Let $M$, $N$ be the midpoints of the sides $AB$, $AC$, respectively. Since $\angle AMO = \angle ANO = 90^\circ = \angle AIO$, we see that the points $A$, $M$, $N$, $I$ lie on the circle having $AO$ as a diameter. In particular, the quadrilateral $AMIN$ is inscribed in this circle, and hence we have $\angle ANI = \angle BMI$.

Since we have from the triangle inequality that $BC > |AB - AC| = 1$, we can take a point $D$ on the side $BC$ in such a way that $BM = BD$ holds. From $BM = BD$ and $\angle MBI = \angle DBI$, we get the fact that the triangles $MBI$ and $DBI$ are congruent, and hence that $\angle BMI = \angle BDI$.

Thus we have $\angle BDI = \angle ANI$, from which it follows that $\angle INC = \angle IDC$. Since we also have $\angle NCI = \angle DCI$, as $I$ is the incenter of the triangle $ABC$, we conclude that the triangles $NCI$ and $DCI$ are congruent. Hence, we have $CD = CN = \frac{3}{2}$, and therefore, $BC = BD + DC = BM + CN = 1 + \frac{3}{2} = \frac{5}{2}$.

Denote by $H$ the foot of perpendicular line drawn from $A$ to the line $BC$. From $2^2 + (\frac{5}{2})^2 > 3^2$, we see that the point $H$ lies on the side $BC$. If we let $BD = x$, then we get from the Pythagorean theorem that $2^2 = x^2 = 3^2 - (\frac{5}{2} - x)^2$. Solving this, we get $x = \frac{1}{4}$. We then have $AH = \sqrt{2^2 - \frac{1}{16}} = \frac{3\sqrt{7}}{4}$ and finally, we obtain that the area of the triangle $ABC = \frac{1}{2} BC \cdot AH = \frac{15\sqrt{7}}{16}$.