Acme Posted April 3, 2014 Share Posted April 3, 2014 Janus, As I recall, the tetrahedra on the left do not complete the sphere. snip... Regards, TAR You are correct. 20 tetrahedra do not close-pack to form an icosahedron. I don't have a ready reference, but investigated this with a friend some years ago when discussing Fuller's Synergetics: Explorations In The Geometry of Thinking. Link to comment Share on other sites More sharing options...

Janus Posted April 3, 2014 Share Posted April 3, 2014 Janus, As I recall, the tetrahedra on the left do not complete the sphere. That is, if you make 12 regular ones and fit them together, they "look" like they are going to complete a "sphere", but they are a "little" off. Resulting in the reality, that if you are to join all the edges on the outside...you can't. The radii going toward the center are just a little too long...and you wind up with spaces. This would mean that twelve balls would not exactly fit around a center sphere of the same radius, in this arrangement. However, it appears that twelve balls do fit exactly around the figure on the right, and the edges shown are exactly one radii long. As are the internal edges of the six pyramids and eight tetrahedra, that point toward the center (not shown in the wire frame). This results in the fact that placing a ball at each vertices, each with a diameter equal to the length of an edge, and one in the center of the figure will result in the ping pong ball arrangement, and each ball will use exactly half its diameter to reach the outside edge of each of the twelve balls that surround it. Being that this positioning of balls is exactly the same as taking a regular cube and placing the appropriate size ball exactly at the midpoint of each of the twelve edges of the cube, and one of the same size in the exact center of the cube, the division of the cube into twelve equal solid angles, from its center point, is assured by this arrangement. As well, the division of the sphere in the center, into twelve equal solid angles is likewise insured. The diamond shapes in my original figure, are the result of forming the flat perpendiculars exactly half way between the adjacent pen holes(vertices). So in this regard, my close packing, wins, and space can be discribed, from any one point, as either 6 square areas and 8 triangular ones, in this arrangement, or the twelve identical diamond shapes that result from this arrangement. Regards, TAR regular tetrahedra won't fit snugly, but irregular ones can. The one on the left still allows you to fit 13 balls in the same sphere as the one on the right, the balls just won't all fit snugly to each other. IF that is a consideration, then the one one on the right is the choice you want. (However, as a D&D die, the shape on the left is better as the one of the right would be "loaded". ) Returning to your original topic, here's another way of dividing up a sphere into 12 identical sections, shown next to the first method. Link to comment Share on other sites More sharing options...

tar Posted April 3, 2014 Author Share Posted April 3, 2014 Janus, But isn't the one on the right, our pentagonal arrangement, with the problems? I don't think you can use the one on the right, to seed a matrix with the "dual" characteristics of the one on the left. I know I had considered the one on the right. Probably could find a construction paper model like that (with flat faces) to prove it, if I had to, in some box or drawer. It just doesn't have the possibilities and is not quite so "loaded" with symmeties and nice finds, and "forced" qualities, as the one on the left, offers. My quess is, if pentagons were better than hexagons, we would not have honeycombs and snowflakes taking the hexagon route. All we have to back up the pentagon is starfish...well probably other things, but they do not come to mind. I was looking at some images on the web the other day, of crystals and electron microscope renderings and we have a lot of hexagonal stuff. I am trying to see what the geometric space rules are, that things tend to mind. To answer questions such as why electrons have the shell rules they have, for instance. Regards, TAR oh, and by the way, calling the figure "mine" is just a figure of speech, it is obviously "ours", and rather natural and evident and previously discovered, at that Link to comment Share on other sites More sharing options...

Janus Posted April 4, 2014 Share Posted April 4, 2014 Janus, But isn't the one on the right, our pentagonal arrangement, with the problems? Here are two side by side: While the original one does have pentagons formed of 5 triangles each, the pentagons don't all fit edge to edge. You can have only two pentagons that do not share any triangles and those pentagons are joined by triangular faces. The one on the right is made up of 12 pentagons that only share edges. if you take it one step further and form these pentagons from triangles like om the left shape, you end up with a solid with 60 faces. Link to comment Share on other sites More sharing options...

