Talk:Hypercycle (geometry)

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An entry from Hypercycle (geometry) appeared on Wikipedia's Main Page in the Did you know? column on October 24, 2006.

[edit] List of properties copied?

It seems that the bulleted list of properties were copied directly from http://www.math.uncc.edu/~droyster/math3181/notes/hyprgeom/node68.html, which is referenced in the article. -- SilverStar 05:30, 25 October 2006 (UTC)

I started from that list, but I reformulated the properties, changed the order and added some. The proofs are not taken from there: I did them myself, partially following Martin's book. Eubulide 07:36, 25 October 2006 (UTC)

[edit] Circle Limit III

The text says that the Escher print contains hypercycles, but it doesn't explain why. Where's the straight line that one of those hypercycles is based on, and what defines the distance between them? 171.64.71.123 00:01, 11 January 2007 (UTC)

In the Poincaré model, a circle represents

* a circle if it does not meet the horizon;

* a horocycle if it is tangent to the horizon;

* a straight line if it cuts the horizon at right angles;

* a hypercycle if it cuts the horizon at any other angle.

Since Escher's arcs are not perpendicular to the horizon, they must be hypercycles. The axis of a hypercycle is represented by the perpendicular circle that meets the horizon in the same pair of points. Sorry I can't help more than that. —Tamfang 08:39, 28 February 2007 (UTC)

I can see that in the actual woodcut, the lines seem to be at slight angles to perpendicular where they meet the horizon. The question is whether this is intended or not. I always thought of it as just a representational imperfection, just as how pencil strokes of finite width usually represent ideally widthless geometrical objects.

I started to write this comment with the intention of showing that Escher must have intended straight lines, but along the way I became convinced that he couldn't have. At first I thought that their being hypercycles would spoil the bilateral symmetry of the fish, but they are not symmetric to begin with - angle of the back corner of the fin is always 90° for right fins but 120° for left fins.

Even if each fish is individually non-symmetric we must, however, assume that each fish is congruent with every other fish; otherwise the entire artistic point of the woodcut is lost. This means by necessity that the white lines always meet each other at 60° angles (each angle in a six-way meet is alternately the right part of a head plus the right part of a tail, or the left part of a head plus the left part of a tail, but these two possibilities must be equal because they are opposite). Consider now the two different quadrilaterals centered on the middle of the disk. One, standing on its corner, has spines of yellow and green fish as its sides and a side length of one fish, and another, lying on a side, has spines of red and blue fish as its sides and a side length of two fish. If the white lines are straight, we have two regular quadrilaterals, both with 60° corners but one with straight sides twice as long as the other. This is an impossible configuration in hyperbolic geometry; there is no such thing as figures that are similar but have diferent sizes. Thus we're forced to abandon the assumption that the spines are straight lines. And having seen that, they do indeed look like hypercycles. –Henning Makholm 14:56, 6 May 2007 (UTC)

The recent biography of Coxeter says the arcs are "branches of equidistant curves that cut through corresponding vertices of the octagons of the underlying tessellation." The "underlying tessellation" must be {8,3}. Its vertices are the intersections of the white lines plus the triple points of the left fins. —Tamfang 17:06, 6 May 2007 (UTC)

That makes sense. I'll try to brush up a bit on hyperbolic trigonometry, and see if I can compute the answer to the original question, i.e. where the axes are with respect to the fish pattern. –Henning Makholm 20:36, 6 May 2007 (UTC)