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From the 1951 novel The Universe Between by Alan E. Nourse.

Bob Benedict is one of the few scientists able to make contact with the invisible, dangerous world of The Thresholders and return—sane! For years he has tried to transport—and receive—matter by transmitting it through the mysterious, parallel Threshold.

[...]

Incredibly, something changed. A pause, a sag, as though some terrible pressure had suddenly been released. Their fear was still there, biting into him, but there was something else. He was aware of his body around him in its curious configuration of orderly disorder, its fragments whirling about him like sections of a crazy quilt. Two concentric circles of different radii intersecting each other at three different points. Twisting cubic masses interlacing themselves into the jumbled incredibility of a geometric nightmare.

The author might be just throwing some terms together to give the reader a sense of awe, but maybe there's some non-euclidean geometry where this is possible.

Yly
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Mindwin Remember Monica
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    I failed to find a solution but I can prove that this universe doesn't have conservation of linear or angular momentum. Therefore, it has an infinite power source or power sink. (probably both) – Joshua Aug 20 '19 at 17:53
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    Your question is unclear because you didn't say how a circle is defined in geometry with nonuniform curvature. People should have to read the whole book to know what your question is. Geometry with nonuniform curvature is possible so you have to break some assumptions from which you can deduce that the geometry has uniform curvature. What are the assumptions you want to make that allows for a geometry with nonuniform curvature? – Timothy Aug 21 '19 at 01:17
  • Wouldn't a cylinder permit intersecting concentric circles? Always an even number of intersections, so not the space that this passage is describing, but it meets the other conditions... – Stobor Aug 21 '19 at 03:25
  • @Timothy: I would define a "circle" as "the set of all points which are an arbitrary fixed distance (the radius) from some arbitrary point (the center) under the given geometry's metric." Did you have some other definition in mind? – Kevin Aug 21 '19 at 04:54
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    This violates assumption (3) of Tanner's answer, but consider latitudes of the Earth, with the North Pole as the centre. Let the radius be the distance travelled to the latitude from the south pole by an aircraft that initially flies south. The equator can be defined as having a radius equal to 90 + 360*n degrees of latitude for integer n. Thus concentric circles of different 'radius' occupying the same degree of latitude (infinitely many intersections). Obviously these circles are identical, but this may not be known to the aircraft, which may be enough to convince you they're distinct. – Hugh Aug 21 '19 at 05:59
  • Woa... I read that book like 40 years ago when I was a kid. – Almo Aug 21 '19 at 20:38
  • @Kevin And for that definition the answer is obviously "doesn't exist", if you assume that for two points A and B, there is only one distance between them. So another definition is needed - or defining what the word distance means. – JiK Aug 22 '19 at 21:15

7 Answers7

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Yes, with the appropriate definition of "circle". Namely, define a circle of radius $R$ centered at $x$ on manifold $M$ to be the set of points which can be reached by a geodesic of length $R$ starting at $x$. This seems pretty reasonable, and reproduces the usual definition in Euclidean space.

It's not hard to see that concentric circles on a torus or cylinder can have four intersection points.

enter image description here

(Here's how to interpret this picture: The larger circle has been wrapped in the y-direction, reflecting a torus or cylinder topology. Coming soon: A picture of this embedded in 3D.)

By flattening one side of the torus a bit, you can make one side of the larger circle intersect the smaller circle at two points, while the other side just grazes at a single point*. Thus you get three intersections.

*As a technical point, this can definitely be accomplished in Finsler geometry, though I'm not sure if it can be done in Riemannian geometry.

