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

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\textbf{(E)}\ 90^{\circ} </math>
 
\textbf{(E)}\ 90^{\circ} </math>
  
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== Solution ==
+
== Solution 1 (MAA) ==
 +
 
 +
By the Inscribed Angle Theorem, <cmath>\angle BOC = 2\angle BAC = 120^\circ .</cmath>Let <math>D</math> and <math>E</math> be the feet of the altitudes of <math>\triangle ABC</math> from <math>B</math> and <math>C</math>, respectively. In <math>\triangle ACE</math> we get <math>\angle ACE = 30^\circ</math>, and as exterior angle <cmath>\angle BHC = 90^\circ + \angle ACE = 120^\circ .</cmath>Because the lines <math>BI</math> and <math>CI</math> are bisectors of <math>\angle CBA</math> and <math>\angle ACB</math>, respectively, it follows that<cmath>\angle BIC = 90^\circ + \tfrac 12\angle A = 120^\circ .</cmath>Thus the points <math>B, C, O, I</math>, and <math>H</math> are all on a circle.
 +
<asy>
 +
import geometry;
 +
size(200);
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defaultpen(fontsize(12)+0.8);
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pair O,A,B,C,D,E,I,H;
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real h=2*sqrt(3);
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O=(0,1/h); B=(-0.5,0); C=(0.5,0);
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path c1=CR(O,length(O-B));
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// A=IP(c1,O--O+5*dir(120));
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A=IP(c1,B--B+5*dir(80));
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I=incenter(A,B,C);
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H=orthocenter(A,B,C);
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D=extension(A,C,B,H); E=extension(A,B,C,H);
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path c2=circumcircle(O,I,H);
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pair o2=circumcenter(O,I,H);
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draw(c1, royalblue); draw(A--B--C--A); draw(B--D^^C--E, dotted);
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draw(arc(o2,length(O-o2),10,170), dotted+red); draw(B--I--C--O--B, black+0.3);
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pen p =black+3.25;
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dot("$O$", O, N,p);  dot("$A$", A, dir(110),p); dot("$B$", B, dir(210),p); dot("$C$", C, dir(-30),p); dot("$I$", I, N,p); dot("$H$", H, N,p); dot("$D$", D, D-H,p); dot("$E$", E, (E-C),p);
 +
</asy>
 +
Further, since
 +
<cmath>\angle OCI = \angle OCB - \angle ICB = 30^\circ - \tfrac 12\angle C </cmath>
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<cmath>\angle ICH = \angle ACE - \angle ACI = 30^\circ - \tfrac 12\angle C </cmath>
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we have <math>OI=IH</math>.
 +
 
 +
Because <math>[BCOIH]=[BCO]+[BOIH]</math>, it is sufficient to maximize the area of quadrilateral <math>BOIH</math>. If <math>P_1</math>, <math>P_2</math> are two points in an arc of circle <math>BO</math> with <math>BP_1<BP_2</math>, then the maximum area of <math>BOP_1P_2</math> occurs when <math>BP_1=P_1P_2=P_2O</math>. Indeed, if <math>BP_1\neq P_1P_2</math>, then replacing <math>P_1</math> by the point <math>P_1’</math> located halfway in the arc of the circle <math>BP_2</math> yields a triangle <math>BP_1’P_2</math> with larger area than <math>\triangle BP_1P_2</math>, and the area of <math>\triangle BOP_2</math> remains the same. Similarly, if <math>P_1P_2\neq P_2O</math>.
 +
 
 +
Therefore the maximum is achieved when <math>OI=IH=HB</math>, that is, when <cmath>\angle OCI = \angle ICH = \angle HCB = \tfrac 13 \angle OCB = 10^\circ.</cmath>Thus <math>\angle ACB = 40^\circ</math> and <math>\angle CBA = 80^\circ</math>.
 +
 
 +
== Solution 2==
  
 
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> (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=(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>  
+
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.
 
Since  <math>\angle IBH=\angle IBO</math>,  <math>IH=IO</math> by Inscribed Angles.
  
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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,
−
<cmath>\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}</cmath>  
+
<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}}

Latest revision as of 12:05, 28 June 2022

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 1 (MAA)

By the Inscribed Angle Theorem, \[\angle BOC = 2\angle BAC = 120^\circ .\]Let $D$ and $E$ be the feet of the altitudes of $\triangle ABC$ from $B$ and $C$, respectively. In $\triangle ACE$ we get $\angle ACE = 30^\circ$, and as exterior angle \[\angle BHC = 90^\circ + \angle ACE = 120^\circ .\]Because the lines $BI$ and $CI$ are bisectors of $\angle CBA$ and $\angle ACB$, respectively, it follows that\[\angle BIC = 90^\circ + \tfrac 12\angle A = 120^\circ .\]Thus the points $B, C, O, I$, and $H$ are all on a circle. [asy] import geometry; size(200); defaultpen(fontsize(12)+0.8); pair O,A,B,C,D,E,I,H; real h=2*sqrt(3); O=(0,1/h); B=(-0.5,0); C=(0.5,0); path c1=CR(O,length(O-B)); // A=IP(c1,O--O+5*dir(120)); A=IP(c1,B--B+5*dir(80)); I=incenter(A,B,C); H=orthocenter(A,B,C); D=extension(A,C,B,H); E=extension(A,B,C,H); path c2=circumcircle(O,I,H); pair o2=circumcenter(O,I,H); draw(c1, royalblue); draw(A--B--C--A); draw(B--D^^C--E, dotted); draw(arc(o2,length(O-o2),10,170), dotted+red); draw(B--I--C--O--B, black+0.3); pen p =black+3.25; dot("$O$", O, N,p); dot("$A$", A, dir(110),p); dot("$B$", B, dir(210),p); dot("$C$", C, dir(-30),p); dot("$I$", I, N,p); dot("$H$", H, N,p); dot("$D$", D, D-H,p); dot("$E$", E, (E-C),p); [/asy] Further, since \[\angle OCI = \angle OCB - \angle ICB = 30^\circ - \tfrac 12\angle C\] \[\angle ICH = \angle ACE - \angle ACI = 30^\circ - \tfrac 12\angle C\] we have $OI=IH$.

Because $[BCOIH]=[BCO]+[BOIH]$, it is sufficient to maximize the area of quadrilateral $BOIH$. If $P_1$, $P_2$ are two points in an arc of circle $BO$ with $BP_1<BP_2$, then the maximum area of $BOP_1P_2$ occurs when $BP_1=P_1P_2=P_2O$. Indeed, if $BP_1\neq P_1P_2$, then replacing $P_1$ by the point $P_1’$ located halfway in the arc of the circle $BP_2$ yields a triangle $BP_1’P_2$ with larger area than $\triangle BP_1P_2$, and the area of $\triangle BOP_2$ remains the same. Similarly, if $P_1P_2\neq P_2O$.

Therefore the maximum is achieved when $OI=IH=HB$, that is, when \[\angle OCI = \angle ICH = \angle HCB = \tfrac 13 \angle OCB = 10^\circ.\]Thus $\angle ACB = 40^\circ$ and $\angle CBA = 80^\circ$.

Solution 2

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 (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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