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

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