Showing why a doughnut and a mug are topologically equivalent…
A. Demanet devised an interesting method of solution of trinomial equations which depends on the use of communicating vessels of convenient forms.
To solve an equation of the third degree of the form:
x3 + x = c
where c is a constant, an inverted cone and a cylinder, joined together by means of a tube, are taken.
As shown below.
The radius r of the cone and its height h are in the ratio:
r/h = √3/√π
while the base of the cylinder is taken as 1 cm2
A polyhedron compound of two cubes is obtained by allowing two cubes to share opposite polyhedron vertices, and then rotating one a sixth of a turn about the axis that joins the two opposite vertices (see fig. 1 below).
As you can see from fig. 2, the two-cube compound is made up of 12 pyramidal modules. Each pyramidal module is composed of two right triangles with ratio 2:1 and one isosceles right triangle.
Print the PDF with the paper model (shown in fig. 3) to make your own compound of two cubes. Continue reading “Cube in a Cube or the Intersecting Tetrahedra”