Note on regular polygons of rational order Peter Kahlig Ludwig Reich 75 A Tribute by Students, Colleagues, and Friends Harald Fripertinger, Wolfgang Prager, Jens Schwaiger, and Jörg Tomaschek (Eds.) Grazer Math. Ber., ISSN 1016-7692 Bericht Nr. 363 (2015), 106-131 This article: CORRECTED VERSION (2018) Abstract. Exploiting the Dido functional equation, regular polygons of integer order (from 3 to 10, convex polygons) and of rational order (like 5/2 or 8/3, star polygons) are treated. Connections with in…nite products yield representations for . [Corrected version with extensions.]
1. Introduction Regular polygons of integer order have been a long-standing topic of geometry ([1, 6, 10]). Regular polygons of rational order lead to star-shaped …gures (self-intersecting polygons, so-called star polygons). The emerging patterns are aesthetically pleasing; some of them became respected symbols (for instance pentagram and hexagram, cf. [1]). –The theory of regular polygons (of integer order) is intimately connected with the theory of cyclotomic …elds, and dividing a circle into equal parts is a well-known task in complex function theory (cf. [4, 5, 11, 12]). (For brevity, we often denote regular polygons simply as polygons.) Convex polygons n and star polygons may be characterized by the Schlä‡i symbol f m g (cf. [5]) where the integer n is the number of points on the outer circle, and the integer m describes the condition that every mth neighbor point on the outer circle is connected by a straight line n (1 m 2 ) called stroke line (or draw line). For instance, f5g denotes a pentagon (5 points on the outer circle, every neighbor point is connected by a stroke line), while f 52 g denotes a pentagram (5 points on the outer circle [forming an auxiliary “mother” n pentagon], every second neighbor point is connected by a stroke line). Within f m g, n and m must not be divided by a common factor. (The same holds for our index notation, n n n := s n e.g. s m f m g for brevity.) The fraction m (where 1 < m 2 ) may be interpreted n as rational order of a star polygon. By de…nition, f 1 g = fng, i.e. n is (in case m = 1) n the order of a convex polygon. – Use and interpretation of the Schlä‡i symbol f m g in 0
Mathematics Subject Classi…cation 2010: Primary: 51M20, 39B12; secondary: 40A20. Keywords and phrases: Dido functional equation, regular polygons (of rational order), regular star polygons, regular …gures, Schlä‡i symbol, in…nite products for . 1
2
PETER KAHLIG
connection with polygons (either convex or star) is critically documented on some web sites, for instance
https://en.wikipedia.org/wiki/Regular_polygon Speci…cally, following [5], convex polygons are denoted by fng; n = 2; 3; 4; :::; and star n n polygons by f m g; n; m coprime. (If n; m are not coprime, f m g denotes a compound 5 star polygon.) For instance: f5g pentagon, f6g hexagon, f 2 g pentagram, f 26 g hexagram (6 and 2 are not coprime, hence f 26 g =2 f 13 g by so-called Schlä‡i factoring, denoting a regular compound of 2 triangles). In Section 2 we recall some basic formulas and present partially known results from an apparently new point of view. In Section 3 we illustrate the …rst twenty-…ve regular polygons: f2g digon f3g triangle (equilateral) f4g square f 42 g tetragram f5g pentagon f 52 g pentagram f6g hexagon f 62 g hexagram A f 36 g hexagram B f7g heptagon f 72 g heptagram A f 73 g heptagram B f8g octagon f 82 g octagram A f 83 g octagram B f 84 g octagram C 9 f9g nonagon f 2 g nonagram A f 93 g nonagram B f 49 g nonagram C 10 10 10 f10g decagon f 10 2 g decagram A f 3 g decagram B f 4 g decagram C f 5 g decagr.D Strictly, it would be consistent to denote the nonagon by enneagon, but both names appear to be in general use (cf. [2]). – Polygons of order n 2 are constructible (by ruler and compass) if and only if one of the following conditions is met: r (1) Either n = 2k ; k 2 N; or n = 2k product of di¤erent Fermat primes 22 +1 (young Carl Friedrich Gauss 1796, cf. [10, 3, 1]). Presently known r 2 f0; 1; 2; 3; 4g: (2) Dido value f (n) is a constructible algebraic number (cf. [12, 4, 9]). (3) Euler function value '(n) is a power of 2 (cf. [3]). – This criterion is convenient to apply (constructible n-gons are marked by *; for extreme case n = 2 see Example 2):
n '(n)
2 20
3 21
4 21
5 22
6 21
7 6
8 22
n '(n)
11 10
12 22
13 12
14 6
15 23
16 23
17 24
9 6 18 6
10 22 19 18
20 23
n n Polygons of rational order m are constructible (for integers 1 < m 2 ) if and only if the corresponding “mother” polygon of integer order n is constructible. In Section 4 we compile some in…nite products for .
2. Some basic theory A concise notation for powers of functions is used, for instance f (x)2 := [f (x)]2 . – The Dido functional equation permits a convenient approach to regular polygons ([7, 8]). It reads r 1 2f (2x) = f (x) + f (x)2 + 2 ; x 2; x
REGULAR POLYGONS OF RATIONAL ORDER
3
where f : [2; 1) ! [0; 1) is a uniquely de…ned continuous function. Under a certain regularity condition (suitable asymptotic behavior at in…nity) the solution is 1 f (x) = cot( ); x 2 [2; 1): x x Denote by sn the half-side of a regular convex polygon of order n, and by f (n) the Dido value f (n) = n1 cot( n ) for n = 2; 3; 4; ::: [f (2) = 0 is valid for the extreme case “2-gon” (digon), see Example 2]. Proposition 1. The perimeter Pn of a convex polygon of order n = 2; 3; 4; ::: is Pn = 2n sn : The stroke distance dn of a convex polygon of order n is dn =
1 2 Rn = 2 sin( )Rn = 2sn ; Pn = p n n [nf (n)]2 + 1
i.e. for convex polygons the stroke distance is simply the polygon side (twice the half-side). The draw length Dn of a convex polygon of order n is the sum of all stroke distances, Dn = ndn = 2nsn = Pn ; i.e. for convex polygons the draw length is simply the perimeter. The inner radius rn of a convex polygon of order n is rn = f (n)
1 Pn = nf (n)sn = cot( ) sn : 2 n
The outer radius Rn of a convex polygon of order n is r p 1 1 1 Pn = [nf (n)]2 + 1sn = sn : Rn = f (n)2 + 2 n 2 sin( n )
Inner radius rn and outer radius Rn are related by rn f (n) =q Rn f (n)2 +
1 n2
Rn2
= cos( ); n
rn2 = s2n :
The area An of a convex polygon of order n is 1 1 Pn = rn sn n = f (n) ( Pn )2 = n cot( ) s2n 2 2 n n2 n 1 1 = f (n)d2n = cot( )d2n = f (n) ( Dn )2 = cot( )Dn2 4 4 n 2 4n n 1 2 1 2 = r = n tan( )rn = q rn Rn = n sin( ) rn Rn f (n) n n n f (n)2 + 1
An = rn
n2
=
f (n) f (n)2 +
1 n2
1 Rn2 = q f ( n2 )2 +
22 n2
Rn2 =
n 2 sin( )Rn2 : 2 n
4
PETER KAHLIG
n n of a star polygon of order Proposition 2. The perimeter P m m is (with n n = 4; 5; 6; ::: and 1 m 2 ) v #2 " u r n n u nf (n) m f(m ) (m 1) 2 1 t n = 2n s n = Pn = 1 + tan( Pm Pn 1 + n2 ) Pn = m n (m 1) n f (n)f ( ) + 1 cos( ) m m n
=
2n
cos( (m n1)
)
sn =
2n tan( n )
cos( (m n1)
)
rn =
2n sin( n )
cos( (m n1) )
n of a star polygon of order The half-side s m n = sm
n m
Rn :
is
tan( n ) sin( n ) 1 1 sn = rn = Rn : Pn = (m 1) (m 1) 2n m cos( n ) cos( n ) cos( (m n1) )
n n of a star polygon of order The stroke distance d m m is s (m 1) sin( mn ) 2 sin( m 2 sin( m ) [nf (n)]2 + 1 n ) n ) cos( n n = n dm d = d = s = sm n n n n n 2 [m sin( n ) sin( n ) sin( n ) f(m )] + 1 s 2 sin( mn ) 2 [nf (n)]2 + 1 2m m n n = r = rn = )r m n n n 2 n r m = 2 tan( nf (n) [ m f ( m )] + 1 cos( n ) nf ( m ) n
=p
2 m )Rn : Rn = 2 sin( n n 2 n [m f(m )] + 1
n of a star polygon of order The draw length D m n = nd n = Dm m
sin( mn
n m
is the sum of all stroke distances,
sin( mn
) ) 2n sin( mn ) Dn = Pn = sn sin( n ) sin( n ) sin( n )
(m 1) 2n sin( m ) sin( mn ) cos( (m n1) ) n ) cos( n n = n Pm sm sin( n ) sin( n ) s 2n sin( mn ) 2 [nf (n)]2 + 1 2m m n n = rn = rn = )r m n n 2 n r m = 2n tan( f (n) [ m f ( m )] + 1 cos( n ) f(m ) n
=
=p
2n n 2 n f(m )] [m
+1
Rn = 2n sin(
m )Rn : n
n n of a star polygon of order The inner radius r m m is s cos( mn ) 1 cos( mn ) f (n)2 + n12 1 cos( mn ) cos( (m n1) ) n n = f( n rm ) P = P = s = sm n n n 2 n 2 m 2 n sin( ) 2 sin( ) sin( ) f(m ) +m 2 n n n n s n n 2 f(m) f (n) + n12 cos( mn ) f(m ) m rn = q = Rn = cos( )Rn : 2 rn = n m f (n) f ( m )2 + n2 cos( n ) n n 2 m2 f( ) + m
The k
n m #k
n2
n n middle radius m #k (see Proposition 6) of a star polygon of order m is r 1 n k2 1 n = = f ( )2 + 2 r m rn n f( k ) k n cos( nk ) m s 2 n n rm f(m ) f ( nk )2 + nk 2 cos( m n ) Rn ; k 2 f1; 2; :::; m 1g: = Rn = R = n 2 n n m 2 r nk f ( k ) f ( m ) + n2 cos( nk )
th
REGULAR POLYGONS OF RATIONAL ORDER
In particular, the last middle radius n m #(m
= =
1)
=
f (n)2 + n12 f (n)+ n2 fm ( n )
1 2 Pn
m
f (n)+ n2 f1(n)
is
1)
cos( m n ) n sin( n ) cos(
(m
1) n
)
1 2 Pn
=
cos( m n ) sin( n ) cos(
(m
1) n
)
sn
1
f (n)+ n2 f (n) cos( m n ) rn r = n (m 1) f (n)+ n2 fm cos( ) cos( n ) ( n ) n m q q f (n)2 + 12 cos( m n ) Rn : f (n)2 + n12 12 Pn = f (n)+ mnn Rn = (m 1) cos( ) n2 f ( ) n m n is star polygon of order m
f (n) 12 Pn =
f (n)+ n2 fm ( n ) m q f (n)2 + n12 f (n)+ n2 fm ( n ) m
n of a The area A m
1 n n = rm P n = ns m 2 m
n = r n Am m
=
n m #(m
5
= n sin( )Rn n
n m #(m
1)
=
n m #(m
1)
1 Pn = nsn 2
n m #(m
1)
cos( mn ) R2 Rn2 = n sin( ) n cos( (m 1) ) n n n sin( ) n n rn2 = r2m (m 1) ) cos( m ) cos( ) n n
n f(m ) n f (n)f ( m ) + nm2
cos( mn ) = n tan( ) n cos( ) cos( (m 1) n n =
cos( mn ) n [nf (n)]2 + 1 n n2 2 f ( ) n2 d = d2 n n 4m m m f (n)f ( m 4 sin( ) cos( (m 1) ) n )+1 n n
=
cos( mn ) n [nf (n)]2 + 1 1 1 2 f ( ) n2 D = D2 n n 4m m m f (n)f ( m 4n sin( ) cos( (m 1) ) n )+1 n n
=
cos( mn ) 2 n2 n [ n f ( n )]2 + 1 2 n sin( n ) n = dm dn f ( ) n2 m m n m 4m m m f (n)f ( m ) + 1 4 sin( n )2 cos( (m 1) ) m n
cos( mn ) 1 n [ n f ( n )]2 + 1 1 sin( n ) n = D2n f ( ) n2 m m n D2m m 4m m m f (n)f ( m ) + 1 4n sin( n )2 cos( (m 1) ) m n n cos( mn ) cos( mn ) 1 2 2 ( Pn ) = sn = n sin( ) cos( (m 1) ) 2 sin( ) cos( (m 1) ) =
n
2
=
=
f (n) f (n) +
n
+ n12 m n n2 f ( m )
f (n)2 +
= f(
1 ( Pn )2 = 2
1 n2 mf (n) n n2 f ( m )
f (n)2 +
n
An =
n
f (n) + n2 f1(n) f (n) f (n) + n2 fm( n ) m
f (n) + f (n) +
1 n2 f (n) m n n2 f ( m )
An =
f (n)2 + n12 n 1 1 2 ( Pn )2 = f ( ) n m ( Pn ) 2 m f (n)f ( m ) + n2 2
n f(m cos( mn ) ) f (n)2 + n12 An m An = n f (n) f (n)f ( m ) + n2 cos( n ) cos( (m n1) )
n n f(m ) + n2 m ) + nm2 1 n f (n)f ( m f (n) 1 2 2 n ) = f (n) n ) ( ( Pm ) P 2 m n 2 n m2 m f(m 2 2 ) +m f ( ) + n n2 m n2 f ( ) m
cos( mn ) cos( (m n1) ) 1 n cos( mn ) cos( (m n1) ) 2 2 n ) = n : = ( Pm sm n sin( n ) 2 sin( n ) Remark 1. Some particular cases of m (valid for n 2m) are the following. m = 1 convex polygons fng (n = 2; 3; 4; :::) 2 Pn = 2nsn = f (n) rn = 2n tan ( n )rn = q 22 1 Rn = 2n sin ( n )Rn , 1 sn = 2n Pn =
dn = 2sn =
1 nf (n) rn
2 nf (n) rn
f (n) + n2 1 Rn [nf (n)]2 +1 2 tan( n )rn = p 2 2 2 Rn n f (n) +1
= tan ( n )rn = p
=
= sin ( n )Rn ,
= 2 sin ( n )Rn ,
6
PETER KAHLIG
Dn = ndn = 2nsn = P n =
2 f (n) rn
= 2n tan( n )rn = q
rn = f (n) 12 Pn = f (n) nsn = cot ( n )sn =
q
2
f (n)
f (n)2 + n12
2 f (n)2 + n12
Rn = 2n sin ( n )Rn ,
Rn = cos ( n )Rn ,
2
2
2
2
1 An = rn 21 Pn = f (n) ( 12 Pn ) = n cot ( n )sn = n4 cot ( n )dn = 14 f (n) Dn = 4n cot ( n )Dn 1 = f (n) rn2 = n tan( n )rn2 f (n) = f (n) 2+
2
1 n2
Rn2 = n cos ( n ) sin ( n )Rn = q
2
1 2
2 2 f( n 2 ) + n2
Rn2 = n2 sin ( 2n )Rn .