Yly
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    But can this be done with exactly three points of intersection? – Novak Aug 20 '19 at 16:13
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    @Novak You could probably use a manifold of genus 2 and have the circles only intersect at tangents (a torus is genus 1). – Sam Benner Aug 20 '19 at 17:03
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    @Novak yes, as shown in the last paragraph. If you require x-style intersections, this becomes impossible in 2D - each circle has an outside and an inside and you must exit any circle you enter, where being inside means there's a radius passing through that point. Note that the number of such radii can only change discontinuously, can only do so at an intersection, and that is the property of that (point, circle) pair. . – John Dvorak Aug 20 '19 at 17:04
  • @novak Even in 3D to get exactly three intersections you'll need space that isn't centrally symmetric, while at the same time the presence of human observers demands that the space is locally euclidean enough not to interfere with basic biology. Note that adding discontinuities into the space will break the definition of a circle (the set of points reachable through a geodesic of length x becomes non-manifold). – John Dvorak Aug 20 '19 at 17:15
  • @wizzwizz4 Yes, I've fixed it now. Thanks. – Yly Aug 21 '19 at 00:54
  • Note that if you have four intersections, you also have three intersections (plus one left over). Nothing in OP's quote rules that out. – Kevin Aug 21 '19 at 04:58
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    Nice redefinition of "circle" using a different metric which isn't really a metric. – Ethan Bolker Aug 21 '19 at 11:41
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    @kevin nothing except the rules of colloquial English. That statement is a prose description from a short novel, not a formal statement in a theorem. – Novak Aug 21 '19 at 20:13
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    Amazing. The next challenge: Simulate with raytracing what it would look like if you were near this formation in a three-dimensional analogue of the space (if not $\mathbb{T}^3$ then perhaps $\mathbb{T}^2 \times \mathbb{R}$ or similar w/suitable Riemannian metric). My imagination is that it will look something like an infinite hall of mirrors (due to the repeating toroidal geometry causing the "light" to wrap around over and over) with interlocking circles all throughout below you - very strange. But that's a conjecture esp. given that simply tiling the given doesn't result in circles only. – The_Sympathizer Aug 22 '19 at 00:12
  • And make the rings made of little fragmentary bits, too :g: – The_Sympathizer Aug 22 '19 at 00:15
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    Would it look a bit like this? https://i.sstatic.net/zgtHl.png Certainly not mathematically correct, but maybe good enough for imagination. – Thomas Weller Aug 22 '19 at 22:29
  • I don't get how flattening the torus or cylinder can produce exactly three intersections. It seems to me that the point of tangency of the two circles would have to lie on a single geodesic running from the center to the tangency point to the center. To have three intersections I think you would have to be able to mark off two non-overlapping sections of $r_1$ and $r_2$ one way along that geodesic, and two overlapping sections of $r_1$ and $r_2$ another way. – David K Aug 23 '19 at 12:57
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    Your definition of circle here, with center $x\in M$, corresponds to first considering the Euclidean circle inside the tangent room $T_xM$ (which is just a copy of the Euclidean plane) with center in the origin and radius $R$. Then use the exponential map $\exp: T_xM \to M$ to map this subset of the tangent room to a subset of the Riemannian manifold $M$, and call the image "a circle". When $R$ exceeds the injectivity radius at $x$, one such "circle" can intersect itself. – Jeppe Stig Nielsen Aug 23 '19 at 15:13
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    (continued) To imagine this, take $M$ as a cylindrical surface in $\mathbb{R^3}$. Draw a big circle on a piece of transparent paper, and wrap the paper around the cylinder. The "circle" will intersect (even) itself. It is easy to find a circle with another radius on another transparent paper and wrap that around the cylinder as well, with coincident centers, in a way that the two circles intersect. – Jeppe Stig Nielsen Aug 23 '19 at 15:13
  • @DavidK Yeah, that's why I said I wasn't sure it could be done in Riemannian geometry. In Finsler geometry though the length of a path can be different if you traverse it backwards, so the situation you describe is possible. – Yly Aug 23 '19 at 16:15
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The situation is impossible if we make the following assumptions:

  1. Each circle has exactly one center, which is a point.
  2. Concentric circles have the same center.
  3. Each circle has exactly one radius, which is a number. (We make no assumptions, besides those listed, about the meaning of the word "number.")
  4. If two circles intersect each other at a point, then that point lies on both circles.
  5. Given any unordered pair of points, there is exactly one distance between those points, which is a number.
  6. If a point $p$ lies on a circle, then the distance between $p$ and the center of the circle is the radius of the circle.

