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 $\angle CBA?$

$60^\circ$

$72^\circ$

$75^\circ$

$80^\circ$

$90^\circ$

Solution:

When $\angle BAC = 60^\circ,$ a classical fact is that $B,$ $C,$ $O,$ $I,$ and $H$ all lie on a common circle, so $BCOIH$ is a convex cyclic pentagon whose vertices depend only on the shape of the triangle.

Fixing $BC = 1$ and $\angle A = 60^\circ,$ the circumradius is $R = \tfrac{1}{\sqrt3},$ and $O, I, H$ are determined by $\angle CBA = B$ (with $\angle BCA = 120^\circ - B$ ). Writing the pentagon area as a function of $B$ on the allowed range $60^\circ \le B \le 90^\circ$ and maximizing gives an interior maximum at $B = 80^\circ.$

So the maximizing angle is $\angle CBA = 80^\circ.$

Thus, the correct answer is D.

2011 AMC 12A Problem 25

Solution:

Problem 25 in Other Years