Prove that for one vertex of a convex pentagon, the sum of distances to the other four is greater than the perimeter

The problem is from the journal 'Crux Mathematicorum', originally proposed by Paul Erdős and Esther Szekeres for the case of a convex $n$-gon with $n > 5$, and can be found here together with a proof for that case (pdf page 20) . Unfortunately they leave the case $n = 5$ open to the reader, so I would like to know how to prove:

Any convex pentagon has a vertex whose sum of distances to the other four vertices is greater than the perimeter of the pentagon.

I was not able to extend the method from the pdf above to the $n = 5$ case, because it relies on comparing the perimeter of the pentagon to that of a regular $n$-gon centered on the centroid of the original polygon and then using the inequality $\sin(\frac \pi n) \leq \frac 1 2 $, which is not true for $n = 5$.

Thanks in advance for any help!

EDIT: I would prefer a proof that would be feasible in the context of a math competition like the IMO or Putnam, but any kind of result is appreciated.

One more result that might be helpful: If for two distinct vertices $U$ and $V$ we denote by $s_U$ and $s_V$ the sum of distances from $U$ and from $V$ respectively, one can show that

$s_U + s_V > 3\vert UV\vert + p$,

where $p$ is the perimeter of the pentagon and $\vert UV\vert$ is the distance from $U$ to $V$. Therefore if we had $\vert UV\vert \geq \frac{p}{3}$, this would give us a proof, so we can assume wlog that the distance between any two vertices is at most $\frac{p}{3}$.

EDIT 2: If we could prove the inequality in this post, we would have a proof using the inequality from the pdf.

Solution 1:

This is nothing like a solution, but since the question asks for any help, here are some observations.

Notation: The pentagon vertices are $ABCDE$ in counterclockwise order; the perimeter is $p$, the sum of distances from vertex $V$ is $s_V$, the maximum sum is $m = \max\{s_V\}$, and the excess is $e=m-p$. The claim to prove is that $m>p$.

Without loss of generality we may assume that the longest side has length $1$. Then $p \le 5$. Using this we can isolate at least some cases.

Observation 1. If $D$ is far from $A$ and $B$, more precisely, if $|AD|+|BD| \ge 3.5$, then $|CD|+|ED| > 1.5$ (because $|AE|\le 1$ and $|BD|\le 1$), thus $m > 5 \ge p$. So, in any possible counterexample, from any vertex, the sum of the two diagonals must be less than $3.5$.

Corollary 1. In any counterexample, at least three diagonals are short (= has length smaller than $1.75$). This is because from any vertex we have at least one short diagonal, and there are five vertices (two short diagonals are not enough because they can have at most four endpoints).

Corollary 2. In any counterexample, there is a vertex with two short diagonals. This is because there are three short diagonals and at least two of them must have a common endpoint.

That makes the problem somehow bounded (we do not need to care about extremely elongated pentagons). However, any possible proof cannot rely solely on the fact that $p \le 5$, because we can easily find pentagons where $m < 5$. Also, the excess can be made smaller than $0.206$.

Numerical example. This pentagon was found by a simple grid search, with $A$ fixed at $(0,0)$ and $B$ at $(1,0)$, and the six coordinates of $C,D,E$ searched for in a grid with successive refinements. It is mirror-symmetric but this was not enforced in the search.

$$ A=(0,0)\\ B=(1,0)\\ C=(1.1699928358529263, 0.5584839529924609)\\ D=(0.5, 1.076262271881104)\\ E=(-0.1699928358529265, 0.5584839529924609) $$

With this we have perimeter $p \approx 3.861064$ and maximum sum $m \approx 4.066970$, thus excess $e \approx 0.205906$. In fact, here we have $s_V \approx m$ for all five vertices. The distance matrix is: $$ \begin{bmatrix} 0 & 1.0000 & 1.2965 & 1.1867 & 0.5838 \\ 1.0000 & 0 & 0.5838 & 1.1867 & 1.2965 \\ 1.2965 & 0.5838 & 0 & 0.8467 & 1.3400 \\ 1.1867 & 1.1867 & 0.8467 & 0 & 0.8467 \\ 0.5838 & 1.2965 & 1.3400 & 0.8467 & 0 \end{bmatrix} $$ and all its column sums are roughly equal.

Thoughts towards a proof: One may need to divide the parameter space into a few cases (like Observation 1 above), and handle each case with a different proof. The fact that the numerical minimal-excess solution has all $s_V$ equal is also suggestive; perhaps there is a local differential argument for this.

In fact, one could probably produce a kind of a proof by running a meticulous grid search (which I did not do); if the grid is fine enough, one could then appeal to the fact that $p$ and $m$ cannot vary too much within a grid cell. Such a proof would settle the truth but would not be geometrically very appealing.