hapter 22 The regular pentagon 22.1 The golden ratio The construction of the regular pentagon is based on the division of a segment in the golden ratio. 1 Given a segment, to divide it in the golden ratio is to construct a point P on it so that the area of the square on P is the same as that of the rectangle with sides P and, i.e., P 2 = P. construction of P is shown in the second diagram. M P P Suppose P has unit length. The length ϕ of P satisfies ϕ 2 = ϕ +1. 1 In uclid s lements, this is called division into the extreme and mean ratio.
802 The regular pentagon This equation can be rearranged as ( ϕ 1 ) 2 = 5 2 4. Since ϕ>1, wehave Note that P = ϕ = 1 2 This explains the construction above. ( 5+1 ). ϕ ϕ +1 = 1 ϕ = 2 5 1 =. 5+1 2 22.2 The diagonal-side ratio of a regular pentagon onsider a regular pentagon. It is clear that the five diagonals all have equal lengths. Note that (1) = 108, (2) triangle is isosceles, and (3) = = (180 108 ) 2=36. In fact, each diagonal makes a 36 angle with one side, and a 72 angle with another. P It follows that (4) triangle P is isosceles with P = P =36, (5) P = 180 2 36 = 108, and (6) triangles and P are similar.
22.3 onstruction of a regular pentagon with a given diagonal 803 Note that triangle P is also isosceles since (7) P = P =72. This means that P =. Now, from the similarity of and P,wehave : = : P. In other words P = P P,orP 2 = P. This means that P divides in the golden ratio. 22.3 onstruction of a regular pentagon with a given diagonal Given a segment, we construct a regular pentagon with as a diagonal. (1) ivide in the golden ratio at P. (2) onstruct the circles (P ) and P (), and let be an intersection of these two circles. (3) onstruct the circles () and () to intersect at a point on the same side of as. (4) onstruct the circles (P ) and (P ) to intersect at. Then is a regular pentagon with as a diameter. P
804 The regular pentagon 22.4 Some interesting constructions of the golden ratio 22.4.1 onstruction of 36, 54, and 72 angles ach of the following constructions begins with the division of a segment in the golden ratio at P. 36 36 36 P 54 72 36 P 54 72 P 22.4.2 Odom s construction Let and be the midpoints of the sides and of an equilateral triangle. If the line intersects the circumcircle of at, then divides in the golden ratio.
22.4 Some interesting constructions of the golden ratio 805 22.4.3 ϕ and arctan 2 Given a segment, erect a square on it, and an adjacent one with base. If is the vertex above, construct the bisector of angle to intersect at P. Then P divides in the golden ratio. Proof. If =1, then P P : P =1 : 5, P : =1 : 5+1 :2 =1 : 5+1 : P = 5+1:2 =ϕ :1. This means that P divides in the golden ratio.
806 The regular pentagon
hapter 23 onstruction of regular polygons regular polygon of n sides can be constructed (with ruler and compass) if it is possible to divides the circle into n equal parts by dividing the 360 at the center into the same number of equal parts. This can be easily done for n =6(hexagon) and 8 (octagon). 1 We give some easy alternatives for the regular hexagon, octagon and dodecagon. 23.1 The regular hexagon regular hexagon can be easily constructed by successively cutting out chords of length equal to the radius of a given circle. 1 Note that if a regular n-gon can be constructed by ruler and compass, then a regular 2n-gon can also be easily constructed.
808 onstruction of regular polygons 23.2 The regular octagon (1) Successive completion of rhombi beginning with three adjacent 45 - rhombi. (2) We construct a regular octagon by cutting from each corner of a given square (of side length 2) an isosceles right triangle of (shorter) side x. This means 2 2x : x = 2:1, and 2 2x = 2x; (2 + 2)x =2, x = 2 2+ 2 = 2(2 2) (2 + 2)(2 2) =2 2. O Q x Q x 2 2x P x P Therefore P =2 x = 2, which is half of the diagonal of the square. The point P, and the other vertices, can be easily constructed by intersecting the sides of the square with quadrants of circles with centers at the vertices of the square and passing through the center O.
23.3 The regular dodecagon 809 23.3 The regular dodecagon (1) Trisection of right angles. (2) regular dodecagon can be formed from 4 equilateral triangles inside a square. 8 of its vertices are the intersections of the sides of these equilateral triangles, while the remaining 4 are the midpoints of the sides of the squares formed by the vertices of the equilateral triangles inside the square. n easy dissection shows that the area of the regular dodecagon is 3 4 of that of the (smaller) square containing it.
