This is a variation on an earlier question resolved by user35353: Can a tangle of arcs interlock? In that question, the arcs were restricted to circular arcs, and user35353's proof that one arc can be removed without disturbing the others relies on the circularity of the arcs. That proof fails with elliptical arcs. The length of the major and minor axes are arbitrary, and, just as in the previous question, the arcs must leave a positive gap, i.e., they cannot be complete ellipses.
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asked Aug 11, 2013 at 1:10
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7 Answers
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Three tangled ellipses that can't be unlocked. The major radii of the smaller ellipses have to be a bit smaller than the minor radius of the large ellipse. The minor radii of the small ellipses have to allow them to be linked, but of course smaller than the large radius, so that they can't rotate.
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Let's consider an ellipse with radii $A,B$ $A\gt B$, and two identical ellipses with radii $a,b$, $a\gt b$. Assume that $a$ is very close to $B$, but still $a\lt B$.
Because $a\lt B$, there is a minimal distance $d$, so that, if we introduce the large ellipse through a small one, the distance between their centers cannot be smaller than $d$.
Because $a\gt b$, there is a maximal distance $D$, so that, if we introduce the large ellipse through a small one, and the distance between their centers is smaller than $D$, the small ellipse can't be rotated around the large one.
To find a configuration as in the figure which is locked, we look to satisfy the following conditions.
$b\gt d$, because we want the two small ellipses to be tangled.
$b\lt D$, because we want to prevent the rotation of the small ellipses around the large one.
To find the values satisfying these conditions, start from a configuration with $b\lt a\lt B\lt A$. Then, if the condition is not satisfied, replace $a\mapsto (a+B)/2$. Then, $d$ decreases, but $D$ remains unchanged, since it depends only on $b$. If we repeat this, $d\to 0$, so at some point $d\lt D$ and $d\lt b$. Then, $b$ can be replaced by a smaller value, which is larger than $d$, but smaller than $D$.
answered Aug 12, 2013 at 5:28
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$\begingroup$ Bravo! This ultimate example ends the competition. $\endgroup$ Commented Aug 12, 2013 at 10:35
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$\begingroup$ This is the most convincing example. I echo Wlodzimierz's Bravo! $\endgroup$ Commented Aug 12, 2013 at 13:19
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$\begingroup$ OK, I think I buy this one--very nice! One minor comment--the word "larger" in the second sentence should be "smaller." $\endgroup$ Commented Aug 12, 2013 at 21:03
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Four tangled ellipses that cannot be unlocked.
answered Aug 11, 2013 at 22:11
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$\begingroup$ Plausible, but I don't see how to prove that any of these constructions work. In particular, here's an issue you have to balance: you want the green rings to be very oblong, so that you can't rotate them and unlock them from the blue ring; but if they are very oblong, it seems easier to slide the red ring along them and escape that way... $\endgroup$ Commented Aug 12, 2013 at 3:04
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$\begingroup$ @Daniel Litt: There is no need that the greens are very oblong. If you drag the upper part of the left green ellipse toward right, there will be a maximal inclination of it. If is in this position, and is not circle, it can't rotate. Do similar for the right green. Now, the red has to be tight enough, so that it can't rotate along the greens. This is not difficult, since the green grows wider toward the center, and since the initial position of the red is at the end of the green, which is the closest end. $\endgroup$ Commented Aug 12, 2013 at 5:20
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$\begingroup$ I still don't see a proof, Cristi. There are many other motions one can imagine; for example, one could try rotating the greens along a vertical axis, and then slipping the red free. A proof dealing with all such examples seems quite difficult--in particular, these movements do put constraints on the parameters of the ellipses, and since you haven't specified any parameters, it's hard to tell whether any will work. $\endgroup$ Commented Aug 12, 2013 at 6:09
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$\begingroup$ @Daniel Litt: Whether the green is more horizontal or vertical, the red still has to elongate to slide along the green. Of course not any ellipses can be locked in this configuration, but the end where the red is placed is minimal for more reasons. First, it is smaller there. And the green is smaller there too. And the dimensions are chosen so that the red is tied at the closer green end. You can chose it so that the angle between the major axes of red and green exceeds 90. In this case, even if the green is a line, the red would have to increase the length to slide. $\endgroup$ Commented Aug 12, 2013 at 6:25
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$\begingroup$ Hmm--well, I think you're missing my point, which is that even if your remarks address my comments correctly (which is not obvious, to me at least), there's still no proof that these configurations are locked... $\endgroup$ Commented Aug 12, 2013 at 6:45
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This is not an answer, but I cannot write short comments.
