Calculating coordinates of vertices of regular convex polygon given number of sides and the radius of circumscribing circle

[glevik]

Hello,

I am working on a program where I need to draw a shape which is a
regular convex polygon. The parameters I am given is the number of
sides and the radius of the of the circumscribing circle. Let's say
the bottom side of the polygon is to be horizontal, I am trying to
calculate coordinates of its vertices so that I can draw it. I can
calculate many of polygon's properties like the length of its side or
its area, but I am a little stuck calculating the coordinates of its
vertices. Any advice is highly appreciated.

Leo

[Avni Pllana]

Hi Leo,

the vertices of a regular n-gon with circumradius R are:

for n even:

Xn(k)=R*[-sin(k*2*pi/n-pi/n) cos(k*2*pi/n-pi/n)] ,

and for n odd:

Xn(k)=R*[-sin(k*2*pi/n-2*pi/n) cos(k*2*pi/n-2*pi/n)];

where k = 0, 1, ..., n-1 .

Best regards,
Avni

[Ben Saucer]

Let n be the number of sides, and r be the circumradius. Then solve in
the complex plane all of the nth roots of r^n (there will be n such
roots). One of the points will be at r + 0i. The rest of the roots will
also be on the circle centered about 0 + 0i, and having radius r. The
roots are r * cos (2pi * k / n) + i * r * sin (2pi * k / n) where k is
integer from 0 to n-1.

Ben

[mark]

While Avni and Ben both gave what I'm sure are excellent answers, I'm a little dumber than that mathematically. When I read the above I pictured x and y coordinates as needed in the form: (x,y) for the center and (x+a, y+b) for the vertices.
I'm not sure Leo is as helpless as me, but could someone get me from Avni or Ben's formulae to xy coordinates?

[Philippe 92]

mark wrote :
>> ...regular convex polygon. ...
>> calculate coordinates of its vertices...
>> Leo

>
> could someone get me from Avni or Ben's formulae to xy coordinates?

Hi,

They do give the coordinates, although not explicitely for you.

The key is that the coordinates x, y of points on a unit circle
centered at origin are x = cos(t), y = sin(t)
That is for any circle centered in (a, b) with radius r :
x = a + r*cos(t), y = b + r*sin(t).
I'll generally discard the r factor and the offset (a, b) for
the sequel.

For vertices of a regular n-gon inscribed in the circle, the angles
are 2pi/n apart from each other (2pi is in radian, that is 360 degree
all further angles are in radian).

That is x[k] = cos(t0 + k*2pi/n) , y[k] = sin(t0 + k*2pi/n)
the t0 being the angle of some [given] vertex.

Ben has choosen t0 = 0, and has written the coordinates (x, y) as
a complex number z = x + i*y, a usual way to represent points on
a plane, but totally equivallent to (x, y), just easing some
calculations. For instance writing cos(t) + i*sin(t) as exp(i*t)
and using rules for exponentiation gives "at once" the formulas
for cos(n*t), cos(a+b) etc...

However the requirement was :

"the bottom side of the polygon is to be horizontal"

Hence Ben formulas is false, as it has choosen its first vertex (t0)
at (1, 0) or introducing r at (r, 0), written as z0 = r + 0*i.
The bottom side is then not always horizontal, depending on n.

Avni has choosen an other t0, introducing two separate cases
n odd : then a vertex is at (0, R), angle t0 = pi/2
n even : then the upper side is also horizontal and a vertex is at
angle t0 = pi/2 - pi/n
As cos(pi/2 + t) = - sin(t) and sin(pi/2 - t) = cos(t),
this results into his set of formulas.
Coordinates of a point X being written as [x y] (the space to
separate coordinates is not a so good choice however)

I would prefer choosing the initial vertex at one end of this
bottom side ! this results into a formula which doesn't depend
on the parity of n.
That is t0 = -pi/2 + pi/n whatever n.
Then apply cos(-pi/2 + t) = sin(t) and sin(-pi/2 + t) = -cos(t),
this results into the coordinates of the k-th vertex :

x[k] = a + r*sin(pi/n + k*2pi/n)
y[k] = b - r*cos(pi/n + k*2pi/n)

Best Regards.

--
Philippe Ch., mail : chephi...@free.fr
site : http://mathafou.free.fr/ (recreational mathematics)

[mark]

Thanks Philippe 92, That's better.

Mark

[gudi]

You can conveniently use polar coordinate form of equation of circle

passing through origin to obtain the vertices. a is length of a
polygon side,

R is Circum-radius. Using Law of Sines for each triangle in the
polygon,

where al is angle contained in the circle segment looking at origin,

al = 2 Pi/n = t step for polygon of n sides to get coordinates of
vertices.

r/ sin(t + al) = a/ sin(al) = 2R, sin(al) = a / 2R,

(x,y) = 2 R (cos(t), sin(t) ) sin ( t + arcsin ( a / 2R ) )

t = 0,2 Pi, with step al.

Narasimham