imatfaal Posted April 4, 2014 Share Posted April 4, 2014 Here are two side by side: While the original one does have pentagons formed of 5 triangles each, the pentagons don't all fit edge to edge. You can have only two pentagons that do not share any triangles and those pentagons are joined by triangular faces. The one on the right is made up of 12 pentagons that only share edges. if you take it one step further and form these pentagons from triangles like om the left shape, you end up with a solid with 60 faces. Tar Janus Those two shapes are connected as duals. If you take the one made of triangles (20 of them thus called an icosahedron) and cut the vertices off you will be able to visualise that you will get a little pentagon emerging as the cut face. As you carry on shaving off the vertex the pentagon gets bigger, then starts overlapping its neighbour, and finally you are left with the figure on the right made from 12 pentagons (the dodecahedron). Similarly with the dodecahedron will become the icosahedron when the vertices are shaved off. dodecahedron - pentagons - 12 faces - 20 vertices - 30 edges icosahedron - triangles - 20 faces - 12 vertices - 30 edges Link to comment Share on other sites More sharing options...

tar Posted April 4, 2014 Author Share Posted April 4, 2014 imatfaal, Shaving off the vertices, in a manner normal to the center, til the new surfaces meet is the same way you get to my original figures arrangement, starting with a cube. I have not taken angle measurements on my clay bar, for obvious reasons, but noticed just before, a rather familiar looking diamond in the middle of the pentagonal arrangement on the right, above. Was wondering if anybody knows the angles of the diamonds in my original figure. The measurement of the obtuse angle, and of the acute, and what the ratio of the length of one diagonal is, to the other. Regards, TAR Link to comment Share on other sites More sharing options...

imatfaal Posted April 4, 2014 Share Posted April 4, 2014 Ratio of long diagonal to short is sqrt(2) (this is for the flat edged polyhedron not for the spheroid). If the short diagonal is 2 units long, the long diagonal is 2.sqrt(2), and very happily the length of the side is sqrt(3). The small angle is 70.53 degrees. The rhombic dodecahedron is the dual of the cubic octahedron (a nice mixture of equitriangles and squares). The rhombic dodecah can be formed in a really neat way; if you arrange 7 cubes with one in the centre and one each of each of its six faces you get a sort of cross in 3d. Draw a set of lines linking the centre of each of the 6 outer cubes and you have the edges of a rhombic dodecah. If you draw lines along all the short diagonals of a rhombic dodecah you get a cube and along the long diagonals you get an octahedron (the cube and the octahedron being duals of each other) Great shape! 1 Link to comment Share on other sites More sharing options...

Janus Posted April 4, 2014 Share Posted April 4, 2014 (edited) Tar Janus Those two shapes are connected as duals. If you take the one made of triangles (20 of them thus called an icosahedron) and cut the vertices off you will be able to visualise that you will get a little pentagon emerging as the cut face. As you carry on shaving off the vertex the pentagon gets bigger, then starts overlapping its neighbour, and finally you are left with the figure on the right made from 12 pentagons (the dodecahedron). Similarly with the dodecahedron will become the icosahedron when the vertices are shaved off. You mean like this: On its way you can note that it passes through this shape: As well as the pentagon-hexagon pattern familiar from a soccer ball. Edited April 4, 2014 by Janus 2 Link to comment Share on other sites More sharing options...

tar Posted April 5, 2014 Author Share Posted April 5, 2014 (edited) Imatfaal,Was wondering a few things. Looking at your profile figure, I noticed the four points at the top frontish seem to describe a square and the two of these, along with a point on the left describe a triangle. Is your figure, the same arrangement as the cubic octahedron? It looks like it is made of intersecting tetrahedra, but I would be interested to know how it is put together.And after looking through this link on the cubic octahedron http://kjmaclean.com/Geometry/Cubeoctahedron.htmlwith the vector equilibrium idea, and the fact that it fit more snugly into a unit sphere, than the cube, I was wondering if we could come up with a regular strategy, based on this arrangement, to derive a figure that fit even more snugly into the unit sphere. If so, it might be possible, not to square the circle, but to sphere the cube, and thereby derive a function whereby the increasingly small diamond shapes of the same ratio, would have, at each iteration an exact number of which would complete the figure. If such a function could be described, when taken to the limit, to its smallest integral, approaching an infinite number of sides/diamonds, of increasingly small dimension, but the same ratio of diagonals...we would have a function that could sphere a cube. Leading me to wonder if we could find Pi in the ratio of the diagonals of the diamonds derived and depicted in my original clay figure.Regards, TARor perhaps, not that we could, but that we should be able to find Pi in the ratio of the diagonals Janus,Can you do MD65536's divisions in quarters of the diamonds, keeping all the vertices on the surface of the unit sphere?And then divide each of the resulting diamonds in the same quarters, again keeping all the vertices on the surface of a unit sphere?We might already have the strategy. Conceptionally use the TARadian, whose solid angle is 1/12 of a sphere's, and the surface area of a Janus section which is 1/12th the surface area of a sphere, and divide the section in quarters, which will simultaneously divide both the solid angle and the surface area into quarters.Regards, TAR Edited April 5, 2014 by tar Link to comment Share on other sites More sharing options...

