Euclidean Construction of Regular Hyperbolic Polygons

Abstract We show that the hyperbolic regular polygons with all right angles and five or more sides which can be constructed with Euclidean compass and straightedge are exactly the same family as the constructible Euclidean regular polygons with five or more sides. A few handy tricks for constructions in the Poincaré disk help us reach this conclusion fairly quickly.

The Polygons Are All Right The constructions in hyperbolic geometry, even though they reside in the Euclidean plane, take on a life of their own. The hyperbolic disk model consists of all the Euclidean points inside a boundary circle @. The hyperbolic lines in the disk are diameters of the disk and arcs of circles perpendicular to the disk. We can have polygons with five or more sides and all the angles 90E. (Angles are measured using Euclidean tangent lines at the vertex of the angle.) One characteristic of this geometry is the lack of similarity among triangles: if two triangles have their angles pair-wise congruent, then the triangles themselves must be congruent. In this sense, if a specific all right, regular polygon exists, it is the only one. We now need to consider some constructions. If the Euclidean ray OA (where O is the center of @Þ twice intersects the Euclidean circle containing a hyperbolic line, then the two points of intersection are related by inversion across @. Conversely, if a Euclidean circle passes through a point and its inversion across @, then the Euclidean circle must be perpendicular to @. In the figure below, the thick circle is the boundary, @, and all other solid circles contain hyperbolic lines. (As seen in a 2001 Monthly article, [2], and hinted at in various texts, [3], [4], there are several ways to take advantage of the Euclidean structure governing the disk model when making constructions.) Figure 1. The All Right Regular Pentagon.