m = 2 “class A” star polygons f n2 g (n = 4; 5; 6; :::) q q 1 1 1 2 2 P n2 = 2ns n2 = f (n) f (n) + n2 Pn = cos( ) Pn = f (n) f (n)2 + 2 =
2 1 2 f (n) + n2 2 f( n ) f (n) 2
q
1 s n2 = 2n P n2 =
n
tan( )
r n2 = 2n cos( 2n ) r n2 = n
tan( )
1 n2 rn
= 2n cos ( n ) rn n
2 f (n) Rn
= 2n tan ( n )Rn , p 1 [nf (n)]2 + 1rn ) sn = [nf (n)]2
1 f (n)2 + n12 sn = cos(1 f (n) n tan( ) sin( ) sin( ) sin( ) = cos( n ) rn = [cos( n)]2 rn = 1 [sin(n )]2 rn = [1+sin( )][1n sin( )] rn n n n n n f (n)2 + 1 tan( ) = nf 1( n ) f (n)2n2 r n2 = cos( 2n ) r n2 = nf1(n) Rn = tan ( n )Rn , 2 n 4f (n) 4f (n)2 q d n2 = q 2f (n) d = 2 cos( sn = 4 cos( n )sn = f (n) )d = n n 2+ 1 n 1 1 2 2 f (n) + f (n) + n2 n2
= q n2
n2
2
s n2 = 4 cos( n )2 s n2
Rn = 2 sin ( 2n )Rn = 4 sin( n ) cos( n )Rn ,
2 f( n 2 ) +1 2f (n) nd n2 = q f (n)2 + n12 22
D n2 =
=
= r n2 = q
2
4nf (n) f (n)2 + n12 2 q
2
r =
n f (n)2 + n12 n f ( )f (n) q 2 f (n)2 + n12
=
cos( 2n ) cos( n ) rn f ( n )f (n) 1 q 2 2 Pn = f (n)2 + 1
f( n )
2 = f (n) rn = q 1 f (n)2 + = f (n) n 2 #1
2
Rn = 2n sin ( 2n )Rn = 4n sin( n ) cos( n )Rn ,
f (n) = f ( n2 ) f (n) 2+
n 2
2
s n2 = 2 cos ( n ) P n2 = 4n cos ( n ) s n2
2 2 f( n 2 ) + n2
f( n 2)
Dn = 2 cos ( n )Dn = 2 cos( n )Pn = 4n cos ( n )sn
n2 2
1 n2
(n)] 1 Rn = [nf [nf (n)]2 +1 Rn =
f (n)2 n12 f (n)2
rn = (1
1 n 1 n n2 r 2 = cos( n ) r 2
n f( n ) 2 q 22 n 2 f( 2 ) + n 2
n
ns n2
Rn = cos ( 2n )Rn ,
2 1 tan( n ) )rn [nf (n)]2 )r n = (1 cos ( 2 ) f( n cos( 2 ) 2) = cos( n)2 rn = q Rn = cos( n ) Rn , n n f (n)2 + 1 n2
f( n f( n 1 1 1 2) 2) cos( n ) 2 Pn = f (n) rn 2 Pn = f (n) An 2 1 = (1 [nf (n)] tan( n ) )An 2 )An = (1 2 2 = f ( n2 ) ( 12 Pn ) [an “analogy” to An = f (n) ( 21 Pn ) , see warning below] 2 2 = f ( n2 ) sn n2 = 2n cot ( 2n )sn 2 1 2 2 n2 n 2 n 2 1 = 4 f ( 2 )dn = 2 cot ( n )dn = 4 f ( n2 )Dn = 2n cot ( 2n )Dn q 2 2 2 f( n ) 2 cot( 2 ) +1 2 1 2 = 16 f ( n2 ) n ff(n) d n = n16 f (n) f ( n2 )2 + n2 2 d2n = n8 cos( n)2 d2n (n)2 2 2 2 n n 2 2 f( n ) 2 2 2 2 1 = n8 2f (n)22 (f (n) + n12 )d n = n16 [f (n) n2 f1(n) ][1+ n2 f1(n)2 ]d n = n16 f (n)[1 [nf (n)] 4 ]d n 2 2 2 4 2 2 1 1 [nf (n)] 1 2 = 16 [n f (n) n2 f1(n)3 ]d n = 16 d n 2 3 2 2 q n f (n) 2 1 n 2 1 n f (n) + n2 1 f( 2 ) n 2 22 1 cot( n ) 2 2 = 16 f ( 2 ) f (n)2 D n = 16 f (n) f ( 2 ) + n2 D2n = 8n 2 Dn cos( 2 2 2 n)
A n2 = r n2
1 n 2P 2
=
f( n 2) f (n)
f ( n )f (n) f ( n )f (n) 1 P n2 = f (n)2 2 + 1 1 2 2 2
nsn = f (n)2 2 +
cos( n )rn
REGULAR POLYGONS OF RATIONAL ORDER n n 2 f( 2 ) f (n)2
2 2 1 1 1 1 f (n)[1 [nf (n)] = 16 4 ]D n n2 f (n) ][1+ n2 f (n)2 ]D n 2 2 4 1 4 2 1 [nf (n)] 1 2 1 f (n) 1 2 n4 [f (n) n4 f1(n)3 ]D n = 16 D D = 16 n= n 4 3 3 n f (n) 16 f (n) 2 2 2 f( n ) 2 = f (n)22+ 1 Rn2 = n tan ( n ) cos ( 2n )Rn n2 = n2 12 Pn = cos(1 ) r n2 12 P n2 cos( n ) =r n2 12 P n2 = A n2 . n n 1 n 6= f ( n WARNING. We have rn = f (n) 12 Pn ,but r m m ) 2 Pn = m #(m 1) for m 6= 1[in par1 n ticular for m = 2: r n2 6= f ( 2 ) 2 Pn = n ]. Nevertheless the surprising relation A n2 = 2 2 f ( n2 ) ( 12 Pn )2 , being formally analogous to the m = 1 case An = f (n) ( 21 Pn ) , does 2 hold for m = 2. At …rst sight it seems to suggest a generalization of An = f (n) ( 21 Pn ) 2 n n 1 n =? f ( (with integer n) to rational m , namely something like A m m ) ( 2 Pn ) , which is, 1 = 8n
2
7
2
1 (f (n) + n12 )D n = 16 [f (n) 2
n above in however, NOT true for general m 6= 1 and m 6= 2 (cf. the true formula for A m Proposition 2). – Sketch of proof for the surprising simpli…cation in the m = 2 case: an
f (n)2 +
1 2
n 1 2 n n (from Proposition 2), A n = f ( area formula for A m m ) f (n)f ( n )+ m ( 2 Pn ) , gives for m m
m = 2 the expression
n2
1 n f (n)2 + n12 ( Pn )2 ; A n2 = f ( ) n 2 2 f (n)f ( 2 ) + n2 2 but we know from Remark 2 (below) the identity f ( n2 ) = f (n)
1 n2 f (n) ,
resembling the 2
2 cot ( 2n
identity ) = cot ( n ) tan ( n ); therefore, the denominator is f (n)f ( n2 )+ n22 = f (n) + n12 , and the last expression for A n2 in fact reduces to
n 1 A n2 = f ( )( Pn )2 2 2 f( n )
2 [complying with A n2 = f ( n2 ) 12 Pn 21 Pn = n 21 Pn .]. A consequence is the relation A n2 = f (n) An 2 (being so simple only in the present m = 2 case of “class A” star polygons). –See also the deeper reconsideration in Proposition 7.
m = 3 “class B” star polygons f n3 g (n = 6; 7; 8; :::) r q 22 2 f( n 2 sin ( n ) 2 ) + n2 1 n 2 2 1 2 n n P 3 = 2ns 3 = f ( n ) f ( 2 ) + n2 Pn = cos( 2 ) Pn = f ( n ) f (n) 2 + 1 Rn = 2n cos( 2 ) Rn , 2 2 2 n n n r q n 2 22 f ( ) + 2 sin ( n ) 2 1 n2 s n3 = 2n P n3 = f (1n ) f ( n2 )2 + n2 2 sn = cos(12 ) sn = nf 1( n ) f (n) 2 + 1 Rn = cos( 2 ) Rn , 2
d n3 = q n2 32
2
2 f( n 3 ) +1
2
n
Rn =
2 sin ( 3n
)Rn =
n2
n
sin ( 3n ) sin ( n ) dn ,
sin ( 3 )
D n3 = nd n3 = q n2 2nn Rn = 2n sin ( 3n )Rn = sin ( n ) Dn , 2 +1 n f ( ) 2 3 r 3 1 f (n)2 + n2 1 cos ( 3 ) f( n ) cos ( 3 ) Pn = sin ( n ) sn = q n 3 32 Rn = cos ( 3n )Rn = cos( n ) rn , r n3 =f ( n3 ) n 2 32 2 f ( 3 ) + n2
f ( 3 )2 + n2
n
n
cos( 3 ) cos( 3 ) n = cos( n ) Rn = cos(1 ) r n3 = [cos( n)]2 rn , 3 #1 n n n cos( 3n ) cos( 3n ) 1 n= n = R = r r , 2 2 #2 n 3 cos( n ) cos( n ) 3 cos( 2n ) cos( n ) n
A =r n 3
n 3
= n sin ( n ) cos ( =
r
1 n n 2 P 3 =f ( 3 )
n 3 #2
1 2 Pn :
f (n)2 + n12
2
3 2 f( n 3 ) + n2
2 1 2 Pn
cos ( 3 )
n sin ( n ) r2n 3 n ) cos ( n ) 3
= n sin ( n ) cos ( 2n ) Rn2 = cos ( 2 n
cos ( 3n ) cos ( 3n ) cos ( 3n ) 2 rn2 = cos ( ) cos n tan ( n )rn = cos ( ) cos A 2 2 2 ( ) ( 2n ) n ) cos ( ) n n n n n
m = 4 “class C” star polygons f n4 g
(n = 8; 9; 10; :::)
8
PETER KAHLIG
P = 2ns = n 4
n 4
1 f( n 3)
q
f ( n3 )2
+
q 1 P n4 = f (1n ) f ( n3 )2 + s n4 = 2n 3
d n4 = q n2 42
2
2 f( n 4 ) +1
32 n2 Pn 32 n2
= cos(13 ) Pn = f (2n ) 3 n
sn = cos(13 ) sn = n sin ( 4n
Rn = 2 sin ( 4n )Rn = sin (
)
n)
1 nf ( n 3)
dn ,
r
r
2
3 2 f( n 3 ) + n2
f (n)2 + n12 2
3 2 f( n 3 ) + n2
f (n)2 + n12
sin ( )
Rn = 2n cos( 3n ) Rn , n
sin ( ) Rn = cos( 3n ) Rn , n
sin ( 4 )
D n4 = nd n4 = q n2 2nn Rn = 2n sin ( 4n )Rn = sin ( n ) Dn , 2 n 2 f ( 4 ) +1 4 r 1 2 4 n f (n) + n2 1 cos ( ) f( ) cos ( 4 ) r n4 =f ( n4 ) Pn = sin ( n ) sn = q n 4 42 Rn = cos ( 4n )Rn = cos( n ) rn , n 2 42 2 f ( 4 ) + n2
f ( 4 )2 + n2
n
n
cos( 4 ) cos( 4 ) n = cos( n ) Rn = cos(1 ) r n4 = [cos( n)]2 rn , 4 #1 n n n cos( 4n ) cos( 4n ) 1 n n= = R = r r , 2 2 #2 n 4 cos( n ) cos( n ) 4 cos( 2n ) cos( n ) n 4 ) cos( 4n ) cos( n n = cos( 3 ) Rn = cos(13 ) r n4 = cos( 3 ) cos( rn , 4 #3 n n n n)
A n4 = r n4
r
1 n n 2 P 4 =f ( 4 )
2
2 1 2 Pn
cos ( 4 )
cos ( 4n ) An 3 n ) cos ( n )
= n sin ( n ) cos ( 3n ) Rn2 = cos ( n
sin ( ) cos ( 4 ) =r = cos( 4n )Rn n cos( 3n ) Rn = n sin ( n ) cos ( 3n ) Rn2 n n cos( 4 ) cos ( 4 ) = n4 #3 21 Pn = cos( 3n ) Rn n sin ( n )Rn = n sin ( n ) cos ( 3n ) Rn2 n n cos( 4 ) cos( 4 ) n sin( ) cos( 4 ) = n sin(n ) 12 Pn 12 P n4 = n sin(n ) n sin ( n )Rn cos( 3 n ) Rn = n sin ( n ) cos( n3 ) Rn2 n n n n 4 cos( 4n ) cot( 4 ) n sin( n ) cos( n ) 2 n 2 = 4 sin( ) cos( 3 ) dn = 4 sin( 4 )2 cos( 3 ) d n = n4 cos( 3n ) dn d n4 4 n n n n n n = n=2 collapsed star polygons f n=2 g (asterisks) (n = 4; 6; 8; :::) n 4
m
f (n)2 + n12
4 2 f( n 4 ) + n2
1 n 2P 4
These are interesting degenerate cases of regular star polygons: the Dido value is the n n smallest possible, f ( m ) = f ( n=2 ) = f (2) = 0, leading geometrically to collapsed star polygons (“asterisks”) with vanishing inner radius, vanishing middle radii, and vanishing area. Here, n and m = n=2 are not coprime, hence Schlä‡i factoring is applicable and n gives f n=2 g = (n/2)f 21 g, denoting a compound star polygon, consisting of n/2 digons
mutually rotated through 2 =n = =(n=2). Also interpretable as n spikes (half-digons) forming an “n-spike asterisk” (for short “n-asterisk”) where n is an even integer, for instance 6-asterisk (with 6 spikes, the most common asterisk). [Odd-spiked asterisks (like 5-asterisk) cannot be represented as collapsed star polygons, but may be imitated like ? (pentagram).] q 2
n = P n=2 [nf (n)] +1Pn = sin1( ) Pn = 2nRn , n q 2 1 1 n = n = s n=2 P [nf (n)] +1s n = sin ( ) sn = Rn , 2n n=2 n = 2s n d n=2 = sin1( n=2
n
n)
dn = 2Rn ,
1 n = nd n = D n=2 sin ( n ) Dn = 2nRn , n=2 n = 0, r n=2 n #k n=2
cos( )
n = 0 = cos( k2 ) Rn = cos(1k ) r n=2
n = r n A n=2 n=2
n
1 n 2 P n=2 =
n
0.
for all k = 1; 2; 3; :::; n2 1,
REGULAR POLYGONS OF RATIONAL ORDER
9
Remark 2. Some identities, implied by the Dido functional equation and/or its solution f (n) = n1 cot( n ):
1 f (2n) = 2
r
f (n) +
n f ( ) = f (n) 2
r
1 f (n)2 + 2 n
1 n2 f (n)
=
!
=
1 cot( ); n = 2; 3; 4; :::; 2n 2n
2 2 cot( ); n = 4; 5; 6; :::; n n
1 1 = 2f (2n) f (n) = ; n = 2; 3; 4; :::; n2 n sin( n ) 1 f (2n) q = = n sin( ); n = 2; 3; 4; :::; 1 2+ n 1 f (2n) 2 (2n)2 f (n) + f (n)2 +
n2
1 1 1 1 f (2n) p = sin( ); n = 2; 3; 4; :::; = = 1 n 2f (2n) f (n) n f (2n)2 + (2n) n n2 f (n)2 + 1 2 f (n)
1 f (n)
q f (n)2 + r
1 n2
f (2n)2 f (n) = = 2f (2n) f (n) f (2n)2 +
1 (2n)2 1 (2n)2
= cos( ); n = 2; 3; 4; :::; n
1 2f (2n) 1 ; n = 3; 4; 5; :::; = 1= 2 n f (n) cos( n ) f (n) 1 n 2 =q = n sin( ) cos( ) = sin( ); n = 4; 5; 6; :::; n n 2 n f (n)2 + n12 n 2 22 f ( 2 ) + n2
f (n)2 +
f ( n2 )f (n) f ( n2 ) q = 1 f (n)2 + n2 f ( n2 )2 +
f ( n2 ) q f (n)2 +
= 1 n2
22 n2
=
f ( n2 ) f (n)2 n = 2f (n) f ( 2 ) f (n)2 +
1 n2 1 n2
= cos(
2 ); n = 4; 5; 6; :::; n
f ( n2 ) cos( 2n ) = ; n = 4; 5; 6; :::; 2f (2n) f (n) cos( n )
f ( n )f (n) f ( n2 )f (n) cos( 2n ) q 2 = = ; n = 4; 5; 6; :::; 2f (2n) f (n) n sin( n ) f (n)2 + n12 f ( n2 ) f (n)2 +
1 n2 f ( n2 )
r
f (n)
2 = n tan( ) cos( ); n = 4; 5; 6; :::; n n =1
1 =1 [nf (n)]2
cos( 2n ) tan( )2 = ; n = 4; 5; 6; :::; n cos( n )2
n 22 n 2 f ( )2 + 2 = f ( ) + 2 = 2f (n) 2 n 2 n f (n)
n 2 ; n = 4; 5; 6; :::; f( ) = 2 n sin( 2n )
10
PETER KAHLIG
Further identities are the following, valid for n = 2; 3; 4; ::: and m = 1; 2; :::; f (n) + n2 f1(n) f (n) + n2 fm( n ) m f(
=
n 2
:
n f(m ) f (n)2 + n12 cos( mn ) n m = f (n) f (n)f ( m ) + n2 cos( n ) cos( (m n1) )
n n f(m ) + n2 m cos( mn ) cos( (m n1) ) ) + nm2 n f (n)f ( m f (n) = f (n) = ) 2 2 n 2 n m f(m n sin( n ) ) +m ) + n2 m f(m n2 f( n ) m
n ) f(m n f (n)f ( m ) +
cos( mn )
m n2
= n sin( ) n cos( (m
1) n
)
cos( mn ) n n [nf (n)]2 + 1 f ( ) n2 = n m m m f (n)f ( m )+1 sin( n ) cos( (m n1) ) sin( n ) cos( mn ) n n [ n f ( n )]2 + 1 f ( ) n2 m m n = 2 (m 1) m m m f (n)f ( m ) + 1 sin( m ) n ) cos( n q n n f(m ) f (n)2 + n12 )[2f (2n) f (n)] f(m cos( mn ) = = n n m m f (n)f ( m ) + n2 f (n)f ( m ) + n2 cos( (m n1) ) v " #2 u p n 2 n n n f(m )] + 1g u f[nf (n)]2 + 1gf[ m f(m ) nf (n) m t = 1 + n2 n2 n n m f (n)f ( m ) + 1 m f (n)f ( m ) + 1 r 1 (m 1) 2 ) = = 1 + tan( (m 1) n cos( ) n
Proposition 3. An in…nite alternating product for f (n) is (cf. [9]) f (n) =
1 Y
(1
(jn=2)
2 ( 1)j+1
)
; n = 3; 4; 5; :::;
j=1
f (n) is the ratio of inner circle perimeter 2 rn and polygon perimeter Pn , i.e. f (n) = (2 rn )=Pn . This quantity admits representation as in…nite product not only for integer n, but also for rational n (like 25 or 82 , providing a connection to star polygons). As shown in [9], the in…nite products for f (n), n > 2, arise by considering the identity f (n) = ( =n) cot( =n) = cos( =n)=[(n= ) sin( =n)] and substituting well-known product representations of cosine and sine, then replacing the resulting standard product by an (equivalent) alternating product. Note on middle circles. For geometric interpretation and application, it is essential to know certain auxiliary circles relevant for the shape of regular polygons. For convex polygons it is well known that inner and outer circle are important (and no others). But for star polygons, there also exist middle circles, located between inner and outer circle and connecting points of self-intersection. (These middle circles may serve as part of a polar grid to …x the shape in geometric drawings of star polygons.) For instance, a pentagon has inner and outer circle (and no middle circle) since there is no self-intersection, while a pentagram has inner, outer and a middle circle
REGULAR POLYGONS OF RATIONAL ORDER
11
(connecting points of self-intersection). Heptagram B has two middle circles (each one connecting points of self-intersection), in their notation distinguished simply by numbering (#1, #2). Proposition 4. A convex polygon of order n = 2; 3; 4; ::: has one inner and one outer circle. Given the radius Rn of the outer circle, the radius of the inner circle is f (n) Rn = cos( ) Rn : rn = q n f (n)2 + n12
n g with n = 4; 5; 6; ::: and m 2 Proposition 5. A star polygon of order f m n f1; 2; :::; 2 g has one inner circle and one outer circle. Given the radius Rn of the outer circle, the radius of the inner circle is s n n f(m ) ) f (n)2 + n12 f(m cos( m m n ) n = q ) Rn = Rn = cos( rn : rm r = 2 n n m 2 2 n f (n) f ( m ) + n2 cos( n ) f ( n )2 + m2 m
n
n ) 0, implying Remark 3. Upper bound n2 for m follows from condition f ( m n n 2 , hence m or (to emphasize that this is an inequality among integers) m m 2 b n2 c. Half-brackets bxc denote the ‡oor function (greatest integer x).
