Some years ago, I asked some 'famous' people in an advanced Plane Geometry forum about the following:

> *Let $ABC$ be arbitrary triangle, how can one construct a point $P$ in the plane such that $P$ is the circumcenter of the cevian triangle of $P$ with respect to $ABC?$*

The answer was negative, even in the sense of calculations by a Computer (can't construct by rule and compass, even can not calculations by computer). 

> **My conjecture:** *Let $ABC$ be arbitrary triangle, then there exist a point $P$ such that $P$ is the circumcenter of the cevian triangle of $P$ with respect to $ABC$*. 

> **Question:** How can construct this point?

I propose this conjecture to you for a clear proof.

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/a4W0L.png