Another solution to the problem:

Draw the line XP and the altitude, ha, at X. Draw the circle with
center P passing through O and intersecting ha at R. Let Q be the
midpoint of OR and draw the line PQ intersecting ha at A.

The solution is based on the fact that AP is the angle bisector of
angle XAO.

Regards,
Harold Connelly


--- In Hyacinthos@yahoogroups.com, "Paul Yiu" <yiu@f...> wrote:
>
> Dear Ricardo,
> [RB]: To construct the ABC triangle knowing:
> O, cicuncenter,
> X [your Ha] the foot of the height from vertex A, and
> P [your Wa] the foot of bisecting line from vertex A.
>
> [PY]: A necessary and sufficient condition for the existence of the
> triangle is that O does not lie inside the circle (C), center P,
> passing through X. ...
>
> *** After posting this construction, I checked with Wernick's
article
> in Math. Magazine, volume 55 (1982) number 4. This problem was one
> among his list, and Wernick had a solution. He posed it as Problem
> 1149(b) in the same issue. A solution by J. M. Stark appeared in
> volume 57, number 1. It was the same construction as I gave, with a
> very simple proof. I, however, designed the construction by first
> solving the problem analytically and then interpreted the resulting
> expressions geometrically.
>
> Best regards
> Sincerely
> Paul