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Volumes of Regular Icosahedron and Regular Tetrahedron 
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#1
Jul1403, 05:40 AM

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Please teach me.
Is one Regular Icosahedron equal to twenty Regular Tetrahedrons ? If the edgelength of both Regular Polyhedras is 1, What would be their volumes ? Can we prove (or disprove) the equation below ? volume of Regular Icosahedron = 20 * volume of Regular Tetrahedron Thank you. 


#2
Jul1403, 11:17 AM

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now I have gotten interested. So maybe you could help. the icosahedron has 20 faces, so it looks like it might be made of 20 tetrahedrons fitted together (so your estimate of volume would be right) But I wanted to check this and I fetched a formula for the vol of tetrahedron and when I calculated it out it said 0.11785 Is that right? 


#3
Jul1403, 11:29 AM

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Something is eluding me about this, or I am doing something wrong. Here's a formul for tet volume in terms of edgelengths (on web from Berkeley prof Bill Kahan)
sqrt( 4u^2 v^2 w^2 – u^2 (v^2 +w^2 –U^2 )^2 – v^2 (w^2 +u^2 –V^2 )^2 – w^2 (u^2 +v^2 –W^2 )^2 + + (v^2 +w^2 –U^2 )(w^2 +u^2 –V^2 )(u^2 +v^2 –W^2 ) )/12 . In the case you propose, all these edgelengths u,v,w,U,V,W are equal to one. So it should be routine to get the volume 


#4
Jul1403, 11:34 AM

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Volumes of Regular Icosahedron and Regular Tetrahedron
sqrt 2 divided by 12 Damn, it comes out 0.11785 again! I am having a hard time comprehending how the vol of the icosahedron can be only 20 times the vol of the tet if the tet vol is so small. 


#5
Jul1403, 12:01 PM

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it certainly looks like the icos is made of 20 tets fitted together
because the icosahedron certainly has 20 faces but we should check volumes just to make sure and if I calculated right the tet volume is 0.11785 which would make the icosahedron volume 2.357 .....(edit)... And so 20 times the tet volume2.357is looking good. Plus it is what one naturally visualizes because of the 20 faces. WARNING (edited in later) I checked on google and was told that the ratio of volume is not 20 but 18.51229586..... so I am confused about this. Have to go for now, but will come back to it later. 


#6
Jul1403, 12:14 PM

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Thank marcus.
I am not a mathematician, so I really need the teaching. I had visited askgeeve.com a few days ago, this site gave me many formulas of the volumes of Polyhedras, but I could not understand them. I am mostly interested to confirm my assumption  A Regular Icosahedron can be made by 20 Regular Tetrahedrons. I had made many pieces of regular tetrahedron & regular octahedron, when I put these pieces together, I found something interesting. We can make larger tetrahedron or octahedron with just the tetrahedron and octahedron units. For example  Higher order Tetrahedron may be built by t * Tetrahedron pieces + o * Octahedron pieces Order 2 : t = 4, o = 1 Order 3 : t = 11, o = 4 Order 4 : t = 24, o = 10 Order 5 : t = 45, o = 20 etc. Higher order Octahedron may be built by t * Tetrahedron pieces + o * Octahedron pieces Order 2 : t = 8, o = 6 Order 3 : t = 32, o = 19 etc. I also put 20 tetrahedron pieces together and got a icosahedron shape, but this model has some gaps, not the perfect regular one. I guess if all of my prototypes of tetrahedron pieces are Regular (standard), I would get the Regular Icosahedron when 20 of them were connected together. My assumption: The volume of a regular Icosahedron is equal to the volumes of 20 Regular Tetrahedrons. Is there any proof or disproof of this before ? 


#7
Jul1403, 01:00 PM

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who has some expertise will step in and reply. I am pretty sure the vol of tet is 0.11785 so that your estimate of isos is 2.357 Let us check that against somebody's calculation of the actual volume of the icosahedron. Is it actually 2.357 or not? You said you asked jeeves and got volume formulas so what volume formula for the icosahedron did you get? It should be easy to evaluate and see if it is 2.357. OUCH!!!!! I just looked on google and it said the ratio of volume is not 20 but 18.51229586!!!!! http://euch3i.chem.emory.edu/proposa...d/volumes.html 


#8
Jul1403, 01:15 PM

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Josdavi and others,
this seemingly simple problem has me quite baffled I think my figure for the tetrahedron volume must be wrong although I calculated it using a formula posted by the highlyesteemed Prof. Kahan So all bets are off and moreover I have to be off for a few hours and cannot clear this up! Maybe someone else can step in and resolve this. Be back later, m 


#9
Jul1403, 03:45 PM

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Well, we can compute the volume of a regular tetrahedron via the volume formula for any old pyramid:
The base would have vertices: (0, 0, 0) (1, 0, 0) (1/2, sqrt(3)/2, 0) Due to the symmetry, the apex would have to lie over the centroid at: (1/2, sqrt(3)/6, x) To have a unit edge length, x has to be sqrt(6)/3 The area of the base triangle is one half base height: A = (1/2) * 1 * (sqrt(3) / 2) = sqrt(3) / 4 The volume of the pyramid is one third base height (1/3) * (sqrt(3) / 4) * (sqrt(6) / 3) = sqrt(2) / 12 = .11785 This is certainly correct. My hypothesis is that the 20 tetrahedra making the icosohedron are not regular; I can't see any clear reason why they should be in the first place. Regular tetrahedra packed together like to form hexagonallike structures, but the icosahedron is pentagonallike. 


