# Fibonacci series directly relates the tetrahedron to the pentagon | GEOMETRY: Platonic Solids & the Symmetries of Space | Sacred Geometry Web | Forum

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7:36 am

October 17, 2019

11:04 am

April 3, 2018

That is interesting, but I’m not sure I understand fully.

Do you mean that, for instance, if the length of three edges is in proportions 3 : 5 : 8

then the new face would be a point of a 5 pointed star?

That seems counter-intuitive to me.

I can imagine that if the lengths were 3 : 3 : 5

that the new face might approximate a star point,

but “any three terms of the Fibonacci series” seems like it would often generate a new face

that was not an isosceles triangle, let alone a golden ratio one.

Perhaps I’m misunderstanding what you mean?

12:20 pm

October 17, 2019

I actually do mean 3:5:8 but the phi ratio gets closer the higher in the sequence one goes; the 3:5 triangle has the shorter side of the star( the side of a pentagon) while the 3:8 and 5:8 triangles -dont ask me HOW…wind up having equal lengths for the third sides of their triangles- the two diagonals of the pentagon, or the sides that form the “point” of the star

12:37 pm

October 17, 2019

10:22 am

July 2, 2016

I’m going to actually try this with sticks and glue.

Make a tetrahedron.

Mark points 3 5 and 8 units down three edges.

Join them up with other sticks.

Examine the resulting triangle.

I agree with @Cindy that intuitively I would think you need two lengths the same

in order to get an isosceles triangle, so Im intrigued 🙂

I find it fascinating that once you have phi ratios in a form

then they tend to appear everywhere within it.

So if the lengths were phi ratio instead of Fibonacci then I guess the triangle

would be a perfect star point?

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