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Fibonacci series directly relates the tetrahedron to the pentagon
October 17, 2019
7:36 am
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If you take a tetrahedron and cut it with a plane keeping the three 60 degree angles at the top vertex but changing the lengths to any three terms of Fibonacci series the new fourth face is one point of the five-pointed star...all you need to solve are sine and cosine laws.

November 18, 2019
11:04 am
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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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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and I should have clarified- the terms of the series must be in order, no "jumping around"

March 7, 2021
10:22 am
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
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
Byron Bay NSW Australia
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Interesting explanation @d0b123

Definitely still awake,
(and counting tetrahedrons as they jump the fence).

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