Problem:
Triangle ABC has β BAC=60β,β CBAβ€90β,BC=1, and ACβ₯AB. Let H,I, and O be the orthocenter, incenter, and circumcenter of β³ABC, respectively. Assume that the area of the pentagon BCOIH is the maximum possible. What is β CBA ?
Answer Choices:
A. 60β
B. 72β
C. 75β
D. 80β
E. 90β Solution:
By the Inscribed Angle Theorem, β BOC=2β BAC=120β. Let D and E be the feet of the altitudes of β³ABC from B and C, respectively. Because CE and BD intersect at H,
and β ICH=β ACHββ ACI=(90βββ EAC)β21ββ ACB=30ββ21ββ ACB. Thus OI=IH. Because [BCOIH]=[BCO]+[BOIH] and BCO is an isosceles triangle with BC=1 and OB=OC=3β1β, it is sufficient to maximize the area of quadrilateral BOIH. If P1β,P2β are two points in an arc of circle BO with BP1β<BP2β, then the maximum area of BOP1βP2β occurs when BP1β=P1βP2β=P2βO. Indeed, if BP1βξ =P1βP2β, then replacing P1β by the point P1β²β located halfway in the arc of circle BP2β yields a triangle BP1β²βP2β with larger area than β³BP1βP2β, and the area of β³BOP2β remains the same. Similarly, if P1βP2βξ =P2βO, then replacing P2β by the midpoint P2β²β of the arc P1βO causes the area of β³P1βP2β²βO to increase and the area of β³BP1βO to remain the same.
Therefore the maximum is achieved when OI=IH=HB, that is, when β OCI=β ICH=β HCB=31ββ OCB=10β. Thus 30ββ21ββ ACB=10β, so β ACB=40β and β CBA=80ββ.
