Hyperbolic polygonal spirals

Transcription

1 Rose- Hulma Udergraduate Mathematics Joural Hyperbolic polygoal spirals Jillia Russo a Volume 11, No., Fall, 010 Sposored by Rose-Hulma Istitute of Techology Departmet of Mathematics Terre Haute, IN mathjoural@rose-hulma.edu a Aquias College

2 Rose-Hulma Udergraduate Mathematics Joural Volume 11, No., Fall, 010 Hyperbolic polygoal spirals Jillia Russo Abstract. This article is based o the costructio of Nested Hyperbolic Polygoal Spirals. The costructio uses costructible Euclidea agles to create hyperbolic polygos of five or more sides. The ested polygos are formed by coectig the midpoits of the sides of the origial polygo, thus creatig a spiral. The costructio is icluded for the readers to be able to costruct oe for themselves as they read alog. This costructio, alog with hyperbolic trigoometric formulas, led to the results: measures of the agles, side legths ad areas of all the parts of the spiral. Furthermore, the costructio is used to prove the costructible hyperbolic regular polygos have the same umber of sides as the costructible Euclidea polygos. Ackowledgemets: We thak the Deiso ad Marguerite Mohler Fud for creatig a summer research experiece for my advisor, Dr. Michael McDaiel, ad me. Dr. Aaro Cizori s talk o geometric spirals at our Math Club ispired our hyperbolic versio. We appreciate the referee s directios, which helped clarify the article.

3 RHIT Udergrad. Math. J., Vol. 11, No. Page Itroductio I this paper we costruct ad examie what we have amed ested hyperbolic polygoal spirals withi the Poicaré Disk (Figure 1). The costructio is easy to follow, so the readers will be able to costruct oe for themselves as they read alog. My advisor ad I studied the areas, agles ad side legths of the parts of the spiral. Also, the costructio leads to a proof describig which regular hyperbolic polygos with all right agles are costructible. Figure 1. Nested hyperbolic petagos. Hyperbolic geometry dates back to the ieteeth cetury with Bolyai ad Lobachevsky beig credited as the cofouders. This geometry comes from a chage i the parallel postulate. Istead of the Euclidea requiremet that for every give lie ad a give poit ot o that lie, there is oe parallel lie through the give poit, i hyperbolic geometry there are may parallel lies through a give poit (P) ot o a give lie (l) (Figure.) There are differet models for represetig this type of geometry, but for our purposes we focused o the Poicaré Disk Model.

4 Page 198 RHIT Udergrad. Math. J., Vol. 11, No. Figure. Hyperbolic disk properties. I the disk model, the hyperbolic space is cofied by a boudary circle. There are two kids of hyperbolic lies: diameters ad arcs of circles orthogoal to the boudary. I this space, hyperbolic lies may appear bet i the Euclidea sese, but i hyperbolic space they are straight lies. Also, agles withi this geometry are measured usig their tagets. This creates the characteristic property that the sum of the agles of a hyperbolic triagle is betwee 0 ad π. Furthermore, all similar triagles are cogruet ad o-itersectig lies have exactly oe mutual perpedicular lie. We ca ow examie Figure 1 with the vocabulary of hyperbolic geometry. The complete circle is the boudary of the space. The outermost petago is the level 1 hyperbolic petago. It has all right agles. Oe side has bee draw with the hyperbolic lie cotaiig it. The other lies have bee hidde i order to keep the drawig simple. Ay ested petago has the midpoits of the sides of the previous petago for vertices, givig us the k levels we use throughout the paper. The ested petagos form hyperbolic triagles; we will fid properties of oe triagle from each level. The shadig is to emphasize choosig oe triagle from each level. Here are a few more hyperbolic geometry facts we will use without proof. We ca have a hyperbolic polygo with all right agles ad as may sides as we wat, as log as we wat more tha four. The ested polygos have a limitig case as they are costructed iward: the Euclidea regular polygo of that may sides. The hyperbolic segmets get Euclidea straighter as the polygos get smaller.

