Iterated golden rectangles in detail

Keep in mind that these diagonals cut the innermost rectangular shape in the exact same method as the outer triangle. So whenever we repeat the 4 actions above, we will constantly have 2 diagonals that converge in the same location. So the limit of the procedure assembles on the point that is the intersection of the two diagonal lines.

Weve cut off left, top, right, and bottom. Were entrusted a rectangular shape with one side of each color above. We could repeat the procedure above on this rectangle, cutting off left, leading, right, and bottom, and so on ad infinitum.

Start with a golden rectangle in landscape mode. Well outline our rectangle with the lower left corner at the origin and the upper right corner at (φ, 1) where φ is the golden ratio. No imaging cutting off the largest square you can on the left side, displayed in blue.

Where does this process end in the limitation? Draw two diagonal lines as revealed below.

Ive seen the illustration of nesting golden rectangular shapes lot of times, but Ive never ever seen a presentation enter into much detail. This post will go into more information than normal, consisting of Python code.

Lastly, focus on the staying rectangular shape and picture cutting off the square on bottom, displayed in red.

Now focus on the staying rectangular shape and imagine cutting off the square on the right, revealed in orange.

Next, concentrate on the staying rectangular shape and picture cutting off the square on top, displayed in green.

The two diagonals lines have equations

y = -( 1/ φ) x + 1

and

y = φx– φ

therefore we can calculate their intersection to be at

plotrect(( 0, 0), (phi, 1), “black”).
plotrect(( 0, 0), (1, 1), “blue”).
plotrect(( 1, 2-phi), (phi, 1), “green”).
plotrect(( 2 * phi – 2,0), (phi, 2-phi), “orange”).
plotrect(( 1,0), (2 * phi-2,2 * phi-3), “red”).
, “–“, color=” gray”)
plt.savefig(” golden_rect5.

, “–“, color=” gray”)
plt.savefig(” golden_rect5.

and.

Heres Python code to product the last plot above.

I intend to find the formula of the spiral that touches two corners of each square in the next post.

def plotrect( lowerleft, upperright, color):.
x0, y0 = lowerleft.
x1, y1 = upperright.
plt.plot( [x0, x1], [y0, y0], c= color).
, c= color).
plt.plot( [x0, x0], [y0, y1], c= color).
, c= color).

phi = (1 + 5 ** 0.5)/ 2.

import numpy as np.
import matplotlib.pyplot as plt.

, c= color).
, c= color).
, c= color).
, c= color).

y0 = 1/( φ + 2).

plt.gca(). set_aspect(” equal”).

The building and construction process, in the limit, converges to the point (x0, y0) where the diagonals cross.

x0 = (2φ + 1)/( φ + 2).

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