hp41programs

polygon

Area of a Polygon for the HP-41


Overview
 

 1°)  General Case
 2°)  Brahmagupta's Formula

   a)  Focal Programs
   b)  M-Code Routines ( Heron & Brahmagupta Formulae )

 3°)  Convex Cyclic Polygons

   a)  Area ( 2 programs )
   b)  Diagonals Lengths

 4°)  Regular Star Polygons
 

-> The latest addition is the second program in paragraph 3-a)
 

1°)  General Case
 

-Here, we assume that the coordinates of all the vertices are given.

-The following program calculates the area of a polygon  A1 A2 ...... An    where  Ai(xi,yi)  are the n vertices.
-The result is positive if the n points are stored counterclockwise.
 

Data Registers:       •  R00 = n                                 ( All these  registers are to be initialized before executing "PGA" )

                                  •  R01 = x1   •  R03 = x2   ..........   •  R2n-1 = xn
                                  •  R02 = y1   •  R04 = y2   ..........   •  R2n = yn
Flags: /
Subroutines: /

-Synthetic registers M N O may be replaced by any unused data registers.
 
 

 01  LBL "PGA"
 02  RCL 00
 03  ST+ X
 04  STO M
 05  RCL 02
 06  RCL IND Y
 07  STO O
 08  -
 09  DSE M
 10  RCL 01
 11  RCL IND M
 12  STO N
 13  +
 14  *
 15  DSE M
 16  LBL 01
 17  RCL IND M
 18  ENTER^
 19  X<> O
 20  -
 21  DSE M
 22  RCL IND M
 23  ENTER^
 24  X<> N
 25  +
 26  *
 27  -
 28  DSE M
 29  GTO 01
 30  2
 31  /
 32  CLA
 33  END

 
( 56 bytes / SIZE 2n+1 )
 
 

      STACK        INPUT      OUTPUT
           X             /         Area

 
Example:      The 6 vertices of an hexagon are defined by their coordinates (x,y)
 
 

     xi      -2      -1      2      3      2     -1
     yi       0      -2     -1      1      2      3

 
-We store these 12 numbers in registers

      R01     R03     R05    R07    R09    R11
      R02     R04     R06    R08    R10    R12            respectively

-Then,       6  STO 00     XEQ "PGA"  >>>>   Area =  16.0000   ( in 2.7 seconds )

Note:

-Execution time is of the order of 43 seconds if n = 100
 

2°)  Brahmagupta's Formula
 

     a)  Focal Programs
 

-Brahmagupta's formula generalizes Heron's formula for a convex cyclic quadrilateral ( i-e that may be inscribed in a circle )

-Let  a , b , c , d  its sides lengths, the area A of the cyclic quadrilateral is

   A =  ( ( s-a ).( s-b ).( s-c ).( s-d ) )1/2   where  s = semi-perimeter = ( a + b + c + d )/2
 

Data Registers: /
Flags: /
Subroutines: /
 
 

 01  LBL "BRM"
 02  STO M
 03  RDN
 04  ST+ M
 05  RDN
 06  ST+ M
 07  X<>Y
 08  ST+ M
 09  SIGN
 10  ST+ X
 11  ST/ M
 12  CLX
 13  X<> M
 14  ST- Y
 15  ST- Z
 16  ST- T
 17  ST- L
 18  X<> L
 19  *
 20  *
 21  *
 22  SQRT
 23  END

 
( 43 bytes / SIZE 000 )
 
 

      STACK        INPUTS      OUTPUTS
           T             a             /
           Z             b             /
           Y             c             /
           X             d         Area

 
Example:    a = 4 , b = 5 , c = 6 , d = 7

   4  ENTER^
   5  ENTER^
   6  ENTER^
   7  XEQ "BRM"  >>>>  Area = 28.98275349

Notes:

-This formula may also be applied for a triangle with d = 0
-If we also need the radius R of the circumcircle, it may be computed by

   R = (1/4) [ ( a.b + c.d ) ( a.c + b.d ) ( a.d + b.c ) ]1/2 / Area
 

Data Registers:   R00 = Area , R01 = a , R02 = b , R03 = c , R04 = d
Flags: /
Subroutines: /
 
 

