Answer: The three points divide the line into four parts, two of which have finite length. If the ratio of the larger of these parts to the smaller is greater than $4$ then there are $3$ such triangles, otherwise their are only $2$. Let the triangle be $ABC$ with $AB=AC$ and label the points $I$ and $I_A$ (If it is another excenter then it will not lie on the line created by the incenter and circumcenter). Notice $B$ and $C$ lie on the circle with diameter $I_AI$ and are also the tangents from $A$ to this circle. Let this circle have center $M$. Notice that $A$ must lie outside the circle. The point $O$ is the midpoint of $AM$ and thus can take on all positions to the same side of the midpoint of $IM$ as $I$. This is sufficient to lead to the above answer. (Given the label of each point we can uniquely reconstruct the triangle).

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