In Euclidean geometry, Fermat point (X(13) in ETC) is constructed as follows. Let ABC be a triangle. Construct the triangle BA'C having base BC, angles BCA’ and CBA’ = Pi/3 and vertex A' on the negative side of BC; similarly construct triangles CB'A and AC'B based on the other two sides. AA’, BB’ and CC’ concur in a point P. Fermat point (X(13)) has a property such that the angles of BPC, CPA, APB = 2Pi/3.
In hyperbolic geometry, we can construct a point Q in similar manner, though it is not always possible. The angles of BQC, CQA, AQB do not have the exact value 2Pi/3, but near it. It is very likely that there is a point R such that the angles of BRC, CRA, ARB = 2Pi/3. In order to confirm the title, I would like to know how to construct R, if possible.