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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.

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?

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.

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 $ABC$

I propose this conjecture to you for a clear proof.

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.

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.

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 $ABC$

I propose this conjecture to you for a clear proof.

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.

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 $ABC$

I propose this conjecture to you for a clear proof.