imatfaal Posted April 5, 2014 Share Posted April 5, 2014 Imatfaal, Was wondering a few things. Looking at your profile figure, I noticed the four points at the top frontish seem to describe a square and the two of these, along with a point on the left describe a triangle. Is your figure, the same arrangement as the cubic octahedron? It looks like it is made of intersecting tetrahedra, but I would be interested to know how it is put together. And after looking through this link on the cubic octahedron http://kjmaclean.com/Geometry/Cubeoctahedron.html with the vector equilibrium idea, and the fact that it fit more snugly into a unit sphere, than the cube, I was wondering if we could come up with a regular strategy, based on this arrangement, to derive a figure that fit even more snugly into the unit sphere. If so, it might be possible, not to square the circle, but to sphere the cube, and thereby derive a function whereby the increasingly small diamond shapes of the same ratio, would have, at each iteration an exact number of which would complete the figure. If such a function could be described, when taken to the limit, to its smallest integral, approaching an infinite number of sides/diamonds, of increasingly small dimension, but the same ratio of diagonals...we would have a function that could sphere a cube. Leading me to wonder if we could find Pi in the ratio of the diagonals of the diamonds derived and depicted in my original clay figure. Regards, TAR or perhaps, not that we could, but that we should be able to find Pi in the ratio of the diagonals.... Not sure what you are looking at Tar. The points are in exactly the same places as the vertices of a dodecahedron. I don't have the model in front of me but I don't think there are any instances of 4 way symmetry. It is enantiamorphic though - which means it has a flipped mirror version! Link to comment Share on other sites More sharing options...

Acme Posted April 5, 2014 Share Posted April 5, 2014 Not sure what you are looking at Tar. The points are in exactly the same places as the vertices of a dodecahedron. I don't have the model in front of me but I don't think there are any instances of 4 way symmetry. It is enantiamorphic though - which means it has a flipped mirror version! Perhaps I'm mistaken, but I think TAR was referring to your avatar. Link to comment Share on other sites More sharing options...

imatfaal Posted April 5, 2014 Share Posted April 5, 2014 (edited) Perhaps I'm mistaken, but I think TAR was referring to your avatar. I guessed - but I should have made it clear I was working on that assumptionl; I meant I could not see where on my avatar Tar was seeing a square. My avatar is indeed 5 interlocking tetrahedra. It is made from 30 pieces of origami paper (well ten squares cut into 3 strips each) - it's a construction for the modular origami masochist. I really cannot describe how the interlacing of the origami version works in practice - that one photo'd took about 3 hours of a very boring teleconference to assemble! I once made a very expensive one with Benjamin Franklins face on each strip - I was my boss's 3 grand unfortunately - I still have the Greenback version I made as proof of concept. Mathematically I believe it is an inscribed tetrahedron in each of the cubes that can be inscribed in an dodecahedron - as each cube could have two tetrahedron inscribed that is where you get the enantiamorphism (one set of inscriptions is left handed the other right handed) How it fits together - well If Janus still has his modelling gloves on he could probably whip one up on his cad programme. Take a dodecahedron with a vertex straight at your eye - draw a straight line through the shape to another vertex, you should miss two on the way (one to the left and one to the right - or vice versa). Do this another two times always going away from you and missing two other vertices to reach almost the "other side" of the dodecahedron (it isn't quite the other side). You should be able to join the far ends of the three lines you have made into an equilateral triangle. There is the first tetrahedron. Rinse Repeat four times. I quite like the idea of the TaRadian;; pretty sure it wont catch on tho. There is another interesting thing about the rhombic dodeca - it fills space; unlike the standard dodeca, you can pack the rhombic so that it leaves no spaces Edited April 5, 2014 by imatfaal to put in an important "not" Link to comment Share on other sites More sharing options...

