In 1937, mathematician Victor Thébault found that squares constructed on a parallelogram’s sides yield a square when their centers are connected.

Read more: https://en.wikipedia.org/wiki/Th%C3%A9bault’s_theorem
In 1937, mathematician Victor Thébault found that squares constructed on a parallelogram’s sides yield a square when their centers are connected.
Read more: https://en.wikipedia.org/wiki/Th%C3%A9bault’s_theorem
The Fibonacci Zoetropes are sculptures by John Edmark. The spirals in the sculptures follow the Fibonacci sequence. When filmed at 24 frames per second and spun at 550 revolutions per minute, each frame represents a 137.5 degree rotation, which is equivalent to the Golden Angle.
This is a simple linkage-mechanism for converting binary numbers to decimal numbers.
This is one of my earliest color optical illusions. There is no yellow or green in the diamond shapes, just vertical black lines! (If you don’t believe it, use a eyedropper tool to check it.)
Is it possible to 3D print an impossible cube ? Here is a way to do it… After all, it’s all about perspective!
Source: Wolfram Community
As you maybe know, I am an expert in optical illusions… So, I would like to show you one of my oldest illusions I created in the 90s. In the picture you may see ghost-like dark radial beams. This illusion is a variant of the Herman’s scintillating grid illusion. I designed this illusion just by turning 45 degrees the Herman grid and then by applying a polar transformation.
Continue reading “A Neat Geometrical Illusion: The Scintillating Starburst”
In a polygon, an exterior angle is formed by a side and an extension of an adjacent side. The sum of exterior angles in any convex polygon always adds up to 360 degrees, as shown in the 2 visual proofs below. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon.
Source: https://twitter.com/panlepan/status/1138686590216298497?s=20 by @panlepan
Perpetual Escher’s waterfall
Take 6 points on a circle such that every second edge (green chords) has length equal to the radius of the circle. Then the midpoints of the other three sides of the cyclic hexagon form an equilateral triangle.
Read more: https://www.archimedes-lab.org/2-Sunday_puzzle_43.html