n Proposition 6. A star polygon of order f m g; n 4; m 2 f1; 2; :::; n2 g; has m 1 middle circles. Radius of the k th middle circle (k 2 f1; 2; :::; m 1g): r 1 n 2 k2 1 n n = = rn f ( ) + 2 rm #k n m f( k ) k n cos( nk ) m s 2 n n rm f(m ) f ( nk )2 + nk 2 cos( m n ) Rn ; k 2 f1; 2; :::; m 1g: = Rn = R = n 2 n n m r nk f ( k ) f ( m )2 + n2 cos( nk )
Remark 4. The case n = 3 requires m 2 f1; :::;
3 2
g = f1g, a singleton, i.e. m = 1 only, implying that there are no middle circles and (equivalently) no star polygons of order 3 fm g; m > 1. Therefore, star polygons exist only for n 4. n g, inner and outer circles may be interpreted as Remark 5. For a star polygon f m n n (inner extreme cases “#0” and “#m” of middle circles: “k = 0” leads to m #0 = r m n n n radius of f m g), and “k = m” leads to m #m = Rn (outer radius of f m g). – The inner n )] carries npoints of tangency with the polygon of order n (or circle [radius rn (or r m n m ). The outer circle (radius Rn ) carries n edge points (or spike points) of the polygon n n of order n (or m ). A middle circle (radius m #k ) carries n points of self-intersection
n of the polygon of order m . – The ratio of two successive middle radii of a star polygon of n order m is (with n = 4; 5; 6; :::and m 2 f1; 2; :::; n2 g) n m #k n m #(k
1)
f (n) + n2 fk( n ) cot( ) + tan( kn ) k k = cos( )[1 + tan( ) tan( )] = q = pn 2 n n n cot( n ) + 1 f (n)2 + n12 =
cos( k n 1 ) cos( kn )
;
k 2 f1; 2; :::; m
1g:
12
PETER KAHLIG
Putting “k = m” as above (and taking reciprocals) yields an expression for the last middle n circle m #(m 1) , n m #(m
Rn
1)
=
q f (n)2 +
f (n) +
1 n2 m n n2 f ( m )
=
p
cos( m [nf (n)]2 + 1 n ) : = 1 (m 1) nf (n) + n f ( n ) cos( ) n m m
Remark 6. An overview of the radii of inner circle r, middle circle(s) circle R for the …rst twenty-…ve polygons may be given as a list: f2g digon: f3g triangle (equilat.): f4g square: f 42 g tetragram: f5g pentagon: f 52 g pentagram: f6g hexagon: f 26 g hexagram A: f 73 g hexagram B: f7g heptagon: f 72 g heptagram A: f 37 g heptagram B: f8g octagon: f 28 g octagram A: f 83 g octagram B: f 84 g octagram C: f9g nonagon: f 92 g nonagram A: f 93 g nonagram B: f 94 g nonagram C: f10g decagon: f 10 2 g decagram A: f 10 3 g decagram B: 10 f 4 g decagram C: f 10 5 g decagram D:
and outer
r2 = 0; R2 extrem e (convex) p olygon r3 ; R 3 r4 ; R 4 r 42 = 24 #1 = 0; R4 collapsed star p olygon (4-asterisk) r5 ; R 5 star p olygon, 1 m iddle circle r 25 ; 52 #1 ; R5 r6 ; R 6 r 26 ; 62 #1 ; R6 star p olygon, 1 m iddle circle collapsed star p olygon (6-asterisk) r 63 = 62 #1;2 = 0; R6 r7 ; R 7 r 27 ; 72 #1 ; R7 star p olygon, 1 m iddle circle r 73 ; 73 #1 ; 37 #2 ; R7 star p olygon, 2 m iddle circles r8 ; R 8 star p olygon, 1 m iddle circle r 82 ; 28 #1 ; R8 r 83 ; 38 #1 ; 38 #2 ; R8 star p olygon, 2 m iddle circles collapsed star p olygon (8-asterisk) r 84 = 48 #1;2;3 = 0; R8 r9 ; R 9 star p olygon, 1 m iddle circle r 29 ; 29 #1 ; R9 r 93 ; 39 #1 ; 39 #2 ; R9 star p olygon, 2 m iddle circles star p olygon, 3 m iddle circles r 94 ; 94 #1 ; 49 #2 ; 49 #3 ; R9 r10 ; R10 r 10 ; 10 ; R10 star p olygon, 1 m iddle circle 2 2 #1 10 10 r 10 ; ; ; R star p olygon, 2 m iddle circles 10 3 3 #1 3 #2 r 10 ; 10 ; 10 ; 10 ; R10 star p olygon, 3 m iddle circles 4 4 #1 4 #2 4 #3 10 r 10 = = 0; R collapsed star p olygon (10-asterisk) 10 5 5 #1;2;3;4
Remark 7. Star polygons f n2 g (“class A” star polygons) have only one middle circle; thus, for brevity, its radius n2 #1 may be denoted simply and uniquely by n2 (n 4). In the list above, this pertains to f 24 g (tetragram), f 52 g (pentagram), f 62 g (hexagram A), f 72 g (heptagram A), f 28 g (octagram A), f 92 g (nonagram A), f 10 2 g (decagram A). – In the illustrations below, one can observe that a star polygon n g produces by self-intersection all its predecessors fng; f n2 g; f n3 g; :::; f mn 1 g in fm n miniature, resulting in a rich geometric structure. [In asterisks f n=2 g, this rich geometric structure is unobservable since it collapsed to the center point.]
REGULAR POLYGONS OF RATIONAL ORDER
13
3. Applications Example 1. Elements of a convex polygon and a star polygon
half-side s5 (side 2s5 ), inner radius r5 , perimeter P5 = 10s5 , outer radius R5 area A5
half-side s 5 , 2 inner radius r 5 , 2 middle radius 5 , 2 outer radius R5
perimeter P 5 = 10s 5 , 2 2 area A 5 2
Polygon fng (n-gon, n 2) is enclosed between its inner circle (radius rn ) and outer circle n (radius Rn ). Star polygon f m g (m th n-gram, m 2 f1; 2; :::; n2 g; n 4) is enclosed n ) and outer circle (radius R ). between its inner circle (radius r m n
Example 2. Polygon f2g (Digon)
Digon Polygon f2g (digon) is enclosed between its inner circle (radius r2 = 0) and outer circle (radius R2 ). It is to be considered as having two coincident sides (cf. [5]). I f2gA (regular convex) polygon of order 2 (digon) has half-side s2 and Dido value f (2) = 21 cot( 2 ) = 0, a constructible algebraic number (containing only rationals and/or square roots), so the digon is constructible (by ruler and compass). The digon is an extreme case of a (convex) polygon: a double straight line (of …nite length), also interpretable as twin line. Each twin has length 2s2 , so this 2s2 is the side of the digon (half-side s2 ), and the perimeter of the digon is P2 = 4s2 . Inner radius r2 and area A2 of the digon vanish (r2 = 0, A2 = 0), outer radius R2 = s2 . Stroke distance of digon: d2 = 2s2 (side of digon). Draw length of digon: D2 = 2d2 = 4s2 = P 2 (perimeter of digon). (There is no star polygon.)
Expressions in terms of outer radius R2 f2g (digon): 1 1 q R2 = 2 sin ( 2 )R2 = 2R2 ; 2 P2 = 2s2 = 1 2
f (2) + 22 1 1 s2 = 2 ( 2 P2 )= sin ( 2 )R2 =
R2 ;
14
PETER KAHLIG
d2 = 2s2 = 12 P2 = 2R2 ; D2 = 2d2 = 4s2 = P 2 = 4R2 ; 2 2 f (2) 2 R2 = cos ( 2 ) R2 = 0; A2 = f (2) r2 = q f (2) 2 + 1 R2 = 2 sin ( 2 ) cos ( 2 )R2 = sin ( )R2 = 0: 1 2 f (2) + 22
22
Example 3. Polygon f3g (Equilateral Triangle) [Trigon]
Equilateral triangle Polygon f3g (equilateral triangle) is enclosed between its inner circle (radius r3 ) and outer circle (radius R3 ).
I f3g A convex polygon of order 3 (equilateral triangle) has half-side s3 and Dido value p 1 f (3) = 13 cot( 3 ) = 3p = 19 3 (t 0:1925), 3 a constructible algebraic number (containing only rationals and/or square roots). Perimeter of equilateral triangle: P3 = 6s3 . Half-side of equilateral triangle: s3 = 61 P3 . Stroke distance of equilateral triangle: d3 = 13 P3 = 2s3 . Draw length of equilateral triangle: D3 = 3d3 = P3 . Inner radius of equilateral triangle: p r3 = f (3) 12 P3 = cot( 3 )s3 = 13 3s3 (t 0:5774 s3 ). Outerq radius of equilateral triangle: p R3 = f (3)2 + 312 12 P3 = sin(1 ) s3 = 32 3s3 (t 1:1547 s3 ). 3 p Area of equilateral triangle: A3 = r3 21 P3 = f (3) ( 21 P3 )2 = 3s23 (t 1:7321 s23 ). Pythagorean relationship: s23 = R32 r32 . (There is no star polygon.) Expressions in terms of outer radius R3 f3g (equilateral triangle): p 1 1 q R3 = 3 sin( 3 )R3 = 23 3R3 (t 2: 5981R3 ), 2 P3 = 3s3 = 1 2 f (3) + 32 p s3 = 31 ( 12 P3 ) = p 2 1 2 R3 = sin( 3 )R3 = 21 3R3 (t 0:8660R3 ), 3 f (3) +1 p d3 = 2s3 = 2 sin( 3 )R3 = 3R3 (t 1:7321R3 ), p D3 = 3d3 = 6s3 = P3 = 6 sin( 3 )R3 = 3 3R3 (t 5:1962R3 ), r3 = q f (3) R3 = cos( 3 ) R3 = 12 R3 , f (3)2 + 312 p 2 f (3) 3 2 3 2 2 A3 = f (3) 3R3 (t 1:2990 R32 ). 2 + 1 R3 = 2 sin( 3 )R3 = 4 33
Example 4. Polygon f4g (Square) [Tetragon], Star Polygon f 24 g (FourSpike Asterisk) [Tetragram]
REGULAR POLYGONS OF RATIONAL ORDER
Square
15
Tetragram
Polygon f4g (square) is enclosed between its inner circle (radius r4 ) and outer circle (radius R4 ). Star polygon f 42 g (four-spike asterisk) is enclosed by its outer circle (radius R4 ). All points of self-intersection of the four-spike asterisk lie in the center. Inner circle (radius r 4 = 0) and middle circle (radius 4 = 0) collapsed to zero in the center. 2
2
I f4g A convex polygon of order 4 (tetragon, square) has half-side s4 and Dido value f (4) = 14 cot( 4 ) = 14 , a constructible algebraic number (containing only rationals and/or square roots). Perimeter of square: P4 = 8s4 . Half-side of square: s4 = 18 P4 . Stroke distance of square: d4 = 41 P4 = 2s4 . Draw length of square: D4 = 4d4 = P4 = 8s4 . Inner radius of square: r4 = f (4) 12 P4 = 14 4s4 = cot( 4 )s4 = s4 . q p Outer radius of square: R4 = f (4)2 + 412 12 P4 = sin(1 ) s4 = 2s4 (t 1:4142 s4 ). 4
Area of square: A4 = r4 21 P4 = f (4) ( 12 P4 )2 = 14 (4s4 )2 = 4s24 . There is one star polygon (the four-spike asterisk) [tetragram]: I f 24 g A star polygon of order 24 (tetragram, four-spike asterisk ) has half-side s 42 and
Dido value
f ( 42 ) = f (2) = 0, a constructible algebraic number (containing only rationals and/or square roots), so the tetragram is constructible (by ruler and compass). – Interpretation of Schlä‡i factoring f 42 g =2 f 21 g: the tetragram is a compound star polygon, consisting of 2 digons (mutually rotated through 2 =4 = =2); digons and rotation angle are constructible, hence the tetragram is constructible. q
p 2 [4f (4)] +1P4 = sin1( ) P4 = 2P4 = 8R4 . 4 p p Half-side of tetragram: s 4 = 18 P 4 = 81 2P4 = 2s4 = R4 . 2 2 p p p Stroke distance of tetragram: d 4 = 2d4 = 2s 4 = 2 2s4 = 41 P 4 = 14 2P 4 = 2R4 . 2 2 p p 2 p Draw length of tetragram: D 4 = 4d 4 = 2D 4 = 8s 4 = 8 2s4 = P 4 = 2P 4 = 8R4 . 2 2 2 2
Perimeter of tetragram: P 4 = 2
The tetragram (four-spike asterisk) is a collapsed star polygon; inner radius, middle radius and area of tetragram vanish: r 4 = 0, 4 = 0, A 4 = 0. 2 2 2 (Tetragram and its mother square have the same outer radius R4 .)
Pythagorean relationships: s24 = R42 r42 ; s24 = s24 + r42 . 2 Expressions in terms of outer radius R4 f4g (square):
16 1 2 P4
PETER KAHLIG
= 4s4 =
q
p R4 = 4 sin( 4 )R4 = 2 2R4 (t 2:8284 R4 ), p R4 = sin( 4 )R4 = 21 2R4 (t 0:7071R4 ), 2
1 f (4)2 + 412
s4 = 41 ( 12 P4 ) = p
1 42 f (4) +1
p d4 = 2s4 = 41 P4 = 2R4 (t 1:4142 R4 ), p D4 = 4d4 = P4 = 4 2R4 (t 5:6569 R4 ), p r4 = q f (4) R4 = cos( 4 )R4 = 12 2R4 (t 0:7071 R4 ), 1 2 f (4) + 42
A4 =
f (4) f (4)2 + 412
R42 = 2 sin( 2 )R42 = 2R42 .
f 42 g (tetragram, four-spike asterisk ): p 1 1 1 1 1 4 4 2 P 2 = 4s 2 = cos( ) 2 P4 = 2 2 P4 = f (4) R4 = 4 tan ( 4 )R4 = 4R4 ; 4
s 24 = 14 ( 12 P 42 ) = cos(1 ) s4 = 4f1(4) R4 = tan ( 4 )R4 = R4 ; 4 d 24 = 2R4 ; D 42 = 4d 42 = 8R4 ; 4 = 0; r 42 = 0; A 42 = r 42 21 P 42 = 42 21 P4 = 0: 2 Example 5. Polygon f5g (Pentagon), Star Polygon f 52 g (Pentagram)
Pentagon
Pentagram
Polygon f5g (pentagon) is enclosed between its inner circle (radius r5 ) and outer circle (radius R5 ). Star polygon f 25 g (pentagram) is enclosed between its inner circle (radius r 25 ) and outer circle (radius R5 ); middle circle (radius 52 ) dashed. All points of self-intersection of the pentagram lie on the middle circle (radius 5 ). 2 Self-intersection of the pentagram produces a miniature pentagon, enclosed between inner circle (radius r 5 ) and middle circle (radius 5 ). 2
2
I f5g A convex polygonq of order 5 (pentagon) has half-side s5 and Dido value p p p 1 f (5) = 51 cot( 5 ) = 15 1 + 25 5 = 25 25 + 10 5 = p 1 p (t 0:2753), 5
5 2 5
a constructible algebraic number (containing only rationals and/or square roots). Perimeter of pentagon: P5 = 10s5 . 1 Half-side of pentagon: s5 = 10 P5 . Stroke distance of pentagon: d5 = 15 P5 = 2s5 . Draw length of pentagon: D5 = 5d5 = P5 . Inner radius of pentagon: p p r5 = f (5) 21 P5 = cot( 5 )s5 = 15 25 + 10 5s5 (t 1:3764 s5 ). Outerq radius of pentagon: p p R5 = f (5)2 + 512 12 P5 = sin(1 ) s5 = 15 50 + 10 5s5 (t 1:7013 s5 ). 5 Area of pentagon: p p A5 = r5 21 P5 = f (5) ( 12 P5 )2 = 25 + 10 5s25 (t 6:8819 s25 ). There is one star polygon (the pentagram):
REGULAR POLYGONS OF RATIONAL ORDER
17
I f 52 g A star polygon of order 52 (pentagram) has half-side s 52 and Dido value p p 2 f ( 52 ) = 52 cot( 25 ) = 25 25 10 5 = 52 p 1 p (t 0:1300), 5+2 5
a constructible algebraic number (containing only rationals and/or square roots). Perimeter of pentagram: q p 1 P 25 = 10s 52 = f (5) f (5)2 + 512 P5 = cos(1 ) P5 = ( 5 1)P5 (t 1:2361 P5 ). 5 p 1 Half-side of pentagram: s 52 = 10 P 52 = cos(1 ) s5 = ( 5 1)s5 (t 1: 236 1s5 ). 5 Stroke distance of pentagram: q p
f (5) 5+ 5 2 R5 = 45 f (5) 2 + 1 R5 = 2 sin( 5 )R5 = 2 R5 52 p p = 12 10 + 2 5R5 (t 1:9021R5 ) p sin( 2 ) = 4 q f (5) s5 = 2 sin( 5 ) s5 = 4 cos( 5 )s5 = ( 5 + 1)s5 (t 3:2361s5 ) 1 2 5 f (5) + 52 p 1 2 5 = 4[cos( 5 )] s 2 = 2 ( 5 + 3)s 52 (t 2:6180s 25 ). Draw length of pentagram: q
d 25 =
4q 5
1
2 f ( 52 )2 + 522
p
f (5) 5+ 5 2 D 25 = 5d 52 = 4 q 51 22 R5 = 4 f (5) 2 + 1 R5 = 10 sin( 5 )R5 = 5 2 R5 f ( 2 )2 + 52 52 p p = 52 10 + 2 5R5 (t 9:5106R5 ): Inner radius of pentagram: cos( 2 ) f ( 5 )f (5) 1 f (5)2 1 1 q 2 r 25 = f ( 52 ) q f (5) P5 = 5 sin(5 ) 12 P5 = f ( 52 ) f (5) P 52 2+ 1 2 P 5 = 2 2 1 2 2 2 5 f (5) + 52 f ( 52 )2 + 252 52 p p cos( 2 ) 1 = sin( 5 ) s5 = 10 50 10 5s5 (t 0:5257s5 ). 5 (One and only) middle radius of pentagram: q p 1 1 2 + 1 r5 = 5 5 = ( 5 = f (5) r 5 1)r 25 (t 1:2361r 52 ) 2 #1 ) f (5) 5 cos( 2 2 2 2 5 p p = f ( 52 ) 12 P5 = f ( 52 )5s5 = 2 cot( 25 )s5 = 52 25 10 5s5 (t 0:6498s5 )
=
f ( 5 )f (5) q 2 f (5)2 + 512
1 5 2P2 .
