REGULAR PENTAGON

* Area of Pentagon =r2n*tan(180/n) or  [R2n*sin(360/n)]/ 2 or a2*n /(4*tan 180/n)

*Δ GTU & Δ GTA are congruent. Δ GAC & Δ GTU are congruent. Δ GAC & Δ GAT are congruent. K is the center.

* If we draw the diagonals, another pentagon U1C1A1G1T1 with side a1 emerges. If you put the length of this side in stead of AG, another pentagon with side a2 emerges and so on so forth symmetric around the center K.

* There are 10 triangles which are sum of 2 triangles:

1+2=11; 2+3=12; 3+4=13; 4+5=14; 5+6=15; 6+7=16; 7+8=17; 8+9=18; 9+10=19;10+1=20;

* There are 5 triangles which are sum of 3 triangles:

2+3+4=21(GTU); 4+5+6=22(TUC); 6+7+8=23(UCA);8+9+10=24(CAG);10+1+2=25(AGT);

The combinations above are cyclical in nature.

* There are 5 triangles which are sum of 2 triangles + X

1+X+7=26; 1+X+5=27;  3+X+9=28; 3+X+7=29;   5+X+9=30;

* There are 5 triangles which are sum of 4 triangles +X

1+X+5+6+7=31(CGU) ; 3+X+7+8+9=32(ATC);  5+X+1+9+10=33(UAG) ; 7+X+1+2+3=34(CGT) ;  9+X+3+4+5=35(ATU);

The combinations above are non-cyclical in nature.

* In total, there are 35 no. of triangles.

* There are 5 parallelograms :

1+2+3+4+5+X ; 3+4+5+6+7+X; 5+6+7+8+9+X; 7+8+9+10+1+X; 9+10+1+2+3+X ;

* For a crystal, it is impossible to find a dihedral symmetry for n=5,6 & n>6.

* We take r(k) as rotation ( conventionally anti-clockwise) to make the pentagon invariant. Hence value of r(k) = k2π/n= 2πk/n where k=0,1,2,3,....(n-1). r(0)=0,r(1)=2π/n,r(2)=4π/n,r(3)=6π/n, r(4)=8π/n

* Similarly, We take s(k) as rotation of axis of reflection to make pentagon variant under reflection. Hence s(k)= kπ/n= πk/n where k=0,1,2,3,....(n-1). s(0)=0,s(1)=π/n,s(2)=2π/n,s(3)=3π/n, s(4)=4π/n

. This means that reflection axis are spaced 36 degree to each other.

Dn={r(0),r(1),r(2),r(3),r(4),s(0),s(1),s(2),s(3),s(4)} The set having 10 elements i.e order=10. The elements of Dn   can be thought of as linear transformation of the plane(2-D) leaving the pentagon invariant. Thus each element of Dn    can be represented as 2x2 matrix with group operation                 ( composition ) as matrix multiplication.

Dn ={T(0),T(1),T(2),T(3),T(4),S(0),S(1),S(2),S(3),S(4)}

Cn ={T(0),T(1),T(2),T(3),T(4)} is a subset of  Dn  & forms a cyclical group with respect to composition of matrix multiplication. T(0) is the identity element, T(1) is the Generator & T2(1)= T(2),   T3(1)= T(3),  T4(1)= T(4),  T5(0)= T(0). This is an Abelian group since the elements obey the commutation law with respect to composition of matrix multiplication. T(1)-1 = T(4) ; T(2)-1 = T(3) and T(0)-1 = T(0) ;

Symmetric transformations that form the Dihedral group consist of rotation & reflection and are isometric or distance preserving in nature where distance between any 2 points does not change after the transformation.

The center of group Dn     as well as Cn  is{T(0)}which is a sub-group of both the groups. By definition, center is a subset of the group which commutes with every element of the group.

Cn is a cyclic group of order 5 and by definition, the order of the sub-groups are divisors of the order of the group. This group has 2 sub groups of order 1,5 respectively - one being the identity element and the other is the group which is its own sub group.

The only finite plane symmetry groups are Dn    and  Zn   . Thus D5   and permutation group Z5   are finite plane symmetry groups.

Now T(i).T(j)=T(i+j) ;

T(i).S(j)=S(i+j) ;

S(i).T(j)=S(i-j) ;

S(i).S(j)=T(i-j) ;

where i,j <=0 and <=n-1 and i,j are computed modulo n. T(0) is the identity & T-1(i) =T(n-i)  and   U-1(i) =U(i)

* Since in general, TS is not equal to ST (i.e. Matrix multiplication is not commutative), the group is a Non-Abelian Group.

Dn :Elements

If  all these elements are positive numbers, the group will be a symmetrical group (Check up )

Every permutation of a finite set can be written  as a cycle or as a product of disjoint cyclical groups. And disjoint cycles commute.

