# Sum of the lengths of all diagonals in a regular polygon

I believe that the sum of the lengths of diagonals of a regular polygon ($$n$$-gon) is always greater than or equal to any other irregular polygon ($$n$$-gon) inscribed in a circle.

For example for a 4-gon, if we consider a regular $$4$$-gon, then the two diagonals pass through the centre of a circle, resulting in the sum of the diagonals' lengths twice the diameter. Considering any other $$4$$-gon (irregular), the sum of the lengths of the diagonals is not greater than $$2\times$$ the diameter.

Case : 5-gon I proved that the sum of the lengths of diagonals in a regular $$5$$-gon is maximum compared to any $$5$$-gon (irregular polygon). I used Lagrange's multiplier to prove it.

Case : $$n$$-gon I tried proving the same for an $$n$$-gon using Lagrange's multiplier, where we need to maximize the sum of the lengths of all diagonals with respect to $$\theta_1+ \theta_2 + \dots + \theta_n =2\pi,$$ but unable to do the calculation as the number of variables increases. Here $$\theta_i$$ is the angle subtended to the centre of a circle by the side $$a_i$$ of the polygon.

I believe that the sum of the lengths of diagonals is maximum if $$\theta_1 = \theta_2 = \dots = \theta_n,$$ which is possible only in a regular polygon.

Edit: same question posted in mathoverflow, https://mathoverflow.net/questions/398248/sum-of-the-lengths-of-all-diagonals-in-a-regular-polygon

• It is not true that the sum for a nonsquare quadrilateral is always less than twice the diameter. If it's a rectangle, the sum is equal to twice the diameter. Also, you have introduced a symbol $\theta$ without telling anyone what it means (but if it means what I think it means, then your last assertion is incorrect). Jul 16, 2021 at 6:48
• OK, then saying the $\theta_i$ are all equal is the same as saying all the sides are of the same length, which is the same as saying it's a regular polygon. Jul 16, 2021 at 12:10
• Yes, using Lagrange's multiplier I have proved that for 5-gon, the sum of the length of diagonals is maximum if 5-gon is a regular polygon. But I am unable to prove it for n-gon. And refering to your first comment, sum of the length of diagonals is maximum for regular 4-gon. Please check. Thank you
– user413988
Jul 16, 2021 at 13:21
• The sum of diagonals for a rectangle is the same as the sum of diagonals for a square. Please check. Jul 16, 2021 at 21:33
• Yeah, it is difficult to get a grip on it. I am struggling with the same. I will explore further and let you know if I get anything.
– user413988
Jul 17, 2021 at 13:58

For simplicity, let us consider the index $$i$$ of $$P_i$$ as the member of $$\mathbb{Z}_n$$.

We aim to show a stronger criterion : given points $$P_1,P_2,...,P_m$$ on the unit circle $$\Omega$$ lying in the counter-clockwise order, for each $$k such that $$gcd(k,m)=1$$, we have $$\sum_{j=1}^m \overline{P_j P_{j+k}}\le \sum_{j=1}^m\overline{Q_j Q_{j+k}}$$ , where $$Q_1,...,Q_m$$ are vertices of a regular $$(m)$$-gon inscribed in $$\Omega$$.

Assuming this, we can easily show the original conjecture observing that the collection of all diagonal lines of the convex $$n$$-gon $$P_1 P_2...P_n$$ is partitioned into the cycles of the form above.

(Explicitly speaking, we can write

$$\sum_{|i-j|\ge 2}\overline{P_iP_j}=\frac{1}{2}\sum_{k=2}^{n-1}\sum_{j=1}^{gcd(k,n)}\left(\sum_{i=1}^{n/gcd(k,n)}\overline{P_{ki+j}P_{k(i+1)+j}}\right)$$ You can check this holds since each segment of LHS is counted at least twice in RHS, and the total count of segments of RHS is $$(n-1)n$$, which is exactly twice of the total number of terms in LHS. )

For better intuition, I post an example of $$n=8$$.