Constructing All Right Polygons

Theorem The hyperbolic regular polygons with all right angles and five or more sides which can be constructed with Euclidean compass and straightedge are exactly the same family as the constructible Euclidean regular polygons with five or more sides. Before we prove the theorem, let’s note a useful construction tool through the following lemma. The figure contains a hyperbolic, all right, regular hexagon (the emphasized polygon with center O and vertices L and H.). Suppose we start with two ultra-parallel hyperbolic lines (no intersection in or on the boundary), for example the circles with centers at E and C. We use our straight edge to construct the two Euclidean secant lines through their points on the boundary. These secants intersect at D in the figure, the center of the unique, mutual perpendicular to our two lines. Now construct the segment between D and E and then construct the circle with that segment as diameter. This circle, (the upper dashed circle in the figure), intersects the top hyperbolic line at the point, L, where the unique, mutual perpendicular line centered at D will pass. We can summarize our trick with a statement. Lemma The Euclidean secant line through the endpoints of a hyperbolic line contains the center of any hyperbolic line perpendicular to the given hyperbolic line. Proof The lemma is clearly true for hyperbolic lines which are diameters, so we will move directly to the case where the hyperbolic line is part of a Euclidean circle. We will use a second hyperbolic line which could be a Euclidean diameter or another Euclidean arc. Again, the diameter case is easy to see, so we work with the second hyperbolic line as a Euclidean arc. To establish the lemma, note that the radii of the circles centered at D and E meet at a 90E angle at the last point constructed, L. Thus we know that our desired mutual perpendicular is indeed perpendicular to the hyperbolic line centered at E. This implies that the points J and G are an inverse pair across the mutual perpendicular circle centered at D. But J and G are also on the boundary circle, implying the boundary circle is perpendicular to the desired mutual perpendicular. So our desired perpendicular is officially a hyperbolic line. Then the two points on the circle centered at C which lie on the boundary are an inverse pair across the desired mutual perpendicular. Therefore the circle centered at D is perpendicular to the circle centered at C, as desired. n The lemma fails to apply when the two hyperbolic lines have Euclidean rays through endpoints which are Euclidean parallel. In this case, there is still exactly one mutual perpendicular: through the Euclidean centers of the arcs. For the small Euclidean investment of three segments, a midpoint and two circles, we have the unique perpendicular to the two given hyperbolic lines. This makes it pretty easy to construct all right hyperbolic polygons with lots of sides, as long as we’re careful how we close the polygon. If the all right polygon has an odd number of sides, the construction of a hyperbolic line perpendicular to a single given line will be necessary. At least the single secant line of the Lemma guides our selection of the center; if we just select a good location on the boundary for a point on our next circle, we can construct a tangent to the boundary to get our next center. (This method is not illustrated in the figure.) So now we can construct all right polygons of as many sides as we want, as long as we want more than four. That’s something we can’t do in Euclidean geometry. While we’re considering hyperbolic projects which are impossible in the Euclidean world, why not try to construct a regular, all right hyperbolic version of a non-constructible regular Euclidean polygon? The quick answer is, that would be logically equivalent to constructing the non-constructible. To see why, we consider the construction above, in a backwards way. The above figure contains the skeleton of our method for constructing a regular, all right hyperbolic polygon. We can start with a regular Euclidean polygon of convenient size. Then we construct the diagonals which skip one vertex. In our figure, the original pentagon is light and the diagonals form the familiar five-pointed star. next, we construct the circle whose diameter has the center O and a vertex of the original polygon P as endpoints. This circle is the lower dotted one in the figure. The dotted circle intersects a diagonal which skips E at the point Q. If we use OQ as the radius of @, we get everything we want. We construct a circle for each vertex with each radius the length EQ. Use the diagonals as the secant lines containing centers of mutually perpendicular circles, all of which are the same Euclidean size, the same Euclidean distance from O. Moreover, these circles all contain hyperbolic lines since they must be perpendicular to @. Each arc which forms a side of our hyperbolic polygon has the same Euclidean size, is the same distance from O and is symmetric across the ray from O through its center. Thus, without reference to the hyperbolic distance formula, we have the sides of the hyperbolic polygon all the same size in hyperbolic measure. (Our hyperbolic pentagon has its five vertices distinguished by lighter ink than the rest and its sides are thickened.) We have applied our first trick, backwards, to get the hyperbolic pentagon. So, if we have a regular Euclidean polygon given, we can construct a regular, hyperbolic all right polygon with the same number of sides. Now suppose we had a construction for a hyperbolic, regular, all right version of a non-constructible regular Euclidean polygon. Maybe one is possible, since we could build off-center. It would be congruent to the one centered at O, but it could look different with our Euclidean eyes. The hyperbolic sides can have equal hyperbolic length and unequal Euclidean measure; maybe there’s room for such a method to exist. The contradiction arrives quickly. We could choose a vertex and construct two perpendicular hyperbolic lines of the same Euclidean radius which intersect at this vertex. The vertex would act as the hyperbolic center of a circle through its adjacent vertices and this circle would cut these two new lines as well. The two new lines would have the same length arc inside this circle as the original two sides of the heptagon. But these two arcs would have the same Euclidean size as well. We could then easily finish constructing this copy of the heptagon in a position where the center of our copy would be the center of the boundary circle. Then we could whip out our straightedge and construct a non-constructible, regular Euclidean polygon. And, since we can also copy hyperbolic angles by construction, no construction can exist in general for the regular, hyperbolic polygon. The construction method above implies that the family of constructible regular polygons in Euclidean geometry is exactly the same family in hyperbolic geometry. REFERENCES

1. B. E. Reynolds and W. E. Fenton, College Geometry Using the Geometer’s Sketchpad, Key College Publishing, 2006. 2. C. Goodman-Strauss, Compass and Straightedge in the Poincaré Disk, The American Mathematical Monthly 108 No. 1 (2001), 38 -49. 3. H. Meschkowski, Noneuclidean Geometry, (Trans. A. Shenitzer), Academic Press, 1964. 4. H. S. M. Coxeter, Non-Euclidean Geometry, University of Toronto, 1965. Author information:

Michael McDaniel, Associate Professor of Mathematics Aquinas College Grand Rapids, MI 49506 [email protected]

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Ph. D. 1997, The George Washington University