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A regular $n$-gon contains a regular $(n+1)$-gon. That is, they are in the same plane, and no part of the regular $(n+1)$-gon is outside of the regular $n$-gon. None of their sides coincide. There are no other restrictions.

Among the integers $n\ge3$, what is the absolute maximum number of points of contact between the inside polygon and the outside polygon?

Context

I made up this question. I think it's a natural question (that I haven't found asked anywhere) with a non-obvious answer.

My attempt

My guess is that the answer is four, and that this maximum can be attained for every $n\ge4$, and that when this maximum is attained the polygons share a line of symmetry. I think any explanation must take into account the fact that $n$ and $n+1$ are coprime, but I don't have a clear idea about this.

I made a desmos graph of a regular $6$-gon containing a regular $7$-gon. You can use the sliders to rotate and expand the hexagon, or translate the heptagon up and down.

EDIT

For example, here is a regular $4$-gon containing a regular $5$-gon, with four contact points. The equations of the lines can be found in this desmos graph.

enter image description here

Dan
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  • "No other restrictions"? Not even on the minimum size or the minimum number of contact points of the (n+1)-gon? Well in that case there are many different answers, but I'll go for the low-hanging fruit: none, if the radius of the (n+1)-gon is below a certain limit and its centre coincides with that of the n-gon! – H. sapiens rex May 17 '23 at 09:12
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    @H.sapiensrex The question asks for the maximum number of contact points. Clearly there are many examples in which there is at least one contact point, so the answer is not zero. – Dan May 17 '23 at 09:17
  • @Dan my point is that it's an ill-posed question. With no restrictions on the size or placement of the (n+1)-gon (other than that it cannot lie outside the n-gon), there is no unique (n+1)-gon to consider and so it is meaningless to ask what the maximum number of contact points are, because that implies that there is only one type of (n+1)-gon under consideration. So I ask you again: are you sure there are no other restrictions? – H. sapiens rex May 17 '23 at 09:25
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    @H.s, for each $n$, it's the maximum over all possible sizes and placements of the two regular polygons (subject to the restrictions mentioned in the question). – Gerry Myerson May 17 '23 at 09:35
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    @H.sapiensrex There are no other restrictions. I'm not sure I follow you: first you said "there is no unique $(n+1)$-gon", then you said "there is only one type of $(n+1)$-gon". Seems contradictory, but perhaps I misunderstand. Anyway, do you think it is possible for a regular $n$-gon to contain a regular $(n+1)$-gon, such that none of their sides coincide, and they have more than four contact points? I think it's not possible. – Dan May 17 '23 at 09:35
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    @Dan I meant that the question seems to imply that there's only one type of (n+1)-gon to consider, but does not actually assert so; instead, it seems to give you the freedom to pick whatever (n+1)-gon you want and asks for the answer based on that. I apologize for confusing you, it was really a tongue-in-cheek interpretation on my part. I am actually in the process of investigating it myself lol – H. sapiens rex May 17 '23 at 09:43
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    Even without being regular, I would usually expect there to be max 4 contact points when you have a convex polygon containing a "more convex" polygon (i.e. larger vertex angles, so that the inner polygon can only touch the sides, not the vertices of the outer poly). For any orientation of the inner polygon, it can be maximally scaled to touch at least 3 sides of the outer one. Generally it will touch 3 sides, but as you rotate it, a contact point will transfer from one edge to another, and momentarily give you 4 contact points. I don't think simultaneous transfers for 5 points will occur here. – Jaap Scherphuis May 17 '23 at 10:00
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    @JaapScherphuis If the polygons are convex but not necessarily regular, there can be more than four contact points. Choose one point on each side of a convex pentagon, and call them $A,B,C,D,E$. Certainly there is a point inside the pentagon, call it $F$, such that hexagon $ABCDEF$ is convex. So we have a convex pentagon containing a convex hexagon, with five contact points. – Dan May 17 '23 at 10:10
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    @JaapScherphuis Without being regular, one can easily imagine a 100-gon inscribed in a 99-gon so that each but one vertex of the 100-gon sits somewhere on one of the 99-gon's edges. Other than that, your reasoning seems sound: we have 4 degrees of freedom for translation/rotation/scaling of the inner polygon, so it is natural to expect that we can match at most 4 contact points. – Ivan Neretin May 17 '23 at 10:10
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    @IvanNeretin If we have an $n$-gon containing an $m$-gon, where $n$ and $m$ can be any integers greater than $2$, could we still apply the reasoning "we have 4 degrees of freedom... at most 4 contact points"? If so, then we have a problem: a regular $5$-gon can contain a regular $10$-gon with five contact points. – Dan May 17 '23 at 10:16