(Pentagram and its mother pentagon have the same outer radius R5 .) Area of pentagram: p p A 52 = r 52 12 P 25 = 52 12 P5 = f ( 52 ) ( 21 P5 )2 = 21 25 10 5s25 (t 0:8123 s25 ) p f ( 5 )f (5) f ( 25 ) f (5)2 = f ( 25 ) f (5) ( 12 P 52 )2 = q 25 ( 12 P 52 )2 = f (5) A5 = 2( 5 2)A5 (t 2+ 1 22 52
f ( 2 )2 + 52
0:4721 A5 ). 2 5) . Pythagorean relationships: s25 = R52 r52 , s25 = s25 + (r5 2 2 Expressions in terms of outer radius R5 f5g (pentagon): p p 1 1 q R5 = 5 sin( 5 )R5 = 45 10 2 5R5 (t 2:9389R5 ), 2 P5 = 5s5 = 1 2 f (5) + 52 p p 1 1 s5 = 5 ( 2 P5 ) = p 2 1 2 R5 = sin( 5 )R5 = 41 10 2 5R5 (t 0:5878R5 ), 5 f (5) +1 p p d5 = 2s5 = 15 P5 = 2 sin( 5 )R5 = 21 10 2 5R5 (t 1:1756R5 ), p p D5 = 5d5 = P5 = 10 sin( 5 )R5 = 52 10 2 5R5 (t 5:8779R5 ), p R5 (t 0:8090 R5 ), r5 = q f (5) R5 = cos( 5 )R5 = 5+1 4 f (5)2 + 512 p p f (5) 5 2 5 2 2 A5 = f (5) 10 + 2 5R52 (t 2:3776 R52 ). 2 + 1 R5 = 2 sin( 5 )R5 = 8 52
18
PETER KAHLIG
f 52 g (pentagram):
p
p
1 1 1 2 5R5 (t 3:6327R5 ), cos( 5 ) 2 P5 = f (5) R5 = 5 tan( 5 )R5 = 5 5 p p 1 1 1 1 s 52 = 5 ( 2 P 25 ) = cos( ) s5 = 5f (5) R5 = tan( 5 )R5 = 5 2 5R5 (t 0:7265R5 ), 5 p p d 25 = 2 sin( 25 )R5 = 12 10 + 2 5R5 (t 1:9021R5 ), p p D 25 = 5d 52 = 10 sin( 25 )R5 = 52 10 + 2 5R5 (t 9:5106R5 ), p f ( 25 ) f (5) q R5 = cos( 25 )R5 = 54 1 R5 (t 0:3090 R5 ), r 25 = f ( 52 ) f (5) 2 + 1 R5 = 5 2 22 2 f ( 2 ) + 52 5 p r5 cos( 25 ) f ( 52 ) 3 5 1 2 5 = q 5 = R = R = r R = 5 5 5 cos( 5 ) 2 r5 cos( 5 ) 2 R5 (t 0:3820R5 ), 2 f (5)2 + 1 1 5 2P2
= 5s 52 =
52
A 25 = r 52
1 5 2P2
=
5 2
1 = r 52 f (5) R5 = f ( 25 ) f (5)21+ 1 R52 52 p p 2 5 2 )R5 = 4 50 22 5R5 (t 1:1226 R52 )
= 5 tan( 5 ) cos( 25 p f ( 52 ) = f (5) A5 = 2( 5
1 2 P5
2)A5 (t 0:4721 A5 ).
Example 6. Polygon f6g (Hexagon), Star Polygon f 62 g (Hexagram A), Star Polygon f 63 g (Hexagram B) [Six-Spike Asterisk]
Hexagon
Hexagram A
Hexagram B
Polygon f6g (hexagon) is enclosed between its inner circle (radius r6 ) and outer circle (radius R6 ). Star polygon f 62 g (hexagram A) is enclosed between its inner circle (radius r 62 ) and outer circle (radius R6 ). (Middle circle dashed.) Star polygon f 36 g (hexagram B) is enclosed by its outer circle (radius R6 ). All points of self-intersection of hexagram A lie on the middle circle (radius 6 ). Self2 intersection of hexagram A produces a miniature hexagon, enclosed between inner circle (radius r 6 ) and middle circle (radius 6 ). 2 2 All points of self-intersection of hexagram B lie in the center. Inner circle (radius r 6 = 0) and all middle circles (radii 6 #1;2 = 0) collapsed to zero in the center. 3
3
I f6g A convex polygon of order p 6 (hexagon) has half-side s6 and Dido value 1 1 f (6) = 16 cot( 6 ) = 2p = 6 3 (t 0:2887), 3 a constructible algebraic number (containing only rationals and/or square roots). Perimeter of hexagon: P6 = 12s6 . 1 Half-side of hexagon: s6 = 12 P6 : Stroke distance of hexagon: d6 = 16 P6 = 2s6 . Draw length of hexagon: D6 = 6d6 = P6 = 12s6 . p Inner radius of hexagon: r6 = f (6) 21 P6 = cot( 6 )s6 = 3s6 (t 1:7321 s6 ). q Outer radius of hexagon: R6 = f (6)2 + 612 21 P6 = sin(1 ) s6 = 2s6 . p 6 Area of hexagon: A6 = r6 21 P6 = f (6) ( 21 P6 )2 = 6 3 s26 (t 10:392 s26 ).
REGULAR POLYGONS OF RATIONAL ORDER
19
There are two star polygons (hexagram A, B): I f 62 g A star polygon of order 62 (hexagram A) has half-side s 62 and Dido value p 1 f ( 62 ) = f (3) = 3p = 19 3 (t 0:1925), 3 a constructible algebraic number (containing only rationals and/or square roots). –Interpretation of Schlä‡i factoring f 26 g = 2 f 31 g: hexagram A is a compound star polygon consisting of 2 triangles (mutually rotated through 2 =6 = =3); triangles and rotation angle are constructible, hence hexagram A is constructible. Perimeter of hexagram: q p 1 f (6)2 + 612 P6 = cos(1 ) P6 = 23 3P6 (t 1:1547 P6 ). P 26 = 12s 62 = f (6) 6 Half-side of hexagram A: p 1 s 26 = 12 P 62 = cos(1 ) s6 = 23 3s6 (t 1:1547 s6 ). 6 Stroke distance of hexagram A: p p d 26 = 4[cos( 6 )]2 s 62 = 3s 62 = 14 P 62 = 4 cos( 6 )s6 = 2 3s6 = 2 cos( 6 )d6 = 3d6 : Draw length of hexagram A: p D 26 = 6d 62 = 12 cos( 6 )d6 = 6 3d6 (t 10:3920d6 ): Inner radius of hexagram A: cos( 26 ) 1 cos( 26 ) 1 1 r 26 = f ( 62 ) q f (6) 2 P6 = 6 sin( ) 2 P6 = sin( ) s6 = s6 = 2 R6 1 2 f (6) + 62
=
f (6)2 f ( 62 ) f (6) 2+ 1 62
6
1 6 2P2
6
f ( 6 )f (6) q 2 2 f ( 26 )2 + 262
= p 1
1 6 2P2
= cos( 26 ) 16 cot( 6 )
1 6 2P2
= cos( 26 ) cot( 6 )s 62 = 2 3s 26 (t 0:8660s 26 ): (One and only) middle radius of hexagram A: q 6 2
6 2 #1
1 f (6)2 + 612 r 62 = cos(1 ) r 62 = p23 r 62 (t 1:1547r 26 ) f (6) 6 p f ( 62 ) 12 P6 = f (3)6s6 = 2 cot( 3 )s6 = 32 3s6 (t 1:1547 s6 ).
=
= s 26 = (Hexagram A and its mother hexagon have the same outer radius R6 .) Area of hexagram A: p A 26 = r 62 12 P 26 = 62 12 P6 = f ( 62 ) ( 21 P6 )2 = f (3) (6s6 )2 = 4 3s26 (t 6:9282s26 ) 2
f (6) = f ( 62 ) f (6) 2+
1 62
( 12 P 62 )2 =
f( 6 )
f ( 6 )f (6) q 2 2 f ( 62 )2 + 262
( 12 P 62 )2 =
f ( 6 )f (6) q 2 f (6)2 + 612
1 6 2P2
1 2 P6
2 2 = f (6) A6 = ff (3) (6) A6 = 3 A6 . 6 I f 3 g A star polygon of order 36 (hexagram B, six-spike asterisk ~ ) has half-side s 36 and
Dido value
f ( 63 ) = f (2) = 0, a constructible algebraic number (containing only rationals and/or square roots), so hexagram B is constructible (by ruler and compass). – Interpretation of Schlä‡i factoring f 63 g =3 f 21 g: hexagram B is a compound star polygon, consisting of 3 digons (mutually rotated through 2 =6 = =3); digons and rotation angle are constructible, hence hexagram B is constructible. q Perimeter of hexagram B: P 6 = 12s 6 =
2
[6f (6)] +1P6 = sin1(
P6 = 2P 6 : 6) 1 1 Half-side of hexagram B: s 6 = 12 P 6 = sin( ) s6 = 2s6 : 3 3 6 Stroke distance of hexagram B: d 6 = sin1( ) d6 = 2d6 = 2s 6 = 4s6 : 3 3 6 Draw length of hexagram B: D 6 = 6d 6 = sin1( ) D6 = 2D 6 = 12s 6 = 24s6 = 3 3 3 6
P 36 = 2P 6 :
Hexagram B (six-spike asterisk) is a collapsed star polygon; inner radius, all middle radii, and area of hexagram B vanish: r 6 = 0;
= 0; A 36 = 0:
3
3
3
6 3 #1;2
20
PETER KAHLIG
(Hexagram B and its mother hexagon have the same outer radius R6 .)
Pythagorean relationships: s26 = R62 r62 , s26 = s26 + (r6 2 Expressions in terms of outer radius R6 f6g (hexagon): 1 1 q R6 = 6 sin( 6 )R6 = 3R6 ; 2 P6 = 6s6 = 1 2 s6 =
1 1 6 ( 2 P6 )
=
f (6) + 62 p 2 1 2 R6 6 f (6) +1
6 2
)2 , s26 = s26 + r62 . 3
= sin( 6 )R6 = 21 R6 ;
d6 = 2s6 = 16 P6 = 2 sin( 6 )R6 = R6 ; p r6 = q f (6) R6 = cos( 6 )R6 = 12 3R6 (t 0:8660 R6 ); f (6)2 + 612 p 2 f (6) 6 2 3 2 2 2 A6 = f (6) 3R6 (t 2:5981R62 ): 2 + 1 R6 = 2 sin( 6 )R6 = 3 sin( 3 )R6 = 2 62
f 62 g (hexagram A): 1 1 1 6 6 2 P 2 = 6s 2 = cos( ) 2 P6 =
p
1 f (6) R6 = 6 tan( 6 )R6 = 2 3R6 (t p tan( 6 )R6 = p13 R6 = 13 3R6 (t 0:5774R6 ); p 6
s 62 =
1 1 6 6(2P2 )
=
3:4641R6 );
d 62 = 2 sin( 3 )R6 = 3R6 (t 1:7321R6 ); p D 62 = 6d 62 = 12 sin( 3 )R6 = 6 3R6 (t 10:3920R6 ); f (6) r 26 = f ( 62 ) f (6) 2+
f ( 26 )
f (3)
R6 = cos( 3 )R6 = 21 R6 ; p cos( 3 ) 6 1 1 6 q 1 6 = f( ) R = 3R6 = s 26 (t 0:5774 R6 ); P = f ( ) R = 6 6 6 ) 2 2 2 cos( 3 1 2 2 6 f (6) + 62 p A 26 = f ( 62 ) f (6)21+ 1 R62 = 6 tan( 6 ) cos( 3 )R62 = 3R62 1 62
R6 =
q
2
f ( 62 )2 + 622
62
= 6 sin( 6 )R6 =
f ( 62 ) f (6)
A6 =
1 2 P6 = f (3) 2 f (6) A6 = 3 A6 = 6 2
=
6 2
6
R6 =
1 6 2P2
2 p
q
f (3)2 + 312
1 R6 cos( 6 ) = r 62 12 P 62 = r 62 f (6)
3R62 (t 1:7321R62 ):
f 36 g (hexagram B, six-spike asterisk ~): 1 6 s 63 = R6 ; d 63 = 2s 63 = 2R6 ; D 36 = 6d 63 = 2 P 3 = 6R6 ; 6 r 36 = 0; = 0; A 63 = r 63 21 P 63 = 36 #2 12 P6 = 0: 3 #1;2
12R6 ;
Example 7. Polygon f7g (Heptagon), Star Polygon f 72 g (Heptagram A), Star Polygon f 37 g (Heptagram B)
Heptagon
Heptagram A
Heptagram B
Polygon f7g (heptagon) is enclosed between its inner circle (radius r7 ) and outer circle (radius R7 ). Star polygon f 27 g (heptagram A) is enclosed between its inner circle (radius r 72 ) and outer circle (radius R7 ). Star polygon f 37 g (heptagram B) is enclosed between its inner circle (radius r 7 ) and outer circle (radius R7 ). Middle circles dashed. 3
REGULAR POLYGONS OF RATIONAL ORDER
21
All points of self-intersection of heptagram A lie on the middle circle (radius 7 ). 2 Self-intersection of heptagram A produces a miniature heptagon, enclosed between inner circle (radius r 7 ) and middle circle (radius 7 ). 2
2
All points of self-intersection of heptagram B lie on the 1st and 2nd middle circle (radius 7 #1 and 7 #2 ). Self-intersection of heptagram B produces a miniature heptagon, 3
3
enclosed between inner circle (radius r 7 ) and 1st middle circle (radius 3
7 3 #1
); moreover,
it produces a miniature heptagram A, enclosed between inner circle (radius r 7 ) and 2nd 3 middle circle (radius 7 #2 ). 3
I f7g A convex polygon of order 7 (heptagon) has half-side s7 and Dido value f (7) = 71 cot( 7 ) (t 0:2966), a non-constructible algebraic number (containing not only rationals or square roots). Perimeter of heptagon: P7 = 14s7 . 1 P7 . Half-side of heptagon: s7 = 14 Stroke distance of heptagon: d7 = 2s7 . Draw length of heptagon: D7 = 7d7 = 14s7 = P7 . Inner radius of heptagon: r7 = f (7) 21 P7 = cot( 7 ) s7 (t 2:0765 s7 ). q Outer radius of heptagon: R7 = f (7)2 + 712 12 P7 = sin(1 ) s7 (t 2:3048 s7 ). 7
Area of heptagon: A7 = f (7) ( 12 P7 )2 = 7 cot( 7 ) s27 (t 14:536 s27 ). There are two star polygons (heptagram A, B): I f 27 g A star polygon of order 27 (heptagram A) has half-side s 72 and Dido value f ( 72 ) = 27 cot( 27 ) (t 0:2278), a non-constructible algebraic number (containing not only rationals or square roots). Perimeter of heptagram A: q P 72 = 14s 72 =
1 f (7)
f (7)2 +
1 72 P7
Half-side of heptagram A: s 27 =
1 cos( 7 ) P7 (t 1:1099P7 ). 1 1 7 14 P 2 = cos( 7 ) s7 (t 1:1099s7 ). d 27 = 2 sin( 27 )R7 (t 1:5637R7 ).
=
Stroke distance of heptagram A: Draw length of heptagram A: D 27 = 7d 72 = 14 sin( 27 )R7 (t 10:9460R7 ). Inner radius of heptagram A: cos( 27 ) 1 1 r 27 = f ( 72 ) q f (7) 2 P7 = 7 sin( ) 2 P7 = 1 2 f (7) + 72 2
f (7) = f ( 72 ) f (7) 2+
1 72
1 7 2P2
7
=
cos( 27 ) sin( 7 ) s7
f ( 7 )f (7) q 2 2 f ( 72 )2 + 272
1 7 2P2
1 7 72 r 2
1 7 cos( 7 ) r 2
(t 1:437 0s7 )
= cos( 27 ) 17 cot( 7 )
1 7 2P2
= cos( 27 ) cot( 7 )s 72 (t 1:2947s 72 ) (One and only) middle radius of heptagram A: q 7 2
7 2 #1
=
1 f (7)
f (7)2 +
=
= f ( 72 ) 12 P7 = f ( 72 )7s7 = 2 cot( 27 )s7 (t 1:5949s7 ): (Heptagram A and its mother heptagon have the same outer radius R7 .) Area of heptagram A: A 72 = r 72 12 P 27 = 72 12 P7 = f ( 72 ) ( 21 P7 )2 = 27 cot( 27 ) (7s7 )2 = 14 cot( 27 ) s27 (t 11:165 s27 ). I f 37 g A star polygon of order 37 (heptagram B) has half-side s 73 and Dido value f ( 73 ) = 73 cot( 37 ) (t 0:0978), a non-constructible algebraic number (containing not only rationals or square roots).
