Computer Aided Drafting, Design and Manufacturing Volume 27, Number 1, March 2017, Page 48 CADDM
A simple method for constructing heptadecagon
Ma Long, Wang Pengfei, Zhang Caiming
School of Computer Science and Technology, Shandong University, Jinan 250101, china.
Abstract: It is proved that among the regular polygons with prime edges, only the regular polygons with Fermat prime edges are constructable with compass and straightedge. As 17 is a Fermat prime number, the construction of heptadecagon has been discussing all the time. Many different construction methods are proposed although they are based on the same theory. Simplification of the construction is still a sensible problem. Here, we propose a simple method for constructing regular heptadecagon with the fewest steps. The accumulation of construction errors is also avoided. This method is more applicable than the previous construction. Key words: heptadecagon; compass and straightedge construction; primitive root; group theory DOI: 10.19583/j.1003-4951.2017.01.0048
1 Introduction In ancient Greece, angle trisection, cube duplication and circle squaring are 3 classical difficult Compass and straightedge construction is an construction problems. In that period, they could not ancient problem in Euclidean Geometry[1-2]. There are be solved. Until the 18th-19th century, with the 5 basic constructions with compass and straightedge: establishment of group theory[3], their constructablity (1) Creating the line through two existing points; was proved to be negative. (2) Creating the circle through one point with Before the solution of the 3 problems, uniform centre another point; division of a circle problem is solved by solving cyclotomic polynomial. It is proved that if a circle can (3) Creating the point which is the intersection of be uniformly divided into p parts, where p is an even two existing, non-parallel lines; prime, then p must be a Fermat prime (4) Creating the one or two points in the 2k intersection of a line and a circle(if they intersect); Fkk 21,0,1,2
(5) Creating the one or two points in the F0 = 3, F1 = 5, F2 = 17, F3 = 257 and F4 = 65537 intersection of two circles(if they intersect). are proved to be primes. But except these 5 primes, no new Fermat primes are discovered by now. The Based on the basic constructions, some much used constructions of regular triangle, regular pentagon are constructions can be derived: simple. K. Gauss gave the earliest construction of regular (1) Constructing the perpendicular bisector from a heptadecagon[4]. The constructions of 257-gon and segment; 65537-gon are also discovered[5]. (2) Finding the midpoint of a segment; Most of methods for constructing heptadecagon are k (3) Drawing a perpendicular line from a point to a troublesome, because the surd expression of cos line; 17 is too complex to construct. But if the best use of (4) Bisecting an angle; intermediate variables during the construction are (5) Mirror a point in a line; made, the construction steps can be reduced. Our method is based on this creative ideal. (6) Construct a line through a point tangent to a circle. Compared to the existing construction method, our
Corresponding author: Zhang Caiming, Male, Ph.D., Professor, E-mail: [email protected]. Ma Long et al., A simple method for constructing heptadecagon 49
16 4 13 2 15 8 9 method can not only reduce the steps, but also avoid 1 accumulative construction errors. 3 14 5 12 10 7 11 6 2 2 Preliminary 16 4 13 1 The vertices of a hetadecagon can be composed of 314512 2 the roots of the cyclotomic polynomial 21589 3 17 z = 1. 611710 4 In order to solve this equation, the roots are According ε17 = 1, it is not difficult to confirm the divided into different groups to construct solvable following relations equations. 1 Let 12 () 21 124 22 cosi sin (1) 17 17 (σk, k = 1, 2 can be solved by geometric construction) and Then εk, k = 1, 2, ···, 16 are the complex roots of
Eq.(1). It is not difficult to verify that 6 is a primitive kk2 k [6] (1,2kk2 ,)k root of 17 (Table 1). kk2 1 Table 1 . (τk, k = 1, 2, 3, 4 can be solved from σk, k = 1, 2 by Index of 6 Remainders geometric construction) 0 1 According to the exponential property of complex 1 6 numbers[7], we have 2 2 2k 3 12 kk17 cos 4 4 17 5 7 2k Then cos ,k 1,2, ,8 can be solved from τk, 6 8 17 7 14 k = 1, 2, 3, 4 from the following relations: 8 16 28 610 9 11 2cos 2cos , 2cos 2cos 17 1712 17 17 10 15 28 610 11 5 4cos cos , 4cos cos 17 1723 17 17 12 13 13 10 4161214 2cos 2cos , 2cos 2cos 14 9 17 1734 17 17 15 3 416 1214 4cos cos , 4cos cos 17 1741 17 17 The complex roots can be grouped according to the remainders (Fig. 1). Every past construction is a solution to the equation in essence. But how to make full use of the intermediate variables to simplify the construction is not simple. In this paper, we will give the answer of this problem.