5 RHIT Udergrad. Math. J., Vol. 11, No. Page 199 As the polygos go iward, the hyperbolic legth of a side approaches zero. The vertex agles grow from π to the Euclidea limit. The ested polygos are ot cogruet ad do ot form a tilig. The iitial vertex agles do ot have to be π ; they could be ay size less tha the Euclidea value of π, where 5. I the secod sectio, we recall a few useful hyperbolic trigoometric idetities. We give istructios for the costructio of this figure i the third sectio. Our fourth sectio has area calculatios, icludig a telescopic series. The proof that the sum of the hyperbolic legths of oe side of each level polygo coverges takes up the fifth sectio. We close with our costructibility proof. Hyperbolic Trigoometry The hyperbolic disk has its ow trigoometry, usig sih x ad cosh x. We oly eed two hyperbolic idetities, available i ay hyperbolic trigoometry text [Keelly]. We use triagle ABC whose sides are hyperbolic segmets with hyperbolic legths a, b, c. Figure 3. Hyperbolic triagle ABC. cosh c = cosh a cosh b sih a sih b cos C (1) The equatio above resembles the Law of Cosies i that C, the oly agle ivolved, correspods with the isolated side o the left had of the equatio. Sice the formula uses cosh a istead of the hyperbolic legth a, we will use cosh a i our calculatios much more tha a, eve though we will ivestigate the hyperbolic legths of the sides of our ested polygos. We will also use the special structure of our polygos to simplify the secod useful idetity.

6 Page 00 RHIT Udergrad. Math. J., Vol. 11, No. Figure 4. Level k sketch of OF D. I Figure 4, the poit O is the ceter of our hyperbolic polygos ad our disk. Euclidea arcs AC ad CE represet two cosecutive sides of a level k hyperbolic polygo ad arc BD represets a side of the hyperbolic polygo at the ext level, k + 1. Usig the otatio ad relatioships i this figure, we have the followig equatio. si OF D si ODF cosh F D = cos DOF +cos OF D cos ODF () The triagle OF D has specific values so that we ca simplify quickly. We will take segmet BD as legth L k, ad the cetral agle coectig cosecutive midpoits of a level k polygo as A k, makig ODF = A k. Sice OF D = π, formula () becomes si A k cosh L k = cos π. (3) With these formulas, we calculated agles, areas, ad legths for as may levels as we wated. We used cosh L k = S k for all our calculatios ad otes ad coverted to hyperbolic legth oly whe we wated actual hyperbolic legths. Our empirical ivestigatio poited the way to the theoretical results i this paper. The level oe side legth, S 1, of the petago is worth lookig at. Lemma 1 For the five-sided, all right-agled, regular hyperbolic petago, cosh L 1 = S 1 = ϕ (the Golde Ratio.) Proof. From equatio () applied to the hyperbolic triagle OGE i Figure 8 (below), we have cosh GE si π si π = cos π + cos π cos π. This implies S = cosh GE = cos π A trig ratio from regular Eulcidea petagos gives us cos π 5 fiishes our proof: cos π + 1 = 5 ϕ + 1 = ϕ. 3 Nested Hyperbolic Polygos = ϕ. A double agle formula We will ow provide a algorithm to costruct a regular, right agled hyperbolic polygo ad create layers of polygos iside it usig the midpoits of the sides. The costructio is

7 RHIT Udergrad. Math. J., Vol. 11, No. Page 01 simple (provided the iitial agle is costructible) ad ca be doe with a series of Euclidea moves usig a compass ad straight edge. This limitatio of the first agle forms a crucial part of our work i the last sectio. Figures 5, 6, ad 7 below will illustrate our procedure i the = 5 case. Regular Right Agled Hyperbolic Polygo Costructio Figure 5. Figure 6. Boudary locatio. Figure 7. Polygo costructio. Before drawig aythig, choose the umber of sides ad calculate ( 180( ) ). Create a segmet AB of ay legth R. Rotate the lie by the agle ( 180( ) ) 90 aroud poit B with legth R, creatig segmet BC. Thus begis Figure 5. Now rotate the segmet BC a egative 90 degrees aroud poit C, creatig segmet CD. We cotiue the first two steps repeatedly, rotatig aroud the ewest created edpoit, util the sides close i o each other, creatig Figure 5. Coect the vertices at the agles of measure ( 180( ) ) 90; it will create a -sided polygo with iterior agles ( 180( ) ). As we see i Figure 6, we have created isosceles right triagles, each with hypoteuse a outer side, like BD. The rotatio ( 180( ) ) 90 is just right to get such a figure for sides. Coect two vertices of the -sided polygo, skippig oe i betwee, like the segmet BH. Now we ca costruct the first side of the hyperbolic regular polygo. Use a vertex, J, of the -sided polygo as the ceter. Use radius of legth R, so the hyperbolic lie will go through the two poits of legth R away from the chose vertex. Now costruct the diagoals of the -sided polygo to fid the ceter. (Lies through O i Figures 6 ad 7)

