Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AMC 12A Problems/Problem 25"

m (→‎Solution)
m (→‎Solution)
Line 15: Line 15:
 
It's well-known that <math>\angle BOC=2A</math>, <math>\angle BIC=90+\frac{A}{2}</math>, and <math>\angle BHC=180-A</math> (verifiable by angle chasing). Then, as <math>A=60</math>, it follows that <math>\angle BOC=\angle BIC=\angle BHC=120</math> and consequently pentagon <math>BCOIH</math> is cyclic. Observe that <math>BC=1</math> is fixed, whence the circumcircle of cyclic pentagon <math>BCOIH</math> is also fixed. Similarly, as <math>OB=OC</math>(both are radii), it follows that <math>O</math> and also <math>[BCO]</math> is fixed. Since <math>[BCOIH]=[BCO]+[BOIH]</math> is maximal, it suffices to maximize <math>[BOIH]</math>.
 
It's well-known that <math>\angle BOC=2A</math>, <math>\angle BIC=90+\frac{A}{2}</math>, and <math>\angle BHC=180-A</math> (verifiable by angle chasing). Then, as <math>A=60</math>, it follows that <math>\angle BOC=\angle BIC=\angle BHC=120</math> and consequently pentagon <math>BCOIH</math> is cyclic. Observe that <math>BC=1</math> is fixed, whence the circumcircle of cyclic pentagon <math>BCOIH</math> is also fixed. Similarly, as <math>OB=OC</math>(both are radii), it follows that <math>O</math> and also <math>[BCO]</math> is fixed. Since <math>[BCOIH]=[BCO]+[BOIH]</math> is maximal, it suffices to maximize <math>[BOIH]</math>.
  
−
Verify that <math>\angle IBC=\frac{B}{2}</math>, <math>\angle HBC=90-C</math> by angle chasing; it follows that <math>\angle IBH=\angle HBC-\angle IBC=90-C-\frac{B}{2}=\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}</math> since <math>A+B+C=180\implies\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90</math> by Triangle Angle Sum. Similarly, <math>\angle OBC=(180-120)/2=30</math> (isosceles base angles are equal), whence <cmath>\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}</cmath> so <math>IH=IO</math> by Inscribed Angles.
+
Verify that <math>\angle IBC=\frac{B}{2}</math>, <math>\angle HBC=90-C</math> by angle chasing; it follows that <math>\angle IBH=\angle HBC-\angle IBC=90-C-\frac{B}{2}=\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}</math> since <math>A+B+C=180\implies\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90</math> by Triangle Angle Sum. Similarly, <math>\angle OBC=(180-120)/2=30</math> (isosceles base angles are equal), whence <cmath>\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}</cmath>  
 +
Since  <math>\angle IBH=</math>\angle IBO<math>,  </math>IH=IO<math> by Inscribed Angles.
  
 
There are two ways to proceed.  
 
There are two ways to proceed.  
  
  
−
Letting <math>O'</math> and <math>R</math> be the circumcenter and circumradius, respectively, of cyclic pentagon <math>BCOIH</math>, the most straightforward is to write <math>[BOIH]=[OO'I]+[IO'H]+[HO'B]-[BO'O]</math>, whence <cmath>[BOIH]=\frac{1}{2}R^2(\sin(60-C)+\sin(60-C)+\sin(2C-60)-\sin(60))</cmath> and, using the fact that <math>R</math> is fixed, maximize <math>2\sin(60-C)+\sin(2C-60)</math> with Jensen's Inequality.  
+
Letting </math>O'<math> and </math>R<math> be the circumcenter and circumradius, respectively, of cyclic pentagon </math>BCOIH<math>, the most straightforward is to write </math>[BOIH]=[OO'I]+[IO'H]+[HO'B]-[BO'O]<math>, whence <cmath>[BOIH]=\frac{1}{2}R^2(\sin(60-C)+\sin(60-C)+\sin(2C-60)-\sin(60))</cmath> and, using the fact that </math>R<math> is fixed, maximize </math>2\sin(60-C)+\sin(2C-60)<math> with Jensen's Inequality.  
  
  
 
A more elegant way is shown below.
 
A more elegant way is shown below.
  
−
'''Lemma:''' <math>[BOIH]</math> is maximized only if <math>HB=HI</math>.
+
'''Lemma:''' </math>[BOIH]<math> is maximized only if </math>HB=HI<math>.
  