 01  LBL "BRMR"
 02  STO 00        
 03  STO 04 
 04  RDN
 05  ST+ 00
 06  STO 03 
 07  RDN
 08  ST+ 00
 09  STO 02
 10  X<>Y
 11  ST+ 00
 12  STO 01        
 13  SIGN
 14  ST+ X
 15  ST/ 00
 16  X<> 00
 17  ST- Y
 18  ST- Z
 19  ST- T
 20  ST- L
 21  X<> L
 22  *
 23  *
 24  *
 25  SQRT
 26  STO 00        
 27  RCL 01 
 28  RCL 02
 29  *
 30  RCL 03
 31  RCL 04
 32  *
 33  +
 34  RCL 01
 35  RCL 03
 36  *
 37  RCL 02        
 38  RCL 04
 39  *
 40  +
 41  *
 42  RCL 01
 43  RCL 04
 44  *
 45  RCL 02
 46  RCL 03        
 47  *
 48  +
 49  *
 50  SQRT
 51  4
 52  / 
 53  RCL 00
 54  ST/ Y
 55  END

 
   ( 76 bytes / SIZE 005 )
 
 

      STACK        INPUTS      OUTPUTS
           T             a             /
           Z             b             /
           Y             c            R
           X             d         Area

 
Example:    a = 4 , b = 5 , c = 6 , d = 7

   4  ENTER^
   5  ENTER^
   6  ENTER^
   7  XEQ "BRM"  >>>>  Area = 28.98275349
                            X<>Y     R   =  3.973161451
 

     b)  M-Code Routines ( Heron & Brahmagupta Formulae )
 

-"HERON" and "BRHM" use the Héron's & Brahmagupta's formulae.
-They are so similar that it is preferable to combine them.

-There is no check for alpha data and synthetic register Q is used.
-13-digit routines will give a better precision...
 

08E  "N"
00F  "O"
012  "R"
005  "E"
008  "H"
104   CLRF 8
033   JNC+06
08D  "M"
008   "H"
012   "R"
002   "B"
108   SETF 8
2A0  SETDEC
0F8  C=X
128   L=C
2BE  C=-C
10E  A=C ALL
0B8  C=Y
01D  C=
060   A+C
078   C=Z
025   C=
060   AB+C
04E   C=0 ALL
10C  ?FSET 8
01B   JNC+03
270   RAMSLCT
038   READATA
0E8   X=C
025   C=
060   AB+C
089   AB
064   STO Q+
0B8  C=Y
2BE  C=-C
10E   A=C ALL
138   C=L
01D  C=
060   A+C
078   C=Z
025   C=
060   AB+C
0F8   C=X
025   C=
060   AB+C
0D1  RCL
064   Q+
149   C=
060   AB*CM
089   AB
064   STO Q+
078   C=Z
2BE  C=-C
10E   A=C ALL
138   C=L
01D  C=
060   A+C
0F8   C=X
025   C=
060   AB+C
0B8  C=Y
025   C=
060   AB+C
0D1  RCL
064   Q+
149   C=
060   AB*CM
089   AB
064   STO Q+
0F8   C=X
2BE   C=-C
10E   A=C ALL
138   C=L
01D  C=
060   A+C
0B8  C=Y
025   C=
060   AB+C
078   C=Z
025   C=
060   AB+C
0D1  RCL
064   Q+
149   C=
060   AB*CM
04E   C
35C
050   =
190
226   16
269   C=
060   AB/C
305   C=
060   sqrt(AB)
0E8   X=C
3E0   RTN

( 96 words )
 
 

      STACK        INPUTS      OUTPUTS
           T             a             a
           Z             b             b
           Y             c             c
           X             d         Area
           L             /            d

 
Example:    a = 4 , b = 5 , c = 6 , d = 7

   4  ENTER^
   5  ENTER^
   6  ENTER^
   7  XEQ "BRHM"  >>>>  Area = 28.98275349

-See "Heron's Formula for the HP-41" for an example of area of a triangle...
 

3°)  Convex Cyclic Polygons
 

     a)  Area
 

-"CPLA" computes the area A of a convex cyclic polygon and the circumradius R, assuming all the sides lengths are known.
-Moreover, we also assume that the center of the circumcircle is inside the polygon.

-If  a1 , a2 , ......... , an  are the sides lengths and   µ1 , µ2 , ......... , µn  are the corresponding central angles,
 we have to solve the system of (n+1) equations:

     2.R sin µ1/2 = a1
     2.R sin µ2/2 = a2
     .........................