tar Posted April 6, 2014 Author Share Posted April 6, 2014 (edited) Imatfaal, I am just giggling rather hard at myself, having made 4 tetrahedron out of color clay balls and toothpicks and was in the process of trying to interlace them. I guess I am easily amused to find clay balls and toothpicks and my inability to sort out the pattern so hilarious. I took a picture, of the source of my amusement, which is definitely NOT your avatar arrangement yet. I'll post it in a bit. Just for fun. Regards, TAR ...WOW... Janus, You really have to do the cutting the diamonds in quarters thing. The new lines trace out the rhombic dodecahedron again. Or so it seems on my masacred clay ball. Regards, TAR well wait, not suprising at all. I drew the lines thru my pen holes. Duh. But I sure would like to see the Janus rendition of the resulting 48 sided figure. We could change the thread title into "dividing the sphere into 48 identical sections" Its neat. You can follow a pair of sides zigzaging all the way round the sphere and landing where you started. Then turn the big diamond 70.53 degrees (or 109.47 the other way) and follow pairs of sides around in the same zigzag. The only big diamonds you miss, doing this, are the two at the acute tips of your diamond. So you can follow pairs of sides around and see 24 little diamonds on each transit. The opposite big diamond (the opposite 4 little ones) you transit on both trips. The eight little ones off the tips of your starting diamond, you never transit following the pairs of sides around. Anyway, neat 48 sided figure, with some possibilities for further exploration. Sure would be nice to see that wireframe...(hint, hint) Here also attached it a first draft of laying the twelve sections of a sphere out where you can see the whole surface at once. Edited April 5, 2014 by tar Link to comment Share on other sites More sharing options...

tar Posted April 6, 2014 Author Share Posted April 6, 2014 Probably best to flip that diagram 90 degrees and put the four lengthwise diamonds along the equator, if one was to use it to divide up the globe, inorder to see the whole thing at once. Nice arrangement, as that it is not a projection, but an actual division. You could even leave the curvature and the terrain on a section, and it would still work to see the whole thing at once. Link to comment Share on other sites More sharing options...

Acme Posted April 6, 2014 Share Posted April 6, 2014 (edited) Probably best to flip that diagram 90 degrees and put the four lengthwise diamonds along the equator, if one was to use it to divide up the globe, inorder to see the whole thing at once. Nice arrangement, as that it is not a projection, but an actual division. You could even leave the curvature and the terrain on a section, and it would still work to see the whole thing at once. What you have drawn is nearly a planar net, were it not for the diamonds 'floating' free and vertex joinings. >> planar net @ Wiki http://en.wikipedia.org/wiki/Net_(polyhedron) Edited April 6, 2014 by Acme Link to comment Share on other sites More sharing options...

tar Posted April 6, 2014 Author Share Posted April 6, 2014 (edited) Acme, I am working on the previous suggestion to try the divisions out on the globe. It worked out really nice in preparation, my globe was about a foot in diameter so I took 6 inch peices of tape brought on down from the North Pole on the Prime Meridian, another on the date line, and another two between. (looks like a cross from the north pole. Did the same from the south. From the ends of the tapes coming from the poles made the wide angles of the four diamonds running lengthwise around the equator by simply putting the one end of the 6in. piece at the end of the tape coming from the pole, and the other end of the tape on the equator. It made the perfect angles. I aborted my effort. I have to take new pictures, minding my lighting and distance from the globe on each shot, and be a lot more careful photoshopping my sections, but here is a "rough draft" beginning of substituting globe sections for the diamonds in my diagram. Nevermind. Its incomplete and lousy to boot. I'll post something when its presentable. Could be a while. Edited April 6, 2014 by tar Link to comment Share on other sites More sharing options...

