2011 AMC 12A Problems/Problem 25

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$ (Error compiling LaTeX. Unknown error_msg)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] $, whence <cmath>[BOIH]=\frac{1}{2}R^2(\sin(60-C)+\sin(60-C)+\sin(2C-60)-\sin(60))</cmath> and, using the fact that$ R $is fixed, maximize$ 2\sin(60-C)+\sin(2C-60)$with Jensen's Inequality.

A more elegant way is shown below.

'''Lemma:'''$ (Error compiling LaTeX. Unknown error_msg)[BOIH] $is maximized only if$ HB=HI$.

'''Proof by contradiction:''' Suppose$ (Error compiling LaTeX. Unknown error_msg)[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: $\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}$

-Solution by thecmd999

2011 AMC 12A (Problems • Answer Key • Resources)
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2011 AMC 12A Problems/Problem 25

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