How to draw octagon inside an octagon where both centered in same point in turtle using Python? This what I did so far:
import turtle
for i in range(8):
turtle.forward(100)
turtle.right(360/8)
turtle.pendown()
turtle.goto(20, -20)
turtle.penup()
for i in range(8):
turtle.forward(40)
turtle.right(360/8)
just put the pen down, trace the first pentagon in the first for loop, travel half the distance of the first line and begin your new smaller octagon there
import turtle
turtle.pendown()
for i in range(8):
turtle.forward(100)
turtle.right(360/8)
turtle.forward(50)
for i in range(8):
turtle.forward(FIGURE OUT WHAT THE NEW DISTANCE IS)
turtle.right(360/8)
if it was important to jump to point 20,-20 and start with octagon 40 length there this would be the version
import turtle
turtle.pendown()
for i in range(8):
turtle.forward(100)
turtle.right(360/8)
turtle.penup()
turtle.goto(20, -20)
turtle.pendown()
for i in range(8):
turtle.forward(40)
turtle.right(360/8)
Remember that points on a circle are given by the formula
x = r * cos(theta)
y = r * sin(theta)
where r is the radius and theta is the angle subtended by the ray extending from the origin to the point and the x-axis. The vertices of an octagon are equally spaced on a circle. So if we choose equally spaced thetas from 0 to 2 pi radians, then we'll have the vertices of an octagon:
import turtle
from math import pi, cos, sin
def ngon(n, dist, phase, shift=(0,0)):
theta = 2 * pi / n
x0, y0 = shift
turtle.penup()
x = dist * cos(phase) + x0
y = dist * sin(phase) + y0
turtle.goto(x, y)
turtle.pendown()
for i in range(1, n+1):
x = dist * cos(i*theta + phase) + x0
y = dist * sin(i*theta + phase) + y0
turtle.goto(x, y)
def demo_ngon():
turtle.speed(0.1)
turtle.width(2.0)
ngon(8, dist=100, phase=0)
ngon(8, dist=68, phase=0)
turtle.mainloop()
demo_ngon()
