This is not surprising because when we go along the cicumference say from P to P'

we encounter twice height of triangle and once side of a triangle. I had so much of trouble writing above in LATEX.

I've inserted spaces using space \ space. Bob Bundy.

]]>The difference looked unrealistic at first sight the cut portion having taken away so much volume.The radius of the sphere which accommodated 20 triangles being less than one side of a triangle also looked odd. But looking around the circumference 2 complete sides of triangle like P'B and heights of 4 triangles i.e about only 6 elements along circumference removed the doubt.

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It consists of 20 identical equilateral triangles joined 5 at each vertex. It has 20 faces, 30 edges and 12 vertices. If you put a dot at the centre of each face and join together adjacent face dots the resulting solid is a dodecahedron, having 12 faces, 30 edges and 20 vertices. These two are 'duals' of each other; a fact I shall make use of later.

If you join opposite vertices, all these lines go through the centre so you can make a sphere that goes through all the vertices. I shall call the centre to vertex distance the radius of the icosahedron. In another thread, Monox D. I-Fly asked about this distance, as it is key to working out the volume of the solid. Here is his diagram:

I shall assume a side length of 1, and show how to work out this distance. Then I'll work out the volume. I also plan to make a separate thread that shows how to use this to get the volume of a dodecahedron.

Outline of my method. (1) Obtain expressions for tan(36), cos(36) and sin(36). I have done that here: http://www.mathisfunforum.com/viewtopic.php?id=22964

(2) Calculate E'C'. (3) prove that E'C'CA is a rectangle. (4) Calculate AC' and hence (5) the radius. (6) Calculate the area of any face. (7) Calculate the volume of a pyramid with such a triangle as its base and O as its vertex. (8) Hence calculate the volume of the solid.

(2) Triangle E'D'C' is isosceles so it can be split in two down its line of symmetry, creating a triangle that is 36-54-90. Thus E'C' is given by

(3) E'C'CA is a quadrilateral; it has E'C' = AC and E'A = C'C so it is a parallelogram. But E'C = AC' = 2.radius so it is a parallelogram with equal diagonals, ie. a rectangle.

(4) So, by Pythagoras

(5)

(6) Area of any equilateral triangle:

(7) I need the perpendicular distance from O to a triangle.

Let M be the midpoint of B'C' and G the centre of triangle PB'C' (third way up median)

(8) every triangle has a pyramid like this and there is no space between them where they meet at O so 20 times the above should give the volume of the icosahedron:

I don't think that is right. So I've got to check to find where I've slipped up.

All errors found and corrected.

Note:

Thus the above simplifies like this:

which is the value given by Monox D. I-Fly here http://www.mathisfunforum.com/viewtopic.php?id=22950

I think I'll rest with that for now.

Bob

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