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

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

@@ Line 9: / Line 9: @@
 \textbf{(E)}\ 90^{\circ} </math>
-== Solution ==
+== Solution 1 (MAA) ==
-) 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>
+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);
+defaultpen(fontsize(12)+0.8);
-) Define the length of <math>BC</math> as a unit length.
+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
+<cmath>\angle OCI = \angle OCB - \angle ICB = 30^\circ - \tfrac 12\angle C </cmath>
+<cmath>\angle ICH = \angle ACE - \angle ACI = 30^\circ - \tfrac 12\angle C </cmath>
+we have <math>OI=IH</math>.
-) 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>.
+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>.
-) Since OMB and XMC are fixed, the area is a constant. Thus, <math>XOIHC</math> also achieved maximum area.
+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>.
-'''Lemma:'''
+== Solution 2==
-<math>m\angle BOC = m \angle BIC = m \angle BHC = 120^{\circ}</math>
+Let <math>\angle CAB=A</math>, <math>\angle ABC=B</math>, <math>\angle BCA=C</math> for convenience.
-For <math>m\angle BOC</math>, we fixed it to <math>120^{\circ}</math> when we drew the diagram.
+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>.
-Let <math>m\angle ABC = \beta</math>, <math>m\angle ACB = \gamma</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), 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.
-Now, let's isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>I</math>.
-<math>m\angle IBC = \frac{\beta}{2}</math>, <math>m\angle ICB = \frac{\gamma}{2}</math>
+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.
-<math>m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - ({180^{\circ} - 120^{\circ})= 120 ^{\circ}</math>
+A more elegant way is shown below.
-Now, lets isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>H</math>.
+'''Lemma:''' <math>[BOIH]</math> is maximized only if <math>HB=HI</math>.
-<math>m\angle HBC = \beta - 30^{\circ}</math>, <math>m\angle HCB = \gamma - 30^{\circ}</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>
-<math>m\angle BHC = 180^{\circ} - \beta - \gamma + 60^{\circ} = 240^{\circ} - 120^{\circ} = 120^{\circ}</math>
+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>
-Lemma proven. The lemma yields that BOIHC is a cyclic pentagon.
+==Video Solution by Osman Nal==
+https://www.youtube.com/watch?v=O3amRG9zEHE&ab_channel=OsmanNal
-Since XOIHC also achieved maximum area,
-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.)
-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>
-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>.
-Let's do some multivariable calculus.
-<math>f_{x_1} = \cos x_1 - \cos (x_3)</math>, <math>f_{x_2} = \cos x_2 - \cos (x_3)</math>
-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).
-Now, we need to verify that such a situation exists and find the angle for this situation.
-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>.
-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.
-Hence, the answer is <math>(D) 80</math>.
 == See also ==
 {{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}}
 {{MAA Notice}}

2011 AMC 12A (Problems • Answer Key • Resources)
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