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.
 A: 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.
A: Let us choose coordinates with origin $P=(0,0)$ and $PA=a,\ PB=b,\ PC=c$ be the fixed distances. Then, $A,B,C$ moves along the circles of radii $a,b,c$ respectively. The area of $\triangle ABC$ is
$$S=\frac{1}{2}ab|\sin\angle APB|+\frac{1}{2}bc|\sin\angle BPC|+\frac{1}{2}ca|\sin\angle CPA|,$$
which is a continuous function on $A,B,C$. Since the domain, the triple product of circles, is compact, we see that a maximum exists.
To see the property of the maximum, let $A,B,C$ be the points where the maximum is attained. To clarify the argument, we separate the cases (i) $P$ is inside $\triangle ABC$, (ii) one vertex, say $B$, is at the opposite to $P$ with regard to the line $AC$. In both cases, we let $\theta=\angle APB$ and $\varphi =\angle BPC$. Then, for (i) we have $\angle CPA=2\pi-\theta-\varphi$ and for (ii) we have $\angle CPA=\theta+\varphi$. By depicting the triangle, we see anyway that the area $S$ is given by
$$S=\frac{1}{2}ab\sin\theta+\frac{1}{2}bc\sin\varphi-\frac{1}{2}ca\sin (\theta+\varphi).$$
Now our $(\theta,\varphi)$ is a local maximum of the two-variable function $S=S(\theta,\varphi).$ So, both the derivatives $\partial S/\partial \theta$ and $\partial S/\partial \varphi$ vanish. Therefore,
$$ab\cos \theta-ca\cos(\theta+\varphi)=bc\cos\varphi-ca\cos(\theta+\varphi)=0.$$
Without loss of generality, we may put $A=(a,0)$ and $A,B,C$ are located counterclockwise. Then, we can compute the inner products as
$$\begin{split}
\vec{OA}\cdot\vec{BC}&=(a,0)\cdot(c\cos(\theta+\varphi)-b\cos\theta,\ast)=0,\\
\vec{OB}\cdot\vec{CA}&=(b\cos\theta,b\sin\theta)\cdot(a-c\cos(\theta+\varphi),-c\sin(\theta+\varphi))=0,\\
\vec{OC}\cdot\vec{AB}&=(c\cos(\theta+\varphi),c\sin(\theta+\varphi))\cdot(b\cos\theta-a,b\sin\theta)=0
\end{split}$$
using the above condition and the addition formula of $\cos$. Therefore, $P$ is the orthocenter of $\triangle ABC$. Maybe we may say that the given argument is correct modulo the existence of the (global) maximum value.
