| * 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.
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