HMMT February February 16, 2019

Solution:

Observe that
$$\angle D B O = \angle D B A + \angle A B O = \angle C B A + \angle B A O = \frac{1}{2}(\angle C B A + \angle B C A) + \frac{1}{2}(\angle B A C) = \frac{1}{2}\left(180^{\circ}\right) = 90^{\circ}.$$
Thus $B C = B D = \sqrt{2}$ by the Pythagorean Theorem on $\triangle D B O$. Then $\angle B O C = 90^{\circ}$, and the distance from $O$ to $B C$ is $\frac{\sqrt{2}}{2}$. Depending on whether $A$ is on the same side of $B C$ as $O$, the height from $A$ to $B C$ is either $1 + \frac{\sqrt{2}}{2}$ or $1 - \frac{\sqrt{2}}{2}$, so the area is $\left(\sqrt{2} \cdot \left(1 \pm \frac{\sqrt{2}}{2}\right)\right) / 2 = \frac{\sqrt{2} \pm 1}{2}$.

Solution:

One can observe that $\angle D B A = \angle C B A = \angle A C B$ by property of reflection and $A B C$ being isosceles, hence $D B$ is tangent to $\Omega$ and Power of a Point (and reflection property) gives $B C = B D = \sqrt{O D^{2} - O B^{2}} = \sqrt{2}$. Proceed as in Solution 1.