3 Our construction method
Our construction method is composed of the following parts: (1) Construct the points with x-coordinates of 1 4 and 2 on the x-axis. Fig. 1 Decomposition of complex roots of zk mod 17. 4
50 Computer Aided Drafting, Design and Manufacturing (CADDM), Vol.27, No.1, Mar. 2017
(8) Construct the intersections C and C between (2) Construct the points with x-coordinates of 1 , 2 4 2 b2 and x-axis, where C2 is on the right, C4 is on the left. 2 , 3 and 4 on the x-axis. Step (5)-(8) are shown in Fig. 3, they make up the 2 2 2 2nd part of the construction. It can be proved that (3) Construct the points with y-coordinates of xkk , 1,2,3,4. 1 1 1 1 Ci 2 1 , 2 , 3 and 4 on the y-axis. 2 2 2 2
1 (4) Construct the points with coordinates , 2
2 1 2 3 1 3 4 1 411 , , , , and , . 2 22 22 22 (5) Construct the points with x-coordinates of 2k 2cos ,k 1,2, ,8 on x-axis. 17 Fig. 3 Solution of l ,l 1,2,3,4. 2 (6) Construct the vertices of a regular heptadecagon on unit circle. (9) Construct D1 above M on the y-axis, such that |MD1| = |OC1|. In a unit circle ⊙O, the concrete construction is shown as followed: (10) Construct D2 above M on the y-axis, such that |MD2| = |OC2|. 1 (1) Construct a point A ,0 on the x-axis. (11) Construct D below M on the y-axis, such that 8 3 |MD3| = |OC3|. 1 (2) Construct the mid-point M 0, of the (12) Construct D4 below M on the y-axis, such that 2 |MD4| = |OC4|. radius along y-axis. Step (9)-(12) are shown in Fig. 4, they make up the (3) Construct a circle a with center A and through M. 3rd part of the construction. It can be proved that (4) Construct the intersections B and B between a 1 1 2 y i . Di and x-axis, where B1 is on the right, B2 is on the left. 2 Step (1)-(4) are shown in Fig. 2, they make up the 1st part of the construction. It can be proved that xkk ,1,2. Bk 4
Fig. 4 Interception on y-axis.
Fig. 2 Solution of k ,1,2k . (13) Construct E1 above C1 and on the right of D2, 4 such that |C1E1| = |OD2| and |D2E1| = |OC1|.
(5) Construct a circle b1 with center B1 and (14) Construct E2 below C2 and on the right of D3, through M. such that |C2E2| = |OD3| and |D3E2| = |OC2|.
(6) Construct a circle b2 with center B2 and (15) Construct E3 below C3 and on the left of D4, through M. such that |C3E3| = |OD4| and |D4E3| = |OC3|.
(7) Construct the intersections C1 and C3 between (16) Construct E4 above C4 and on the left of C4, b1 and x-axis, where C1 is on the right, C3 is on the left. such that |C4E4| = |OD1| and |D1E4| = |OC4|.
Ma Long et al., A simple method for constructing heptadecagon 51
Step (13)-(16) are shown in Fig. 5, they make up the
4th part of the construction. It can be proved that Ek, k = 1, 2, 3, 4 are the points in the 4th part.
Fig. 7 Segments construction.
Fig. 5 New centres of circles.
(17) Construct the intersection N(0, 1) between the y-axis and ⊙O.
(18) Construct circles e1, e2, e3 and e4 respectively with centers E1, E2, E3 and E4 and through point N.
