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

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== Solution ==
 
== Solution ==
  
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1) Let the circumcircle of <math>\triangle ABC</math> have a center <math>O</math>. Since <math>\angle BAC = 60^{\circ}</math>, <math>\overline{BC}</math> is a chord that intercept an arc of <math>120 ^{\circ}</math>
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Let <math>\angle CAB=A</math>, <math>\angle ABC=B</math>, <math>\angle BCA=C</math> for convenience.
  
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2) Define the length of <math>BC</math> as a unit length.
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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>, it follows that <math>O</math> is the midpoint of minor arc <math>BC</math>, so it's fixed as well. This implies that <math>[BCO]</math> is fixed, and since <math>[BCOIH]=[BCO]+[BOIH]</math> is maximal, it suffices to maximize <math>[BOIH]</math>.
  
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3) Draw the diameter <math>\perp</math> to <math>BC</math>. Let's call the interception of the diameter with <math>BC</math> <math>M</math> (because it is the midpoint) and interception with the circle <math>X</math>.
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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>, 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.
  
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4) Since OMB and XMC are fixed, the area is a constant. Thus, <math>XOIHC</math> also achieved maximum area.
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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. A much more elegant way is shown below.
  
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'''Lemma:'''  
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'''Lemma:''' <math>[BOIH]</math> is maximized only if <math>HB=HI</math>.
  
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<math>m\angle BOC = m \angle BIC = m \angle BHC = 120^{\circ}</math>
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'''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>
  
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For <math>m\angle BOC</math>, we fixed it to <math>120^{\circ}</math> when we drew the diagram.
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It's necessary to show the existence of a maximum <math>[BOIH]</math> (although the wording of the problem gives it to you for free), which is not hard. Either way, since <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>
  
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Let <math>m\angle ABC = \beta</math>, <math>m\angle ACB = \gamma</math>
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-Solution by '''thecmd999'''
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Now, let's isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>I</math>.
 
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<math>m\angle IBC = \frac{\beta}{2}</math>, <math>m\angle ICB = \frac{\gamma}{2}</math>
 
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<math>m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - ({180^{\circ} - 120^{\circ})= 120 ^{\circ}</math>
 
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Now, lets isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>H</math>.
 
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<math>m\angle HBC = \beta - 30^{\circ}</math>, <math>m\angle HCB = \gamma - 30^{\circ}</math>
 
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<math>m\angle BHC = 180^{\circ} - \beta - \gamma + 60^{\circ} = 240^{\circ} - 120^{\circ} = 120^{\circ}</math>
 
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Lemma proven. The lemma yields that BOIHC is a cyclic pentagon.
 
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Since XOIHC also achieved maximum area,
 
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Let <math>m\angle XOI = x_1</math>, <math>m\angle OIH = x_2</math>, <math>m\angle IHC = x_3</math>, and the radius is <math>R</math> (which will drop out.)
 
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Then the area = <math>\frac{r^2}{2}(\sin x_1 + \sin x_2 + \sin x_3)</math>, where <math>x_1 + x_2 + x_3 = 60^\circ</math>
 
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So we want to maximize <math>f(x_1, x_2) = \sin x_1 + \sin x_2 + \sin x_3</math>, Note that <math>x_3 = 60 ^\circ - x_1 - x_2</math>.
 
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Let's do some multivariable calculus.
 
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<math>f_{x_1} = \cos x_1 - \cos (x_3)</math>, <math>f_{x_2} = \cos x_2 - \cos (x_3)</math>
 
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If the partial derivatives with respect to <math>x_1</math> and <math>x_2</math> are zero, then <math>x_1 = x_2 = x_3 = 20^\circ</math>, and it is very easy to show that <math>f(x_1, x_2)</math> is the maximum with the second derivative test (left for the reader).
 
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Now, we need to verify that such a situation exists and find the angle for this situation.
 
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Let's extend <math>AI</math> to the direction of <math>X</math>, since <math>AI</math> is the angle bisector, so that <math>AI</math> intersects the midpoint of the arc <math>X</math>. Hence, if such a case exists, <math>m\angle AXB = m \angle ACB = 40 ^\circ</math>, so <math>m\angle CBA = 80 ^\circ</math>.
 
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If the angle is <math>80 ^\circ</math>, it is clear that since <math>I</math> and <math>H</math> are on the second circle (follows from the lemma). <math>I</math> will be at the right place. <math>H</math> can be easily verified too.
 
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Hence, the answer is <math>(D) 80</math>.
 
  
 
== 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 20:16, 22 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 $\angle BHC=180-A$ (indeed, all are 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 $OB=OC$, it follows that $O$ is the midpoint of minor arc $BC$, so it's fixed as well. This implies that $[BCO]$ is fixed, and 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, $\angle OBC=90-A=30$, whence $\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}$ and consequently $IH=IO$ by Inscribed Angles.

There are several 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 much more elegant way is shown below.

Lemma: $[BOIH]$ is maximized only if $HB=HI$.

Proof: Suppose for the sake of contradiction that $[BOIH]$ 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 $[BOIH]$, whence the claim follows. $\blacksquare$

It's necessary to show the existence of a maximum $[BOIH]$ (although the wording of the problem gives it to you for free), which is not hard. Either way, since $HB=HI$ by our lemma and $IH=IO$ from above, it follows that \[\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}\]

-Solution by thecmd999

See also

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

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