    2.R sin µn/2 = an
    µ1 + µ2 + ... + µn = 360°

-Substitutions lead to            Arc sin ( a1 / 2.R ) + Arc sin ( a2 / 2.R ) + ..................... + Arc sin ( an / 2.R )  =  180°      which has only 1 unknown.

-After finding  R , the Area is given by

     A = ( R2/2 ) ( Sin a1 + Sin a2 + ........... + Sin an )
 

Data Registers:           R00 to R03 are used by "SOLVE"            ( Registers Rbb thru Ree are to be initialized before executing "CPLA" )

                                      •  Rbb = a1   •  Rbb+1 = a2   .......................   •  Ree = an         bb > 05       ( R04 & R05: temp )

Flags: /
Subroutine:    "SOLVE"  ( cf "Linear and Non-Linear Systems for the HP-41" or another root-finding program )
 
 

 01  LBL "CPLA"
 02  DEG
 03  STO 04          
 04  "T"
 05  ASTO 00
 06  0
 07  LBL 00
 08  RCL IND Y
 09  X>Y?
 10  X<>Y
 11  RDN
 12  ISG Y
 13  GTO 00
 14  101
 15  %
 16  XEQ "SOLVE"
 17  RCL 04          
 18  0
 19  GTO 02
 20  LBL "T"
 21  STO 05 
 22  RCL 04
 23  0
 24  LBL 01
 25  RCL IND Y
 26  RCL 05          
 27  /
 28  ASIN
 29  +
 30  ISG Y
 31  GTO 01
 32  PI
 33  R-D
 34  -
 35  RTN
 36  LBL 02
 37  RCL IND Y
 38  RCL 01          
 39  /
 40  ASIN
 41  ST+ X
 42  SIN
 43  +
 44  ISG Y
 45  GTO 02
 46  RCL 01          
 47  2
 48  /
 49  STO Z
 50  X^2
 51  *
 52  2
 53  /
 54  CLD
 55  END

 
    ( 90 bytes / SIZE 006+??? )
 
 

      STACK        INPUTS      OUTPUTS
           Y             /             R
           X       bbb.eee          Area

   where   bbb > 005

Example:   Find the area of a convex cyclic polygon with sides  4 , 5 , 6 , 7 , 8 , 9 , 10

-Store these 7 numbers into  R06 thru R12 , then:

    6.012   XEQ "CPLA"  >>>>     Area = 174.6757942                       ---Execution time = 89s---
                                       X<>Y       R   =   8.143816985

Notes:

-The program doesn't work if the center of the circumcircle is outside the polygon.

-Instead of using the secant method, we can also employ Newton's method:
 

Data Registers:              R00 to R03: temp            ( Registers Rbb thru Ree are to be initialized before executing "CPLA" )

                                      •  Rbb = a1   •  Rbb+1 = a2   .......................   •  Ree = an         bb > 03

Flags: /
Subroutines: /
 
 

 01  LBL "CPLA"
 02  DEG
 03  STO 00
 04  X<>Y
 05  STO 01
 06  LBL 01
 07  RCL 01
 08  STO 02
 09  CLX
 10  STO 03
 11  LBL 02
 12  RCL IND 02
 13  RCL 00       
 14  ST+ X
 15  /
 16  ASIN
 17  ST+ 03       
 18  X<> L
 19  ENTER^
 20  X^2
 21  CHS
 22  1
 23  +
 24  SQRT
 25  /
 26  +
 27  ISG 02
 28  GTO 02
 29  RCL 00       
 30  /
 31  R-D
 32  RCL 03
 33  PI
 34  R-D
 35  -
 36  X<>Y
 37  /
 38  ST+ 00
 39  VIEW 00
 40  ABS
 41   E-7
 42  X<Y?
 43  GTO 01
 44  RCL 01
 45  STO 02
 46  CLX
 47  LBL 03
 48  RCL IND 02
 49  RCL 00       
 50  ST+ X
 51  /
 52  ASIN
 53  ST+ X
 54  SIN
 55  +
 56  ISG 02
 57  GTO 03
 58  RCL 00       
 59  STO Z
 60  X^2
 61  *
 62  2
 63  /
 64  CLD
 65  END

 
    ( 91 bytes / SIZE var. )
 
 

      STACK        INPUTS      OUTPUTS
           Y       bbb.eee             R
           X            r          Area

   where   bbb > 003  and  r is an estimation of the radius

Same Example:   Find the area of a convex cyclic polygon with sides  4 , 5 , 6 , 7 , 8 , 9 , 10

-Store these 7 numbers into  R06 thru R12 , then, if you choose  r = 10

    6.012   ENTER^
       10     XEQ "CPLA"  >>>>     Area =  174.6757941                       ---Execution time = 55s---
                                       X<>Y       R   =   8.143816984

Notes:

-A bad guess may lead to a DATA ERROR
-The successive R-approximations are displayed ( line 39 )
 

     b)  Diagonals Lengths
 

              An *        .............