−
'''Proof by contradiction:''' Suppose <math>[BOIH]</math> is maximized when <math>HB\neq HI</math>. Let <math>H'</math> be the midpoint of minor arc <math>BI</math> be and <math>I'</math> the midpoint of minor arc <math>H'O</math>. Then <math>[BOIH']=[IBO]+[IBH']>[IBO]+[IBH]=[BOIH]</math> since the altitude from <math>H'</math> to <math>BI</math> is greater than that from <math>H</math> to <math>BI</math>; similarly <math>[BH'I'O]>[BOIH']>[BOIH]</math>. Taking <math>H'</math>, <math>I'</math> to be the new orthocenter, incenter, respectively, this contradicts the maximality of <math>[BOIH]</math>, so our claim follows. <math>\blacksquare</math>
+
'''Proof by contradiction:''' Suppose </math>[BOIH]<math> is maximized when </math>HB\neq HI<math>. Let </math>H'<math> be the midpoint of minor arc </math>BI<math> be and </math>I'<math> the midpoint of minor arc </math>H'O<math>. Then </math>[BOIH']=[IBO]+[IBH']>[IBO]+[IBH]=[BOIH]<math> since the altitude from </math>H'<math> to </math>BI<math> is greater than that from </math>H<math> to </math>BI<math>; similarly </math>[BH'I'O]>[BOIH']>[BOIH]<math>. Taking </math>H'<math>, </math>I'<math> to be the new orthocenter, incenter, respectively, this contradicts the maximality of </math>[BOIH]<math>, so our claim follows. </math>\blacksquare<math>
  
−
With our lemma(<math>HB=HI</math>) and <math>IH=IO</math> from above: <cmath>\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}</cmath>
+
With our lemma(</math>HB=HI<math>) and </math>IH=IO$ from above: <cmath>\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}</cmath>
  
 
-Solution by '''thecmd999'''
 
-Solution by '''thecmd999'''

Revision as of 16:06, 28 September 2013

Problem

Triangle $ABC$ has $\angle BAC = 60^{\circ}$, $\angle CBA \leq 90^{\circ}$, $BC=1$, and $AC \geq AB$. Let $H$, $I$, and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$, respectively. Assume that the area of pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$?

$\textbf{(A)}\ 60^{\circ} \qquad \textbf{(B)}\ 72^{\circ} \qquad \textbf{(C)}\ 75^{\circ} \qquad \textbf{(D)}\ 80^{\circ} \qquad \textbf{(E)}\ 90^{\circ}$

Solution

Let $\angle CAB=A$, $\angle ABC=B$, $\angle BCA=C$ for convenience.

It's well-known that $\angle BOC=2A$, $\angle BIC=90+\frac{A}{2}$, and $\angle BHC=180-A$ (verifiable by angle chasing). Then, as $A=60$, it follows that $\angle BOC=\angle BIC=\angle BHC=120$ and consequently pentagon $BCOIH$ is cyclic. Observe that $BC=1$ is fixed, whence the circumcircle of cyclic pentagon $BCOIH$ is also fixed. Similarly, as $OB=OC$(both are radii), it follows that $O$ and also $[BCO]$ is fixed. Since $[BCOIH]=[BCO]+[BOIH]$ is maximal, it suffices to maximize $[BOIH]$.

Verify that $\angle IBC=\frac{B}{2}$, $\angle HBC=90-C$ by angle chasing; it follows that $\angle IBH=\angle HBC-\angle IBC=90-C-\frac{B}{2}=\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}$ since $A+B+C=180\implies\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90$ by Triangle Angle Sum. Similarly, $\angle OBC=(180-120)/2=30$ (isosceles base angles are equal), whence \[\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}\] Since $\angle IBH=$\angle IBO$,$IH=IO$by Inscribed Angles.

There are two ways to proceed.


Letting$ (Error compiling LaTeX. Unknown error_msg)O'$and$R$be the circumcenter and circumradius, respectively, of cyclic pentagon$BCOIH$, the most straightforward is to write$[BOIH]=[OO'I]+[IO'H]+[HO'B]-[BO'O]$, whence <cmath>[BOIH]=\frac{1}{2}R^2(\sin(60-C)+\sin(60-C)+\sin(2C-60)-\sin(60))</cmath> and, using the fact that$R$is fixed, maximize$2\sin(60-C)+\sin(2C-60)$with Jensen's Inequality.


A more elegant way is shown below.

'''Lemma:'''$ (Error compiling LaTeX. Unknown error_msg)[BOIH]$is maximized only if$HB=HI$.

'''Proof by contradiction:''' Suppose$ (Error compiling LaTeX. Unknown error_msg)[BOIH]$is maximized when$HB\neq HI$. Let$H'$be the midpoint of minor arc$BI$be and$I'$the midpoint of minor arc$H'O$. Then$[BOIH']=[IBO]+[IBH']>[IBO]+[IBH]=[BOIH]$since the altitude from$H'$to$BI$is greater than that from$H$to$BI$; similarly$[BH'I'O]>[BOIH']>[BOIH]$. Taking$H'$,$I'$to be the new orthocenter, incenter, respectively, this contradicts the maximality of$[BOIH]$, so our claim follows.$\blacksquare$With our lemma($HB=HI$) and$IH=IO$ from above: \[\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}\]

-Solution by thecmd999

See also

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_25&oldid=57167"