An open triangle problem in plane geometry 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 prove conjecture and how construct this point?


 A: This is a comment, but too long for that.  One can try to apply the $p,q$ method, i.e., assume that the three vertices are $(0,0)$,  $(1,0)$ and  $(p,q)$  respectively.  Then if $P$ has barycentric coordinates $(\lambda_1,\lambda_2,\lambda_3)$ with respect to the vertices, the equality of the squares of the lengths from $P$ to the vertices of the cevian triangle, together with fact that they sum to $1$, provides three equations for the $\lambda$‘s.  Messing about suggest that these have precisely one solution for points in the interior—perhaps this can tightened to a rigorous proof by somebody with more facilities than I have available in these pandemic times.
A: Here is a proof that is pretty standard for existence and unicity in such geometric setting by convergence, in the figure the initiale cevians intersect at $P$. Arrange the segments to the sides in increasing order, here  $PF\le PE\le PG$. It is easy to see that the orange parts are smaller than their corresponding segments $(PE,PF)$ and  the green ones are bigger, so now consider $PG$ and $PF$, and move the point $P$ down along the $[AG]$ segment until you get an equal segment to $PG$ but in this way the decay between $PG$ and $PF$ decreases, and so on you can make the same reasoning even with equal part segments to have  a convergence towards one single point inside the triangle.
