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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.

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    $\begingroup$ 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) $\endgroup$ – Joshua Aug 20 at 17:53
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    $\begingroup$ 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? $\endgroup$ – Timothy Aug 21 at 1:17
  • $\begingroup$ 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... $\endgroup$ – Stobor Aug 21 at 3:25
  • $\begingroup$ @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? $\endgroup$ – Kevin Aug 21 at 4:54
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    $\begingroup$ 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. $\endgroup$ – Hugh Aug 21 at 5:59
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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.

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    $\begingroup$ But can this be done with exactly three points of intersection? $\endgroup$ – Novak Aug 20 at 16:13
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    $\begingroup$ @Novak You could probably use a manifold of genus 2 and have the circles only intersect at tangents (a torus is genus 1). $\endgroup$ – Spitemaster Aug 20 at 17:03
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    $\begingroup$ Nice redefinition of "circle" using a different metric which isn't really a metric. $\endgroup$ – Ethan Bolker Aug 21 at 11:41
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    $\begingroup$ @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. $\endgroup$ – Novak Aug 21 at 20:13
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    $\begingroup$ Would it look a bit like this? i.stack.imgur.com/zgtHl.png Certainly not mathematically correct, but maybe good enough for imagination. $\endgroup$ – Thomas Weller Aug 22 at 22:29
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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).

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    $\begingroup$ 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. $\endgroup$ – Cireo Aug 20 at 5:32
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    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Aug 20 at 6:09
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    $\begingroup$ 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. $\endgroup$ – Paul Sinclair Aug 20 at 16:42
  • $\begingroup$ 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". $\endgroup$ – Paul Sinclair Aug 20 at 16:49
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    $\begingroup$ @PaulSinclair and it's an assumption that we can, thankfully, dispute. Any old multiply-connected space will do. $\endgroup$ – John Dvorak Aug 20 at 17:21
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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).

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    $\begingroup$ $(\Bbb Z/4\Bbb Z)^2$ 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 $(0,2)$. $\endgroup$ – Paul Sinclair Aug 20 at 16:33
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    $\begingroup$ @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. $\endgroup$ – The_Sympathizer Aug 20 at 16:38
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    $\begingroup$ I think this answer demonstrates that "squared radius" is a more fundamental concept for circles over arbitrary rings than "radius". $\endgroup$ – Micah Aug 20 at 17:00
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    $\begingroup$ @Micah - but the question is about geometry, not algebra. $\endgroup$ – Paul Sinclair Aug 20 at 17:37
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    $\begingroup$ @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. $\endgroup$ – Ilmari Karonen Aug 20 at 20:33
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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.

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  • $\begingroup$ Right, and there shall be no circle without distance. $\endgroup$ – Michael Hoppe Aug 19 at 18:31
  • $\begingroup$ @darijgrinberg But we'd also have to read "different radii" as "the same radius"... $\endgroup$ – Théophile Aug 19 at 18:40
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    $\begingroup$ @Théophile: Oh, right... $\endgroup$ – darij grinberg Aug 19 at 18:41
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    $\begingroup$ Quote from 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 $\endgroup$ – M.Herzkamp Aug 20 at 10:46
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    $\begingroup$ @M.Herzkamp I hadn't thought of that. The OP is interested in three intersections. $\endgroup$ – Ethan Bolker Aug 20 at 11:53
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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.

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    $\begingroup$ 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. $\endgroup$ – Tanner Swett Aug 20 at 17:06
  • $\begingroup$ @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. $\endgroup$ – jdmc 12 hours ago
  • $\begingroup$ 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. $\endgroup$ – jdmc 12 hours ago
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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).

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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.

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