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    @Dan Yes, of course. That's why your comment about n and n+1 being coprime feels important, only I don't exactly see how to put it all together. – Ivan Neretin May 17 '23 at 10:22
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    @Dan I'm not saying more than 4 cannot occur - I'm saying that more than 4 points will only occur in special circumstances. A max of 4 points is the norm, and will generally always be possible (given the "more convex" condition). To get more than 4 the inner polygon will have to be specially suited to the outer polygon. Regular polygons with a common factor greater than 2 in the number of sides is one example of being specially suited. – Jaap Scherphuis May 17 '23 at 11:24
  • No two points of contact can be on the same side of the larger polygon, so there can be no more than $n$ points of contact. That contradicts your guess of four when $n=3$. – TonyK May 17 '23 at 11:43
  • @TonyK My conjecture is not that the maximum is four each and every value of $n$; my conjecture is that the absolute maximum, when allowing any value of $n$, is four (and that this absolute maximum can be attained for any value of $n$ greater than $3$). – Dan May 17 '23 at 12:15
  • Well that's clear, now that you've edited the question! – TonyK May 17 '23 at 17:26
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    @Dan: After some effort on your problem, I believe that the maximum number of contacts (under the restrictions you propose) is not four but three. The apparent example of four points of contact that you offer with Desmos, could be an optical illusion (when you "see" that there are four points, with the help of the zoom you will be able to see that it is not so in reality). Otherwise, could you give a numerical example of four contact points.? – Piquito May 25 '23 at 23:09
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    @Piquito I've added to the question an example with four contact points. – Dan May 26 '23 at 02:47
  • @Dan: Thanks you very much. I follow trying your nice problem and your example with the pentagon will help me. Thanks again. – Piquito May 26 '23 at 13:03
  • Similar question now posted to MO by OP: https://mathoverflow.net/questions/447594/a-regular-n-gon-contains-a-regular-m-gon-with-n-m-coprime-no-sides-coinc – Gerry Myerson May 28 '23 at 01:55
  • @Gerry Myerson. Be kind, please, with someone whose English is weak: I can't understand your objection to the deleted answer given by Prem. What part of the heptagon is outside of the hexagon? – Piquito May 31 '23 at 20:39
  • @Piq, I think you have to look at the revision history. My recollection is that when I left that comment, parts of the heptagon were outside of the hexagon, and that Prem edited the answer in response to my comment. You will notice that Prem thanked me for my comment. I'm not always kind, but I think that in this instance your criticism misses the mark. – Gerry Myerson May 31 '23 at 21:53
  • @Gerry Myerson: Thanks you very much. I understand now (so is it false that the max is four as the OP supposed? – Piquito May 31 '23 at 22:32
  • @Piq, I don't think we have agreed on an answer to the question of whether it is true or false that the max is four. – Gerry Myerson Jun 01 '23 at 04:31
  • @GerryMyerson I wonder, what is your hunch about whether the max is four? – Dan Jun 01 '23 at 04:33
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    Sorry, Dan, I really don't have one. – Gerry Myerson Jun 01 '23 at 04:42
  • @Dan The hexagon and heptagon given by Prem in his deleted answer, does not show that there are five contact points in this example? – Piquito Jun 01 '23 at 12:25
  • @Piquito I can lo longer see Prem's deleted answer, but if I recall correctly, in his example, when I zoomed in I could see that the so-called contact points were not really contact points. – Dan Jun 01 '23 at 12:32
  • @Dan Aha, very well. Thanks. I feel to have a method to finish and you can explore it: There is a vertex of the (n+1)-gon that is not in any side of the n-gon. Try first with some symmetry (which seems to be necessary) and put this vertex $A_1$ from which start two SMALL SIDES $A_1A_2$ and $A_1A_{n+1}$. The circumcenter of the resulting triangle allows you to draw a small (n+1)-gon inside the n-gon. After this, you could proceed by parallelisms to examine the posible desired contact points. – Piquito Jun 01 '23 at 13:13
  • @Dan Obviously, the angle $\angle{A_{n+1}A_1A_2}$ must be the corresponding angle. – Piquito Jun 01 '23 at 13:22
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    @Gerry Myerson I was thinking that the figure in the deleted answer was correct when I wrote what you reply . Thanks you very much for all. – Piquito Jun 02 '23 at 12:37
  • Clarification: For $n=3$ the answer seems to be 3, right? However, you wrote $n \geq 3$ so I assume you considered this case specifically. Also, by "none of their sides coincide" I assume you mean "no edge from either polygon is collinear with any other such edge"? – geometrian Jun 18 '23 at 15:31
  • @geometrian Yes, for $n=3$ the maximum number of contact points is three. For all $n>3$, I conjecture that the maximum number of contact points is four. And yes, by "none of their sides coincide", I mean that no edge from either polygon is collinear with any other such edge. (If there were coincident sides, then there would be infinite contact points.) – Dan Jun 18 '23 at 21:06
  • Beautiful problem! – Per Alexandersson Nov 29 '23 at 21:04