From the above, suppose that $C$ and $D$ are two concentric circles that intersect at a point. The two circles have the same center, $e$, and call the intersection point $p$. Then the distance between $e$ and $p$ is the radius of $C$, but it is also the radius of $D$, so the two circles cannot have different radii.

We could make the situation possible by discarding some of the axioms, but for the most part, these axioms are so fundamental to the notion of geometry that if you discarded one, the result wouldn't be considered geometry any more (not even non-Euclidean geometry). In particular, axioms 2 and 4 above are essentially just the definitions of the words "concentric" and "intersect," and axioms 1, 3 and 6 essentially constitute the definition of a circle with a given center and radius.

If I had to pick an axiom to discard, I would discard axiom number 5: the statement that given two points, there is only one distance between those points. This is the approach taken in Luca Bressan's answer (a plane based on modular arithmetic, where pairs of points with a distance of $0$ also have a distance of $2$ and vice versa) and in Yly's answer (a cylinder, where a pair of points has infinitely many distances, depending on which direction and how many times you wrap around the cylinder as you measure the distance).

Sophie Swett
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    I don't know why that axiom is hard to discard. I'm just not sure if it helps, because that distance may be isomorphic in that space. Consider a sphere (the earth, perhaps). The distance from the north pole to the equator could be hit by a great circle of radius r*pi/2 as well as r*3pi/2. – Cireo Aug 20 '19 at 05:32
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    It's not necessary to assume the center of a circle is unique, though without that assumption some of the later axioms should be rephrased a little. – Eric Wofsey Aug 20 '19 at 06:09
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    To be more specific about Eric Wofsey's comment, it is possible for circles to have more than one center (great circles in the sphere have two), but there is no need for the uniqueness in your proof. As long as they have at least one common center with different radii, the argument works. – Paul Sinclair Aug 20 '19 at 16:42
  • Many of your axioms are just the definitions of "circle", "center", "radius", "intersect" and "concentric". I.e., if someone's concept doesn't include them, then they are playing silly word games rather than producing interesting geometries. The only necessary geometric assumption is "points have only one distance between them". – Paul Sinclair Aug 20 '19 at 16:49
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    @PaulSinclair and it's an assumption that we can, thankfully, dispute. Any old multiply-connected space will do. – John Dvorak Aug 20 '19 at 17:21
  • @PaulSinclair Yep, that's a good point! I just edit this answer to point out that most of these assumptions are essentially just definitions, and that the statement "points have only one distance between them" is the odd one out. – Sophie Swett Aug 20 '19 at 17:27
  • Regarding your commentary on my answer, I would say that I have violated your assumption 6, rather than 5. There is a well defined metric distance function on a cylinder, but by defining a circle in terms of geodesics which are not necessarily minimum length, the radius of a circle need not coincide with the distance from points on the circumference to the center. – Yly Aug 21 '19 at 00:44
  • Your definition assumes a metric space - in a more general topology, a circle might be any closed curve, say, the image of a circle in $\mathbb{R}^2$ under some continuous function. – j4nd3r53n Aug 21 '19 at 13:27
  • @j4nd3r53n - "circle" means more than just "closed curve". People may occasionally call a closed curve a circle, but only informally. Further, the OP specifically says "geometry", which is stronger than topology, and refers to concentricity and radii, both of which require a metric. – Paul Sinclair Aug 21 '19 at 19:41
  • @PaulSinclair, good points; but let me see if I can save a bit of face :-) Firstly, concentric doesn't necessarily require a metric - it just means having the same centre, which I could simply choose based on my own whim, I think. Next, are you sure the definition of a metric rules out something sufficiently exotic? Finally, I recall, from my introduction to differential geometry, too long ago, that even a potato is a sphere, if you can define you differential structure so that it is. Now, if this isn't a concise, hard-hitting argument ... – j4nd3r53n Aug 22 '19 at 07:56
  • @j4nd3r53n - You don't get to pick the center of a circle on your own. It is part of the definition of the circle. I made no claim at all about the definition of a metric ruling out anything. I just claim that talking about circles and radii implies a metric because of how they are defined. Yes, the surface of a potato is a sphere (unless a worm has gotten at it, in which case it might be a multi-holed torus). The difference between them is not intrinsic to the surface itself, but rather in how someone has forced that surface to live in 3D space. – Paul Sinclair Aug 22 '19 at 14:22
  • This argument cannot be correct, because the assumptions apply to two orthogonal circles in $E^3,$ but the conclusion does not hold. Somehow you need to introduce an assumption that is tantamount to the circles lying in a common plane. In other words, you also need Euclid's axiom that three points determine a plane. – whuber Aug 22 '19 at 18:16
  • @whuber I think my argument is sound for $E^3$. (That's 3-dimensional Euclidean space, right?) Can you be more specific about what the counterexample is? I can't think of a way to arrange two orthogonal circles in such a way that the assumptions hold but the conclusion does not. – Sophie Swett Aug 22 '19 at 21:13
  • Could you point out where you employ this Euclidean axiom of the plane? If not, the issue isn't that there's no counterexample in $E^3,$ but that your logic is incomplete. There are counterexamples in Riemann manifolds. The problem there is that there can be multiple equally short geodesics connecting the center of a circle to any one of its points (which raises thorny questions of what a "circle" might mean in those circumstances). – whuber Aug 22 '19 at 21:38
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Consider the geometry in which the plane is identified with $(\mathbb Z / 4 \mathbb Z)^2$. Define the circle with center $C(a, b)$ and radius $r$ as the locus of all points $P(x, y)$ such that $(x - a)^2 + (y - b)^2 = r^2$.

