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This is a preliminary in the proof of Cauchy-Goursat Theorem for Triangles in complex analysis,

The solid triangle enclosed by $a,b,c$ in the complex plane is defined as $$\Delta(a,b,c):=\{t_1a+t_2b+t_3c: t_1+t_2+t_3=1,t_i\geq0, i=1,2,3 \}$$

I already proved that this is a convex and compact set, now I need to prove that the diameter of this triangle is $$diam(\Delta(a,b,c))=max\{|a-b|,|b-c|,|c-a|\}$$

from basic ideas of metric on $\mathbb{C}$, I can see that the max distance cannot be attained by a straight line joining two interior points and hence the maximum distance of a straight line joining any two points can be attained between two points lying on the boundary of the triangle,

Now I stuck here, how to conclude from here that the max distance is $\max\{|a-b|,|b-a|,|c-a|\}$, Kindly give some suggestions or solutions or ideas.

We can assure that the max is attained by constructing the continuous function $f:T^2\to \mathbb{R}: f(x,y)=|x-y|$, since it is a real valued continuous function on a compact set the maximum value is attained.

Praveen Kumaran P
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    For a point $p$ and a line segment $[a, b]$, can you show that the maximum distance from $p$ to any point on the line is either $|p-a|$ or $|p-b|$? – Calvin Lin Feb 28 '24 at 18:25
  • @CalvinLin follows from Pythagoras theorem? – Praveen Kumaran P Feb 28 '24 at 18:27
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    Sure, that's one way of doing it. Now show that this problem is an immediate corollary (IE We only care about endpoints) – Calvin Lin Feb 28 '24 at 18:29
  • @CalvinLin yes I can see a proof coming, but do you have any other way to prove the claim on your first comment? – Praveen Kumaran P Feb 28 '24 at 18:30
  • The next step might be proving that the maximum cannot be attained by points that are not vertices, i.e. along the lines of your previous step, just now "interior" means interior to the boundary segment, i.e. not a vertex. (You might have to consider a couple of cases separately, together maybe with the fact that triangles are convex.) – Julio Di Egidio Feb 28 '24 at 18:33
  • @PraveenKumaranP Pythagorean theorem is a great simple approach. Love that. No reason to complicate it. $\quad$ But if you insist ... Another approach could be to use the more complicated idea that in a triangle, the largest side is opposite the largest angle. Show that for $ C \in [A, B]$, $\angle PCA$ or $\angle PCB$ is $\geq 90^\circ$ and hence the largest in the corresponding triangle, so $ |PC| \leq \max (|PA|, |PB| )$ – Calvin Lin Feb 28 '24 at 18:37
  • @CalvinLin I edited the question with an argument for this claim, But it feels a bit suspicious for me... is that argument correct? – Praveen Kumaran P Feb 28 '24 at 19:15
  • Your proof looks wrong. I don’t think the sum being less than twice the distance YP is strong enough since that can be true in practice. – Eric Feb 28 '24 at 19:32
  • @Eric yes that's a silly mistake ... – Praveen Kumaran P Feb 28 '24 at 19:55
  • @PraveenKumaranP Your proof doesn't use Pythaorean theorem at all. How I was thinking of using it, is to drop the perpendicular from $P$ to line $AB$ to get point $D$ (which may not be on segment $AB$). Then $|DC| \leq \max (|DA|, |DB|)$ and so $ |PD|^2 + |DC|^2 \leq \max ( |PD|^2 + |DA|^2, |PD|^2 + |DB|^2|$. – Calvin Lin Feb 28 '24 at 20:21
  • Alternatively, just use Triangle inequality: $a|X|+b|Y|=|aX|+|bY|>|aX+bY|$, so |X| is convex, so $f(x,y)=|x-y|$ is convex. – Eric Feb 29 '24 at 02:18

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