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In the icosahedron above, what is the proper way of determining the length of PP'?
My workmate thinks that it is twice the length of P to the midpoint of E'C' plus E'A, since she thinks that (and it does look) ACC'E' is a rectangle.
However, I think that E'A must not be the height of the rectangle, since the height of the rectangle should be the height of the triangle A'BB'. Also, I think the icosahedron's height is twice the length of P to the center of A'B'C'D'E' plus triangle A'BB', though I am not sure if the length of P to the center of A'B'C'D'E' equals the length of P to the midpoint of E'C'.
Which one of us is correct?
Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.
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hi Monox D. I-Fly
To 'see' the true length of a line in 3D you must be looking square on to it. As al the middle triangles are sloping from our perspective, none of those distances is the true one.
A sphere centre O and radius OA will go through all the vertices, so what you need is this radius (then double to get P'P)
This site should help you get there:
http://paulscottinfo.ipage.com/polyhedr … cosah.html
Just before the bottom of the page you will find the coordinates in terms of phi, the golden ratio.
You might also be interested in the angle between to faces:
https://en.wikipedia.org/wiki/Table_of_ … ral_angles
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Sorry, I made a typo. I mean I think the icosahedron's height is twice the length of P to the center of A'B'C'D'E' plus the height of triangle A'BB'. Is it still wrong?
Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.
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hi
If A'B'C'D'E' is horizontal and P'P is vertical then the height of that pyramid is good. But AB'B is not in a vertical plane, so its height is not the same as the vertical distance between A'B'C'D'E' and ABCDE.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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What do you mean that A'B'C'D'E' isn't horizontal? Aren't all five vertices parallel to each other?
Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.
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What do you mean that A'B'C'D'E' isn't horizontal?
I haven't said this. An icosahedron can be rotated to any position. IF we arrange it so that PP' is vertical, then both ABCDE an A'B'C'D'E' are horizontal. And the distance from p' to the pentagon would be part of the answer.
But the missing distance is not the height of A'B'B because that triangle is not vertical. You need the vertical distance between the two pentagons.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I think I get it. It's because the triangles ABA' and BCB' are slanted, right?
Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.
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Yes.
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Someone from My Math Forum gave me the hint that AO/AB = sin 72 degrees. Where did the number 72 come from?
Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.
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ABB'PE is another regular pentagon with O at its centre. So the central angle is split into 5 equal parts, 360/5 = 72.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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ACC'E' is a rectangle, but the entire rectangle is "skewed" in the sense that you mean. Hence the angle ACC' is a right angle and Pythagoras can be used.
Note that sin(108°)/sin(36°) = sin(72°)/sin(36°) = 2sin(36°)cos(36°)/sin(36°) = 2cos(36°).
Thanks to this hint, I understand that cos36° =
, so 2cos36° = .Assuming the length of each triangle's side is a, using Pythagorean Theorem, we have:
Last edited by Monox D. I-Fly (2016-04-06 15:07:21)
Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.
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hi Monox D. I-Fly
I have put a full analysis for the icosahedron here:
http://www.mathisfunforum.com/viewtopic.php?id=22966
The dodecahedron should follow quickly using duality but I won't have a chance to type it up until next week.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hey Monox D. I-Fly,
I don't really know how to help out with the main issue, but I did want to point out that there is something peculiar about the line of equations you wrote.
Namely, the third to seventh lines are really one half of the first two, not equal to them. The correct denominator is 2, not 4.
Last edited by Relentless (2016-04-07 10:21:25)
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Oops, right. Thanks for the correction.
Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.
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