# 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.

• 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). Feb 24 at 15:44
• @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. Feb 24 at 17:24
• @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) Feb 24 at 17:30
• 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 $P$. Feb 25 at 1:42

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.
• 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 $P$. Feb 25 at 1:41
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.