Regular Polygons. by construction and paper folding. Paul Yiu Department of Mathematics Florida Atlantic University.

Transcription

1 Regular Polygons by construction and paper folding Paul Yiu Department of Mathematics Florida tlantic University January 22, 2009

2 1 Regular polygons (a) Triangle (b) Square (c) Pentagon (d) Hexagon (e) Heptagon (f) Octagon 2

3 1.1 onstruction of equilateral triangle with a given side 3

4 1.2 onstruction of a square 4

5 1.3 onstruction of a regular hexagon 5

6 6

7 1.4 onstruction of a regular octagon 7

8 D 1 2x Q x x 1 2x P x B (1 2x) 2 = x 2 + x 2 = x =

9 (1 2x) 2 = x 2 + x 2 = x = 1 D 2 2. O Q P B 9

10 Successive completion of rhombi beginning with three adjacent 45 -rhombi. 10

11 1.5 The regular dodecagon (12 sides) Successive completion of rhombi beginning with five adjacent 30 - rhombi. This construction can be extended to other regular polygons. If an angle of measure θ := 360 2n can be constructed with ruler and compass, beginning with n 1 adjacent θ-rhombi, by succesively completing rhombi, we obtain a regular 2n-gons tessellated by rhombi. (n 1) + (n 2) = 1 n(n 1) 2 11

12 2 The regular pentagon 12

13 2.1 ngle sum = 540 ngle sum of a pentagon = = 540. Each angle of a regular pentagon is =

14

15 P 36 B E D 15

16 2.2 The pentagram 16

17 If each side of the regular pentagon has length s, and each diagonal has length d, then s s s P d s 36 B E D 17

18 s s s d d s d s = s s d = s 5 1 d = 2 It also means that d s = This is called the golden ratio. 18

19 2.3 Dividing a segment in the golden ratio M Q P B 19

20 2.4 onstruction of a regular pentagon with a given diagonal Given a segment B, we construct a regular pentagon BDE with B as a diagonal. (1) Divide B in the golden ratio at P. (2) Divide BB in the golden ratio at P. The points P and P are symmetric with respect to B. (3) onstruct the circles (B), B(), (P) and B(P ), and let (i) be the intersection of (P) and B(P ), (ii) D be the intersection of (B) and B(P ), (iii) E be the intersection of B() and (P). Then BDE is a regular pentagon with B as a diagonal. P P B E D 20

21 2.5 onstruction of a regular 17-gon arl Friedrich Gauss ( ) discovered, at the age of 19, that it is possible to construct a regular 17-gon with ruler and compass. The following construction 1 of a regular 17-gon makes use of Gauss computation. onsider a circle (O) with two perpendicular diameters P Q and RS. (1) On the radius OR, mark a point such that O = 1 4 OP. (2) onstruct the internal and external bisectors of angle OP to intersect the line OP at B and respectively. (3) Mark D and E on PQ such that D = and BE = B. (4) Mark the midpoint M of QD. (5) Mark F on OS such MF = MQ. (6) onstruct the semicircle on OE and mark a point G on it such that OG = OF. (7) Mark H on OP such that EH = EG, and construct the perpendicular to OP at H to intersect the circle at P 1. Then POP 1 = , and P 1 is a vertex of the regular 17-gon adjacent to P on the circle. 1 J. J. allagy, The central angle of the regular 17-gon, Math. Gaz., 67 (1983)

22 R P 1 Q M O D B E H P F G S 22

23 3 The golden ratio 3.1 Let D and E be the midpoints of the sides B and of an equilateral triangle B. If the line DE intersects the circumcircle of B at F, then E divides DF in the golden ratio. D E F O B 23

24 3.2 Two adjacent squares. The bisector of angle D divides B in the golden ratio. D P B 24

25 4 Regular solids regular solid is one whose faces of regular polygons of the same number of sides. t each vertex there are at least three faces, and the sum of the angles must be less than 360. The only possibilities are Number Faces angle of faces solid equilateral 60 3 regular tetrahedron triangles 4 regular octahedron 5? squares 90 3 cube regular pentagons? 25

26 Solid Vertices V Edges E Faces F tetrahedron cube octahedron dodecahedron icosahedron V E + F = 2. 26

27 5 onstruction of the regular dodecahedron 5.1 golden tent Take a square (in horizontal position), and divide a mid-line in golden ratio at B, B. Erect a half-square vertically above BB. P K P B O B 27

28 5.2 pattern for the golden tent 28

29 5.3 onstruction of regular dodecahedron On each of the six faces of a cube, build a golden tent so that faces top and bottom front and rear left and right extra edges east-west vertical north-south Each trapezoid of one tent, together with a triangle of an adjacent tent, forms a regular pentagon. 29

30 P K P X B O Y Q D O In this way we have a solid with twelve faces, each a regular pentagon. This is a regular dodecahedron. 30

31 5 Paper folding 5.1 Folding a square D D Y X Y X B B 31

32 5.2 Folding an equilateral triangle D D X D D B B 32

33 5.3 Folding a regular octagon D D B B 33

34 5.4 Folding a regular pentagon D D 0 B 1 0 B 34

35 D D B 1 2 B 35

36 D D B 1 2 B 36

37 D B 37

38 5.5 Regular pentagon from a paper knot 38