    A1 *                                       A4 *

          A2 *
                                   A3 *

-Let's call   d1 = A1 A3  ,  d2 = A1 A4 , ........................ , dn-3 = A1 An-1           ( n-3 ) diagonals passing through  A1
    and        a1 = A1 A2  ,  a2 = A2 A3 , ........................ , an = An A1                     n    side-lengths   ( n > 3 )

-Applying the law of cosines to triangles like  A1 A2 A3  &  A1 A3 A4  ....................
  and  taking into account   Cos ( A1 A2 A3 ) = - Cos ( A1 A4 A3 )   since  the sum of these angles = 180°
  lead to the following system of ( n-3 ) equations in ( n-3 ) unknowns
 

    dj = SQRT [ { aj+3 dj+1 ( d2j-1 + a2j+1 ) + dj-1 aj+1 ( a2j+3 + d2j+1) } / ( aj+3 dj+1 + dj-1 aj+1 ) ]                      j = 1 , 2 , ..... , n-3
 

  if we define a dummy "d0" = a1 and a dummy "dn-2" = an
 

-"CPLD" solves this system by a successive approximation method.
-The convergence is linear only.
 

Data Registers:           •  R00 = n > 3                                 ( Registers R00 thru Rnn are to be initialized before executing "CPLD" )

                                      •  R01 = a1   •  R02 = a2   .......................   •  Rnn = an

                                          Rn+1 = a1 = "d0" ,  Rn+2 = d1 , Rn+3 = d2 , ................. , R2n-2 = dn-3 , R2n-1 = an = "dn-2"
Flags: /
Subroutines: /
 
 

 01  LBL "CPLD"
 02  CLA
 03  RCL 00        
 04  1
 05  +
 06  STO N
 07  RCL 00 
 08  ST+ X
 09  DSE X
 10  STO Z
 11   E3
 12  /
 13  +
 14  CLRGX
 15  RCL IND 00
 16  STO IND Z
 17  RCL 01 
 18  STO IND Z
 19  RCL 00
 20  2
 21  -
 22   E3
 23  /
 24  2
 25  +
 26  STO M
 27  LBL 01 
 28  RCL N        
 29  2
 30  +
 31  RCL IND X
 32  STO P
 33  RCL M
 34  ISG X
 35  ABS
 36  X<>Y
 37  RCL IND Y
 38  STO Q
 39  *
 40  STO Y
 41  RCL IND M
 42  X^2
 43  RCL IND N
 44  X^2
 45  +
 46  *
 47  RCL IND M
 48  RCL IND N
 49  *
 50  ST+ Z
 51  X<> P
 52  X^2
 53  RCL Q        
 54  X^2
 55  +
 56  RCL P
 57  *
 58  +
 59  X<>Y
 60  /
 61  SQRT
 62  ISG N
 63  CLX
 64  ENTER^
 65  X<> IND N
 66  -
 67  ABS
 68  ST+ O
 69  ISG M
 70  GTO 01
 71  RCL 00        
 72  3
 73  -
 74  ST- M
 75  ST- N
 76  CLX
 77  X<> O
 78  VIEW X
 79  X#0?
 80  GTO 01
 81  RCL 00
 82  ST+ X
 83  2
 84  -
 85   E3
 86  /
 87  RCL 00        
 88  +
 89  2
 90  +
 91  CLA
 92  CLD
 93  END

 
     ( 141 bytes / SIZE 2n )
 
 

      STACK        INPUT      OUTPUTS
           X             /       bbb.eee

   where   bbb.eee  is the control number of the ( n-3 ) diagonals passing through A1

Example:   Find the diagonals lengths of a convex cyclic hexagon with sides  4 , 5 , 6 , 7 , 8 , 9

   n = 6    STO 00    and store   4 , 5 , 6 , 7 , 8 , 9   into   R01 thru R06
 

   XEQ "CPLD"   >>>>    8.010                              ---Execution time = 82s---

