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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.

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  • $\begingroup$ Hint: There is a slight flaw in your proof. The condition that "$AP \perp BC$" isn't entirely correct, you still need something additional (Otherwise you've locally minimized the area instead). $\endgroup$
    – Calvin Lin
    Feb 24 at 15:44
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    $\begingroup$ @CalvinLin I wonder if this proof is good enough, as it's given that "when $\triangle ABC$ takes the minimum value", and so it's given that $A$ both locally and globally maximises the area. But the proof is not a procedure to find one such $\triangle ABC$ from scratch. $\endgroup$
    – peterwhy
    Feb 24 at 17:24
  • $\begingroup$ @peterwhy No, the proof is flawed because it possible that "when $AP \perp BC$, the area is (locally) minimized". $\quad$ What we want is the converse "When the area is at a (possibly local) maximum, then $AP \perp BC"$. From there, we conclude that that "Hence at the global maximum, $AP\perp BC$ and similar, so $P$ is the orthocenter". (And to me, this proof is rigorous, but it seems like OP has some unstated concerns about it) $\endgroup$
    – Calvin Lin
    Feb 24 at 17:30
  • $\begingroup$ I'll change the question a little: is there a maximum value for the area of triangle $ABC$, and if so, indicate the location of point $P$. $\endgroup$
    – Mr.He
    Feb 25 at 1:42

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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.

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  • $\begingroup$ I'll change the question a little: is there a maximum value for the area of triangle $ABC$, and if so, indicate the location of point $P$. $\endgroup$
    – Mr.He
    Feb 25 at 1:41
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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.

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