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The following seems kind of obvious:

Let $\Delta ABC$ be a triangle in the plane and $P, Q$ two points in $\Delta ABC$ (in the interior or on the boundary). Then $|PQ| \le l$ where $l$ is the length of the longest triangle side.

Here $|PQ|$ denotes the Euclidean distance from $P$ to $Q$.

I can prove it (see below), but that proof does not provide much insight why this estimate is true. Is there a (simpler?, perhaps geometric?) proof which makes the inequality really obvious?

The fact that the triangle is convex probably plays a role, but convexity alone is not sufficient: The above is wrong for convex polygons with more than three sides.

My proof: Using barycentric coordinates we have

$$ P = p_1 A + p_2 B + p_3 C \\ Q = q_1 A + q_2 B + q_3 C $$ where the $p_i, q_i$ are non-negative real numbers with $\sum_{i=1}^3 p_i = \sum_{i=1}^3 q_i = 1$. At least two of the three differences $p_i - q_i$ must have the same sign, without loss of generality we can assume that $p_1-q_1$ and $p_2 - q_2$ have the same sign (both are non-negative or both are non-positive).

Using $\Vert \cdot \Vert$ for the Euklidean norm now, we have $$ \begin{align} \Vert P-Q \Vert  &= \Vert (p_1-q_1) A + (p_2-q_2)B + (p_3-q_3)C \Vert \\ &= \Vert (p_1-q_1) (A-C) + (p_2-q_2)(B-C)\Vert \\ &\le |p_1 - q_1| \cdot \Vert A-C \Vert + |p_2 - q_2| \cdot \Vert B-C \Vert \\ &\le \bigl( |p_1 - q_1| + |p_2 - q_2| \bigr) \cdot l \\ &= |(p_1 - q_1) + (p_2 - q_2)| \cdot l \\ &= |p_3 - q_3 | \cdot l \\ &\le l \, . \end{align} $$

Michael Rozenberg
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Martin R
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  • It seems that this has been asked and answered here: https://math.stackexchange.com/q/459357/42969 (which I had not found before). I'll check that and perhaps delete the question. – Martin R Sep 23 '21 at 13:27
  • If we take the projection of $PQ$ on the longest side, we should be able to geometrically show it. – Math Lover Sep 23 '21 at 13:30
  • Consider any $P$ and $Q$. Let $P_1$ and $Q_1$ are two points lying on the boundary of $ABC$ such that $P$ and $Q$ lie on the interval $P_1Q_1$, then $PQ\leq P_1Q_1$. So it is sufficient to prove for the boundary. – Ivan Kaznacheyeu Sep 23 '21 at 13:35
  • @MathLover: But the projection decreases the length, doesn't it? – What I am looking for is a “rigorous, but insightful” proof (if that exists :) – Martin R Sep 23 '21 at 14:08
  • As I mentioned above, this has been asked before. Achille's answer is exactly what I was looking for: $\Vert P-Q \Vert$ is a convex function of both arguments, therefore it attains the maximum at the extremal points of the triangle. – Martin R Sep 24 '21 at 06:52

1 Answers1

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It's enough to solve our problem for $P\equiv A$ and $Q\in BC$.

If $b$ or $c$ is a greatest side, it's obvious.

Let $a$ be a greatest side.

Thus, there are $x\geq0$, $y\geq0$ such that $x+y=1$ for which: $$PQ=|\vec{PQ}|=|x\vec{AB}+y\vec{AC}|\leq xc+yb\leq xa+ya=a.$$

Michael Rozenberg
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