#10
Jul1403, 11:17 PM

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I'm backhad to be away for the day. Ahhh! Hurkyl has taken care of business here. I think after my false starts earlier today that probably that ratio 18.51 is right after all, so since the tetrahedron volume is 0.11785 the icosahedron volume must be 18.51... times that, or 2.1817 


#11
Jul1503, 12:04 PM

P: 9

But if we pack five Regular Tetrahedron pieces together, surely we can get a Pentagonallike structure. I do not know its formal name, therefore I just called it Simplex Star. A Simplex Star formed by 5 regular tetrahedrons, has 7 vertices, 10 faces and 15 edges. Five of the faces have a common vertex, another five of the faces have another common vertex. I have another question here: Let s be the side length of the regular icosahedron and R the circumradius, what is the relationship between s and R ? 


#12
Jul1503, 12:37 PM

P: 192

Well...I think it's impossible to form an icosahedron with tetrahedrons because you cannot put tetrahedrons "around" a point to form a solidangle of 360...unlike triangles...which can be put this way and form a 360 planeangle...



#13
Jul1503, 04:28 PM

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Have you tried to glue 20 regular tetrahedron models to form the regular icosahedron shape ? I did so, and it is definitely possible. I found the relationship between s and R (of previous question) is R = s or The circumdiameter of the regular icosahedron is two times of the edge length of the regular tetrahedron (or the regular icosahedron itself). Would you please disprove this ? Thank you. 


#14
Jul1603, 03:50 AM

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I made a mistake...
http://mathworld.wolfram.com/Icosahedron.html 


#15
Jul1603, 01:31 PM

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It says the icos volume is (5/12)(3+sqrt(5)) which is the volume of 20 pyramids each with unit equilateral base and height of (3sqrt(3) + sqrt(15))/12 and this height is called the "inradius", it is the distance from the center out to the center of a face, or if you prefer it is the radius of an inscribed sphere. Moreover the page gives an elegant discovery of the Greek mathematician Apollonius. Apollonius found that the ratio of surface areas of Icosahedron and Dodecahedron is equal to the ratio of their volumes. Or so this website appears to say. If true it is a new one to me. I seem to recall that Archimedes discovered a similar fact about the sphere and its cylindrical envelope. The picture is of a tomato sitting inside a tin can just big enough to contain it. And the ratio of the surface areas (2/3) is the same as the ratio of volumes. Apollonius of Perga was born around 261 BCsome 25 years after Archimedesand dwelt in Alexandria. 


#16
Jul1603, 04:31 PM

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Ok, short sketch of a proof that a regular icosahedron cannot be made up of regular tetrahedra:
First, consider a regular tetrahedron. In particular, consider the perpendicular bisector of one of its edges (it is a plane). The two opposite vertices of the tetrahedron lie in this plane (i.e. the two that aren't the endpoints of the edge in question). In this plane, one can find the angle the tetrahedron subtends; if A and B are the opposite vertices and M is the midpoint of the edge, then the measure of angle AMB is 70.53 degrees. (Which I computed from the coordinates I computed in a previous post) Now, consider a regular icosahedron. The top 5 faces are five regular triangles meeting at a point. The outer edges of these triangles form a regular pentagon. Find coordinates for the vertices of the pentagon, and then the coordinates of the central point. Now, select two of the adjacent triangles. If the regular icosahedron could be made of 20 regular tetrahedra, then these two adjacent triangles must be faces of one or more of those tetrahedra. Consider the perpendicular bisector of the edge joining these two triangles. Again, we know the coordinates of the opposite vertices and the midpoint, so we can compute the angle in this plane subtended by the icosahedron; it is 138.19 degrees. Since 138.19 is not a multiple of 70.53, we cannot pack the tetrahedra into the icosahedron; one tetrahedron is not enough to form both of these triangles, and if we tried to use two tetrahedra, they'd overlap by about 3 degrees. 


#17
Jul1703, 02:51 AM

P: 9

Thank Hurkyl again.
I just visited a web site having many useful information. http://www.intent.com/sg/polyhedra.html There is a Table of Platonic And Archimedean Solids (Updated: Wed, 26 Mar 2003.) I got some values as below  Platonic Icosahedron total surface area (when edge length = 1) : 9.464101615 total surface area (when circumradius = 1): 9.464101615 Because their total surface areas are the same, I think they are (it is) the identical icosahedron, and it become clear that A regular icosahedron is made of 20 regular tetrahedrons with the same edge length. But I am still confused..... Does any expert know why their volumes are not in the ratio 1:20 (in the same table) ? Volume (* in multiples of edge length cubed ) Regular tetrahedron: 0.117851130 Regular icosahedron: 2.181694991 http://www.intent.com/sg/polyhedra.html 


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