi.</cmath>
+<math>\hspace{10mm} \triangle ABJ:  \hspace{10mm} \angle BAG + \angle ABD = \alpha + \gamma = 180^\circ + 2 \varphi. </math>
+Let arc <math> \overset{\Large\frown} {AB} = 2\psi \implies</math>
+<math>\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.</math>
+<math>\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.</math>
+Therefore <math>\gamma = 180^\circ – 3 \cdot (14^\circ – 4^\circ) = 150^\circ \implies  \alpha = 180^\circ + 2 \cdot 14^\circ – 150^\circ = \boxed{\textbf{058}}.</math>
+==Video Solution==
+https://youtu.be/IuwkX2Dv25s
+~MathProblemSolvingSkills.com
 ==See Also==

2015 AIME I (Problems • Answer Key • Resources)
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