8 Page 0 RHIT Udergrad. Math. J., Vol. 11, No. The ceter of the hyperbolic space will be the ceter of the -sided polygo (O), where all the diagoals meet. To fid the radius of the disk, we follow the segmet BH we coected betwee two vertices (skippig oe), startig at oe of the vertices, B, ad travelig toward the other util we hit the hyperbolic lie. This ulabeled itersectio gives us the disk radius distace. Fiish the polygo by creatig the hyperbolic lies as before. Use the vertices of the -sided polygo with radii of legth R. (Figure 7) To make the first ested polygo, we start by creatig lies perpedicular to the diagoals at all the vertices of the -sided polygo. The itersectios of these perpedicular lies create a ew -sided polygo. (Figure 8 - Petago MNPRS) Note: The midpoits of the arcs of the hyperbolic lies are the poits where the diagoals itersect the arcs. Use the vertices of the ewly created -sided polygo as the ceter of the circles for the sides of the ested hyperbolic polygo. The radius will be the legth from the vertices to the midpoit of the arc of oe of the circles cotaiig that poit. (It will go through both midpoits). Repeat the last step for all vertices to fiish the first ested hyperbolic polygo. Use the same three steps to costruct multiple ested hyperbolic polygos always buildig off the previous step.

9 RHIT Udergrad. Math. J., Vol. 11, No. Page 03 Figure 8. Nested polygo costructio. These Euclidea steps make hyperbolic objects. These steps my be used for ay umber of sides, give that first agle. We will ow justify the key steps i the costructio. Lemma Proof. JAB = π. The level 1 costructio gives a chai of mutually orthogoal circles. We costructed eighborig circles to have perpedicular radii. For example, Lemma 3 The level 1 costructio gives hyperbolic lies. Proof. The circle with ceter J has two (ulabeled) poits o Euclidea segmet HB. Sice the circle J is orthogoal to circle B, these two ulabeled poits are a poit ad its iverse. Ay circle through a poit ad its iverse is orthogoal to the circle of iversio. The circle with ceter O passes through these two poits. Therefore circles J ad B cotai hyperbolic lies i the disk bouded by circle O.

10 Page 04 RHIT Udergrad. Math. J., Vol. 11, No. Lemma 4 The level k costructio gives hyperbolic lies. Proof. The midpoits at ay level lie o alteratig Euclidea rays (of the rays we costruct) through O. Two cosecutive midpoits determie oe hyperbolic lie. The Euclidea ray from O through oe midpoit passes through the circle cotaiig that midpoit agai, at the iverse of that poit. The midpoit, the ceter of the cotaiig circle ad its iverse are Euclidea colliear with O. By costructig the ceter of the ext level usig the itersectio of the Euclidea perpedicular lies at vertices of the previous level Euclidea polygo, we get the ext level circle through the midpoit ad its iverse, makig the level k circle cotai a hyperbolic lie. (I our Figure 8, circle S passes through a midpoit ad its iverse.) We are relyig o a property of orthogoal circles which ca be foud i [Goodma- Strauss], that a circle through a poit ad its iverse must be orthogoal to the circle of iversio. We deped o the diagoals which skip oe vertex, like HB, to fid the ceters of the circles which cotai our hyperbolic lies. The Euclidea segmet HB passes through the itersectio of two circles ad is called the radical axis of circles J ad O. Lemma 5 Whe oe circle is a hyperbolic lie ad the other is the boudary, the the radical axis cotais the ceter of ay hyperbolic lie orthogoal to the give hyperbolic lie (except the diameter, of course.) Proof. Without loss of geerality, cosider the circle cetered at J i Figure 8. We have already built circle J orthogoal to the boudary. We ca chage our poit of view ad see that we have circle O orthogoal to circle J, treatig circle J as the boudary, temporarily, of a hyperbolic disk. The two ulabeled poits form a poit ad its iverse across circle J. This use of the radical axis is also give as a homework problem o the iversio of circles i [Veema]. We wat to emphasize that this costructio works for ay umber of sides, give that first Euclidea agle. 4 The Area of Oe Spiral Figure 1 shows five spirals made from hyperbolic triagles, oe from each level, distiguished by color. The sum of the areas of these triagles is easily calculated. First, we eed the area formula for a hyperbolic triagle ABC. With agles i radias, the area is π (A + B + C). (There is a multiplicatio costat which we will take as 1.) I Figure 1, we see 5 spirals all covergig at O. The picture would look similar for sides, except the level 1 polygo gets bigger as icreases (see Sectio 5 for the limit.) The the spirals fill the iterior of the first level polygo, with o overlap. For sides, the spiral cosists of 1 of the area of the first level polygo. The hyperbolic area formula of the first level is ( ) π π = π π.