22
PETER KAHLIG
Perimeter of heptagram B: q 1 7 2 P 73 = 14s 73 = f ( 7 ) f ( 2 ) + 2
22 72 P7
Half-side of heptagram B: s 37
Stroke distance of heptagram Draw length of heptagram B: Inner radius r of heptagram B: r 37 =
f (7)2 + 712
f ( 73 )
2
f ( 73 )2 + 732
1 P (t 1:6039P7 ). cos( 27 ) 7 1 = 14 P 73 = cos(12 ) s7 (t 1:6039s7 ). 7 B: d 37 = 2 sin( 37 )R7 . D 37 = 7d 73 = 14 sin( 37 )R7 :
=
r
7 1 2 P7 =f ( 3 )
f (7)2 + 712
2
f ( 37 )2 + 372
7s7 =
cos ( 37 ) sin ( 7 ) s7
1st middle radius of heptagram B: q cos ( 37 ) 1 1 2 + 1 r7 = 7 7= r = f (7) 2 #1 f (7) 7 cos( ) sin ( ) cos( 3 3 3 7
7
7
) s7
(t 0:5129s7 ).
(t 0:5692s7 ):
nd
2
middle radius of heptagram B: q 2 cos ( 37 ) 1 7 2 7 = f ( 7 ) f ( 2 ) + 272 r 73 = sin ( ) cos( s7 (t 0:8226s7 ): 2 3 #2 2 7 7 ) (Heptagram B and its mother heptagon have the same outer radius R7 .) Area of heptagram B: r 1 7 2P3
A 37 = r 73
=
7 3 #2
cos ( 37 ) A7 2 7 ) cos ( 7 )
= cos (
1 2 P7
= f ( 73 )
f (7)2 + 712
2 f ( 73 )2 + 372
2
(7s7 ) =
7 sin ( 7 ) cos ( 37 ) 2 cos( 7 ) cos ( 27 ) r7
(t 0:3961A7 ):
Pythagorean relationships: s27 = R72 r72 , s27 = s27 +(r7
7 2
2
)2 , s27 = s27 +(r7 3
7 3 #2
)2 .
Expressions in terms of outer radius R7 f7g (heptagon): 1 1 q R7 = 7 sin( 7 ) R7 (t 3:0372 R7 ), 2 P7 = 7s7 = 1 2 s7 =
1 1 7 ( 2 P7 )
=
f (7) + 72 p 2 1 2 R7 7 f (7) +1
= sin( 7 ) R7 (t 0:4339R7 ),
d7 = 2s7 = 2 sin( 7 ) R7 (t 0:8678 R7 ), D7 = 7d7 = 14s7 = P7 = 14 sin( 7 ) R7 (t 6:0744 R7 ); r7 = q f (7) R7 = cos( 7 ) R7 (t 0:9010 R7 ), 1 2 f (7) + 72
A7 = r7 12 P7 =
f (7) f (7)2 + 712
R72 =
7 2
sin( 27 ) R72 = 7 sin( 7 ) cos( 7 )R72 (t 2:7364 R72 ):
f 27 g (heptagram A): 1 1 1 7 7 2 P 2 = 7s 2 = cos( ) 2 P7 =
1 f (7) R7 = 7 tan( 7 ) R7 (t 3:3710R7 ), 7 1 1 1 s 27 = 7 ( 2 P 27 ) = 7f (7) R7 = tan( 7 ) R7 (t 0:4816R7 ), d 27 = 2 sin( 27 )R7 (t 1:5637R7 ), D 72 = 7d 72 = 14 sin( 27 )R7 (t 10:9460R7 ), f ( 72 ) f (7) q r 27 = f ( 72 ) f (7) R7 = cos( 27 ) R7 (t 0:6235 R7 ); 2 + 1 R7 = 2 2 f ( 72 )2 + 272 7 r7 f ( 72 ) cos( 2 ) 2 q 7 = R7 = cos( 7 ) R7 (t 0:6920 R7 ); r7 R7 = 2 7 f (7)2 + 712 f ( 72 ) 1 f (7) A7 = (1 72 f (7)2 )A7 = [1 f( 7 ) 1 r 72 12 P 27 = r 72 f (7) R7 = f (7)22+ 1 2
A 27 = f ( 72 )( 12 P7 )2 = =
7 2
1 2 P7
=
2:1018R72 ): f 73 g (heptagram B):
7
tan( 7 )2 ]A7 (t 0:7681A7 )
R72 = 7 tan( 7 ) cos( 27 )R72 (t
REGULAR POLYGONS OF RATIONAL ORDER 1 7 2P3
= 7s 73 =
s 73 = 17 ( 12 P 37 )
23
7 sin( 7 ) R (t 4:8713R7 ); cos( 27 ) 7 sin( 7 ) = cos( 2 ) R7 (t 0:6959R7 ); 7
d 37 = 2 sin( 37 )R7 (t 1:9499R7 ), D 73 = 7d 73 = 14 sin( 37 )R7 (t 13:6490R7 ), r 37 =
q
f ( 73 )
2
f ( 37 )2 + 372 r7 3 7 = #1 r 7 R7 = 3 r7 7 = r 37 R7 = 3 #2 2
1 7 2P3
A 73 = r 73
R7 = cos( 37 ) R7 = sin( 14 )R7 (t 0:2225 R7 ), cos( 37 ) cos( 7 ) R7 cos( 37 ) R cos( 27 ) 7
=
= [2 cos( 27 ) (t 0:3569R7 );
1 2 P7
7 3 #2
1]R7 (t 0:2470R7 );
=
cos( 37 ) R cos( 27 ) 7
cos( 3 )
7s7 = 7 sin( 7 ) cos( 27 ) R72 (t 1:0839R72 ): 7
Example 8. Polygon f8g (Octagon), Star Polygon f 82 g (Octagram A), Star Polygon f 83 g (Octagram B), Star Polygon f 84 g (Octagram C) [Eight-Spike Asterisk]
Octagon
Octagram A
Octagram B
Octagram C
Polygon f8g (octagon) is enclosed between its inner circle (radius r8 ) and outer circle (radius R8 ). Star polygon f 28 g (octagram A) is enclosed between its inner circle (radius r 82 ) and outer circle (radius R8 ). Star polygon f 83 g (octagram B) is enclosed between its inner circle (radius r 8 ) and outer circle (radius R8 ). (Middle circles dashed.) Star 3
polygon f 48 g (octagram C) is enclosed by its outer circle (radius R8 ). All points of self-intersection of octagram A lie on the middle circle (radius 8 ). Self2 intersection of octagram A produces a miniature octagon, enclosed between inner circle (radius r 8 ) and middle circle (radius 8 ). 2
2
All points of self-intersection of octagram B lie on the 1st and 2nd middle circle (radius 8 #1 and 8 #2 ). Self-intersection of octagram B produces a miniature octagon, enclosed 3
3
between inner circle (radius r 8 ) and 1st middle circle (radius 3
8 3 #1
); moreover, it produces
a miniature octagram A, enclosed between inner circle (radius r 8 ) and 2nd middle circle 3 (radius 8 #2 ). 3 All points of self-intersection of octagram C lie in the center. Inner circle (radius r 8 = 0) and all middle circles (radii 8 #1;2;3 = 0) collapsed to zero in the center. 4
4
I f8g A convex polygon of order 8 (octagon) has half-side s8 and Dido value p f (8) = 81 cot( 8 ) = 8(p12 1) = 1+8 2 (t 0:3018), a constructible algebraic number (containing only rationals and/or square roots). Perimeter of octagon: P8 = 16s8 . 1 Half-side of octagon: s8 = 16 P8 .
24
PETER KAHLIG
Stroke distance of octagon: d8 = 2s8 . Draw length of octagon: D8 = 8d8 = 16s8 = P8 . p Inner radius of octagon: r8 = f (8) 12 P8 = cot( 8 )s8 = (1 + 2)s8 (t 2:4142s8 ). q p p Outer radius of octagon: R8 = f (8)2 + 812 12 P8 = 4 + 2 2s8 (t 2:6131 s8 ). Area of octagon: p A8 = r8 21 P8 = f (8)( 12 P8 )2 = 8 cot( 8 ) s28 = 8(1 + 2) s28 (t 19:314 s28 ) p p 1 = 8(1 + 2) (1+p r2 = 1+8p2 r82 = 8( 2 1)r82 (t 3:3137r82 ). 2)2 8 There are three star polygons (octagram A, B, C): I f 28 g A star polygon of order 82 (octagram A) has half-side s 82 and Dido value f ( 82 ) = f (4) = 14 , a constructible algebraic number (containing only rationals and/or square roots). –Interpretation of Schlä‡i factoring f 82 g =2 f 14 g: octagram A consists of 2 squares (mutually rotated through 2 =8 = =4); squares and rotation angle are constructible, hence octagram A is constructible. Perimeter of octagram q A: p p 1 P 82 = 16s 82 = f (8) f (8)2 + 812 P8 = cos(1 ) P8 = 4 2 2P8 (t 1:0824P8 ). 8 Half-side of octagram q A: p p 1 1 P 82 = f (8) f (8)2 + 812 s8 = cos(1 ) s8 = 4 2 2s8 (t 1:0824s8 ). s 28 = 16 8 Stroke distance ofp octagram A: p 2 sin( ) d 28 = sin( 8 ) d8 = 2 + 2d8 (t 1:8478d8 ) 8 p = 2 sin( 28 )R8 = 2 sin( 4 )R8 = 2R8 (t 1:4142R8 ): Draw length of octagram p A: p sin( 28 ) D 28 = 8d 82 = sin( ) D8 = 2 + 2D8 (t 1:8478D8 ) 8 p = 16 sin( 28 )R8 = 16 sin( 4 )R8 = 8 2R8 (t 11:3140R8 ): Inner radius of octagram A: p p cos( 28 ) 1 r 28 = f ( 82 ) q f (8) P = )s = s = 2 cos( 2 + 2s8 (t 1:8478s8 ) 8 8 8 2 sin( 8 ) 1 2 f (8) + 82
=
f (8)2 f ( 28 ) f (8) 2+ 1 2
1 8 2P2
8
8
=
f ( 8 )f (8) q 2 2 f ( 82 )2 + 282 p + 12 2)s 82
1 8 2P2
= cos( 28 ) 18 cot( 8 )
1 8 2P2
= cos( 28 ) cot( 8 )s 82 = (1 (t 1:7071s 82 ): (One and only) middle radius of octagram A: q p p 1 1 2 + 1 r8 = 8 8 8 = = f (8) r 2 + 2s 28 2 #1 f (8) 8 cos( ) 2 2 2 2 8
= f ( 82 ) 12 P8 = f (4)8s8 = 2 cot( 4 )s8 = 2s8 . Area of octagram A: A 82 = r 82 12 P 28 = 82 12 P8 = f ( 82 )( 12 P8 )2 = f (4)(8s8 )2 = 16s28 f ( 28 ) f (8) A8
cot( )
f (4) f (8) A8
= (1 82 f1(8)2 )A8 = (1 [tan( 8 )]2 )A8 = 2 cot( 4 ) A8 8 p = 2 tan( 8 )A8 = 2( 2 1)A8 (t 0:8284A8 ). (Octagram A and its mother octagon have the same outer radius R8 .) I f 83 g A star polygon of order 83 (octagram B) has half-side s 83 and Dido value p f ( 83 ) = 38 cot( 38 ) = 38 ( 2 1) (t 0:1553), a constructible algebraic number (containing only rationals and/or square roots). Perimeter of octagram q B: q p 2 1 1 8 8 P 3 = 16s 3 = f ( 8 ) f ( 82 )2 + 282 P8 = f (4) f (4)2 + 412 P8 = cos(1 ) P8 = 2P8 . =
=
2
4
REGULAR POLYGONS OF RATIONAL ORDER
25
Half-side of octagram q B: q p 2 1 1 1 s 83 = 16 P 83 = f ( 8 ) f ( 82 )2 + 282 s8 = f (4) f (4)2 + 412 s8 = cos(1 ) s8 = 2s8 . 4 2 Stroke distance of octagram B: p sin( 3 ) cos( ) d 38 = sin( 8 ) d8 = sin( 8 ) d8 = cot( 8 )d8 = 8f (8)d8 = (1 + 2)d8 (t 2:414 2d8 ) 8 8 p p = 2 sin( 38 )R8 = 2 cos( 8 )R8 = 2 + 2R8 (t 1:847 8R8 ). Draw length of octagram B: p D 38 = 8d 83 = 8f (8)D8 = (1 + 2)D8 (t 2:414 2D8 ): Inner radius r of octagram B: r r 38 = f ( 83 )
f (8)2 + 812 1 P8 =f ( 83 ) 2 f ( 83 )2 + 832 2
f (8)2 + 812
2
f ( 38 )2 + 382
1st middle radius of octagram B: q 1 1 2 + 1 r8 = 8 = f (8) f (8) 82 3 cos( 3 #1
r 8) 3 8
=
8s8 =
p 4
cos ( 38 ) sin ( 8 ) s8
= s8 .
p 2 2r 83 (t 1:0824r 83 ),
2nd middle radius of octagram B: q p p 2 1 8 2 8 = f ( 8 ) f ( 2 ) + 282 r 83 = cos(12 ) r 38 = cos(12 ) s8 = 2s8 = 2r 83 . 3 #2 2 8 8 (Octagram B and its mother octagon have the same outer radius R8 .) Area of octagram B: r r A 83 = r 83 21 P 83 =
cos ( 38 ) 2 8 ) cos ( 8
= cos (
f (8)2 +
1
2
1 82 = f ( 83 ) P8 = f ( 38 ) 2 f ( 83 )2 + 382 2 p 2)A8 (t 0:5858A8 ): A = (2 ) 8 8 3 #2
1 2 P8
f (8)2 + 812
2
f ( 83 )2 + 382
2
(8s8 )
I f 48 g A star polygon of order 48 (octagram C, eight-spike asterisk) has half-side s 48 and
Dido value
f ( 84 ) = f (2) = 0, a constructible algebraic number (containing only rationals and/or square roots), so octagram C is constructible (by ruler and compass). – Interpretation of Schlä‡i factoring f 84 g =4 f 21 g: octagram C is a compound star polygon, consisting of 4 digons (mutually rotated through 2 =8 = =4); digons and rotation angle are constructible, hence octagram C is constructible. Perimeter of q octagram C: q
p P8 = 2(2+ 2)P8 (t 2:6131P 8 ): q p 1 1 8= Half-side of octagram C: s 8 = 16 P s = 2(2+ 2)s8 (t 2:6131s8 ): 8 sin( ) 4 4 8 q p Stroke distance of octagram C: d 8 = sin1( ) d8 = 2(2 + 2)d8 = 2s 8 : 4 2
P 84 = 16s 84 = [8f (8)] +1P8 = sin1(
8
) P8 =
p
2
2
p
8
Draw length of octagramq C:
D 48 = 8d 84 = sin1(
8
) D8 =
2
4
q p p 2(2 + 2)D8 = 16s 8 = P 48 = 2(2 + 2)P8 : 4
Octagram C (eight-spike asterisk) is a collapsed star polygon; inner radius, all middle radii, and area of octagram C vanish: r 8 = 0; 8 #1;2 = 0; A 8 = 0: 4 4 4 (Octagram C and its mother octagon have the same outer radius R8 .)
Pythagorean relationships: s28 = R82 r82 , 2 2 2 8 ) , s 8 = s + (r8 8 s28 = s28 + (r8 )2 ; s28 = s28 + r82 . 8 2 3 #2 2 3 4 Expressions in terms of outer radius R8 f8g (octagon): p p 1 1 q R8 = 8 sin( 8 )R8 = 4 2 2R8 (t 3:0615R8 ), 2 P8 = 8s8 = 1 2 f (8) + 82
26
PETER KAHLIG
p p s8 = 18 ( 12 P8 ) = p 2 1 2 R8 = sin( 8 )R8 = 21 2 2R8 (t 0:3827R8 ), 8 f (8) +1 p p d8 = 2s8 = 2 2R8 (tp 0:7654R8 ), p D8 = 8d8 = 16s8 = P8 = 8 2 2R p8 (tp6:1229R8 ), 1 2 + 2R8 (t 0:9239R8 ), R = cos( r8 = q f (8) ) R = 8 8 8 2 f (8)2 + 812 p 2 f (8) 8 2 2 2 A8 = f (8) 2R8 (t 2:8284R82 ). 2 + 1 R8 = 2 sin( 8 )R8 = 2 82
f 28 g (octagram A): 1 1 1 8 8 2 P 2 = 8s 2 = cos( ) 2 P8 = 8
s 28 =
p
R8 f (8) = 8 tan( 8 ) R8 = 8( 2 p R8 8f (8) = tan( 8 ) R8 = ( 2
1 1 1 8 8 ( 2 P 2 ) = cos( 8 ) s8 = 2 sin( 28 )R8 = 2 sin( 4 )R8 p
p
1)R8 (t 3:3137R8 ), 1)R8 (t 0:4142R8 ),
= 2R8 (t 1:4142R8 ), d 28 = D 28 = 8d 82 = 8 2R8 (t 11:3140R8 ),
p R8 = cos( 28 )R8 = 12 2R8 (t 0:7071R8 ), p r8 p cos( 28 ) f (4) 2 q 8 = R = R = R = 2 2R8 (t 0:7654R8 ); 8 8 8 r8 cos( 8 ) 2 f (8)2 + 812 p f (4) 1 2 2)R82 (t 2:3431R82 ) A 28 = r 82 21 P 82 = r 82 f (8) R8 = f (8) 2 + 1 R8 = 4(2
f (8) r 28 = f (4) f (8) 2+
f ( 82 ) f (8) A8
1 82
R8 =
q
f (4)
f (4)2 + 412
82
cot( )
f (4) f (8) A8
= (1 82 f1(8)2 )A8 = (1 [tan( 8 )]2 )A8 = 2 cot( 4 ) A8 8 p = 2 tan( 8 )A8 = 2( 2 1)A8 (t 0:8284A8 ). f 83 g (octagram B): p p 8 sin( 8 ) 1 8 8 2 2R8 (t 4:3296R8 ); 2 P 3 = 8s 3 = cos( 4 ) R8 = 4 4 p p sin( ) s 38 = 18 ( 12 P 38 ) = cos( 8 ) R8 = 21 4 2 2R8 (t 0:5412R8 ); 4 p p d 38 = 2 sin( 38 )R8 = 2 + 2R8 (t 1:8478R8 ); p p D 38 = 8d 83 = 8 2 + 2R8 (t 14:7820R8 ); p p f( 8 ) r 83 = q 8 3 32 R8 = cos( 38 )R8 = 21 2 2R8 (t 0:3827R8 ); =
=
f ( 3 )2 + 82 r8 cos( 38 ) 3 8 R = = 8 #1 r cos( 8 ) R8 8 3 8 3 #2
A 38 =
=
r8
R8 = 2 r 3
r8
f ( 83 ) f (4)
cos( 38 cos( 28
f (4)2 + 412 2 8 2 [f ( 3 ) + 832 ][f (8)2 + 812
f 48 g (octagram C, 1 8 s 84 = 2 P 4 = 8R8 ; r 48 = 0;
p = tan( 8 )R8 = ( 2 1)R8 (t 0:4142R8 ); p p ) 1 1 R = R = 4 2 2R8 (t 0:5412R8 ); 8 8 2 cos( ) 2 )
8 4 #1;2;3
8
]
p cos ( 3 ) R82 = 8 sin ( 8 ) cos ( 28 ) R82 = 4( 2 1)R82 (t 1:6569R82 ): 8
eight-spike asterisk):
R8 ; d 84 = 2s 84 = 2R8 ; D 48 = 8d 84 = 16R8 ; = 0; A 84 = r 84 21 P 84 = 48 #3 21 P8 = 0:
Example 9. Polygon f9g (Nonagon), Star Polygon f 92 g (Nonagram A), Star Polygon f 93 g (Nonagram B), Star Polygon f 49 g (Nonagram C)
REGULAR POLYGONS OF RATIONAL ORDER
Nonagon
Nonagram A
27
Nonagram B
Nonagram C
Polygon f9g (nonagon) is enclosed between its inner circle (radius r9 ) and outer circle (radius R9 ). Star polygon f 92 g (nonagram A) is enclosed between its inner circle (radius r 9 ) and outer circle (radius R9 ). Star polygon f 39 g (nonagram B) is enclosed between its 2
inner circle (radius r 9 ) and outer circle (radius R9 ). Star polygon f 94 g (nonagram C) is 3 enclosed between its inner circle (radius r 9 ) and outer circle (radius R9 ). Middle circles 4 dashed. All points of self-intersection of nonagram A lie on the middle circle (radius 9 ). 2 Self-intersection of nonagram A produces a miniature nonagon, enclosed between inner circle (radius r 9 ) and middle circle (radius 9 ). 2
2
All points of self-intersection of nonagram B lie on the 1st and 2nd middle circle (radius 9 #1 and 9 #2 ). Self-intersection of nonagram B produces a miniature nonagon, enclosed 3
3
between inner circle (radius r 9 ) and 1st middle circle (radius 3
9 3 #1
); moreover, it produces
a miniature nonagram A, enclosed between inner circle (radius r 9 ) and 2nd middle circle 3 (radius 9 #2 ). 3
All points of self-intersection of nonagram C lie on the 1st , 2nd and 3rd middle circle (radius 9 #1 , 9 #2 and 9 #3 ). Self-intersection of nonagram C produces a miniature 4
4
4
nonagon, enclosed between inner circle (radius r 9 ) and 1st middle circle (radius 9 #1 ); 4 4 moreover, it produces a miniature nonagram A, enclosed between inner circle (radius r 9 ) and 2nd middle circle (radius
4
9 4 #2
); furthermore, it produces a miniature nonagram B,
enclosed between inner circle (radius r 9 ) and 3rd middle circle (radius 4
9 4 #3
).