810 onstruction of regular polygons (3) 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 can be constructed with ruler and compass, beginning with n 1 adjacent θ-rhombi, by succesively completing rhombi, 2n we obtain a regular 2n-gons tesellated by rhombi. (n 1) + (n 2) + +2+1= 1 n(n 1) 2 23.4 onstruction of a regular 15-gon Let 1 2 3 4 5 be a regular pentagon inscribed in a circle (O). onstruct equilateral triangles 1 2 5, 2 3 5, 3 4 1, 4 5 2, 5 1 3 with the extra vertices inside the circle. xtend the sides of these equilateral triangles to intersect the circle again. The 10 intersections, together the vertices of the regular pentagon, form the vertices of a regular 15-gon inscribed in the circle.
23.5 onstruction of a regular 17-gon 811 2 3 3 2 O 1 1 4 5 4 5 23.5 onstruction of a regular 17-gon The following construction 2 of a regular 17-gon makes use of Gauss computation cos 360 17 = 1 ( 17 1+ 16 + 1 8 17 + 3 17 34 2 ) 17 34 2 17 2 34 + 2 17. onsider a circle (O) with two perpendicular diameters PQand 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 and respectively. (3) Mark and on PQsuch that = and =. (4) Mark the midpoint M of Q. (5) Mark on OS such M = MQ. (6) onstruct the semicircle on O and mark a point G on it such that 2 J. J. allagy, The central angle of the regular 17-gon, Math. Gaz., 67 (1983) 290 292.
812 onstruction of regular polygons OG = O. (7) Mark H on OP such that H = G, and construct the perpendicular to OP at H to intersect the circle at P 1. Then POP 1 = 360, and P 17 1 is a vertex of the regular 17-gon adjacent to P on the circle. R P 1 Q M O H P G S
hapter 24 Hofstetter s constructions on the golden section This chapter collects some short papers by Kurt Hofstetter in orum Geometricorum on the golden section. 24.1 simple construction of the golden section We construct the golden section by drawing 5 circular arcs. 1 We denote by P (Q) the circle with P as center and PQ as radius. igure 24.1 shows two circles () and () intersecting at and. The line intersects the circles again at and. The circles ( ) and () intersect at two points X and Y. It is clear that,, X, Y are on a line. It is much more interesting to note that divides the segment X in the golden ratio, i.e., X = 5 1. 2 This is easy to verify. If we assume of length 2, then =2 3 and X = 15 + 3. rom these, X = 2 3 2 5 1 = =. 15 + 3 5+1 2 1 orum Geom., 2 (2002) 65 66.
814 Hofstetter s constructions on the golden section X Y igure 24.1: This shows that to construct three collinear points in golden section, we need four circles and one line. It is possible, however, to replace the line by a circle, say (). See igure 24.2. Thus, the golden section can be constructed with compass only, in 5 steps. X Y igure 24.2: It is interesting to compare this with igure 24.3 which also displays the golden section. 2 Here, is an equilateral triangle. The line joining the midpoints, of two sides intersects the circumcircle at. 2 See. H. owler, The Mathematics of Plato s cademy, Oxford University Press, 1988, p.105, note on 3.5(b); and G. Odom and J. van de raats, lementary Problem 3007, merican Math. Monthly, 90 (1983) 482; solution, 93 (1986) 572. I am indebted to a referee for these references.
24.2 5-step division of a segment in the golden section 815 Then divides in the golden section, i.e., 5 1 =. 2 However, it is unlikely that this diagram can be constructed in fewer than 5 steps, using ruler and compass, or compass alone. igure 24.3: 24.2 5-step division of a segment in the golden section Using ruler and compass only in five steps, we divide a given segment in the golden section. 3 Inasmuch as we have given in 24.1 a construction of the golden section by drawing 5 circular arcs, we present here a very simple division of a given segment in the golden section, in 5 euclidean steps, using ruler and compass only. or two points P and Q, we denote by P (Q) the circle with P as center and PQas radius. onstruction 24.1. Given a segment, construct 1. 1 = (), 2. 2 = (), intersecting 1 at and, 3. 3 = (), intersecting 1 again at, 4. the segment to intersect 3 at, 5. 4 = ( ) to intersect at G. 3 orum Geom., 3 (2003) 205 206.
816 Hofstetter s constructions on the golden section 3 4 G G H 2 1 igure 24.4: The point G divides the segment in the golden section. Proof. Suppose has unit length. Then = 3 and G = = 2. Let H be the orthogonal projection of on the line. Since H = 1 2, and HG2 = G 2 H 2 =2 3 4 = 5 4,wehaveG = HG H = 1 2 ( 5 1). This shows that G divides in the golden section. Remark. The other intersection G of 4 and the line is such that G : = 1 2 ( 5+1):1. 24.3 nother 5-step division of a segment in the golden section We give one more 5-step division of a segment into golden section, using ruler and compass only. 4 4 orum Geom., 4(2004) 21 22.
