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.

EDIT 3: A proof of the result from the first edit was requested: Label the vertices $U_1,\ldots, U_5$ such that consecutive vertices have consecutive indices (mod $5$), then there are two cases, non-consecutive and consecutive vertices.

Case 1:

The vertices are non-consecutive, say $U_1$ and $U_3$. Let $P$ be the point of intersection of the segments $U_1U_4$ and $U_3U_5$. Using the triangle inequality we get

$$ \begin{align} |U_1P| + |PU_3| &> |U_1U_3|,\\ |U_4P| + |PU_5| &> |U_4U_5| \\ \implies |U_1U_4| + |U_3U_5| &> |U_1U_3| + |U_4U_5| \\ \implies s_{U_1} + s_{U_3} &= |U_1U_2| + |U_1U_3| + |U_1U_4| + |U_1U_5| \\ & \hspace{5mm}+ |U_3U_1| + |U_3U_2| + |U_3U_4| + |U_3U_5|\\ &> 3|U_1U_3| + |U_1U_2| + |U_2U_3| + |U_3U_4| + |U_4U_5| + |U_5U_1| \\ &= 3|U_1U_3| + p. \end{align} $$

Case 2:

The vertices are consecutive, say $U_1$ and $U_2$. Just like before we get

$$ \begin{align} |U_1U_3| + |U_2U_4| &> |U_1U_2| + |U_3U_4|,\\ |U_1U_4| + |U_2U_5| &> |U_1U_2| + |U_4U_5| \\ \implies s_{U_1} + s_{U_2} &= |U_1U_2| + |U_1U_3| + |U_1U_4| + |U_1U_5| \\ & \hspace{5mm}+ |U_2U_1| + |U_2U_3| + |U_2U_4| + |U_2U_5|\\ &> 3|U_1U_2| + |U_1U_2| + |U_2U_3| + |U_3U_4| + |U_4U_5| + |U_5U_1| \\ &= 3|U_1U_2| + p. \end{align} $$ $\tag*{$\square$}$

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    $\begingroup$ Cool! Never knew Erdos,Szekeres proposed to Crux $\endgroup$ Sep 7, 2021 at 2:36
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    $\begingroup$ Could you show how to prove that $S_U+S_V>3\mid UV\mid + p$? $\endgroup$ Apr 29 at 15:47
  • 1
    $\begingroup$ @JuanMoreno I just added it under EDIT 3. $\endgroup$ Apr 29 at 20:21
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    $\begingroup$ @user23571113 thanks! $\endgroup$ Apr 29 at 21:57
  • $\begingroup$ @user23571113 I think I have a complete and valid proof, I have posted it as an EDIT to my previous partial answer; check if it looks good for you! $\endgroup$ May 7 at 15:40

2 Answers 2


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.

Pentagon with small excess

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.

  • $\begingroup$ I have added another result at the bottom of my post, it should improve your first observation from $3.5$ to $\frac{10}{3}$ (and $1.75$ to $\frac{5}{3}$), right? $\endgroup$ Sep 15, 2021 at 11:58

If we denote as $S_d$ the sum of the five diagonals, the OP statement is true for every convex pentagon satisfying $$p<\frac{2}{3} S_d$$

This can be easily showed noting that the perimeter (sum of all the edges) and the sum of distances from some vertex to the other four have two edges in common. Thus, the sum of the three non-common edges must be equal or greater than the two diagonals with the common vertex, and this must hold for each vertex of the pentagon, generating five inequalities to hold simultaneously.

Noting that, putting together the inequalities, each edge is counted three times, and each diagonal twice, proves the necessary condition for the OP statement not to hold: $$3p\geq 2S_d$$

Thus, the OP statement is true for every convex pentagon satisfying $$p<\frac{2}{3} S_d$$


After reviewing this problem, I think I got a complete and valid proof.

Without loss of generality, suppose that the perimeter $p$ of the convex pentagon is equal to $1$. Assume that there is no vertex such that the sum of distances to the other four is greater than the perimeter.

Note that the result stated in this answer before the EDIT section implies that $S_d\leq 1.5$, and thus the average of the lengths of the diagonals is at most $0.3$.

Using the Law of Cosines, every side $s_i$ can be calculated as $$s_{i}^2=d_{j}^2+d_{k}^2-2d_{j}d_{k}\cos(\theta_i)$$ Where $\theta_i$ is the interior angle of the star pentagon formed by the diagonals.

The sum of the interior angles of any star pentagon is equal to $180º$; therefore, the average of the interior angles is equal to $36º$. If we calculate the length of some $s_i$ using the average of the interior angles of the star pentagon, and the maximum average length of the diagonals, we get that $$s_{i}^2=(0.3)^2+(0.3)^2-2(0.3)(0.3)\cos(36)$$ $$s_{i}^2=0.18(1-\cos(36))$$ $$s_i\approx0.18541$$

However, the average length of the sides of the convex pentagon, as the perimeter is equal to $1$, equals 0.2. Therefore, there would be needed sides $s_j$ greater than $s_i$ to achieve the perimeter's length, and that could only be achieved (i) with diagonals of greater length than the average, or/and (ii) with interior angles of the star pentagon greater than the average.

Note that, by the result stated in the OP at the EDIT section, every diagonal can be at most equal to $\frac{p}{3}$. If we plug this in the formula for $s_j$ with the average interior angle of the star pentagon, we get that $$s_{j}^2=\frac{2}{9}(1-\cos(36))$$ $$s_j\approx0.206$$

Indeed, $s_j\geq 0.2$ with both diagonals equal to $\frac{p}{3}$ only if $\theta_i>34.9152°$; and if we set $\theta_i=36°$, both diagonals need to be equal or greater than $\frac{1+\sqrt{5}}{10}\approx 0.3236$ to have some $s_j\geq 0.2$. Finally, $s_{j}\geq0.2$ with both diagonals equal to $0.3$ only if $\theta_{i}>38.9424^{\circ}$.

This shows that no matter which greater-than-the-average diagonals we use, or which interior angles of the star pentagon we plug in, we can obtain at most sides such that the sum of the sides $s_i$ obtained with less-or-equal-to-the-average diagonals and interior angles of the star pentagon, and the sides $s_j$ obtained with greater-than-the-average diagonals and/or interior angles of the star pentagon, is less than the perimeter of the convex pentagon; otherwise, the sum of lengths of the diagonals would be greater than the maximum possible of 1.5, and/or the sum of the interior angles of the star pentagon would be greater than 180°.

As we reach a contradiction, the initial assumption that there is no vertex such that the sum of distances to the other four is greater than the perimeter can not be true; therefore, in any convex pentagon there exists at least one vertex such that the sum of distances to the other four is greater than the perimeter.

  • $\begingroup$ I don’t think this proof is complete - it’s missing a lot of detail in the inference from averages and maxima. A bunch of the functions here are very nonlinear, so you can’t just average the maximum possible result with the result from a pointwise average of each parameter and conclude something robustly; it’ll need more precision than that. This angle does seem promising though! $\endgroup$ May 7 at 16:35
  • $\begingroup$ @RavenclawPrefect I have edited the answer to make it clearer and more precise; please let me know if you consider that it is enough to consider it valid, or if not, I would be grateful if you could concrete how to improve it. Thanks! $\endgroup$ May 7 at 23:19

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