I f9g A convex polygon of order 9 (nonagon) has half-side s9 and Dido value f (9) = 91 cot( 9 ) (t 0:3053), a non-constructible algebraic number (containing not only rationals or square roots). Perimeter of nonagon: P9 = 18s9 . 1 Half-side of nonagon: s9 = 18 P9 Stroke distance of nonagon: d9 = 19 P9 = 2s9 . Draw length of nonagon: D9 = 9d9 = 18s9 = P9 . Inner radius of nonagon: r9 = f (9) 21 P9 = cot( 9 )s9 (t 2:7475s9 ). q Outer radius of nonagon: R9 = f (9)2 + 912 12 P9 = sin(1 ) s9 (t 2:9238 s9 ). 9
Area of nonagon: A9 = r9 21 P9 = f (9) ( 12 P9 )2 = 9 cot( 9 ) s29 (t 24:727 s29 ). There are three star polygons (nonagram A, B, C): I f 92 g A star polygon of order 29 (nonagram A) has half-side s 92 and Dido value f ( 92 ) = 92 cot( 29 ) (t 0:2648), a non-constructible algebraic number (containing not only rationals or square roots). q P9 1 Perimeter of nonagram A: P 92 = 18s 29 = 1 + [9f (9)] 2 P9 = cos( ) (t 1:0642P9 ). 9
28
PETER KAHLIG
Half-side of nonagram A: s 92 =
1 9 18 P 2
=
q 1+
1 [9f (9)]2 s9
=
s9 cos( 9 )
(t 1:0642s9 ).
sin( 29
) sin( 9 ) d9 (t 1:8794d9 ): sin( 2 ) 9d 92 = sin( 9 ) D9 (t 1:8794D9 ): 9
Stroke distance of nonagram A: d 92 =
Draw length of nonagram A: D 29 = Inner radius of nonagram A: cos( 29 ) 1 1 r 29 = f ( 92 ) q f (9) 2 P9 = 9 sin( ) 2 P9 = 1 2 f (9) + 92 2
f (9) = f ( 92 ) f (9) 2+
1 92
cos( 29 ) sin( 9 ) s9
9
1 9 2P2
=
f ( 9 )f (9) 1 q 2 P 92 2 2 f ( 29 )2 + 292
(t 2:2398s9 )
= cos( 29 ) cot( 9 )s 92 (t 2:1047s 29 ):
(One q and only) middle radius of nonagram A: 1 2 1 1 9 9 = 1 + [9f (9)] 2 r 9 = cos( ) r 9 = f ( 2 ) 2 P9 = 2 cot( 9 )s9 (t 2:3835s9 ): 2 2 2 9
(Nonagram A and its mother nonagon have the same outer radius R9 :) Area of nonagram A: A 29 = f ( 92 ) ( 12 P9 )2 = 18 cot( 29 ) s29 (t 21:452s29 ) f( 9 )
2 = f (9) A9 = (1 92 f1(9)2 )A9 = [1 tan( 9 )2 ]A9 (t 0:8675A9 ): I f 93 g A star polygon of order 39 (nonagram B) has half-side s 93 and Dido value p f ( 39 ) = f (3) = 19 3 (t 0:1925); although this is a constructible algebraic number (pertaining to each of the 3 component triangles), nonagram B is not constructible since the rotation angle 2 =9 (among the 3 component triangles) is not constructible. (2 =3 would be constructible, but cannot be trisected by ruler and compass. None of the nonagrams is constructible since their mother polygon, the nonagon, is not constructible.) – Interpretation of Schlä‡i factoring f 93 g =3 f 13 g: nonagram B consists of 3 triangles (mutually rotated through 2 =9); triangles are constructible, but not the rotation angle, hence nonagram B is not constructible. Perimeter of nonagram q B:
P 39 = 18s 93 = f (19 ) f ( 92 )2 + 2 Half-side of nonagram q B: s 39 =
1 9 18 P 3
=
1 f ( 92 )
f ( 92 )2 +
Stroke distance of nonagram Draw length of nonagram B: Inner radius r of nonagram B: f (9)2 + 912
22 92 P9
=
22 92 s9
=
1 P cos( 29 ) 9
(t 1:3054P9 ):
1 s (t 1:3054s9 ): cos( 29 ) 9 sin( 39 ) B: d 93 = sin( ) d9 (t 2:5321d9 ): 9 sin( 3 ) D 39 = 9d 93 = sin( 9 ) D9 (t 2:5321D9 ): 9
r
f (9)2 + 912
cos ( 3 )
9s9 = sin ( 9 ) s9 = 2 sins9( ) (t 1:4619s9 ). 9 9 q 1 1 st 1 middle radius of nonagram B: 39 #1 = 1 + 92 f (9)2 r 93 = cos( ) r 93 ; 9 q 22 nd 9 9 2 middle radius of nonagram B: 3 #2 = 1 + 92 f ( 9 )2 r 3 = cos(12 ) r 93 : r 39 = f ( 93 )
2 f ( 93 )2 + 932
1 2 P9 =f (3)
f (3)2 + 312
2
9
(Nonagram B and its mother nonagon have the same outer radius R9 :) Area of nonagram B: r r f (9)2 + 912 f (9)2 + 1 2 2 1 1 9 1 A 39 = r 93 2 P 93 = 39 #2 2 P9 = f ( 3 ) = f (3) f (3)2 + 912 (9s9 ) 9 2 32 2 P9 f ( 3 ) + 92
=
=
9 sin ( 9 ) cos ( 39 ) 2 cos ( 39 ) cos( 9 ) cos ( 29 ) r9 = cos ( 9 ) cos ( 29 ) A9 (t 0:6946A9 ) 3 cos( 39 ) 9 9 sin( 9 ) cos( 9 ) 2 2 2 4 sin( 9 ) cos( 29 ) d9 (t 4:2939d9 ) = 4 sin( 39 )2 cos( 29 ) d 93
32
(t 0:6697d29 ): 3
REGULAR POLYGONS OF RATIONAL ORDER
29
I f 94 g A star polygon of order 49 (nonagram C) has half-side s 94 and Dido value f ( 94 ) = 94 cot( 49 ) (t 0:0784), a non-constructible algebraic number (containing not only rationals or square roots). r 2
Perimeter of nonagram C: P 94 = 18s 94 = Half-side of nonagram C: s 94 =
1 9 18 P 4
=
f ( 94 )2 + 492
r
f (9)2 + 912
P9 =
2
f ( 94 )2 + 492 f (9)2 + 912
s9 =
4 sin( 9 ) P (t sin( 49 ) 9
4 sin( 9 ) s (t sin( 49 ) 9
1:3892P9 ):
1:3892s9 ):
sin( 49
) sin( 9 ) d9 (t 2:8794d9 ): sin( 4 ) 9d 94 = sin( 9 ) D9 (t 2:8794D9 ): 9
Stroke distance of nonagram C: d 49 =
Draw length of nonagram C: D 94 = Inner radius C: r of nonagram r f (9)2 + 912 1 f (9)2 + 912 cos ( 4 ) 9 9 r 94 = f ( 4 ) P =f ( ) 9s9 = sin ( 9 ) s9 (t 0:5077s9 ). 9 9 2 42 2 9 2 42 4 f ( 4 ) + 92 f ( 4 ) + 92 9 q 1 st 1 middle radius of nonagram C: 49 #1 = 1 + [9f (9)]2 r 94 = cos(1 ) r 49 ; 9 q 1 nd 9 9 2 middle radius of nonagram C: 4 #2 = 1 + [ 9 f ( 9 )]2 r 4 = cos(12 ) r 94 ; 2 2 9 q 1 1 9 = 3rd middle radius of nonagram C: 94 #3 = 1 + [3f (3)] r r 9 = 2r 94 : 2 cos( 3 ) 4 4 9
(Nonagram C and its mother nonagon have the same outer radius R9 :) Area of nonagram C: A 49 = r 94 9s 49 = r 94 12 P 49 = 94 #3 9s9 = 49 #3 21 P9 = 9 sin( 9 )R9 94 #3 : Pythagorean relationships: s29 = R92 r92 ; 2 2 2 9 ) ; s 9 = s + (r9 9 9 s29 = s29 + (r9 )2 ; s29 = s29 + (r9 )2 : 9 2 3 #2 4 #3 2 3 4 Expressions in terms of outer radius R9 f9g (nonagon): 1 1 q R9 = 9 sin( 9 ) R9 (t 3:0782R9 ); 2 P9 = 9s9 = 1 2 f (9) + 92 1 R9 92 f (9)2 +1
s9 = 91 ( 12 P9 ) = p
= sin( 9 ) R9 (t 0:3420R9 );
d9 = 2 sin( 9 )R9 (t 0:6840R9 ); D9 = 9d9 = 18 sin( 9 )R9 (t 6:1564R9 ); r9 = q f (9) R9 = cos( 9 ) R9 (t 0:9397 R9 ); 1 2 f (9) + 92
A9 = r9 12 P9 =
f (9) f (9)2 + 912
R92 =
q
1
2
f ( 92 )2 + 292
R92 =
9 2
sin( 29 )R92 (t 2:8925R92 ):
f 29 g (nonagram A): 1 1 1 1 9 9 2 P 2 = 9s 2 = cos( 9 ) 2 P9 = f (9) R9 = 9 tan( 9 )R9 (t 3:2757R9 ); s 29 = 19 ( 12 P 29 ) = cos(1 ) s9 = 9f1(9) R9 = tan( 9 )R9 (t 0:3640R9 ); 9 d 29 = 2 sin( 29 )R9 (t 1:2856R9 ); D 92 = 9d 92 = 18 sin( 29 )R9 (t 11:570R9 ); f ( 29 ) f (9) q r 29 = f ( 92 ) f (9) R9 = cos( 29 )R9 (t 0:7660R9 ); 2 + 1 R9 = 2 f ( 9 )2 + 2 92 9 2
=
r9
r9 R9 = 2
f ( 92 )
f (9)2 + 912
A 92 = f ( 92 )( 12 P9 )2 = = =
92
2
q
R9 =
f ( 92 ) f (9) A9
cos( 29 ) cos( 9 ) R9
(t 0:8152R9 );
1 [9f (9)]2 )A9 f( 9 ) 1 1 9 P9 = r 92 12 P 29 = r 92 f (9) R9 = f (9)22+ 1 2 2 92 9 tan( 9 ) cos( 29 )R92 (t 2:5094R92 ):
= (1
= [1 R92
tan( 9 )2 ]A9 (t 0:8675A9 )
30
PETER KAHLIG
f 93 g (nonagram B):
r q 2 f ( 92 )2 + 292 9 sin( ) 1 1 2 = 9s 93 = R9 = cos( 2 9 ) R9 (t 4:0183R9 ); f (9) + 92 R9 = f ( 9 ) f (9) 1 2+ 2 9 92 r q 2 f ( 92 )2 + 922 sin( ) 1 1 1 1 1 = 9 ( 2 P 39 ) = 9f ( 9 )f (9) f (9)2 + 92 R9 = 9f ( 9 ) f (9)2 + 1 R9 = cos( 29 ) R9 (t 1 f ( 92 )f (9)
1 9 2P3
s 39
2
92
2
9
0:4465R9 ); p d 39 = 2 sin( 39 )R9 = 3R9 (t 1:7321R9 ); p D 39 = 9d 93 = 18 sin( 39 )R9 = 9 3R9 (t 15:588R9 ); r 39 =
f ( 93 )
q
2
f ( 39 )2 + 392 r9 9 = r93 R9 = 3 #1 r9 9 = r 39 R9 = 3 #2
R9 = cos( 39 )R9 = cos( 3 )R9 = 21 R9 ; cos( 39 ) cos( 9 ) R9 cos( 39 ) R cos( 2 ) 9
= [2 cos( 29 ) =
9
2
A 39 = r 93 21 P 93 =
9 3 #2
1 2 P9
f 49 g (nonagram C): 1 2 q 9 9 2 P 4 = 9s 4 = f (9)2 +
s 49 = 19 ( 12 P 49 ) = p
1 92
=
1 1 2 cos 92
cos( 39 cos( 29
R9 =
2 R9 92 f (9)2 +1
1]R9 =
(t 0:5321R9 );
R9 (t 0:6527R9 );
) R ) 9
cos( 3 )
9s9 = 9 sin( 9 ) cos( 29 ) R92 (t 2:0091R92 ):
9 sin( 9 ) cos( 3 ) R9
=
1 1 2 cos( 9 ) R9
9
= 18 sin( 9 )R9 (t 6:1564R9 );
sin( 9 ) cos( 3 ) R9
= 2 sin( 9 )R9 (t 0:6840R9 );
2 sin( 49
)R9 (t 1:9696R9 ); d 49 = D 94 = 9d 94 = 18 sin( 49 )R9 (t 17:727R9 ); r 49 =
f ( 94 )
q
2
f ( 49 )2 + 492 r9 9 = r94 R9 = 4 #1 r9 9 = r 49 R9 = 4 #2 2 r9 4 9 = r 9 R9 = 4 #3
R9 = cos( 49 )R9 (t 0:1736R9 );
3
A 49 =
9 4 #3
= r 49
cos( 49 ) cos( 9 ) R9 cos( 49 ) R cos( 29 ) 9 cos( 49 cos( 39
1 2 P9
1 9 2P4
=
=
) R ) 9
1 f (9)+ 2
(t 0:1848R9 ); (t 0:2267R9 ); = 2 cos( 49 )R9 (t 0:3473R9 );
4 9 f( 9 ) 4
cos( 4 )
R92 = 9 sin( 9 ) cos( 39 ) R92 (t 1:0690R92 )
cos( 49 ) A cos( 9 ) cos( 39 ) 9
9
(t 0:3696A9 ):
Example 10. Polygon f10g (Decagon), Star Polygon f 10 2 g (Decagram A), 10 Star Polygon f 10 g (Decagram B), Star Polygon f g (Decagram C), Star Poly3 4 g (Decagram D) [Ten-Spike Asterisk] gon f 10 5
Decagon
Decagram A
Decagram B
Decagram C
Decagram D
REGULAR POLYGONS OF RATIONAL ORDER
31
Polygon f10g (decagon) is enclosed between its inner circle (radius r10 ) and outer circle (radius R10 ). Star polygon f 10 2 g (decagram A) is enclosed between its inner circle (radius r 10 ) and outer circle (radius R10 ). Star polygon f 10 3 g (decagram B) is enclosed between 2
its inner circle (radius r 10 ) and outer circle (radius R10 ). Star polygon f 10 4 g (decagram 3 C) is enclosed between its inner circle (radius r 10 ) and outer circle (radius R10 ). (Middle 4
circles dashed.) Star polygon f 10 5 g (decagram D) is enclosed by its outer circle (radius R10 ). All points of self-intersection of decagram A lie on the middle circle (radius 10 ). 2 Self-intersection of decagram A produces a miniature decagon, enclosed between inner circle (radius r 10 ) and middle circle (radius 10 ). 2
2
All points of self-intersection of decagram B lie on the 1st and 2nd middle circle (radius 10 #1 and 10 #2 ). Self-intersection of decagram B produces a miniature decagon, en3
3
closed between inner circle (radius r 10 ) and 1st middle circle (radius 3
10 3 #1
); moreover,
it produces a miniature decagram A, enclosed between inner circle (radius r 10 ) and 2nd 3 middle circle (radius 10 #2 ). 3
All points of self-intersection of decagram C lie on the 1st , 2nd and 3rd middle circle (radius 10 #1 , 10 #2 and 10 #3 ). Self-intersection of decagram C produces a miniature 4
4
4
decagon, enclosed between inner circle (radius r 10 ) and 1st middle circle (radius 10 #1 ); 4 4 moreover, it produces a miniature decagram A, enclosed between inner circle (radius r 10 ) and 2nd middle circle (radius
4
10 4 #2
); furthermore, it produces a miniature decagram B,
enclosed between inner circle (radius r 10 ) and 3rd middle circle (radius 10 #3 ). 4 4 All points of self-intersection of decagram D lie in the center. Inner circle (radius r 10 = 0) and all middle circles (radii 10 #1;2;3;4 = 0) collapsed to zero in the center. 5
5
I f10g A convex polygon ofp order 10 (decagon) has half-side s10 and Dido value p 1 1 f (10) = 10 cot( 10 ) = 10 5 + 2 5 (t 0:3078), a constructible algebraic number (containing only rationals and/or square roots). Perimeter of decagon: P10 = 20s10 . 1 Half-side of decagon: s10 = 20 P10 . Stroke distance of decagon: d10 = 2s10 . Draw length of decagon: D10 = 10d10 = 20s10 = P10 . Inner radius of decagon: p p r10 = f (10) 12 P10 = cot( 10 ) s10 = 5 + 2 5s10 (t 3:0777 s10 ): Outer radius of decagon: q p R10 = f (10)2 + 1012 21 P10 = sin(1 ) s10 = (1 + 5)s10 (t 3:2361 s10 ): 10 Area of decagon: p p A10 = f (10) ( 21 P10 )2 = 10 cot( 10 )s210 = 10 5 + 2 5 s210 (t 30:777 s210 ). There are four star polygons (decagram A, B, C, D): 10 I{ 10 of order 10 2 } A star polygon p 2 (decagram A) has half-side s 2 and Dido value p 1 f ( 10 25 + 10 5 (t 0:2753), 2 ) = f (5) = 25 a constructible algebraic number (containing only rationals and/or square roots). 5 – Interpretation of Schlä‡i factoring f 10 2 g =2 f 1 g: decagram A consists of 2 pentagons (mutually rotated through 2 =10 = =5); pentagons and rotation angle are constructible, hence decagram A is constructible. Perimeter of decagram A: q p p P10 1 1 10 = P 10 = 20s 1 + 50 10 5P10 (t 1:0515P10 ): [10f (10)]2 P10 = cos( ) = 5 2 2 10
32
PETER KAHLIG
Half-side of decagram A: q p p s10 1 1 1 10 = s 10 = P 1 + = 50 10 5s10 (t 1:0515s10 ). 2 s10 = cos( 20 [10f (10)] ) 5 2 2 10 Stroke distance of decagram A: p p sin( 210 ) 1 d 10 = d = 2 cos( )d = 10 + 2 5d10 (t 1:0921d10 ): 10 10 sin( 10 ) 10 2 2 Draw length of decagram A: p p sin( 210 ) 1 10 = D 10 = 10d D = 2 cos( )D = 10 + 2 5D10 (t 1:0921D10 ): 10 10 sin( 10 ) 10 2 2 2 Inner radius of decagram A: cos( 210 ) f (10) 1 2 q = f ( 10 r 10 2 ) f (10)2 + 1 2 P10 = sin( 10 ) s10 = 2 cot( 10 ) cos( 10 )s10 2 102 p = 12 (3 + 5)s10 (t 2:6180s10 ) f ( 10 )f (10)