Janus Posted April 6, 2014 Share Posted April 6, 2014 Janus, Can you do MD65536's divisions in quarters of the diamonds, keeping all the vertices on the surface of the unit sphere? And then divide each of the resulting diamonds in the same quarters, again keeping all the vertices on the surface of a unit sphere? Regards, TAR It can't be done by dividing each section into diamonds,and have all the divisions have identical shapes. Look at the sphere on the left in post #27. Look at where the orange, blue and red sections meet, the interior angles of each at that point is 120 degrees. Now look at the left point of the orange section. It meets with a blue, red, and yellow section at this point and the interior angles are 90 degrees. If you draw lines along the surface from the midpoint of each side of the section to the midpoint of the other side, and then do the same for the other two sides, they cross in the middle at right angles to each other and are at a right angle were they meet the sides. You have divided the section into four pieces. Call them Starting at the top and going clockwise, call them North, East, South and West. The East and West pieces both have 90 degree interior angle at each corner. However, the North and South pieces have 3 90 degree corners and 1 120 degree corner. Not only that but the four corners of the curved section with the 120 degree interior angle don't lie in the same plane, so you can't even flat surface for a polygon side out of it. Changing how you slice up the section doesn't help if you keep the existing interior angles intact. Each piece would have to have at least 1 90 degree and 1 120 degree interior angle, and there is no way to get four identical pieces with this requirement to fit the existing section. I know that it looks like it should be possible at first glance, but you have to take into consideration that you are working with a curved surface. You can take another route, which is divide the section up from the corners. Draw a line from each corner to the opposite corner. You have now divided the section into 4 triangular pieces that each have a 90, 45, and 60 degree internal angle. And since the pieces are triangular, there is no issue with all the corners being the same plane. However, even these new sections aren't identical. Using the same nomenclature as before, the Northwest and Southeast sections are mirror images of the Northeast and Southwest ones. Link to comment Share on other sites More sharing options...

tar Posted April 6, 2014 Author Share Posted April 6, 2014 (edited) Janus, Got it. I think my clay was too forgiving and the slivers could have fit anyway I wanted them to. Thanks for trying it out with real angles. Guess that brings the TaRadian to a screaching halt. If they don't divide into identical sections then the volume and surface areas can not be both 1/4ed with each successive division and the plan is poop. I was considering that if the solid angle at the center was quartered and the diamond shape was quartered, the interior angles would take care of themselves. Looks like I thought wrong. Anyway I really do appreciate you building out the ping pong ball arrangement, I have been trying to do that for a really long time. Still an amazing arrangement, just not as amazing as I was thinking it might be. Guess I will abandon the globe splitting quest. Without being able to reliably divide the solid angles in a manner that would keep the sections identical, there is no hope of using it as a measure of solid angle. Looks like we have the steradian for a good reason. Thanks again Janus. Was sort of exciting there for a moment. Regards, TAR Edited April 6, 2014 by tar 1 Link to comment Share on other sites More sharing options...

md65536 Posted April 7, 2014 Share Posted April 7, 2014 Guess that brings the TaRadian to a screaching halt. If they don't divide into identical sections then the volume and surface areas can not be both 1/4ed with each successive division and the plan is poop. I was considering that if the solid angle at the center was quartered and the diamond shape was quartered, the interior angles would take care of themselves. Looks like I thought wrong.Oops, I blame myself for an earlier suggestion along those lines. I should have realized something was wrong. If you take a regular tetrahedron, you can successively divide each triangle into 4 equilateral triangles, recursively. Then you have a bunch of equal-sized triangles. If however the tetrahedron is inscribed in a sphere, and you extend all the interior triangle vertices out to the surface of the sphere, the points will not all be translated the same distance. So the triangles on the sphere would not all be equal. I should have realized that equal size on a such a polyhedron wouldn't mean equal size on a sphere, but I didn't until told. Link to comment Share on other sites More sharing options...

tar Posted April 7, 2014 Author Share Posted April 7, 2014 MD65536 and Janus, But wait. The diamonds forever are out, but you both suggested the diagnoal route to four triangular faces, from the diamond. I did a little diamond cutting, trying to cut the diamond face flat, while staying true to the internal angles and attempting to stay normal to the center of the sphere, and like Janus says, the vertices are not on the same plane, so you wind up with a "cut" diamond, looking a little like a 4 faceted jewel. Regards, TAR I cut that one a little too deep, not knowing what to expect. It appears, to stay true to form, the exact center of the diamond should still be there as the only point on the original surface. That, and the four vertices of the original diamond shape. So the diagonal lines, leaving four triangles seems like it is the way to go to 48 divisions. But it doesn't look like the pieces with be identical. They appear to be, like Janus said, mirror images of each other. Same dimensions and angles, but one shape is lefthanded and the other right. So of the four, the catty corners are identical to each other, but the neighbors are mirror images, along the diagonals. Don't know if you can split these guys in 4, or 2 or 3. Maybe I'll wait to see if this way to 48 facets, with all vertices on the unit sphere, works. Link to comment Share on other sites More sharing options...