Here I'll try to describe a counterexample, i.e. a construction of elliptical arcs that cannot be separated mechanically from each other. Take 12 equal ellipses with eccentricity close to 1, so the ellipses may be considered "close" to a segment. Now arrange them in a cubical construction, each ellipse corresponding to an edge of the cube in a way that any two ellipses meeting in a "vertex" are linked. Make small gaps in each ellipse near its "middle" (in a point of minimal curvature). I have the feeling that this construction is "solid" and the arcs cannot be separated - if we try to make profit of the gaps we should destroy the "vertices", but then the ellipses will touch very soon... Of course, this is not a proof, maybe it is easier to make a corresponding model and to try to disassemble the construction.
On the other hand, one may replace the cubical model by an arbitrary solid model (with elliptical edges) and it is hard to believe that any such construction is demountable.
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2$\begingroup$ Nice idea! Perhaps simpler: Make a regular tetrahedron from six thin ellipses. $\endgroup$ Commented Aug 11, 2013 at 11:43
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2$\begingroup$ Or just a nearly degenerate triangle from three of length $1$, $1$, $1.99$? $\endgroup$ Commented Aug 11, 2013 at 13:13
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2$\begingroup$ Um, on second thought the smaller ellipses' attachment points can still slide towards each other. $\endgroup$ Commented Aug 11, 2013 at 13:58
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$\begingroup$ @Noam D. Elkies - O.K. but I suppose a triangle of (thin) elliptical arcs is not enough - we may always unlink two of them and then it seems easy to disassemble all the construction. EDIT: Now I saw your self-edit. $\endgroup$– simeon2Commented Aug 11, 2013 at 14:09
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1$\begingroup$ For cube: the upper face of the cube can be made to slide down along the vertical edges. For tetrahedron and cube: three edges meeting in one point can be made to slide through the vertex. $\endgroup$ Commented Aug 11, 2013 at 18:02
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Here is a construction of five interlocking elliptical arcs.
In Stage 1, a large elliptical arc $a$ is braced by two very narrow elliptical arcs $b$ and $c$. The distance between the braces can be very small. The braces can be moved outwards and slip off of a, but they can be moved towards the middle only a little bit.
In Stage 2, the first two braces are braced quite tightly by secondary braces $d$ and $e$, very close to the arc $a$ and much narrower than the first ones. Now the configuration is locked.
answered Aug 11, 2013 at 20:33
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$\begingroup$ Does your ellipse d hook onto b and c on opposite sides of the plane of a, or on the same side? $\endgroup$ Commented Aug 11, 2013 at 20:42
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$\begingroup$ @YoavKallus On opposite sides. $\endgroup$ Commented Aug 11, 2013 at 20:50
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$\begingroup$ I was thinking to the edge. But then I thought that maybe they embrace both sides of b and c, and this will lock them $\endgroup$ Commented Aug 11, 2013 at 20:53
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$\begingroup$ Yes, they do as you say: embrace both sides of b and c. $\endgroup$ Commented Aug 11, 2013 at 21:01
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$\begingroup$ I am not sure, but maybe $e$ can be deleted. $\endgroup$ Commented Aug 11, 2013 at 21:08
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The Borromean rings have an elliptical form in $\mathbb R^3$ given by the union of the three:
$$ x^2 + y^2/2 = 1, z=0 $$ $$ y^2 + z^2/2 = 1, x=0 $$ $$ z^2 + x^2/2 = 1, y=0 $$
I think if you cut out small chunks from these ellipses appropriately, they will still be locked. If I find a proof I'll edit this later.
edit: this is not an answer.
answered Aug 11, 2013 at 2:12
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1$\begingroup$ I don't know if I believe this. Call the ellipses $A, B, C$. It seems like we're free to slide $B$ and $C$ towards each other until one of them is near enough to the chunk in $A$ to slip it free, at which point we're done. $\endgroup$ Commented Aug 11, 2013 at 5:58
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$\begingroup$ I also think that your picture is a bit misleading, since the ellipses should be infinitely thin. $\endgroup$ Commented Aug 11, 2013 at 6:32
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$\begingroup$ I'm starting to agree with you. But I think if the ellipses are given thickness (which is not the question) then this example could work. If they're genuine ellipses as in the formula above, I can see a way to untangle them regardless of the size of the cut -- in fact, you only need to cut one component, it seems. $\endgroup$ Commented Aug 11, 2013 at 6:33
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I think that this configuration cannot be untangled.
answered Aug 11, 2013 at 20:27
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Here's another example of four interlocking ellipses: The large ones are in planes parallel to the screen; the small ones lie in planes perpendicular to it, and they are very narrow.
answered Aug 11, 2013 at 23:22
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1$\begingroup$ Also looks like a strange object in this picture image.dhgate.com/albu_106878897_00-1.0x0/…. $\endgroup$ Commented Aug 12, 2013 at 13:21
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$\begingroup$ I see this construction has been immediately applied in "real life" :-) $\endgroup$ Commented Aug 12, 2013 at 18:00
