Maximizing the area of a cyclic pentagon made up of triangle centres.

Question

Triangle $ABC$ has $\angle BAC=60^\circ$, $\angle CBA \le 90^\circ$, $BC=1$, and $AC \ge AB$. Let $H$, $I$, and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$, respectively. Assume that the area of the pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$?

This is problem 25 from the 2011 AMC12A. My friend sent me the following solution:

Angle formulas give that $BCOIH$ is cyclic. Further $I$ and $H$ are both on the arc to the "left" of $O$, with $H$ further "left" than $I$. To maximize the area, we want to distribute $I$ and $H$ evenly over arc $BO$. This gives that $\angle{IBC}=\angle{IBO}+\angle{OBC}=10+30=40$. Multiplying by two gives that the answer is $80$.

I agree with this solution up to the point where they claim they want to distribute evenly abut BO. Why not distribute evenly over the arc BC, doesn't this make the area even greater? If anyone could explain why we would only distribute on arc BO that would be great.

Answer 1

Note that $O$ is fixed as midpoint of arc $BC$. Since the perpendicular bisector of chord $BC$ passes through $O$.

Answer 2

Begin with a circle with center $O$, and the three triangle points $A$, $B$, and $C$ have to lie on the circle. How are you going to get the incenter point $I$ on the arc $BC$? The incenter is formed by the bisectors of the angles of the triangle, and thus you would have to have a degenerate null triangle to have the bisector ray moving along one of the triangle edges. And if you mean the circular arc $BC$, that doesn't work either, since the incenter will always be in the triangle's interior.

Could you give an example triangle to consider that illustrates better what you have in mind?

Answer 3

Observe that, points $I$ and $H$ will always stay on one side of point $O$$[$ Drop perpendiculars from point $A$, $O$ and $I$ on side $BC$ and then calculate the distances from point $B$ to the feet of these perpendiculars.$]$ and thereafter stay on the arc $BO$.

Hence, to maximize the area, we need to evenly distribute points $I$ and $H$ over the arc $BO$.