5. Isosceles triangle $A B C$ with $A B=A C$ is inscribed in a unit circle $\Omega$ with center $O$. Point $D$ is the reflection of $C$ across $A B$. Given that $D O=\sqrt{3}$, find the area of triangle $A B C$.
Answer: $\frac{\sqrt{2}+1}{2}$ OR $\frac{\sqrt{2}-1}{2}$
Solution 1. Observe that
$$
\angle D B O=\angle D B A+\angle A B O=\angle C B A+\angle B A O=\frac{1}{2}(\angle C B A+\angle B C A)+\frac{1}{2}(\angle B A C)=\frac{1}{2}\left(180^{\circ}\right)=90^{\circ} \text {. }
$$

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 2. 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 .

Note. It was intended, but not specified in the problem statement that triangle $A B C$ is acute, so we accepted either of the two possible answers.