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 $\angle CBA$?
$\textbf{(A)}\ 60^{\circ} \qquad \textbf{(B)}\ 72^{\circ} \qquad \textbf{(C)}\ 75^{\circ} \qquad \textbf{(D)}\ 80^{\circ} \qquad \textbf{(E)}\ 90^{\circ}$
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. 

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 $\angle CBA = 80^\circ$.