Let $C = (0, 0)$ and consider the two circles centered at $C$ with radii $0$ and $2$. Since $2^2 = 0$ in $\mathbb Z / 4 \mathbb Z$, the two equations are clearly equivalent, and they both define the set $\{ (0, 0), (2, 0), (0, 2), (2, 2) \}$. Therefore we can say that there are two concentric circles with different radii that intersect at three different points (other than their center).

Luca Bressan
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    $(\Bbb Z/4\Bbb Z)^2$ is just 16 discrete points, so it really defies the idea of "circles". You could change it to $(\Bbb R/4\Bbb Z)^2$, which would be a torus. But then your strictly $\Bbb Z_4$ concept of radius would need adapted to something continuous, and the circle of radius $0$ would be just $(0,0)$, while the circle of radius $2$ would be a circle tangent to itself at the points $(2,0)$ and $(0,2)$. – Paul Sinclair Aug 20 '19 at 16:33
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    @Paul Sinclair : Yes. In particular, given the quoted passage, one can reasonably argue (from the need to support "whirling" and many different pieces surrounding a view point) that the space needs to be a continuum. – The_Sympathizer Aug 20 '19 at 16:38
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    I think this answer demonstrates that "squared radius" is a more fundamental concept for circles over arbitrary rings than "radius". – Micah Aug 20 '19 at 17:00
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    @Micah - but the question is about geometry, not algebra. – Paul Sinclair Aug 20 '19 at 17:37
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    @Micah: Indeed. In particular, in such spaces you can have circles that don't have a radius, because their "squared radius" has no square root. The circle $x^2 + y^2 = 2$, $(x,y) \in (\Bbb Z/4\Bbb Z)^2$ (containing the points (1,1), (1,3), (3,1) and (3,3)) is one such example. – Ilmari Karonen Aug 20 '19 at 20:33
  • +1 changing the meaning of "circle" as in other answers is easy. My first thought was also changing the shape of the space to a torus, a sphere or a rod. – rexkogitans Aug 21 '19 at 04:47
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Not with the usual definition of a circle as the set of points at a fixed distance $r$ from a center $C$. If circles are concentric that means they have the same center. If they intersect at one point then they have the same radius. That means they are the same circle.