2 Answers2

4

This is a fascinating problem! I was able to come up with a partial proof, proving some results and bounds. Specifically, when $n = 3$, the maximum is exactly $3$. For $n>3$, the maximum is at least $4$ (proven constructively). For all $n$, the number of intersections is no more than $n$. And because of that, when $n = 4$, the maximum is exactly $4$. In sum:

  • $n=3$: max is $m=3$
  • $n=4$: max is $m=4$
  • $n \geq 5$: max is $4 \leq m \leq n$

Upper bound (medium): There are at most $n$ intersections.

For any "outer" polygon $A$ (regular with sides $n$) and inner polygon $B$ (regular with sides $n+1$), consider the bounding circle of $B$. This circle passes through $A$ at most $2n$ times (twice on each edge), and these are the only places where $B$ might intersect $A$. Since $B$ cannot select both intersections on the same edge of $A$ (else that would create a collinear edge), there are at most $n$ intersections.

Note that this contradicts your conjecture when $n < 4$. Specifically, when $n = 3$, we can only have $3$ intersections. Actually, we can achieve this upper bound:

square inscribed in triangle according to below description

The point at the bottom is a midpoint, which allows the other two vertices to touch exactly (there are other ways to do it, but that's easiest to prove).


Corollary (near-trivial): For $n = 3$, the maximum number of intersections is $3$.

The diagram above constructively shows a solution with three vertices, and per the upper bound it cannot be higher than this. It follows that the true maximum is $3$.


Lower bound (easy): There are at least $2$ intersections.

We will get a stronger bound later, but it's much easier to lower-bound it with $2$ by doing something like the following prototypical construction:

tiny octagon inscribed in heptagon according to below description

Basically just make $B$ smaller than a side of $A$ and shove it in a corner where it touches two sides in two places. The one vertex of $A$ turns by $2\pi/n$ and the two vertices of $B$ together turn by $4\pi/(n+1)$, which is always (i.e. for $n>1$) more, so $B$ never intersects those edges in more than two points, and it can't intersect any other edges because it's smaller than a side of $A$.


Lemma (easy): All intersections lie on the interior of edges.

Suppose by way of contradiction that an intersection $p$ could lie on a vertex of $A$. Because the turning angle of $B$ is smaller than that of $A$, at least one of the connected edges in $B$ would fall outside of $A$. Thus, $p$ must lie on the interior of some edge $e_1$ of $A$.


Lemma (easy): Given a neighbor of an intersection, it can only intersect $A$'s next edge.

From the previous lemma, we know that a given intersection $p$ lies on the interior of $A$'s edge $e_1$. Consider neighbor $q$ of $p$ in $B$.

If $q$ intersects $A$, then it cannot be on the same edge $e_1$, because $p$ is already on it, and this would create a collinear edge. Also note that the side lengths of $B$ must be strictly shorter than those of $A$ (else $B$ cannot fit inside $A$). Hence, by the triangle inequality, $q$ cannot lie on anything but neighbor $e_2$ of $e_1$, if it lies on $A$ at all.


Lower bound (harder): For $n \geq 4$, there are at least $4$ intersections.

Here is a constructive proof that for $n \geq 4$, there are at least four points.

Consider the following prototypical diagram:

general diagram, used below

We have the outer polygon in gray and the inner polygon in blue. The polygons are both aligned so that their rightmost vertex is on the $x$-axis. WLOG $A$ has radius $1$, so $a = \langle 1, 0 \rangle$. We can also see that $b = \langle \cos(2\pi/n), \sin(2\pi/n) \rangle$. The inner polygon $B$ has center $C = \langle x_C, 0 \rangle$ and radius $R<1-x_C$ (note that this means that its rightmost vertex does not intersect $A$).

