On the maximum area of a triangle with the fixed distances from its vertices to a given point. Let $\triangle ABC$ be a triangle with the fixed distances from its vertices to a given point $P$. Prove that when the area of $\triangle ABC$ takes the maximum value, $P$ must be the orthocenter.
I think this question is not easy, and some proof methods are incorrect. For example, fixed points $B, C$, we can easily get when $AP$ is perpendicular to $BC$, the area can be maximized. The same reason we can get when $BP$ is perpendicular to $AC$, the area can be maximized. So when the area of $\triangle ABC$ takes the maximum value, $P$ must be the orthocenter.
But I think the above proof process is not rigorous, and  whether there is a rigorous proof method.

The vertices of $\triangle ABC$ are on three concentric circles sharing one center which is point $P$.  I'll give a simple rigorous proof that point $P$ must be the orthocenter of $\triangle ABC$.  The proof is by contradiction.  Suppose one gives you a triangle where $PA$ is not perpendicular to $BC$, and claims that this is the "maximum area possible" triangle, then, clearly, one can obtain a triangle with a greater area by moving point $A$ on the circle to which it belongs such that $A$ is farthest from segment $BC$ (segment $BC$ here is kept fixed).  This shows immediately, that in the maximum area triangle, segments $PA, PB, PC$ MUST be perpendicular to segments $BC, AC, AB$ respectively.  This completes the proof by contradiction.  This is a rigorous proof.
