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If $T\subseteq \mathbb R^2$ is a generic plane triangle, I want to find its diameter $$d=\sup\{\lvert| x-y \rvert|: x,y\in T\}$$ Intuitively I think that $d$ is the length of the longest edge of $T$. How can I formally prove this?

Adriano
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Dubious
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2 Answers2

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Given any triangle $T$, in fact any bounded closed convex subset of $\mathbb{R}^2$. The map

$$T^2 \ni (x,y) \mapsto |x-y|^2 \in \mathbb{R}$$

is a continuous function on $T^2$ bounded from above. Since $T^2$ is compact, the map achieves its maximum on some $(u, v) \in T^2$. i.e.

$$\sup \{\;|x-y|^2 : x, y \in T\;\} = | u - v |^2.$$

If either $u$ or $v$ is not an extremal point of $T$, say $u = \frac12 ( u_1 + u_2 )$ where $u_1, u_2 \in T$, then by substituting $x_1$ by $u - v$ and $x_2$ by $\frac12 ( u_1 - u_2 )$ into parallelogram identity: $$| x_1 + x_2 |^2 + |x_1 - x_2|^2 = 2 ( |x_1|^2 + |x_2|^2 )$$ one find at least one of $|u_1 - v|^2$ and $|u_2 - v|^2$ is greater than $|u-v|^2$. This contradicts with the role of $u, v$ that maximize $|u - v|^2$.

As a result, the $u, v$ that maximize $|u-v|^2$ are both extremal points and

$$\sup\{\;|x-y|^2 : x,y \in T\;\} = \max\{\;|u-v|^2 : u, v \in T, \text{ both extremal}\;\}$$

A triangle has 3 extremal points, i.e. its 3 vertices, and hence its diameter is the length of its longest edge.

achille hui
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    This is very elegant! If I interpret it correctly then you are essentially using that $\Vert x-y \Vert^2$ is a convex function of both arguments, and therefore attains its maximum at extremal points on a convex set. – Martin R Sep 24 '21 at 07:04
  • @MartinR yup. that's the idea. – achille hui Sep 24 '21 at 07:04
  • One could also argue that $\Vert x-y \Vert$ (without the square) is convex, because of the triangle inequality. – Martin R Sep 24 '21 at 07:06
  • @MartinR yes, you can. However, for Euclidean spaces, the squared distance is slightly better because it is strictly convex. – achille hui Sep 24 '21 at 07:54
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[intuitive answer]

If $y$ is fixed and if $x$ is a point in interior, we draw a small ball $B(x,r)$.

Then there exists a point $x'\in \partial B(x,r)$ such that

$$ d(x',y) =d(x,y)+r $$

HK Lee
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