The angle (marked) at $a$ from the vertical to the polygon is $\pi/n$ (that plus the same angle on the other side have to add up to $2\pi/n$, the amount turned at each vertex). Thus, the slope of polygon $A$'s edge $e_1$ is $-\cot(\pi/n)$. With point $a$ and the slope, we can use the point-slope form of a line to get: $$ (y-0) = (-\cot\theta) (x-1)\text{,}\hspace{1cm}\text{where }\theta:=\frac{\pi}{n} $$ We would like $p$ to lie on this line and further, it must be a vertex of $B$ as per: \begin{alignat*}{2} x &= x_C ~+~ && R \cos(2\phi)\text{,}\hspace{1cm}\text{where }\phi:=\frac{\pi}{n+1}\\ y &= && R \sin(2\phi) \end{alignat*} Substituting the latter into the former, we get a linear equation in $R$ and $x_C$: $$ R \sin(2\phi) = -\cot\theta( x_C + R \cos(2\phi) - 1 ) $$

We now proceed similarly. The slope of $e_2$ can now be found as a rotation of the vector defining $e_1$. Its slope works out to be $-\cot(3\theta)$ (remember your trig identities, kids). Combined with the point $b$ and this slope, we can define the line $e_2$ from point-slope form. We know that $q$ is on the inner polygon $B$, and we can intersect these to get a second linear equation in $R$ and $x_C$. $$ R \sin(4\phi) = -\cot(3\theta)( x_C + R \cos(4\phi) ) + \cot(3\theta)\cos(2\theta) + \sin(2\theta) $$

We now have two linear equations in $R$ and $x_C$, so we can combine them and solve for their intersection (i.e., the values of $R$ and $x_C$). This appears to be unavoidably messy, but ultimately it's closed-form and always possible.

Here are some values up through $n = 10$ to demonstrate, rounded to 6 sigfigs:

n $x_C$ $R$
3 0.0490381 0.549038
4 0.0514622 0.752764
5 0.0486795 0.842470
6 0.0451871 0.888297
7 0.0418499 0.914587
8 0.0388455 0.931084
9 0.0361838 0.942190
10 0.0338318 0.950094
def calc_params(n):
    theta = pi /  n
    phi   = pi / (n+1)
#Two linear equations, of form   A R = B x_C + C
A1 = sin(2*phi) + cos(2*phi)*cot(theta)
B1 = -cot(theta)
C1 = cot(theta)
A2 = sin(4*phi) + cos(4*phi)*cot(3*theta)
B2 = -cot(3*theta)
C2 = cos(2*theta)*cot(3*theta) + sin(2*theta)

#Solve (divide through by first coeff and sub)
B1/=A1; C1/=A1
B2/=A2; C2/=A2
x_C = (C2-C1)/(B1-B2)
R = B1*x_C + C1

return x_C, R

Here's $n=3$ through $n=8$, just for an intuition:

diagrams using these parameters

Thus we have two points $p$ and $q$ that are intersections of $A$. Because of the symmetry over the $x$-axis, since $p$ and $q$ are intersections, $p'$ and $q'$ are also. Together, that is four intersections. Because we can solve these equations always, we have at least $4$ intersections (except in the case $n = 3$, where $q = q'$, so there are only three unique intersections).


Corollary (near-trivial): For $n = 4$, the maximum number of intersections is $4$.

Per the former lower bound and the above upper bound, we have exactly $4$ instersections for $n=4$.

geometrian
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  • Assuming the n-gons intersect, isnt the Lower bound 1? Example: a regular pentagon inside a square with the pentagon side smaller than the square and one vertex shared between the two figures. – vvg Jun 19 '23 at 02:34
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    @vvg We are asked to find the maximum number of intersections. The lower-bounds derived give the minimum value of that maximum. – geometrian Jun 19 '23 at 02:39
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COMMENT:- An example illustrating the suggested method to try to solve the `problem with parallels (whoever knows how to use Geogebra, I think can act easily).

With the same side of the red hexagon (HEX) draw a blue heptagon (HEP) as suggested in the attached imperfectly drawn figure. Next, try to draw a heptagon inscribed in the hexagon, using parallels to the sides of HEP. It would be a way of verifying geometrically that the figure given by PREM is incorrect and that DAN's conjecture still stands.

I think the method would be capable of being generalized for polygons of n and n+1 sides.

enter image description here

Piquito
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