Solution of the problem #25 in AMC 12 A 2011

-

Triangle ABC has angle BAC = 60°, angle CBA ≤ 90°, BC = 1, and AC ≥ AB. Let H, I, and O be the orthocenter, incenter, and circumcenter of triangle ABC, respectively. Assume that the area of the pentagon BCOIH is the maximum possible. What is angle CBA?
(A) 60° (B) 72° (C) 75° (D) 80° (E) 90°

Solution:

The range of possible positions of vertex B on the side AB of angle BAC is restricted by the requirements AC ≥ AB and angle CBA ≤ 90°.

In the extreme case when AC = AB, triangle ABC is equilateral and all three center points H, I and O are at the same point. The pentagon BCOIH is reduced to the triangle BCO in this case.

The other extreme position of vertex B is where angle CBA = 90°. In this case, vertex B is the orthocenter of the right triangle ABC and the circumcenter O is the midpoint of its hypotenuse AC. The pentagon BCOIH is reduced to the quadrilateral BCOI in this case.

So, when vertex B moves within its allowed range along the AB side of angle BAC, the following angles vary accordingly:

60° <= angle CBA <= 90°

and

30° ≥ angle OCA ≥ 0°.

Angle BOC = 120° since it's a central angle of the circumcircle of ABC whose inscribed angle BAC = 60°. Let O’ be the point of intersection of perpendicular bisector of BC with the minor arc BC. Since triangle BOO’ is isosceles and angle BOO’ = 120° / 2 = 60°, it’s actually an equilateral triangle. The same is true for triangle COO’. Thus, the radii of the circumcircles of triangles ABC and BCO  are congruent.

So, arc BO always has constant radius R and constant arc measurement 60°, regardless of the position of vertex B and the corresponding position of the circumcircle O.

Let's prove that both orthocenter H and incenter I also always lie on arc BO. For that, it suffices to prove that angles BHC and BIC have measurement 120° each, since angle BOC has measurement 120° and is subtended by the same chord BC from the same side.

Let's prolong line CH until it crosses side BA at point P and line BH until it crosses side AC at point Q. Since H is the orthocenter of triangle ABC, BQ and CP are altitudes of this triangle. Since angle BAC = 60°, angle ACP = 90° - 60° = 30° and angle CHQ = 60°, so that angle BHC = 180° - 60° = 120°.

Further, I is the incenter, so BI and CI are bisectors of angles B and C of triangle ABC. Since angle A = 60°, the sum of angles B and C is 120° and the sum of angles IBC and ICB is 120° / 2 = 60°, so that angle BIC = 180° - 60° = 120°. We have proved that both the orthocenter H and the incenter I always lie on arc BO.

Let BOMN be a quadrilateral where M and N are two arbitrary points in this order on the arc BO.

Based on the Maximum Area of a Cyclic Polygon Theorem (see its proof in this paper: 
https://advancinginmath-349b7.firebaseapp.com/Papers/CyclicPolygonsTheorem1.pdf), the area of the cyclic quadrilateral BOMN is the largest of all such quadrilaterals when points M and N divide the arc BO = 60° into three equal arcs 20° each (and, accordingly, divide angle BCO into three equal angles 10° each). 

Note that the area of the corresponding pentagon BCOMN is also maximum possible in this case, because the area of the isosceles triangle BCO is fixed (all measurements of its sides and angles are constant: BC = 1, angle BOC = 120° and both angles OBC and BCO are 30°).

If we slide vertex B along the side BA within its allowed range to the point where angle OCA = 10° (which is somewhere inside its range: 30° ≥ angle OCA ≥ 0°), then line CM bisects angle BCA. Since incenter I lies on this angle bisector and also on the arc OB, points I and M are located at the same point.

If angle OCA = 10°, then angle BCA = 40° and angle CBA = 180° - 60° - 40° = 80°. Then, angle CPB (where CP passes through point N) = 180° - 80° - 10° = 90°. Since orthocenter H lies on this altitude and also on the arc OB, points H and N are located at the same point.

The answer is: angle CBA = 80° (D)