A Check digit scheme based on D5 :

Let us take Dn ={T(0),T(1),T(2),T(3),T(4),S(0),S(1),S(2),S(3),S(4)}={0,1,2,3,4,5,6,7,8,9}. This finite set can be written as a product of disjoint cycles. Let us take the permutation- σ = From (05782419)o(36). The idea is to view the digits 0 through 9as the elements of group D5 and to replace ordinary addition with calculation done in D5. From LHS, to every digit,a1,a2,a3,a4,a5,a6,a7,..a(n-1), an- We append a σ so that σ(a1)*σ2(a2)*σ3(a3)*........*σn(an)*checksum =0;In this way, all single digit errors and transposition errors involving adjacent digits are detected.

Example: Suppose the unique identification no. of a bank note is 0285368277. What should be the check digit ?

Ans: σ(a1)*σ2(a2)*σ3(a3)*σ4(a4)*σ5(a5)*σ6(a6)*σ7(a7)*σ8(a8)*σ9(a9)*σ10(a10)=σ(0)*σ2(2)*σ3(8)*σ4(5)*σ5(3)*σ6(6)*σ7(8)*σ8(2)*σ9(7)*σ10(7)=5*1*1*4*6*6*7*2*8*2=9*8*9*3*9*2*4*7 =5.  Now 5* check sum = 0 => check sum=5 from Caley's table.

Symmetry :

(1)The pentagon has a 5-fold rotational symmetry in the plane of pentagon when rotated about an axis passing through its center and perpendicular to it. The angle of rotation are 0,72,144,216,288,360 degrees or 0*2π/5,1*2π/5,2*2π/5,3*2π/5,4*2π/5, 5*2π/5 or by integral multiple of 2π/n. The general formula for such rotation is (5-n)2π/n where n=0,1,2,3,4,5. Since rotation by zero degree & 360 degree bring pentagon to the original position, we call it identity transformation. rest are 4 so that total is 5. Nomenclature wise, these are cyclical. For exa,2π/5 clockwise makes pentagon  AGTUC from GTUCA. Such rotation is carried out entirely in 2-D plane of the pentagon. The symmetry is known as cyclic rotation symmetry.

(2)Pentagon also has symmetry when rotated by π about the axis joining the vertex and mid-point of the opposite side (known as axis of reflection). Since there are 5 such axes, such type of rotational symmetry is 5 in 3-D space. Nomenclature wise, the vertex remains same & other symmetrical positions about the axis are flipped.For example, rotation π about GM clockwise results in GACUT from GTUCA. The symmetry is known as reflection symmetry.

Total no. of coincidental rotations of a pentagon is 10. This is the special case of generalized statement that a polygon of n equal sides has 2n coincidental rotations. Set of All coincidental rotations of a regular polygon of side n form what is known as a Dihedral group of degree n and order 2n represented by Dn

Example: Suppose, we want to find out the location of a point after 4 rotations and 3 reflections in anti-clockwise direction, put k=4,k1=3, give a value of (x,y), click.

Websites : http://www.calculatorsoup.com/calculators/geometry-plane/polygon.php

Side AG : a
No. of Sides : n
Value of k ( any integral value starting from zero) for rotation (T-matrix)
Value of ka ( any integral value starting from zero) for subsequent rotation (T'-matrix)
Value of k i.e k1 ( any integral value starting from zero) for reflection 1 (S-matrix)
Value of k i.e k1a ( any integral value starting from zero) for subsequent reflection  (S'-matrix)
Value of k, i.e k2 (------do-------------------------------) for reflection 2 (U matrix)
Assign any numerical value to X-coordinate of G
Assign any numerical value to Y-coordinate of G
Find out
interior angle: AGT (180 - 360/n)
Angle : GAT
Angle : AGU1
Angle : AU1G
Angle : GTA
Angle : GU1C1 , GC1U1 &
Angle : U1GC1 & C1GT &
AU1 : b1= (a*sin GAU1 / sin AU1G)
U1C1: a1=(b1*sin U1GC1 / sin GU1C1)
AT: 2b1+a1
AT : (a * sin AGT / sin GAT )
U1T
GY
YK
U1T/AU1(τ =Golden ratio )[ diagonals intersect each other in Golden Ratio ]
AT/AG (diagonal : side = τ =Golden ratio= [1+√5]/2)
τ2τ -1 =0
Area : A =a2*n /(4*tan 180/n)
Perimeter : Peri (na)
Area of Inner Pentagon, Ai of U1C1A1G1T1[Ai=a12*n /(4*tan 180/n) ]
Perimeter of Inner pentagon: perii(na1)
No. of Diagonals : D= (n/2)*(n-3)
Sum of interior Angles : 180(n-2) degree or radian
Apothem (In-Radius) : r=(a/2) *cot (180/n)
Circum Radius : R = (a/2) *cosec(180/n)
r+R (GM)
R-r
Perimeter of inner pentagon/perimeter of original pentagon :a1/a = (1/2cos 36)2
Area of inner pentagon/original pentagon (U1C1A1G1T1/UCAGT=a12 /a2)
(a1/a)2
Perimeter triangle 1 (2b1+a1)
Area of triangle 1
Perimeter triangle 2 (2b1+a)
Area of triangle 2
area triangle1/area triangle2 ( τ -1 )
perimeter triangle1/perimeter triangle2 [(2b1+a1)/(2b1+a)]
area of inner pentagon/ area of triangle1
area of inner pentagon/ area of triangle2
Find X-co-ordinate  of G after k th rotation
Find Y-co-ordinate  of G after k th rotation
Find X-co-ordinate  of G after k2 th reflection1
Find Y-co-ordinate  of G after k2 th reflection1
T ()  matrix (k rotation)