Mike Smith Cosmos Posted April 7, 2014 Share Posted April 7, 2014 (edited) What happens to me ? I am standing here with a pot of white paint, a compass and a 3 meter rod, and a ticket to a volcanic island in the middle of the Atlantic Ocean .? Sorry the rod more exactly is........hummm..... [ Radius of Earth 6271000 meters and kept dividing by two until I got to 3.0379 meters.] I am off to catch the plane , ! De dum de dum! .... Look out ,or in , from space ,with google earth for the white line appearing ...hopefully in a NW SE direction. ..in the middle of Gran Canaria , wait a minute have I got that wrong ? SW to NE I mean . Got it . " Houston ! We are go for launch " ---------/ yeahhhh! Mike Edited April 7, 2014 by Mike Smith Cosmos Link to comment Share on other sites More sharing options...

Janus Posted April 7, 2014 Share Posted April 7, 2014 jeweled diamond 1.jpg MD65536 and Janus, But wait. The diamonds forever are out, but you both suggested the diagnoal route to four triangular faces, from the diamond. I did a little diamond cutting, trying to cut the diamond face flat, while staying true to the internal angles and attempting to stay normal to the center of the sphere, and like Janus says, the vertices are not on the same plane, so you wind up with a "cut" diamond, looking a little like a 4 faceted jewel. Regards, TAR Something interesting happens when you try and fit these pieces to make a faceted solid, you get this: The facets from each piece line up with the facets of adjacent pieces, forming 1 side. So you end up with a 24 identical, if oddly shaped, faces. The same type of thing happens with tetrahedra. Take 5 regular ones, start with one and fit the other four to each of its sides. You end up with a cube. I cut that one a little too deep, not knowing what to expect. It appears, to stay true to form, the exact center of the diamond should still be there as the only point on the original surface. That, and the four vertices of the original diamond shape. So the diagonal lines, leaving four triangles seems like it is the way to go to 48 divisions. But it doesn't look like the pieces with be identical. They appear to be, like Janus said, mirror images of each other. Same dimensions and angles, but one shape is lefthanded and the other right. So of the four, the catty corners are identical to each other, but the neighbors are mirror images, along the diagonals. Don't know if you can split these guys in 4, or 2 or 3. Maybe I'll wait to see if this way to 48 facets, with all vertices on the unit sphere, works. Link to comment Share on other sites More sharing options...

vampares Posted April 7, 2014 Share Posted April 7, 2014 I thought the question was connect the 20 dots and get 30 simple lines. Is this the sphere with lines, cut to the center. There are 12 pieces. Looking at the tetrahedron of the dodecapentahedron, at what angle are the adjacent non-members? Link to comment Share on other sites More sharing options...

tar Posted April 7, 2014 Author Share Posted April 7, 2014 Janus, That is really interesting. Perhaps the four hexagonal planes (and the three square planes) at just those right angles to each other, conspire to equalize the vectors as Mr. Fuller noticed. Perhaps it really is as "magical" (meaning really really neat) a figure as I have sensed. Imagine what you did there. Cut twelve in 1//4s to get 24. Interesting indeed. On a parallel note, I was rethinking the cutting of the diamond into smaller diamonds. Granted it does not work so neatly if you are trying to make faces, but what if you stay on the surface? I remembered on the way home today, that when I drew my 48 and then 192 diamonds on my clay ball, I never broke the surface. The lines I was drawing where always exactly one radius away from the center of the sphere (cause they where on the surface). It is not a matter of figuring was surface area the shape is subtending, because its ON the surface and the area is exactly what the area is. In the case of the 192 diamonds, that would be 1/192nd the surface area of a sphere. Regards, TAR Link to comment Share on other sites More sharing options...

vampares Posted April 7, 2014 Share Posted April 7, 2014 twelve quarters on a cue ball is the pummel of a sword as much to blame as the man who welds it? Link to comment Share on other sites More sharing options...

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