Let $$P$$ and $$Q$$ be the boundary segments of two planar simple polygons. View these boundaries as rigid wires. Fix $$Q$$ in, say, the $$xy$$-plane, and imagine $$P$$ arranged in $$\mathbb{R}^3$$ so that $$P$$ and $$Q$$ are disjoint and not topologically linked. Say that $$P$$ and $$Q$$ are interlocked if it is impossible to separate $$P$$ to infinity via rigid motions that at all times avoids $$P$$ touching $$Q$$. The polygons cannot flex or distort; each is rigid.

Two unlinked but interlocked polygons are shown below ($$Q$$ red):

My question is:

Q. What is the fewest number of vertices $$|P|+|Q|$$ that achieves unlinked interlocking?

Reducing the red rectangle $$Q$$ to a triangle, and reducing $$P$$ to a hexagon, achieves $$9$$ vertices. Can interlocking be achieved with $$8$$ total vertices?

• It seems fairly clear we should be able to make your "P" into a (non-convex) pentagon to save another vertex. Jan 22 at 22:55
• @SamHopkins: Plausible, but not so clear to me. Jan 22 at 23:43

As Sam Hopkins commented, 8 vertices are enough. Let $$Q$$ be the pentagon from the picture and let $$\pi$$ be the plane containing it. Now we can define the triangle $$P$$ as a triangle of less diameter than the black segment and intersecting $$\pi$$ at two points: one point $$a_0$$ in the open blue region $$B$$ and one point $$b_0$$ in the open green region $$G$$. $$P$$ and $$Q$$ are not linked, and it is intuitively clear that they are interlocked, but let´s prove it in detail. Suppose there is a path of triangles $$P_t=i_t(P)$$ going to infinity, such that $$P_t$$ is disjoint with $$Q$$ for all $$t\in[0,\infty)$$ and $$i_t:P\to\mathbb{R}^3$$ is a continuous path of isometries (continuous in the supremum norm for functions $$P\to\mathbb{R}^3$$) with $$i_0=Id_P$$. Let´s try to derive a contradiction.

Lemma 1: Let $$i$$ be an isometry such that $$i(P)$$ intersects $$\pi$$ at exactly two points $$a,b$$. Then $$\forall\varepsilon>0\;\exists\delta>0$$ such that for any isometry $$j$$ with $$d(i,j)<\delta$$, $$j(P)$$ intersects $$\pi$$ at two points $$a',b'$$, with $$d(a,a')<\varepsilon$$ and $$d(b,b')<\varepsilon$$.

Proof: First of all, if $$d(i,j)$$ is small enough, $$j(P)$$ will intersect $$\pi$$ in two points, because the images by $$j$$ of at least two vertices of the triangle will be outside $$\pi$$.

Now consider the function $$F(x,y)=$$ intersection of the line passing through $$x,y$$ and $$\pi$$, defined in an open subset of $$\mathbb{R}^6$$. This function is continuous in its domain.

Suppose $$a$$ is not a vertex of the triangle. Then it is in the edge formed by two vertices $$i(v_1),i(v_2)$$, each in one half space of $$\pi$$. By continuity of $$F$$, if $$d(i,j)$$ is small enough we have a point $$a'=F(j(v_1),j(v_2))$$ in $$j(P)$$ at distance $$<\varepsilon$$ of $$a$$.

So the lemma works when $$a,b$$ are not vertices of the triangle. $$a$$ and $$b$$ cannot both be vertices of the triangle, so suppose $$a$$ is a vertex, $$a=i(v_1)$$ with $$v_1$$ a vertex of $$P$$, and $$b$$ is not. Then for $$d(i,j)$$ small enough both $$F(j(v_1),j(v_2))$$ and $$F(j(v_1),j(v_3))$$ will be at distance $$<\varepsilon$$ from $$a$$, so the one which is in $$j(P)$$ will be the point $$a'$$. $$\square$$

Now that we are done with the lemma 1, we can define

$$k=\sup\{t>0;P_s\textit{ intersects \pi at two points, one in B and one in G, }\forall s\in[0,t)\}.$$

Lemma 2: $$k>0$$, and in $$[0,k)$$ there are paths $$a_t$$, $$b_t$$ such that $$\{a_t,b_t\}=P_t\cap A\;\forall t.

Proof: By lemma 1, there is some $$\varepsilon>0$$ such that $$\forall t\in [0,\varepsilon)$$, $$P_t\cap A$$ consists on two points, one in $$G$$ and one in $$B$$, so $$k>0$$. Let $$a_t$$ be the point in $$B$$ and $$b_t$$ be the point in $$B$$. By lemma 1 again, $$a_t$$ and $$b_t$$ are continuous in $$[0,\varepsilon)$$. Now consider the maximum $$\varepsilon$$ such that the $$a_t$$ and $$b_t$$ are defined in $$[0,\varepsilon)$$. Then $$\varepsilon=k$$, because if not, by the same argument using lemma 1, $$a_t$$ and $$b_t$$ will be defined in a neighborhood of $$\varepsilon$$.$$\square$$

Now let´s see that $$P_k$$ has to intersect $$Q$$, leading to a contradiction. First of all, $$P_k$$ has to intersect $$Q$$ in at least two points, one point $$a_k\in\overline{B}$$ and one point $$b_k\in\overline{D}$$. To prove this, let $$t_n$$ be an ascending sequence, $$t_n\to k$$, such that $$a_{t_n}$$ is convergent. Call $$p_n=i_{t_n}^{-1}(a_n)$$, we can suppose that $$p_n$$ converges to some point $$p\in P$$ after taking a subsequence. Finally, we can let $$a_k$$ be $$i_k(p)=\lim_n i_{t_n}(p_n)$$, and similarly with $$b_k$$. Now we can consider two cases:

If $$P_k$$ intersects $$\pi$$ only at $$a_k$$ and $$b_k$$ we cannot have $$a_k\in A$$ and $$b_k\in B$$: that would contradict the definition of $$k$$. So either $$a_k\in\partial B\setminus Q$$ and $$b_k\in\overline{G}$$ or $$b_k\in\partial G\setminus Q$$ and $$a_k\in\overline{B}$$. Both cases are impossible, because $$d(\partial B\setminus Q, G)$$ and $$d(\partial G\setminus Q, B)$$ are both bigger than the diameter of $$P$$.

