Problem:
Triangle ABC has ∠BAC=60∘,∠CBA≤90∘,BC=1, and AC≥AB. Let H,I, and O be the orthocenter, incenter, and circumcenter of △ABC, respectively. Assume that the area of the pentagon BCOIH is the maximum possible. What is ∠CBA ?
Answer Choices:
A. 60∘
B. 72∘
C. 75∘
D. 80∘
E. 90∘ Solution:
By the Inscribed Angle Theorem, ∠BOC=2∠BAC=120∘. Let D and E be the feet of the altitudes of △ABC from B and C, respectively. Because CE and BD intersect at H,
and ∠ICH=∠ACH−∠ACI=(90∘−∠EAC)−21​∠ACB=30∘−21​∠ACB. Thus OI=IH. Because [BCOIH]=[BCO]+[BOIH] and BCO is an isosceles triangle with BC=1 and OB=OC=3​1​, it is sufficient to maximize the area of quadrilateral BOIH. If P1​,P2​ are two points in an arc of circle BO with BP1​<BP2​, then the maximum area of BOP1​P2​ occurs when BP1​=P1​P2​=P2​O. Indeed, if BP1â€‹î€ =P1​P2​, then replacing P1​ by the point P1′​ located halfway in the arc of circle BP2​ yields a triangle BP1′​P2​ with larger area than â–³BP1​P2​, and the area of â–³BOP2​ remains the same. Similarly, if P1​P2â€‹î€ =P2​O, then replacing P2​ by the midpoint P2′​ of the arc P1​O causes the area of â–³P1​P2′​O to increase and the area of â–³BP1​O to remain the same.
Therefore the maximum is achieved when OI=IH=HB, that is, when ∠OCI=∠ICH=∠HCB=31​∠OCB=10∘. Thus 30∘−21​∠ACB=10∘, so ∠ACB=40∘ and ∠CBA=80∘​.