-And we get

     R08 = d1 =  8.462784374
     R09 = d2 = 12.12358501
     R10 = d3 = 12.97690534

-There are in fact  n(n-3)/2  diagonals whose lengths may be obtained by "rotating" the sides lengths in registers R01 to R06 and we have similarly:

     d4 =  9.998827970                    d7 = 11.30861231
     d5 = 13.01214483                     d8 = 13.06010803                   n(n-3)/2 = 9  if  n = 6
     d6 = 11.49035918                     d9 = 12.32872367

Notes:

-The successive sums of the differences between 2 consecutive approximations ( in absolute values ) are displayed.
-They should tend to zero.
-However, the termination criterion, line 79  X#0?  may lead to an infinte loop.
-So, it could be replaced for instance by something like:  RCL 00   RCL 01  *   E9  /   X<Y?

-The iteration starts with all diagonals lengths = 0 ( line 14 ) which is very uninspired !
-Modify the first lines as you like provided they contain the following instructions before  LBL 01

  CLA            RCL 01           RCL IND 00        E3
  RCL 00       STO IND Y     STO IND Y         /
  1                  RCL 00            RCL 00               2
  +                  ST+ X               2                        +
  STO N         DSE X              -                        STO M

-Since there are 319 registers at most, "CPLD" can find the diagonals lengths of a 159-gon
-The execution time will not be small without an emulator.

-We can also write an M-code routine to perform the calculations of lines 39 to 61 without disturbing registers M , N , O.
-"DIAG" computes

       SQRT [ { x.y ( z2 + t2 ) + z.t ( x2 + y2 ) } / ( x.y + z.t ) ]     assuming  x , y , z , t  are  in  registers  X , Y , Z , T  upon entry
 

087  "G"
001  "A"
009  "I"
004  "D"
2A0  SETDEC
0F8  C=X
10E  A=C ALL
0B8  C=Y
135   C=
060   A*C
268   Q=C
046   C
270   =
038   T
128   L=C
10E   A=C ALL
135   C=
060   A*C
070   N=C ALL
078   C=Z
10E  A=C ALL
135   C=
060   A*C
0B0  C=N ALL
025   C=
060   AB+C
278   C=Q
13D  C=
060   AB*C
228   P=C
138  C=L
10E  A=C ALL
078  C=Z
135   C=
060   A*C
128   L=C
278   C=Q
025   C=
060   AB+C
268   Q=C
0F8   C=X
10E   A=C ALL
135   C=
060   A*C
070   N=C ALL
0B8  C=Y
10E   A=C ALL
135   C=
060   A*C
0B0  C=N ALL
025   C=
060   AB+C
138   C=L
13D  C=
060   AB*C
238   C=P
025   C=
060   AB+C
278   C=Q
269   C=
060   AB/C
305   C=
060   sqrt(AB)
0E8   X=C
3E0   RTN

( 65 words )
 

-The listing becomes:
 
 

 01  LBL "CPLD"
 02  CLA
 03  RCL 00
 04  1
 05  +
 06  STO N
 07  RCL 00        
 08  ST+ X
 09  DSE X
 10  STO Z
 11   E3
 12  /
 13  +
 14  CLRGX
 15  RCL IND 00
 16  STO IND Z
 17  RCL 01 
 18  STO IND Z
 19  RCL 00        
 20  2
 21  -
 22   E3
 23  /
 24  2
 25  +
 26  STO M
 27  LBL 01
 28  RCL N
 29  2
 30  +
 31  RCL IND X
 32  RCL M        
 33  1
 34  +
 35  X<>Y
 36  RCL IND Y
 37  RCL IND M
 38  RCL IND N
 39  DIAG
 40  ISG N
 41  CLX
 42  ENTER^
 43  X<> IND N
 44  -
 45  ABS
 46  ST+ O
 47  ISG M
 48  GTO 01
 49  RCL 00        
 50  3
 51  -
 52  ST- M
 53  ST- N
 54  CLX
 55  X<> O
 56  VIEW X
 57  X#0?
 58  GTO 01
 59  RCL 00        
 60  ENTER^
 61  ST+ X
 62  2
 63  ST+ Z
 64  -
 65   E3
 66  /
 67  +
 68  CLA
 69  CLD
 70  END

 
    ( 110 bytes / SIZE 2n )
 