So $$P_k$$ has to intersect $$\pi$$ in an edge. For the last time, we will consider two cases:

If $$a_k$$ and $$b_k$$ are both contained in the edge, then the edge intersects $$Q$$, which is a contradiction.

If not, the whole $$P$$ has to be contained in $$\pi$$. Consider an edge of the triangle containing the point $$a_k$$. As the edge doesn´t intersect $$Q$$, one of its vertices must be in $$B$$. There is also other vertex in $$G$$ (one of the vertices of the edge containing $$b_k$$), so the edge joining those two vertices must intersect $$Q$$, a contradiction.$$\\[20pt]$$

P.S: I had some comments regarding whether there are $$P,Q$$ as in the question with $$7$$ total vertices but they were wrong (I had not considered one case). An argument that they do not exist is given in Del's answer.

• Convincing argument re the pentagon. Jan 23 at 0:15
• Yes, it would probably be painful to do it in detail (same as the case of 7 vertices). You have to prove that the two points of intersection of the triangle and the plane are continuous in time and some more things regarding the first moment at which there are not two intersection points in the green and blue regions. I will change the affirmative answer to a "it seems" just in case Jan 23 at 0:38
• @JosephO'Rourke btw how would the proof of the case with an hexagon and a triangle go? Jan 23 at 9:04
• Saúl, I did not yet try to prove it formally. I've worked on other interlocking results and the proofs are rarely easy. Jan 23 at 12:36
• @JosephO'Rourke It should work now. I suppose my definition of "path" of triangles coincides with the usual one. Jan 23 at 19:18

It is not possible with 7 (i.e., with a triangle $$T$$ and a quadrilateral $$Q$$). I write a rough proof.

First, any quadrilateral $$Q$$ lying in a plane $$\pi$$ can be partitioned in two triangles $$Q_1$$ and $$Q_2$$, whose common edge is a diagonal $$d$$ of $$Q$$. Now the intersection of the triangle $$T$$ with $$\pi$$ consists of two points (otherwise they are coplanar and the conclusion is trivial). Since the polygons are not linked, there are two cases: either both points lie outside $$Q$$, or they both lie inside $$Q$$.

Case 1: both lie outside. An inspection of the cases shows that we can just translate $$T$$ in direction parallel to one of the bisectors of the triangles $$Q_1$$ or $$Q_2$$ starting from one of the vertices not belonging to $$d$$.

Case 2: both lie inside. We will consider a modified problem: given a quadrilateral $$Q$$ and two initial points $$x(0),y(0)$$ inside it, find two continuous curves $$x(t),y(t)$$ such that the distance between the two is decreasing in time to 0, and they never lie in $$Q$$. The solution is, e.g., a linear homotopy that sends both points to the midpoint of $$d$$. Now we only have to realize $$x(t)$$ and $$y(t)$$ as the intersections of $$T$$ and the plane $$\pi$$. One can convince themselves that this is possible by pulling $$T$$ in direction orthogonal to $$\pi$$ while translating it.

• In Case 1, did you mean $Q_1$ and $Q_2$ instead of $T_1$ and $T_2$? Jan 24 at 15:39
• @JosephO'Rourke Ah yes, thanks.
– Del
Jan 25 at 17:58
• This is a very nice proof! Jan 25 at 20:54

Here is another example with 8 vertices: a short fat Star Trek symbol and a square in orthogonal planes. Since the distance between the base points of the red figure is greater than its height, one cannot rotate the square to take it out.

Addendum: As Saúl Rodríguez Martín mentions, this example may not work. If we assume, however, that the links are physical, that is they have some thickness, then I think that it should work.

• It's more convincing with 3D tubes. :-) Jan 23 at 17:13
• Seems like they can be taken apart: you can translate the green square parallel to its two sides pointing up until it almost hits the red one. Then you can rotate the square a bit so that three of its vertices are in the same side of the plane of the star trek symbol, and from there it´s not difficult to imagine how to separate them. Jan 23 at 23:39
• @Saúl Rodríguez Martín. Maybe you are right. If we assume that the loops have no thickness one might be able to separate them, as you say. But they might interlock if we have some thickness. Jan 24 at 0:45
• I’m half tempted to pull out a roll of brass wire and try it out. But then, I’m half tempted to do a lot of things. Jan 24 at 19:04

I think that 8 might be possible, by interlocking two Star Trek symbols as shown below. Adendum: This candidate may not work, as quarague points out, but I leave it as a potential "how not to" example. There are other ways to cross the two figures, while they remain unlinked, and one of these variations could be more promising. Also I think it is important to specify whether the loops have any thickness or not. Some configurations may interlock assuming there is nonzero thickness.

• Are those two linked? Jan 22 at 23:55
• No they are not linked. Jan 22 at 23:56
• Ah, I see now. Nice! Jan 22 at 23:58
• This looks to me like they can be taken apart. Rotate the red figure so that the 2 spikes point downwards and the singlespike upwards. Then move it north east and downwards until it is fully separated from the blue one. Jan 23 at 8:13