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**Monox D. I-Fly****Member**- Registered: 2015-12-02
- Posts: 860

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?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,139

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

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**Monox D. I-Fly****Member**- Registered: 2015-12-02
- Posts: 860

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?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,139

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

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**Monox D. I-Fly****Member**- Registered: 2015-12-02
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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?

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**bob bundy****Administrator**- Registered: 2010-06-20
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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

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**Monox D. I-Fly****Member**- Registered: 2015-12-02
- Posts: 860

I think I get it. It's because the triangles ABA' and BCB' are slanted, right?

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**bob bundy****Administrator**- Registered: 2010-06-20
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Yes.

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Monox D. I-Fly****Member**- Registered: 2015-12-02
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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?

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**bob bundy****Administrator**- Registered: 2010-06-20
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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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Monox D. I-Fly****Member**- Registered: 2015-12-02
- Posts: 860

skipjack wrote:

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:

Because AC' = PP', which is the icosahedron's height, the height of the icosahedron is .

*Last edited by Monox D. I-Fly (2016-04-06 15:07:21)*

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,139

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

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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**Monox D. I-Fly****Member**- Registered: 2015-12-02
- Posts: 860

Oops, right. Thanks for the correction.

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