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

(Solution)
(Added in the TERRIFIC video solution.)
(16 intermediate revisions by 5 users not shown)
Line 13: Line 13:
 
Let <math>\angle CAB=A</math>, <math>\angle ABC=B</math>, <math>\angle BCA=C</math> for convenience.
 
Let <math>\angle CAB=A</math>, <math>\angle ABC=B</math>, <math>\angle BCA=C</math> for convenience.
  
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> (indeed, all are 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, hence 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=90-A=30</math> (isosceles base angles are equal), whence <math>\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}</math> and consequently <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), hence <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=\angle IBO</math><math>IH=IO</math> by Inscribed Angles.
 +
 
 +
There are two ways to proceed.  
  
There are several 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 much 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:''' Suppose for the sake of contradiction that <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>, whence the 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>
  
We assume there is a maximum <math>[BOIH]</math> since the problem says so. Then <math>HB=HI</math> by our lemma and <math>IH=IO</math> from above, it follows that <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</math> from above, along with the fact that inscribed angles that intersect the same length chords are equal,
 +
<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'''
+
==Video Solution by Osman Nal==
 +
https://www.youtube.com/watch?v=O3amRG9zEHE&ab_channel=OsmanNal
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}}
 
{{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 08:29, 19 April 2021

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, hence 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), hence \[\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}\] Since $\angle IBH=\angle IBO$, $IH=IO$ by Inscribed Angles.

There are two ways to proceed.


Letting $O'$ 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 \[[BOIH]=\frac{1}{2}R^2(\sin(60-C)+\sin(60-C)+\sin(2C-60)-\sin(60))\] 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: $[BOIH]$ is maximized only if $HB=HI$.

Proof by contradiction: Suppose $[BOIH]$ 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, along with the fact that inscribed angles that intersect the same length chords are equal, \[\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}\]

Video Solution by Osman Nal

https://www.youtube.com/watch?v=O3amRG9zEHE&ab_channel=OsmanNal

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
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