2
f (10) 1 2 1 q 2 10 10 10 = f ( 10 2 ) f (10)2 + 12 2P 2 = 2 P 2 = cos( 10 ) cot( 10 )s 2 22 2 f ( 10 10 2 ) + 102 p p = 14 50 + 22 5s 10 (t 2:4899s 10 ): 2 2 (One and only)q middle radius of decagram A: 1 1 1 2 10 10 = 1 + [10f (10)] = f ( 10 r 10 2 r 10 = cos( 2 ) 2 P10 = 2 cot( 10 )s10 2 2 #1 2 2 10 ) p p = 25 25 + 10 5s10 (t 2:7528s10 ): (Decagram A and its mother decagon have the same outer radius R10 :) Area of decagram A: p p 10 1 2 2 2 = f ( A 10 ) ( P ) = 20 cot( ) s = 4 25 + 10 5s210 (t 27:528s210 ) 10 10 2 2 10 2 f ( 10 )
2 = f (10) A10 = (1 102 f1(10)2 )A10 = [1 tan( 10 )2 ]A10 (t 0:8944A10 ) 1 1 10 1 2 10 10 = r 10 2P 2 = 2 P10 = f ( 2 )( 2 P10 ) : 2 2 10 10 I f 3 g A star polygon of order 3 (decagram B) has half-side s 10 and Dido value 3 p p 3 3 3 f ( 10 5 2 5 (t 0:2180), 3 ) = 10 cot( 10 ) = 10 a constructible algebraic number (containing only rationals and/or square roots). Perimeter of decagram B: q p 1 1 10 = P 10 = 20s 1 + 1)P10 (t 1:2361P10 ): [5f (5)]2 P10 = cos( 5 ) P10 = ( 5 3 3 Half-side of decagram B: q p 1 1 1 10 = s 10 P 1 + 5 1)s10 (t 1:2361s10 ): = 2 s10 = cos( ) s10 = ( 20 [5f (5)] 3 3 5 Stroke distance of decagram B: p p sin( 3 ) = sin( 10 ) d10 = 1 pd510 1 ( 14 5 + 14 ) = 12 (3 + 5)d10 (t 2: 618d10 ): d 10 3 10 4 4 Draw length of decagram B: p sin( 3 ) D 10 = 10d 10 = sin( 10 ) D10 = 12 (3 + 5)D10 (t 2: 618D10 ): 3 3 10 Inner radiusr of decagram B: r
r 10 = f ( 10 3 ) 3
f (10)2 + 1012
32 2 f ( 10 3 ) + 102
1 10 2 P10 =f ( 3 )
f (10)2 + 1012 2
3 2 f ( 10 3 ) + 102
cos ( 3 )
10s10 = sin ( 10 ) s10 = 10
sin ( 210 ) sin ( 10 ) s10
p p p p p = 14 (1 + 5) 10 2 5s10 = 12 10 + 2 5s10 (t 1: 902 1s10 ). 1st middle qradius of decagram B: p p 1 1 1 10 10 = 10 = = 1 + r r 50 10 5r 10 (t 1:0515r 10 ): 2 #1 [10f (10)] cos( ) 3 5 3 3 3 3 10
2nd middle q radius of decagram B: p 1 1 10 10 = ( = 1 + r 5 10 10 2 r 10 = 2 #2 [ f ( )] cos( ) 3 3 3 2
2
10
1)r 10 (t 1:2361r 10 ): 3 3
(Decagram B and its mother decagon have the same outer radius R10 .) Area of decagram B:
REGULAR POLYGONS OF RATIONAL ORDER
33
r f (10)2 + 1012 2 1 = f ( 10 ) 32 3 2 P10 2 f ( 10 3 ) + 102 p p 2 cos ( 3 ) 2 2 = 2 650 290 5r10 = 10 sin ( 10 ) [ cos ( )]210cos ( 2 ) r10 (t 2:4822 r10 ) 10 10 p p 1 2 2 2 = 10 sin ( 10 ) cos ( 2 ) cos ( 3 ) r 10 = 4 25 10 5r 10 (t 6:4984 r 10 ):
1 A 10 = r 10 P 10 = 3 3 2 3
10 3 #2
1 2 P10
10
10
I f 10 4 g A star polygon of order f ( 10 4 )
=
f ( 52 )
=
2 5
cot( 25
)
3
3
3
10 C) has half-side s 10 and Dido 4 (decagram 4 p p 2p 1 2 = 25 25 10 5 = 5 p (t 0:1300), 5+2 5
value
a constructible algebraic number (containing only rationals and/or square roots). 5 – Interpretation of Schlä‡i factoring f 10 4 g =2 f 2 g: decagram C consists of 2 pentagrams (mutually rotated through 2 =10 = =5); pentagrams and rotation angle are constructible, hence decagram C is constructible. Perimeter of decagram q C: p p P10 1 10 2 32 1 10 = = 20s f ( ) + = 50 + 10 5P10 (t 1:7013P10 ): P 10 10 3 2 P10 = 3 10 5 f ( ) cos( ) 4 4 3 10 Half-side of decagram q C: p p 1 1 10 2 32 1 1 10 = = s 10 P f ( ) + s = s = 50 + 10 5s10 (t 1:7013s10 ): 10 3 2 10 10 20 3 10 5 f( 3 ) cos( 10 ) 4 4 Stroke distance of p decagram C: p sin ( 410 ) = sin ( ) d10 = 5 + 2 5d10 (t 3:0777d10 ). d 10 4 10 Draw length of decagram C:p p sin ( 4 ) = 10d 10 = sin ( 10 ) D10 = 5 + 2 5D10 (t 3:0777D10 ). D 10 4 4 10 Inner radius r of decagram C: r r 10 =f ( 10 4 ) 4
f (10)2 + 1012 1 10 42 2 P10 =f ( 4 ) 2 f ( 10 4 ) + 102
cos ( 4 ) = cos( 10 ) r10 10
=
1 5
p
25
p
f (10)2 + 1012 2
4 2 f ( 10 4 ) + 102
cos ( 4 )
10s10 = sin ( 10 ) s10 = s10 10
10 5r10 (t 0:3249r10 ): q r 10 cos( 4 ) 1 4 = 1st middle radius of decagram C: 10 1 + [10f (10)] = [cos( 10)]2 r10 ; 2 r 10 = cos( 4 #1 4 10 ) 10 q r 10 cos( 410 ) 1 nd 4 = 1 + [ 10 f ( 10 )]2 r 10 r10 ; 2 middle radius of decagram C: 10 = cos( 2 ) = cos( 2 ) cos( 4 #2 4 2 2 10 10 10 ) q r 10 4 cos( ) 10 3rd middle radius of decagram C: 10 = 1 + [ 10 f (110 )]2 r 10 = cos(43 ) = cos( 3 ) cos( r : ) 10 4 #3 4 3
3
10
10
10
(Decagram C and its mother decagon have the same outer radius R10 .) Area of decagram C: r p p f (10)2 + 1012 2 10 cos( 410 ) 10 1 A 10 =f ( 4 ) P10 = sin( ) cos( s210 = 2 50 + 10 5s210 (t 17:013s210 ) 3 10 2 42 2 ) 4 f ( 4 ) + 102 10 10 p p 2 cos( 410 ) 1 10 2 = 4 sin( ) cos( 3 ) d10 = 2 50 + 10 5d10 (t 4:2533d210 ) 10 10 p p sin( 10 ) cos( 410 ) 2 = 10 = 12 50 22 5d210 (t 0:4490d210 ) 4 [sin( 410 )]2 cos( 310 ) d 10 4 4 4 p cos ( 410 ) 1 = cos ( ) cos A = (1 5)A (t 0:5528A ): 3 10 10 10 5 ( ) 10
10
I f 10 5 g A star polygon of order s 10 and Dido value 5 10 f ( 5 ) = f (2) = 0;
10 5
(decagram D, ten-spike asterisk) has half-side
a constructible algebraic number (containing only rationals and/or square roots), so decagram D is constructible (by ruler and compass). – Interpretation of Schlä‡i factoring 2 f 10 5 g =5 f 1 g: decagram D is a compound star polygon, consisting of 5 digons (mutually rotated through 2 =10 = =5); digons and rotation angle are constructible, hence decagram D is constructible.
34
PETER KAHLIG
Perimeter of decagram D: q
p P10 = ( 5+1)P 10 (t 3:2361P 10 ): p 1 Half-side of decagram D: s 10 = 20 P 10 = sin(1 ) s10 = ( 5 + 1)s10 : 5 5 10 p Stroke distance of decagram D: d 10 = sin (1 ) d10 = ( 5 + 1)d10 = 2s 10 : 5 2
P 10 = 20s 10 = [10f (10)] +1P10 = sin (1 5 5
10 )
D 10 = 10d 10 = sin ( 5 5
10 )
5
10
Draw length of decagram D: p 1
D10 = ( 5 + 1)D10
p p = 20s 10 = 20( 5 + 1)s10 = P 10 = ( 5 + 1)P10 : 5
5
Decagram D (ten-spike asterisk) is a collapsed star polygon; inner radius, all middle radii, and area of decagram D vanish: r 10 = 0; 10 #1;2;3;4 = 0; A 10 = 0: 5 5 5 (Decagram D and its mother decagon have the same outer radius R10 .) 2 Pythagorean relationships: s210 = R10
2 r10 ; s210 = s210 + (r10 2
10 2
)2 ;
2 10 10 s210 = s210 + (r10 )2 ; s210 = s210 + (r10 )2 ; s210 = s210 + r10 : 3 #2 4 #3 3 4 5 Expressions in terms of outer radius R10 f10g (decagon): p 1 1 q R10 = 10 sin( 10 ) R10 = 52 ( 5 1)R10 (t 3:0902R10 ); 2 P10 = 10s10 = 1 2 f (10) + 102 p 1 1 s10 = 10 ( 2 P10 ) = p 2 1 2 R10 = sin( 10 ) R10 = 14 ( 5 1)R10 (t 0:3090R10 ); 10 f (10) +1 p p f (10) 1 q r10 = 10 + 2 5R10 (t 0:9511 R10 ); R = cos( )R = 10 10 10 4 1 f (10)2 + 102 p p 2 f (10) 10 2 5 2 2 2 A10 = r10 21 P10 = f (10) 10 2 5R10 (t 2:9389R10 ): 2 + 1 R10 = 2 sin( 10 )R10 = 4
f 10 2 g (decagram A):
1 10 2P 2
p
p 1 R10 2 P10 10 5R10 (t 3:2492R10 ); cos( 10 ) = f (10) = 10 tan( 10 )R10 = 2 25 p p R10 1 1 1 10 ) = ( P = tan( )R = 25 10 5R10 (t 0:3249R10 ); 10 10 2 10f (10) 10 5 2 p f (10) f (5) q R10 = cos( 5 )R10 = 41 (1+ 5)R10 (t 0:8090R10 ); f (5) f (10) 2 + 1 R10 = 2+ 1 2 f (5) 10
= 10s 10 = 2
s 10 = 2 r 10 = 2 10 2
102
=
A 10 = 2 =
r 10
r10 R10 = 2
52
f (5) q f (10)2 + 1012
f (5) f (10) A10 = (1 f (5)( 12 P10 )2 =
R10 =
p p 50 + 10 5R10 (t 0:8507R10 ); p = (1 [tan( 10 )]2 )A10 = 25 5A10 (t 0:8944A10 )
cos( 210 ) cos( 10 ) R10
1 [10f (10)]2 )A10 1 10 10 2 P10 = r 2 2
1 10 2P 2
=
1 10
= r 10 2
1 f (10) R10
=
f (5) f (10)2 + 1012
2 R10
p p 2 2 2 = 10 tan( 10 ) cos( 210 )R10 = 21 50 10 5R10 (t 2:6287R10 ): 10 f 3 g (decagram B): r q 22 2 f ( 10 2 ) + 102 1 1 1 2+ 1 R 10 = 10s 10 = P f (10) = 10 10 2 10 2 + 1 R10 2 10 f ( )f (10) f ( ) f (10) 3 3 2 2 102 p 10 sin( ) = cos( 2 10) R10 = 5(3 5)R10 (t 3:8197R10 ); 10 r q 22 2 f ( 10 2 ) + 102 1 1 1 1 1 2 s 10 = 10 ( 2 P 10 ) = 10f ( 10 )f (10) f (10) + 102 R10 = 10f ( 10 ) f (10) 2 + 1 R10 3 3 2 2 102 p sin( ) = cos( 210 ) R10 = 12 (3 5)R10 (t 0:3820R10 ); 10 p 3 1 = 2 sin( d 10 )R = (1 + 5)R10 (t 1:6180R10 ); 10 10 2 3 p 3 D 10 = 10d 10 = 20 sin( 10 )R10 = 5(1 + 5)R10 (t 16:180R10 ); 3 3 p p f ( 10 3 1 3 ) q r 10 = R = cos( )R = 10 2 5R10 (t 0:5878R10 ); 10 10 2 10 4 10 3 3 f ( 3 )2 + 102 r 10 cos( 3 ) 3 10 = r10 R10 = cos( 10 ) R10 3 #1 10
p = 12 ( 5
1)R10 (t 0:6180R10 );
REGULAR POLYGONS OF RATIONAL ORDER
10 3 #2
=
r 10 3
r 10
cos( 310 ) R cos( 210 ) 10
R10 =
2
=
p 5
35
p 2 5R10 (t 0:7265R10 );
cos( 3 )
cos( 3 )
1 1 2 10 10 10 10 = r 10 A 10 2P 3 = 2 P10 = cos( 210 ) R10 10s10 = 10 sin( 10 ) cos( 210 ) R10 3 3 3 #2 p p 2 2 = 25 50 22 5R10 (t 2:2451R10 ):
f 10 C): 4 g (decagram r
1 10 2P 4
=
2
3 2 f ( 10 3 ) + 102
1 f ( 10 3 )
f (10)2 + 1012
r
sin (
)
R10 = 10 cos( 310 ) R10 = 10
p 50
p 10 5R10 (t 5:2573R10 ),
p p sin ( ) 1 R10 = cos( 310 ) R10 = 10 50 10 5R10 (t 0:5257 R10 ), 10 p p 2 4 1 q d 10 = R = 2 sin ( )R = 10 + 2 5R10 (t 1:9021R10 ), 10 2 10 10 2 10 10 2 4 f ( ) +1 4 42 p p 20 10 = q = 10d D 10 R10 = 20 sin ( 410 )R10 = 5 10 + 2 5R10 (t 19:021R10 ), 102 10 4 4 1 1 10 s 10 = 10 2P 4 = 4
1 10f ( 10 3 )
f(
42
4
2
3 2 f ( 10 3 ) + 102
f (10)2 + 1012
)2 +1
p
f ( 10 4 )
q R10 = cos( 410 )R10 = 41 ( 5 1)R10 (t 0:3090R10 ); 42 2 f ( 10 4 ) + 102 p r 10 p cos( 4 ) 4 10 R10 = cos( 10 ) R10 = 15 25 10 5R10 (t 0:3249R10 ); = r10 4 #1 10 r 10 p cos( 410 ) 1 4 10 = 5 R10 (t 0:3820R10 ); #2 r 10 R10 = cos( 210 ) R10 = 2 3 4 2 p r 10 p 4 cos( ) 1 4 10 = r 10 R10 = cos( 310 ) R10 = 10 50 10 5R10 (t 0:5257R10 ); 4 #3
r 10 = 4
10
3
= r 10 A 10 4 4
1 10 2P 4 =
r
1 10 2 P10 =f ( 4 )
10 4 #3
cos ( 4 )
2 = 10 sin ( 10 ) cos ( 310 ) R10 = 10
p 25
f (10)2 + 1012 2
4 2 f ( 10 4 ) + 102
2 1 2 P10
p 2 2 10 5R10 (t 1:6246R10 ):
f 10 5 g (decagram D, ten-spike asterisk): 1 10 s 10 = R10 ; d 10 = 2s 10 = 2R10 ; D 10 = 10d 10 = 20R10 ; 2 P 5 = 10R10 ; 5 5 5 5 5 1 1 10 10 = r 10 10 10 = = 0; = 0; A P P = 0: r 10 10 5 5 #1;2;3;4 5 5 2 5 5 #4 2 4. Some in…nite products for The connection of regular polygons with in…nite products leads to a multitude of in…nite product representations for (cf. [9] where representations of integer order 20 are listed). Here we add …rst instances of rational order. In each case, we have a representation of as a product of an algebraic scale factor (of magnitude of ) and a transcendental correction factor (of magnitude 1, an in…nite product). [Our constructible cases have simple algebraic scale factors (containing only rationals and/or square roots).] Using Proposition 3 we get the following. Ratio of middle circle perimeter 2 also, ratio of middle circle area
Q1
2
5 2
5 2
to polygon perimeter P5 is 2
to star polygon area A 5 is
2 ( 1)j+1
p
p
2
2 2
5 2
2
= f (15 ) 2
j=1 (1
( 54 j)
= 52
1 (15=4) 1 5 2
2
2
1 (25=4) ) 5+2 5 2 1 (15=2) 2 ::: Ratio of inner circle perimeter 2 r 3 to polygon perimeter P3 is 2 r3 =P 3 = f (3), also, ratio of inner circle area r 23 to polygon area A3 is r 23 =A3 = f (3), also, ratio of middle circle perimeter 2 6 to polygon perimeter P6 is 2 6 =P 6 = f ( 26 ) = f (3), 2 2 also, ratio of middle circle area 26 to star polygon area A 6 is 26 =A 6 = f ( 26 ) = f (3) ) 2 2 2 2 p 2 Q1 j+1 2 1 (9=2) 2 1 (15=2) 2 1 order 3: = f (3) ( 32 j) )( 1) = 3 3 1 1(3=2) ::: 2 j=1 (1 3 1 6 2 1 9 2 5 order 2 :
1 (5=4) 1 (5=2)
=P 5 = f ( 52 ), f ( 52 ) ) 5 =A 5 =
2
36
PETER KAHLIG
Ratio of inner circle perimeter 2 r 4 to polygon perimeter P4 is 2 r4 =P 4 = f (4), also, ratio of inner circle area r 24 to polygon area A4 is r 24 =A4 = f (4), also, ratio of middle circle perimeter 2 8 to polygon perimeter P8 is 2 8 =P 8 = f ( 28 ) = f (4), also, ratio of middle circle area
2
2
Q1
2
to star polygon area A 8 is
8 2
8 2
2
2
j+1
2
2
2
=A 82 = f ( 28 ) = f (4) ) 2
