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

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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,
 
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,
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<cmath>\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2((30-\frac{C}{2})+(30+\frac{C}{2}+\frac{1}{3}\angle OCB))=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}</cmath>  
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<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 18:05, 26 December 2019

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 $O'$ 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, \[\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}\]

-Solution by thecmd999

See also

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AMC 12 Problems and Solutions

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