Triangle $ABC$ has $\angle BAC=60^\circ$, $\angle CBA\le90^\circ$, $BC=1$, and $AC\ge AB$. 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 $\angle BCA$?

1. This is the problem's triangle, so by the argument there the maximum occurs at $\angle CBA=80^\circ$, with $\angle BAC=60^\circ$ fixed.
2. $\angle BCA=180^\circ-60^\circ-80^\circ=\boxed{40^\circ}$.