(2j) )( 1) = 4 11 24 2 11 68 2 11 10 12 2 ::: Ratio of inner circle perimeter 2 r 5 to polygon perimeter P5 is 2 r5 =P 5 = f (5), also, ratio of inner circle area r 25 to polygon area A5 is r 25 =A5 = f (5), also, ratio of middle circle perimeter 2 10 to polygon perimeter P10 is 2 10 =P 10 = f ( 10 2 ) = f (5), 2 2 also, ratio of middle circle area 210 to star polygon area A 10 is 210 =A 10 = f ( 10 ) = f (5) ) 2 2 2 2 2 p p 2 2 2 Q j+1 1 2 1 (15=2) 1 (25=2) 1 ( 52 j) )( 1) = 5 5 2 5 1 1(5=2) order 5: = f (5) j=1 (1 5 2 1 10 2 1 15 2 ::: Ratio of inner circle perimeter 2 r 10 to polygon perimeter P10 is 2 r10 =P 10 = f (10), also, ratio of inner circle area r 210 to polygon area A10 is r 210 =A10 = f (10) ) p p Q1 2 ( 1)j+1 5 2 1 15 2 1 25 2 1 25 10 5 11 10 (1 (5j) ) order 10: = f (10) = 2 2 1 20 2 1 30 2 ::: j=1 1 [With the Dido equation we calculate f (20) from f (10) to obtain the scale factor f (20) :] Ratio of inner circle perimeter 2 r 20 to polygon perimeter P20 is 2 r20 =P 20 = f (20), also, ratio of inner circle area r 220 to polygon area A20 is r 220 =A20 = f (20) ) p p p Q1 j+1 2 1 10 2 1 30 2 1 50 2 order 20: = f (20) (10j) )( 1) = 20(1+ 5 5 + 2 5) 11 20 2 1 40 2 1 60 2 ::: j=1 (1 order
4:
1 = f (4)
j=1 (1
Numerical values of scale factor
1/f
in dependence on order:
8 10 3 4 5 10 20 1 2 2 —————————————————————————————————— 1/f 7:69 5:20 5:20 4:00 4:00 3:63 3:63 3:25 3:17
order
6 2
5 2
Remark 8. When the scale factor 1=f is of su¢ ciently high integer or rational order (so that 1=f itself is already “close” to ) then only few terms of the in…nite product are needed to calculate an approximation to of su¢ cient accuracy. Remark 9. Let us reconsider the “surprising result” from Remark 1 more n from Proposition 2: the area A n of deeply, starting with an area formula for A m m n a regular polygon of order m is n = Am
f (n)2 + n12 f (n) + n2 fm( n )
1 Pn 2
2
:
m
Now the question arises: for what values of m does the last equation reduce to 2 n 1 n = f( ? [This holds, as we have observed in Remark 1, for m = 1 Am m) 2 Pn (convex polygons), but surprisingly also for m = 2 (“class A” star polygons).] The answer is given by the following. n Proposition 7. The two area formulas A m =
f (n)2 + n12 f (n)+ n2 fm ( n )
2 1 2 Pn
and
m
2
1 n n = f( are equivalent in two cases only, namely m = 1 and m = 2. Am m) 2 Pn In all cases m 3, only the …rst formula is valid. (The second formula holds not only for m = 1, but surprisingly for m = 2 too.) – The same applies to related n formulas. For instance, expressing area by the last middle circle radius m #(m 1) n = in the form A m
n m #(m
1)
1 2 Pn
, we have either
n m #(m
1)
=
f (n)2 + n12 f (n)+ n2 fm ( n ) m
1 2 Pn
REGULAR POLYGONS OF RATIONAL ORDER
37
n 1 n or (on trial) m #(m 1) = f ( m ) 2 Pn , being equivalent only in the two cases m = 1 and m = 2. Another consequence (for m = 2) is A n2 =An = f ( n2 )=f (n) = n2 =rn .
Sketch of proof. Obviously the two formulas are equivalent if f (n)2 +
f (n)2 + n12 f (n)+ n2 fm ( n )
1
n = f(m )
m
n2 (an implicit equation to determine m). Rewrite is as f (n)f ( n )+ m = 1, yielding the m n2 “test equation” m 1 n : f ( ) = f (n) m n2 f (n) We are looking for positive integer values of the solution m, e.g. m = 1; 2; 3; :::, so let us test one after the other: case m = 1 gives f ( n1 ) = f (n) 0, which is the trivial identity f (n) = f (n), thus the test equation is satis…ed; case m = 2 gives f ( n2 ) = f (n) n2 f1(n) , which is (from Remark 2) a familiar identity, thus the test equation is satis…ed; cases m = 3; 4; 5; ::: cannot ful…ll the test equation (hence there are no other solutions besides m = 1 and m = 2). In detail: from the multiplication formula for the cotangent function we can produce a “division formula” for the Dido function f (n) = n1 cot( n ), a quite general functional equation for f , PK m jmk ( 1)k 2k [nf (n)]m 2k n n f ( ) = PK k=0 ; where K = m k m m 2 [nf (n)]m 2k 1 k=0 ( 1)
2k+1
(
m 2
greatest integer
m 2 ).
This gives for m = 1 the trivial identity nf (n) =
i.e. f (n) = f (n), and for m = 2 we get
n n 2 f( 2 )
=
[nf (n)] 1 2nf (n) ,
1
n f ( ) = f (n) 2
2
n2 f (n)
nf (n) 1 ,
i.e.
;
(which is in fact our familiar identity, complying with the test equation). For m = 3 (n)]3 3nf (n) we obtain n3 f ( n3 ) = [nf3[nf (n)]2 1 , i.e. [nf (n)]2 n f ( ) = 3f (n) 3 3[nf (n)]2 and for m = 4 we get
n n 4 f( 4 )
=
3 1
f (n)
1 n2 f (n)
[nf (n)]4 6[nf (n)]2 +1 4[nf (n)]3 4nf (n) ,
n 1 [nf (n)]4 6[nf (n)]2 + 1 f( ) = 2 4 n f (n) [nf (n)]2 1
f (n)
1 5[nf (n)]2 + 1 ; n2 f (n) 3[nf (n)]2 1 i.e. 1 n2 f (n)
4f (n) ; [nf (n)]2 1
showing that the more complicated structure of the last two division formulas cannot comply with the simple structure of the test equation. (Similarly for m > 4.) Remark 10. In this presentation we chose to have equal outer radius of any n n , so we may (convex) polygon fng and related star polygons f m g, i.e. Rn = R m n ) even in the case of star polygons, and we can simply write Rn (instead of R m n interpret fng as (convex) “mother” polygon, containing related star polygons f m g. –Alternatively, one could choose other elements to be equal for reference, e.g. inner n , now paying attention to the fact that R n (producing equivradius rn = r m n 6= R m alent graphics, but with …gure size increasing with m). [It is clear that this option n = 0.] would not work for collapsed star polygons m = n=2 (asterisks) due to r m n Another interesting choice for illustrative purposes could be equal half-side sn = s m
38
PETER KAHLIG
n and r n ), or equal stroke distance d n (then, of course, with Rn 6= R m n 6= r m n = dm etc.
Remark 11 (Compound star polygons are genuine polygons, due to existence of continuous draw line with constant length.). Some scientists are reluctant to accept compound star polygons as genuine polygons by arguing that “they cannot be drawn by one line without interruption”, like hexagram A with its two separate triangles. However, this argumentation is incorrect; in fact, any polygon can be drawn by one continuous line simply by joining suitable paths. For instance hexagram A: …rst draw the perimeter, then join it continuously with the inner hexagon. There are many other options of line drawing by continuous switching of line sections between perimeter and inner hexagon at any point of self-intersection; the draw length is the same in each case. The false argumentation stems from focussing merely on the obvious “direct” strokes along each single triangle from corner to corner, requiring a “jump” in going from one triangle to the other. (Interestingly, the draw length is still the same, since a “jump” does not contribute to the length of the draw line, yet the draw line has the “blemish” of discontinuity. By calling it an “equivalent discontinuous” draw line with equal draw length, the dispute could be settled.) Of course, the fact that any polygon can be drawn by one continuous line simply by joining suitable paths, also holds for the popular drawing of a pentagon: luckily, the obvious “direct” strokes from corner to corner readily permit a continuous line drawing, but there are many other options of performing the continuous line drawing by changing the path at any point of self-intersection (always resulting in a continuous draw line with the same constant length). Interpreting a pentagon geometrically as consisting of …ve joining isosceles triangles, the draw length of the pentagon may be calculated by adding the perimeters of the …ve “separate” isosceles triangles (again an instance of an “equivalent discontinuous” draw line). Asterisks are another example for the fact that any polygon can be drawn by one continuous line simply by joining suitable paths: being a compound of digons, they can be drawn by one continuous line, obviously in di¤ erent ways, by drawing in di¤ erent order the spikes emerging from the center. Remark 12. The Dido function f (n) simpli…es calculation of inner radius, middle circle radius, area of convex polygons fng and of class A star polygons f n2 g by taking the argument of f directly from the Schlä‡i symbol: 1 rn = f (n) 12 Pn ; An = rn 21 Pn = f (n) ( 12 Pn )2 = f (n) rn2 ; n 1 n = f( ) A n2 = n2 12 Pn = f ( n2 ) ( 12 Pn )2 = f (1n ) 2n : 2 2 Pn ; 2 2 2 With R := limn!1 rn = limn!1 n2 , circle perimeter P and circle area A become 1 1 P := limn!1 Pn = 2 lim( f (n) rn ) = 2 lim( f (1n ) n2 ) = 2R lim f (n) = 2 R; 2
1 1 A := limn!1 An = lim( f (n) rn2 ) = lim A n2 = lim( f (1n ) 2n ) = R2 lim f (n) = R2 : 2 2 From Proposition 1 q we know that the outer radius Rn of a convex poly-
gon of order n is Rn = f (n) f (n)2 + n12
f (n)2 +
1 1 n2 2 Pn ,
and the area is An = f (n)( 12 Pn )2 =
Rn2 .
(a) If Rn = R (…xed for all n) as preferred in this presentation, we get in the limit n ! 1 the circle perimeter 2 R, 1 limn!1 Pn = 2R limn!1 q 12 1 = 2R limn!1 f (n) = 2 R; f (n) + n2
and the circle area R2 ,
REGULAR POLYGONS OF RATIONAL ORDER
limn!1 An = R2 limn!1
f (n) f (n)2 + n12
= R2 limn!1
1 f (n)
39
= R2 :
(b) If Pn = P (…xed for all n) as in isoperimetric problems, we get in the limit n ! 1 the circle radius 2P , q limn!1 Rn = 12 P limn!1 f (n)2 + n12 = 12 P limn!1 f (n) = 2P ; 2
and the circle area P4 , limn!1 An = ( 21 P )2 limn!1 f (n) =
P2 4 :
Remark 13. The isoperimetric inequality may be formulated for (convex) polygons via Dido’s function of integer argument f (n); n 2 (cf. [7, 8, 9]) as 1 (where equality holds for n ! 1), or equivalently f (n) rn f (n) = 2Pnrn 1, 1 2 Pn interpretable as ratio of inner circle perimeter (2 rn ) to polygon perimeter (Pn ), r 1P r2 r2 1, with consequence f (n) = 1rPn = 1 P n r = Ann = (n1 P2 )n2 = ( 1APn )2 2
n
2
n
n
2
n
n
2
interpretable as ratio of inner circle area ( rn2 ) to polygon area (An ). There are similar relations for Class A star polygons via Dido’s function of 1 half-integer argument f ( n2 ); n 4, namely f ( n2 ) (where equality holds for n ! 1), or equivalently 2
n
n
2 f ( n2 ) = Pn2 1, 1 2 Pn interpretable as ratio of (star polygon’s) middle circle perimeter (2 polygon perimeter (Pn ),
with consequence f ( n2 ) =
n 2 1 2 Pn
=
2 n 2 1 n 2 Pn 2
interpretable as ratio of middle circle area (
= 2
n 2
2 n 2 An 2
=
n 1 Pn 2 2 ( 12 Pn )2
=
n 2
) to (mother) An 2
( 12 Pn )2
1,
) to star polygon area (A n2 ).
Remark 14. There are many cases where polygons come into play. Road signs (e.g. octagon for STOP), cross section of jars or buildings (e.g. Pentagon), national emblems (small pentagrams as stars in many ‡ags, also heptagrams e.g. in the Australian ‡ag), models of re‡ections between circular walls, work of arts and crafts (stitchwork, badges, medals, threads or wires around nail patterns), plants (e.g. Hoya carnosa with two-fold pentagram in each blossom). Remark 15. The original version (2015) of this article is a book chapter: Kahlig P., Note on regular polygons of rational order. In: Fripertinger H., Prager W., Schwaiger J., and Tomaschek J. (eds.), Ludwig Reich 75, Grazer Math. Ber., ISSN 1016-7692, Bericht Nr. 363 (2015), 106-131. New in the present corrected version (2018): aspects connected with area of regular ( convex and star) polygons of any order. Again, Dido functional equation n and Schlä‡i symbol f m g are helpful in calculation and interpretation of properties n of general star polygons f m g. The exceptional role of “class A” star polygons n f 2 g is emphasized, and some surprising simpli…cations are explained via functional equations. Asterisks are treated as collapsed star polygons. Compound star polygons are justi…ed as genuine polygons. Dedicated to Professor Ludwig Reich on the occasion of his 75th birthday.
40
PETER KAHLIG
QUINDECIM
AD
LUSTRIS
MULTOS
PERACTIS:
ANNOS
SERENOS!
Acknowledgement. Stimulating conversation with L.R. is appreciated. Support of M.K. and J.S. is gratefully acknowledged. The referees’ constructive comments helped to improve the paper.
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[email protected] 