24.3 nother 5-step division of a segment in the golden section 817 Inasmuch as we have given in 24.1,24.2 5-step constructions of the golden section we present here another very simple method using ruler and compass only. It is fascinating to discover how simple the golden section appears. or two points P and Q, we denote by P (Q) the circle with P as center and PQas radius. 3 2 1 G igure 24.5: onstruction 24.2. Given a segment, construct 1. 1 = (), 2. 2 = (), intersecting 1 at and, 3. the line to intersect 1 at (apart from ), 4. 3 = () to intersect 2 at (so that and are on opposite sides of ), 5. the segment to intersect at G. The point G divides the segment in the golden section. Proof. Suppose has unit length. It is enough to show that G = 1 2 ( 5 1). xtend to intersect 3 at H. Let intersect at I, and let J be the orthogonal projection of on. In the right triangle H, H =4, =1. Since 2 = J H, J = 1 4. Therefore, IJ = 1 4. It also follows that J = 1 4 15.
818 Hofstetter s constructions on the golden section G H I G J igure 24.6: 5 2 Now, IG GJ = I = 1 2 3 J 1 4 15 = 2 5. It follows that IG = 2 5+2 IJ =, and G = 1 + IG = 5 1. This shows that G divides in the 2 2 2 golden section. Remark. If is extended to intersect H at G, then G is such that G : = 1( 5+1):1. 2 fter the publication of 24.2, ick Klingens and Marcello Tarquini have kindly written to point out that the same construction had appeared almost one century ago in the following references. (i). Lemoine, Géométrographie ou rt des onstructions Géométriques,. Naud, Paris, 1902; p.51. (ii) J. Reusch, Planimetricsche Konstruktionen in Geometrographischer usführung, Teubner, Leipzig, 1904; S.37. 24.4 ivison of a segment in the golden section with ruler and rusty compass We give a simple 5-step division of a segment into golden section, using ruler and rusty compass. 5 5 orum Geom., 5 (2005) 135 136.
24.4 ivison of a segment in the golden section with ruler and rusty compass 819 In 24.3 we have given a 5-step division of a segment in the golden section with ruler and compass. We modify the construction by using a rusty compass, i.e., one when set at a particular opening, is not permitted to change. or a point P and a segment, we denote by P () the circle with P as center and as radius. 2 1 3 M G igure 24.7: onstruction 24.3. Given a segment, construct 1. 1 = (), 2. 2 = (), intersecting 1 at and, 3. the line to intersect at its midpoint M, 4. 3 = M() to intersect 2 at (so that and are on opposite sides of ), 5. the segment to intersect at G. The point G divides the segment in the golden section. Proof. xtend to intersect 1 at. ccording to 24.3, it is enough to show that =2. Let be the orthogonal projection of on. It is the midpoint of M. Without loss of generality, assume = 4, so that M = =1and =2 =7. pplying the Pythagorean theorem to the right triangles and M,wehave
820 Hofstetter s constructions on the golden section 2 1 M G 3 igure 24.8: 2 = 2 + 2 = 2 + M 2 M 2 =7 2 +4 2 1 2 =64. This shows that =8=2. 24.5 4-step construction of the golden ratio We construct, in 4 steps using ruler and compass, three points two of the distances among which bear the golden ratio. 6 We present here a 4-step construction of the golden ratio using ruler and compass only. More precisely, we construct, in 4 steps using ruler and compass, three points two of the distances among which bear the golden ratio. It is fascinating to discover how simple the golden ratio appears. We denote by P (Q) the circle with center P, passing through Q, and by P (XY ) that with center P and radius XY. onstruction 24.4. Given two points and, construct 6 orum Geom., 6 (2006) 179 180.
24.5 4-step construction of the golden ratio 821 H 3 G 1 2 igure 24.9: 1. the circle 1 = (), 2. the line to intersect 1 again at and extend it long enough to intersect 3. the circle 2 = () at and, 4. the circle 3 = () to intersect 1 at and G, and 2 at H and I. Then GH = 5+1. H 2 Proof. Without loss of generality let =1, so that = = H = H =2. Triangle H is equilateral. Let 4 = H(), intersecting 1 at J. y symmetry, GJ is an equilateral triangle. Let 5 = J() = J(G) = J(), intersecting 1 at K. inally, let 6 = J(H) =J(). See igure 24.10. With 1, 5, 2, 6, following [1], K divides GH in the golden section. It suffices to prove H = GK = 3. This is clear for GK since the equilateral triangles GK and JK have sides of length 1. On the other hand, in the right triangle H, =1and H =2. y the Pythogorean theorem H = 3.
822 Hofstetter s constructions on the golden section 4 H K J 1 3 G 2 6 5 igure 24.10: Remark. G H = 2 2.