, 2014

Solution:
Answer: $\pi\left(\frac{3-\sqrt{3}}{3}\right)-1$

We will make use of the following lemmas.

Lemma 1: If $ABC$ is a triangle with incenter $I$, then $\angle BIC = 90 + \frac{A}{2}$.

Proof: Consider triangle $BIC$. Since $I$ is the intersection of the angle bisectors, $\angle IBC = \frac{B}{2}$ and $\angle ICB = \frac{C}{2}$. It follows that
$$\angle BIC = 180 - \frac{B}{2} - \frac{C}{2} = 90 + \frac{A}{2}.$$

Lemma 2: If $A$ is on major arc $BC$, then the circumcenter of $\triangle BIC$ is the midpoint of minor arc $BC$, and vice-versa.

Proof: Let $M$ be the midpoint of minor arc $BC$. It suffices to show that $\angle BMC + 2\angle BIC = 360^\circ$, since $BM = MC$. This follows from Lemma 1 and the fact that $\angle BMC = 180 - \angle A$. The other case is similar.

Let $O$ be the center of $\omega$. Since $BC$ has the same length as a radius, $\triangle OBC$ is equilateral. We now break the problem into cases depending on the location of $A$.

Case 1: If $A$ is on major arc $BC$, then $\angle A = 30^\circ$ by inscribed angles. If $M$ is the midpoint of minor arc $BC$, then $\angle BMC = 150^\circ$. Therefore, if $I$ is the incenter of $\triangle ABC$, then $I$ traces out a circular segment bounded by $BC$ with central angle $150^\circ$, on the same side of $BC$ as $A$.

Case 2: A similar analysis shows that $I$ traces out a circular segment bounded by $BC$ with central angle $30^\circ$, on the other side of $BC$.

The area of a circular segment of angle $\theta$ (in radians) is given by $\frac{1}{2} \theta R^2 - \frac{1}{2} R^2 \sin \theta$, where $R$ is the radius of the circular segment. By the Law of Cosines, since $BC = 1$, we also have that $2R^2 - 2R^2 \cos \theta = 1$. Computation now gives the desired answer.