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?
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
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?
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.
Interesting explanation @d0b123
Definitely still awake,
(and counting tetrahedrons as they jump the fence).
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