11 RHIT Udergrad. Math. J., Vol. 11, No. Page 05 We ca divide by to see the spiral has area π π. A more iterestig calculatio ca be made for this area, this time applyig the hyperbolic area formula at each level ad addig. We will get a hoest-to-goodess telescopic series whose sum is, agai, π π. A telescopic sum has terms which cacel throughout the series, usually leavig a frot term ad a back term, as we will see i our example. We begi with a figure of two cosecutive levels i a polygo of sides. Figure 9. Supplemetary agles at D. Usig the area formula for a hyperbolic triagle agai, we fid the area of triagle BCD is π (γ + β) = π (γ + (π δ) ) = δ γ. (4) Now, these two agles γ ad δ are just the vertex agles of isosceles hyperbolic triagles from two cosecutive levels. The sum of the areas of hyperbolic triagles i the spiral would be a telescopic sum where the survivig terms would be the limit of the iermost agle mius the outermost agle. Luckily, both of these are available. The ested polygos approach a Euclidea regular polygo of sides. The outermost agle is a right agle because we started with all right agles. We get the same area of the spiral: π π = π π. From this example, we see that we ca fix the starter agle at size α ad the area of the spiral becomes π α π because the telescopic series will still happe. If we use o-midpoits for the ested polygo vertices, we lose the ice isosceles structure. Yet, we keep the telescopig series! To see this i Figure 8, thik of the two β s as β 1 ad β. The calculatio for the area of triagle BCD, from (4), becomes π (γ + β 1 + β ) = γ + (π β 1 β )) = δ γ. The limit of the series is π α = π α π, the same aswer as usig midpoits. This makes sese because, eve though we chose vertices off the midpoits, we still geerate a spiral which stads for 1 of the area iside the first polygo.

12 Page 06 RHIT Udergrad. Math. J., Vol. 11, No. 5 Sum of side legths For hyperbolic triagle ABC, the hyperbolic trigoometric formulas (1) ad () work icely with our ested polygos. We ca show that the hyperbolic legth of oe side of the first level polygo is bouded as icreases. Lemma 6 For the -sided, all right agled, regular, hyperbolic polygo, lim S 1 = 3. Proof. From formula () applied to triagle OCE, si π si π cosh L = cos π + cos π cos π. Solvig for cosh L = S 1 ad takig the limit gives the desired result. To traslate ito our levels, we use A k for the agle C. We will use L k for the hyperbolic legth of oe side at level k. More ofte, we will use S k = cosh L k. The isosceles structure of the triagle makes a = b =.5 cosh 1 (S k ). The side opposite agle A k already has a ame: S k+1. (The reaso cosh 1 S k turs up so much is that S k is ot hyperbolic legth. We have to take cosh 1 S k to get hyperbolic legth.) Substitutio ito formula (1) gives us our recursive formula for the hyperbolic legth of a side of oe of our ested hyperbolic polygos: S k+1 = [cosh (.5 cosh 1 (S k )) (cos A k sih (.5 cosh 1 (S k ))]. Simplificatios for the compositios cosh (.5 cosh 1 (S k )) = S k + 1 ad sih (.5 cosh 1 (S k ) = S k 1 have bee kow sice at least 1707 [Keelly]. (The substitutios sih 1 x = l(x + x + 1) ad cosh 1 x = l(x + x 1) from ay calculus text provide the key step.) The simplificatios lead to a shorter recursive formula: S k+1 = 1[S k(1 cos A k ) + (1 + cos A k )]. (5) We are ow ready to cosider the sum of oe side from each level. Theorem 1 cosh 1 (S k ) < for fixed, the umber of sides i the polygo. k=1 Proof. Startig with the ratio test, we apply our recursive formula (5) to the umerator. Sice all the subscripts have become k, we ca suppress the subscripts. We are goig to treat both S ad A as fuctios of k. So, for example, S = ds. dk cosh 1 (S k+1 ) lim k cosh 1 (S k ) cosh 1 ( 1 [S(1 cos A) + (1 + cos A)]) = lim k cosh 1 (S) Sice the limitig case is the ceter poit O, we kow the umerator ad deomiator go to 0. So we apply L Hopital s rule with k as the idepedet variable. lim k cosh 1 ( 1 [S(1 cos A) + (1 + cos A)]) cosh 1 (S).