The collection of the diagonal segments of the octagon above is the disjoint union of the red, green, blue-colored chains plus the four main diagonals. Assuming our criterion is correct, we can argue that the total length of the segments of each color (including the main diagonals) is maximized when the polygon is regular, which concludes our proof.

Now, to show the criterion, we first exclude the trivial case $$m=2$$, and allow some of $$P_j$$'s to be the same. (but not in a switched order)

Writing $$P_j=e^{i\theta_j}$$, one can easily observe that the set of such $$(\theta_1,...,\theta_m)\in \mathbb{T}^m$$ is compact.

(Here, $$\mathbb{T}^m$$ is the $$m$$-dimensional torus $$\mathbb{R}^m/\mathbb{Z}^m$$ with the natural metric)

Thus, there exists a pair $$(\theta_1,...,\theta_m)$$ that maximizes the sum.

Now, observe that for each $$j$$, $$P_j$$ contributes only to two distinct segments : $$\overline{P_jP_{j+k}}$$ and $$\overline{P_jP_{j-k}}$$.

By the sine addition formula $$\sin \alpha+\sin \beta=2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2})$$, this can be maximized only if $$\overline{P_jP_{j+k}}=\overline{P_jP_{j-k}}$$.

It follows that the lengths $$\overline{P_jP_{j+k}}$$ are all equal, which requires $$P_1...P_m$$ to be vertices of a regular $$(m)$$-gon, ending the proof of the desired criterion.

P.S. This is a formalization of the great idea suggested by Toni Mhax.

• By this "collection of all diagonal lines of the convex n-gon P1, P2, ..., Pn is partitioned into the cycles of the form above", you mean to say all diagonals will be partitioned, but I doubt that some diagonals will not be taken into the account. Please make it is easy for me. I did not get the first part.
– user413988
Jul 25, 2021 at 5:46
• You should pay attention when you say each $j$ contributes to two distinct segments this works only in $n=5$... Jul 25, 2021 at 5:56
• @PawanMishra - I added some more explanation. Hope this one helps. Jul 25, 2021 at 6:35
• @ToniMhax - It seems that this works for general $n$, could you see this again and tell me if there is anything suspicious? Jul 25, 2021 at 6:36
• @C.Park changing any two diagonals or the position of a point will change the others also wich makes it tricky. You may take $P_jP_{(j+2)}=P_jP_{j-2}$ but maybe this decreases the other distances.... Jul 25, 2021 at 6:44

Yes this is true (presumably but this is not complete), in fact a proof can be done along these lines, sketch. First you prove that for three points $$A,B,C$$ on a circle, the maximum of the distance $$AB+BC$$ for $$A$$ and $$C$$ fixed is for $$B$$ middle of the arc $$\overset{\frown}{AC}$$. (Here is a proof in French A proof), you should prove also that getting away from the middle of the arc will decrease the distance sum. From that a circular (and averaging) reasoning for any $$n$$-gone inscribed in a certain circle gives a convergence towards the regular polygone as maximal for the sum of diagonals.

For example here is the figure for a pentagone, for general $$n\ge 6$$ say $$n=6$$ you pick the middle of the two arcs: points $$P$$ and $$Q$$, take the point that maximises the sum of the three diagonals submerged, it should be inside the arc $$\overset{\frown}{PQ}$$ etc An animation is possible maybe someone could make this...

EDIT i realised this is more complicated than this anyway i'll keep the answer.

• This answer is not very rigorous yet. Jul 25, 2021 at 4:54
• Actually, I made a rigorous version of proof filling in some of the details; hope it matches up Jul 25, 2021 at 5:00
• @ToniMhax- I got your idea. But I feel it is difficult to generalize for all the diagonals. I proved it for 5-gon using Lagrange's multiplier as in 5-gon, for each diagonal only two angles are to be looked for. I mean for diagonal line $P_1P_3$, we need to look $\theta 1, \theta 2$. So, it is easy to solve.
– user413988
Jul 25, 2021 at 6:03
• Yes no proof...(sry) Jul 25, 2021 at 6:18
• I can see the proof (based on your idea), but it is difficult to jot down. Some minor thread is missing in my understanding.
– user413988
Jul 25, 2021 at 12:23