That argument works in any geometry where distance is defined.

If the circles need not be concentric you can imagine a solution. Think of two towns. Consider "time to travel" as a measure of distance. Suppose the towns separated by a range of hills with several low passes. There can be exactly three isolated points each reachable in $10$ minutes from each town.

Ethan Bolker
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  • Right, and there shall be no circle without distance. – Michael Hoppe Aug 19 '19 at 18:31
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    Quote from https://en.wikipedia.org/wiki/Concentric_objects: However, circles in three-dimensional space may be concentric, and have the same radius as each other, but nevertheless be different circles. For example, two different meridians of a terrestrial globe – M.Herzkamp Aug 20 '19 at 10:46
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    @M.Herzkamp I hadn't thought of that. The OP is interested in three intersections. – Ethan Bolker Aug 20 '19 at 11:53
  • @EthanBolker: If my brain was working properly, I might be able to figure out whether there is a Weird Higher Dimensional solution with three intersections. But I'm still not sure how you get around "different radii" unless you use the geodesic approach given in another answer. – Kevin Aug 21 '19 at 05:00
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I'll make a guess: Think about streets, a "circle" is a path in which you can go from point $A$ and return to itself and you can define a "center" in some reasonable way. As the structure of the streets can be very messy (in certain streets, you can go in only one direction, in others, you can go both ways, etc), I guess you can construct different "circles" with the same center that intersect at any "points" you want.

I built this, for example:

$\quad\quad\quad\quad$ enter image description here

The directions with arrow are one way routes, the ones without arrows are two way routes. You can define "distance" like this: The number of different streets (edges) it takes to move from $A$ to $B$. Try to think what are the "circles" with "radius" $2$ and $3$ here. Notice that this is probably a very weird "distance" that may not enjoy all properties of the habitual euclidean space.

Red Banana
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    Do you have an example here of concentric circles of radius 2 and 3 which intersect? I don't quite understand what you're saying. – Sophie Swett Aug 20 '19 at 17:06
  • @Billy: I like your idea of considering "circles" in the context of graphs of vertices and edges, rather than the usual geometric plane, though I think it would be helpful to refine your definition. As you say, "a path in which you can go from point A and return to itself" is simply a closed curve, which doesn't have a "center" in the same sense as a circle. One might derive the "center" of any given closed curve in various computational ways, but a circle is distinctive in that it's a curve derived (in part) from a given center point, rather than the reverse. – jdmc Oct 16 '19 at 16:57
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    In the usual geometric context, a circle can be defined as the set of all points whose distance from a given center point C is equal to a given radius r. We could fashion a definition in terms of a given connected graph G, a given "center" vertex C in that graph, and a given "radius" specified as a certain (integer) number of traversal steps between adjacent vertices. A "circle", then, would be the set of all vertices in G that could be reached in exactly r traversal steps, starting from the "center" vertex C. – jdmc Oct 16 '19 at 17:21
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In hyperbolic geometry, there is the notion of a horocycle, which is essentially a circle of infinite radius. You can have two concentric horocycles which are different but 'meet' (asymptotically) at a single common point p (a so-called ideal point), which also happens to be the center of both horocycles. Everything is nicely illustrated in this Wikipedia enty on horocylces.

There is no contradiction to the answer of @tswett, since the ideal point p is only asymptotically contained in both horocycles (and at infinite distance from all other points of the horocycle).

MKR
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Well, consider two concentric circles, of different radii, in 3D space (you can call them "rings" if you like). That they are concentric does not mean they are coplanar. If two such non-coplanar rings are projected onto 2D space from an appropriate angle, the projected image will show them intersecting at either two points or four. One can, perhaps, conject(ure) that a pair of concentric hyperdimensional "rings" of different radii could be configured in such a way that their projection into 3D space would intersect any arbitrary number of times, including three. They wouldn't look like circles or "rings" in 3D space, of course, but by this reasoning, that doesn't matter.

jdmc
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