(19) Construct the intersections Fk, k = 1, 2, …, 8 between circles $el, l = 1, 2, 3, 4 and x-axis. Step (17)-(19) are shown in Fig. 6, they make up Fig. 8 Final result. the 5th part of the constructions. It can proved that 2k Proof: Ek 2cos ,0 . 17 In Fig. 2, we have
1 xx2 x 12 BB12 A 44 4 1 xx y2 12 BB12 M 444 According to x x and we can derive BB21 21 xkk ,1,2. Bk 4
In Fig. 3, we have Fig. 6 Final solution. xx2 x kkk2 CCkk2 B k22 2 (k 1,2) (20) Construct the intersection S0(1, 0) between the 1 xx y2 kk2 ⊙O and x-axis. CCkk2 M 422
(21) Construct circles fk, k = 1, 2, ···, 8 separately According to xxk , 1, 2 and τk+2 < τk, CCkk2 with centers Fi and radius 1. k we have xkC ,1,2,3,4. (22) Construct the intersections Sk, S17 – k , k = 1, k 2 2, ···, 8 between fk and x-axis. Then Sk, k = 0, 1, 2, ···, 16 are the vertices of the heptadecagon inscribed ⊙O. In Fig. 4 and Fig. 5, we have Step (20)-(22) are shown in Fig. 7 and Fig. 8, the xx14,,x 23, x , EE1 E2 E3 4 make up the 6th part of the constructions. The proof is 222 2 11 1 1 based on the calculation of the horizonal coordinates yy21, yy3 ,,4 EE1 EE234 of Sk, k = 1, 2, 3, ···, 16. 2222
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In Fig. 6, we have References 28 [1] Godfried T. A new look at Euclid’s second proposition [J]. xx22cos2cos x FF14 E 11 17 17 The Mathematical Intelligencer, 1993, 15(3): 12-24.
[2] Gerard A. Foundations of geometry [M]. Upper Saddle 28 xxFF y N(2 y E y N ) 2 (2cos )(2cos ) River: Pearson Prentice Hall, 2006: 1-26. 14 1 17 17 [3] Borel A. Linear algebraic groups [M]. New York: 416 Springer-Verlag, 1991: 50-80. xx2 x 2cos 2cos FF28 E 22 17 17 [4] Herbert W R. A Construction for a regular polygon of 416 seventeen side illustration [J]. The Quarterly Journal of xx y(2 y y ) (2cos )(2cos ) Pure and Applied Mathematics, 2015, 26(4): 206-207. FF28 N E 2 N 3 17 17 [5] Benjamin B. Famous problems of geometry and how to 610 solve them [M]. New York: Dover, 1982: 70-76. xxFF22cos2cos x E 1 [6] Ribenboim P. The new book of prime number 35 3 17 17 records [M]. New York: Springer, 1996: 80-82. 610 xx y(2 y y ) (2cos )(2cos ) [7] Derbyshire J. Unknown quantity [M]. Washington, D.C.: FF35 N E 4 N 2 17 17 Joseph Henry Press, 2006: 58-64.
12 14 xx2 x 2cos 2cos FF67 E 44 17 17 Ma Long is a Ph.D candidite in the School of Computer Science and 12 14 xxFF y N(2 y E y N )1 (2cos )(2cos ) Technology of Shandong University. 67 4 17 17 He got his master degree of From these equations, we can derive mathematics in the School of 2k Mathematics of Shandong University xkF 2cos , 1,2, ,8 in 2011. His main interests are Geometric Modeling, k 17 Computational Geometry and Mesh Construction. In Fig. 7, according the symmetries between ⊙O Wang Pengfei is a Master candidite in and ⊙Fk, k = 1, 2, ···, 8, and x 1, we can derive S0 the School of Computer Science and 2k xxcos , k 0,1, ,8 Technology of Shandong University. SSkk17 17 He got his becholer degree of Computer Science and Technology in This means S , k = 0, 1, 2, ···, 17 are vertices of a k the School of Computer Science and regular heptadecagon inscribed ⊙O. Technology of Shandong University in 2013. His 4 Conclusion main interests are Geometric Modeling, Computational Geometry and Mesh Construction. Compared to the previous constuction method of Zhang Caiming is a professor and regular heptadecagon, our method can doctoral supervisor of the school of simultaneously construct all the vertices of a computer science and technology at the heptadecagon, which have an advantage that the Shandong University. He received a BS error accumulation during the intersection from the and an ME in computer science from original length is avoided. Complex constructions in the Shandong University in 1982 and our methods are also reduced, because they are 1984, respectively, and a Dr. Eng. degree in computer replaced by basic constructions, this can also science from the Tokyo Institute of Technology, Japan, simplify the construction and reduce the error. Our in 1994. From 1997 to 2000, Dr. Zhang has held construction also represent the algebraic variables by visiting position at the University of Kentucky, USA. geometric coordinates intuitively, so it is easy to His research interests include CAGD, CG, information understand and memorize. visualization and medical image processing.