2011 AMC 12A Problem 25

Write $B=\angle CBA$ and $C=\angle BCA=120^\circ-B.$ Since $AC\ge AB,$ we have $B\ge C,$ so $60^\circ\le B\le90^\circ.$ The standard angle formulas give $$\begin{aligned} \angle BOC&=2\angle A=120^\circ, \\ \angle BHC&=180^\circ-\angle A=120^\circ, \\ \angle BIC&=90^\circ+\dfrac{\angle A}{2}=120^\circ. \end{aligned}$$ Hence $B,C,O,I,H$ lie on one circle. Also $BC=1$ and $A=60^\circ$ fix the circumradius $OB=OC=\frac{1}{\sqrt3},$ so $\triangle BCO$ and the circle through $B,C,O$ are fixed. Angle chasing at $C$ gives $$\begin{aligned} \angle OCI&=30^\circ-\dfrac C2, \\ \angle ICH&=30^\circ-\dfrac C2. \end{aligned}$$ Thus the corresponding chords satisfy $OI=IH.$ The pentagon’s area is the fixed area $[BCO]$ plus $[BOIH].$ For two points dividing a fixed arc $OB,$ an inscribed quadrilateral has greatest area when its three consecutive subarcs are equal (equivalently, maximize the sum of their sines). Hence at the maximum $OI=IH=HB.$ In $\triangle BOC,$ $\angle OCB=30^\circ.$ Equal chords make $\angle OCI=\angle ICH,$ $\angle ICH=\angle HCB,$ and each of these angles is $10^\circ.$ Therefore $30^\circ-\tfrac C2=10^\circ,$ so $C=40^\circ$ and $B=80^\circ.$ Thus, the correct answer is D.