13 RHIT Udergrad. Math. J., Vol. 11, No. Page 07 1 = lim [S (1 cos A) + (S 1)A si A] S 1 k S(1 cos A)+(1+cos A) [ ] S 1 [S (1 cos A) + (S 1)A si A] S 1 = lim k [S (1 cos A + cos A) + S(1 cos A) + (1 + cos A + cos A)] 4 S [S (1 cos A) + (S 1)A si A] S 1 = lim k [cos A(S 1) cos A(S 1) + (S + 3)(S 1) S [S (1 cos A) + (S 1)A si A] S + 1 = lim. k [cos A(S 1) cos A(S + 1) + (S + 3) S The lim S = 1; we see the factors of S 1 have caceled by divisio i the last step, k so we ca get ready to take the limit. Before we ca let the (S 1)A si A go to zero, A si A however, we eed to check that lim <. This is worth a look because the umber k S S k approaches 1; so S ca get small quickly. We will ow make sure that the factor A gets small quickly eough to hadle it. We ca rewrite our hyperbolic trigoometry formula (3) for A = (si 1 ( cos ( ) π )) (subscripts suppressed). The we fid da ds = cos( π ) (S + 1) by the Chai Rule. The S + 1 cos ( π ) S = 1 implies that this fractio is fiite for ay fixed. We ca fiish takig our limit. lim k So the difficult-lookig factor A cosh 1 (S k+1 ) si A cacels i the limit ad we fid lim k cosh 1 = (S k ) 1 cos A < 1 for ay fixed. lim k 6 The costructible, regular, all right agled hyperbolic polygos Our costructios give us oe more result. Recall that our costructio method i Sectio 3 begis with a agle. If that agle is costructible, the the resultig regular hyperbolic polygo is costructible. The costructible agles correspod to the costructible regular Euclidea polygos, which are kow. Specifically, a -sided regular Euclidea polygo is costructible if ad oly if is a product of distict Fermat primes ad a power of. A Fermat prime is a prime umber of the form m + 1. May sources give a detailed ivestigatio of the rich history of this result, [Veema] amog them. Theorem The costructible regular hyperbolic polygos with all right agles have the same umber of sides as the costructible Euclidea regular polygos. Proof. We ca show that the family of costructible regular hyperbolic polygos is the same as the family of costructible Euclidea regular polygos. I this case, by family, we S+1

14 Page 08 RHIT Udergrad. Math. J., Vol. 11, No. mea the umber of sides, more tha four. If we ca costruct a regular -sided Euclidea polygo, we ca costruct a regular, right agled hyperbolic polygo from it. We gave its costructio i Sectio 3. I order to show the families are the same, we suppose we could costruct a regular, - sided, right agled hyperbolic polygo where is ot a power of times a product of distict Fermat primes, like = 7. I other words, suppose we have a costructio for a regular, right agled, hyperbolic polygo whose Euclidea versio is ot i the costructible family. If this costructio has O as its ceter, we ca draw Euclidea segmets coectig cosecutive ceters of circles cotaiig hyperbolic sides to costruct a regular Euclidea polygo, which has bee proved impossible. If this hyperbolic costructio is off-ceter, the hyperbolic sides will ot be cogruet Euclidea arcs or segmets. The ceters of the circles cotaiig these Euclidea arcs will ot be spaced symmetrically aroud O. Steps exist, however, which will ceter the drawig, usig compass ad straightedge [Goodma-Strauss]. We ca traslate the off-ceter polygo to the locatio we use i this paper. The we are right back to the previous paragraph, costructig a impossible Euclidea polygo. Therefore, our assumptio that we ca costruct a regular, right agled hyperbolic polygo from outside the Euclidea family leads to a cotradictio. Right agled hyperbolic polygos appear i the study of Riema surfaces, as metioed i the paper by Castro ad Martiez [Castro/Martiez] These authors also metio that oly a few papers exist which study these objects. The Euclidea versios of this problem have received more attetio, i [Emmedorfer et al] ad [Stewart]. 7 Refereces A. E. Keelly, The Applicatio of Hyperbolic Fuctios to Electrical Egieerig Problems, McGraw-Hill (1916). G. Veema, Foudatios of Geometry, Pretice Hall (005). C. Goodma-Strauss, Compass ad straightedge i the Poicaré disk, The America Mathematical Mothly, 108, No. 1 (001), A. F. Costa ad E. Martiez, O hyperbolic right-agled polygos, Geometricae Dedicata, 58, No. 3, (December 1995), D. Emmedorfer, M. Precup, ad A. Warre, Classificatio of Geometric Spirals, Pi Mu Epsilo to appear. J. Stewart, Calculus, 6th editio, (Thomso) 799.