U ()  matrix ( k2 reflection)

TU matrix   (TU11)   (TU12)
(TU21)   (TU22)
UT matrix   (UT11)   (UT12)
(UT21)   (UT22)
Find X-co-ordinate  of G after k1 th reflection
Find Y-co-ordinate  of G after k1 th reflection
S() matrix (k1 reflection )

T().S() matrix = S( ) matrix.              [ Ti.Sj =S(i+j)]   (TS11)   (TS12)
(TS21)   (TS22)
S().T() matrix = S( ) matrix.              [ Si.Tj =S(i-j)]   (ST11)   (ST12)
(ST21)   (ST22)
X-coordinate of G after rotations T, followed by reflections S
Y-coordinate of G after rotations T, followed by reflections S
X-coordinate of G after rotations S, followed by reflections T
Y-coordinate of G after rotations S, followed by reflections T
T' matrix (ka rotation)

S'-matrix (k1a reflection)

T.T'

S.S'

Cayley's Table For Dihedral Group -D5
 T(0)=e T(1) T(2) T(3) T(4) S(0) S(1) S(2) S(3) S(4) T(0)=e T(0) T(1) T(2) T(3) T(4) S(0) S(1) S(2) S(3) S(4) T(1) T(1) T(2) T(3) T(4) T(0) S(1) S(2) S(3) S(4) S(0) T(2) T(2) T(3) T(4) T(0) T(1) S(2) S(3) S(4) S(0) S(1) T(3) T(3) T(4) T(0) T(1) T(2) S(3) S(4) S(0) S(1) S(2) T(4) T(4) T(0) T(1) T(2) T(3) S(4) S(0) S(1) S(2) S(3) S(0) S(0) S(4) S(3) S(2) S(1) T(0) T(4) T(3) T(2) T(1) S(1) S(1) S(0) S(4) S(3) S(2) T(1) T(0) T(4) T(3) T(2) S(2) S(2) S(1) S(0) S(4) S(3) T(2) T(1) T(0) T(4) T(3) S(3) S(3) S(2) S(1) S(0) S(4) T(3) T(2) T(1) T(0) T(4) S(4) S(4) S(3) S(2) S(1) S(0) T(4) T(3) T(2) T(1) T(0)

Cayley's Table For Dihedral Group -D5

[ With re-nomenclature of T(0)=0,T(1)=1,T(2)=2,T(3)=3,T(4)=4,S(0)=5,S(1)=6,S(2)=7,S(3)=8 & S(4)=9 ]

 a     b--> 0 1 2 3 4 5 6 7 8 9 0 0 1 2 3 4 5 6 7 8 9 1 1 2 3 4 0 6 7 8 9 5 2 2 3 4 0 1 7 8 9 5 6 3 3 4 0 1 2 8 9 5 6 7 4 4 0 1 2 3 9 5 6 7 8 5 5 9 8 7 6 0 4 3 2 1 6 6 5 9 8 7 1 0 4 3 2 7 7 6 5 9 8 2 1 0 4 3 8 8 7 6 5 9 3 2 1 0 4 9 9 8 7 6 5 4 3 2 1 0

Cayley's Calculator For Dihedral Group -D5

 a b aob

σ Table For Dihedral Group -D5

σ = (05782419)o(36)

 a     b--> 0 1 2 3 4 5 6 7 8 9 σ 5 9 4 6 1 7 3 8 2 0 σ2 7 0 1 3 9 8 6 2 4 5 σ3 8 5 9 6 0 2 3 4 1 7 σ4 2 7 0 3 5 4 6 1 9 8 σ5 4 8 5 6 7 1 3 9 0 2 σ6 1 2 7 3 8 9 6 0 5 4 σ7 9 4 8 6 2 0 3 5 7 1 σ8 0 1 2 3 4 5 6 7 8 9 σ9 5 9 4 6 1 7 3 8 2 0 σ10 7 0 1 3 9 8 6 2 4 59