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Fibonacci series directly relates the tetrahedron to the pentagon

October 17, 2019

7:36 am

7:36 am

Bradley Grantham

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November 18, 2019

11:04 am

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Cindy

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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?

January 9, 2020

12:20 pm

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Bradley Grantham

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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

January 9, 2020

12:37 pm

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Bradley Grantham

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March 7, 2021

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Shem

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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?

April 15, 2021

7:32 pm

7:32 pm

d0b123

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A logical way to better understand the isoceles triangle concept here is to imagine a tetrahedron of side length 8 standing on sand. Push one corner of the tetrahedron into the sand so that a side of length now = to 3 is standing vertically while one of the other two sides remains touching the surface at it's full length = 8. Push the final side down until it has length 5 above the sand surface. You now have a vertical Pythagorean triangle of sides 3,4,5 with the 3 side at 90 degrees to the horizontal and the 5 side (hypoteneuse) at an angle of 60 degrees to the 3 side (because they are two (part) sides of a tetrahedron, or the equilateral trianglular face of same).

Bisecting the 60 degree angle between the 3 and 5 lines intersects the 4 baseline at it's midpoint. If we draw a line (in the sand ! 😉 ) between the midpoint and the only uncovered side of length 8 where it just touches the sand we can see that there must be 2 equal length sides between the 3-8 and 5-8 points on the horizontal sand surface.

Just for interest (if anyone is still awake at this point?) the midline from the 4 baseline and the 8 side length point has a length of root 51 (square of 2 subtracted from square of root 55) or about 7.141 and the two equal line lengths are root 55 (8 squared - 3 squared) or about 7.416 and the shorter side = 4 as explained before.

The only snag is:

it's NOT a star point! The angle between the equal sides is approx 65 degrees not the 72 degrees in a star (360/5) while the short - long side ratio is approx 1.854 not Phi. 🙁

But it is definitely makes an isoceles triangle in the horizontal plane.

d0b123

May 5, 2021

11:18 am

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Narada Dan Vantari

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Interesting explanation @d0b123

Definitely still awake,

(and counting tetrahedrons as they jump the fence).

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