Diameter of a triangle

Solution 1:

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.

Solution 2:

Let the vertices be $a_1,a_2,a_3$. Then any $x,y \in T$ can be written as convex combinations of these. Now use the triangle inequality.

More generally, this works for any simplex (or even convex polytopes?) in $\mathbb{R}^n$.