 

      STACK        INPUT      OUTPUTS
           X             /       bbb.eee

   where   bbb.eee  is the control number of the ( n-3 ) diagonals passing through A1

Example:   The same one: the diagonals lengths of a convex cyclic hexagon with sides  4 , 5 , 6 , 7 , 8 , 9

   n = 6    STO 00    and store   4 , 5 , 6 , 7 , 8 , 9   into   R01 thru R06
 

   XEQ "CPLD"   >>>>    8.010                              ---Execution time = 65s---

-And we get

     R08 = d1 =  8.462784376
     R09 = d2 = 12.12358502
     R10 = d3 = 12.97690535

Notes:

-We have saved 17 seconds, significant but not a fantastic improvement !
-There are small differences in the last places because "DIAG" employs a few 13-digit routines.
-The modifications suggested for the 1st version also apply with this one.

-After finding the diagonals lengths, we can use Heron's formula to calculate the polygon area by adding the areas of several triangles.
-Unlike "CPLA", this method will also work if the center of the circumcircle is outside the polygon.
 

4°)  Regular Star Polygons
 

-"STLA" computes the area A, the perimeter P, the inradius r and the circumradius R of a regular star polygon { n / k }  from its edge length a

                An-1 *            ........
     A0 *

     A1 *
                   A2 *          A3 *
 

-Assuming  A0 A1 ...................... An-1 An = A0  is a regular convex polygon ( which is not obvious on the drawing above... ),
  the regular star polygon { n / k }  is obtained by connecting  A0 Ak A2k ..............
-In the formulas below,  a = A0 Ak = Ak A2k = .....
 

Formulae:

    A = n R2 Sin (180°/n)  Cos (180°k/n) / Cos [180°(k-1)/n]
    a  = 2.R  Sin (180°k/n)
    r  =   R   Cos (180°k/n)
    P = 2.A / r
 

Data Registers:  R00 to R04: temp
Flags: /
Subroutines: /
 
 

 01  LBL "STLA"
 02  STO 00 
 03  RDN
 04  STO 01 
 05  X<>Y
 06  2
 07  /
 08  STO 03       
 09  STO 04
 10  X^2
 11  *
 12  1
 13  CHS
 14  ACOS
 15  STO 02 
 16  ST* 00
 17  RCL 01       
 18  /
 19  SIN
 20  *
 21  RCL 00
 22  RCL 01       
 23  /
 24  1
 25  P-R
 26  ST* 03
 27  ST* Z
 28  RDN
 29  ST/ 03
 30  ST/ 04
 31  X^2
 32  /
 33  RCL 00       
 34  RCL 02 
 35  -
 36  RCL 01
 37  /
 38  COS
 39  /
 40  ENTER^
 41  ST+ X
 42  RCL 03       
 43  ST/ Y
 44  RCL 04
 45  RDN
 46  X<> Z
 47  END

 
    ( 64 bytes / SIZE 005 )
 
 

      STACK        INPUTS      OUTPUTS
           T             /            R
           Z             a             r
           Y             n             P
           X         k < n/2             A

             ---Execution time = 3.5s---

Examples:

  •  a = 1 , n = 5 , k = 2

   1   ENTER^
   5   ENTER^
   2   XEQ "STLA"   >>>>   A = 0.310270701
                                RDN    P = 3.819660113
                                RDN    r  = 0.162459848
                                RDN   R = 0.525731112

  •  a = 1 , n = 10 , k = 3

   1   ENTER^
  10  ENTER^
   3      R/S    >>>>   A = 0.857567126
                     RDN    P = 4.721359547
                     RDN    r  = 0.363271264
                     RDN   R =  0.618033989

  •  a = PI , n = 41 , k = 13

  PI   ENTER^
  41  ENTER^
  13     R/S    >>>>   A = 9.855571194
                     RDN    P = 19.37713086
                     RDN    r  = 1.017237409
                     RDN   R =  1.871409374

Notes:

-This program works in all angular modes.
-However, DEG mode should be preferable.
-If k = 1, we get the convex regular n-gon. For instance, with  a = 1 , n = 5 , k = 1 ,  "STLA"  returns:

    A = 1.720477401
    P =  5
    r  =  0.688190960
   R  =  0.850650808

 which corresponds to the regular pentagon.
 

Reference:

[1]  http://mathworld.wolfram.com/StarPolygon.html