anonymous
  • anonymous
Suppose a dodecahedron is centered at the origin and two of its twenty vertices are at (0,0,1) and (0,0,-1). Suppose there is a vertex at (a,0,b) with a, b > 0. There are two different possibilities for a and b. Give the exact values for a and b for each possibility.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Loser66
  • Loser66
From (0,0,1) and (0,0,-1), we have the "height" of the weight-shape-"roof" is h =2 Hence the vertexes of the form (a, 0, b) is calculated by \((\pm (1-h^2),0,(1+h))\)
anonymous
  • anonymous
How do you get that? I havent seen anything like that before for problems like these.
Loser66
  • Loser66
hahaha.. it is from https://en.wikipedia.org/wiki/Dodecahedron part "Catesian Coordinate" , read it.

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anonymous
  • anonymous
I see. I would hope that i wouldnt have to use that, because then Id have to make the argument about why thats true, lol. Unless you see a way to explain why thats true. Otherwise we're using using some sort of triangles along with the spherical trigonometry formulas or something.
Loser66
  • Loser66
ok, hihihi.... I don't know how to prove.
anonymous
  • anonymous
Lol, gotcha. No worries then. Yeah, I assumed there was some way to start incorporating triangles into it all.
Loser66
  • Loser66
Question: is it a regular one??
anonymous
  • anonymous
Not specified, so we can't assume that it is without somehow showin git. If it were regular, I couldnt imagine there being more than one possibility for a and b like the problem states, though .
Loser66
  • Loser66
If it is a regular one, then its vertices are on the circle center (0,0,0) and radius 1, right?
Loser66
  • Loser66
the sphere, not circle.
Loser66
  • Loser66
(a,0,b) is the vertex on xz plan, we can find it by vector space, right?
Loser66
  • Loser66
oh, forget it. I am crazy.
anonymous
  • anonymous
Yeah, thatd be too eays becuase yeah, itd be on the x-axis, (1,0,0). SO definitely dont think its regular
Loser66
  • Loser66
@zzr0ck3r any idea?? please.
Loser66
  • Loser66
@Empty
Empty
  • Empty
I like this picture of the orthogonal projection of the regular dodecahedron: https://upload.wikimedia.org/wikipedia/commons/3/31/Dodecahedron_t0_H3.png Now look at the top point and bottom point, these are the points (0,0,1) and (0,0,-1) So to be clear up and down are the z-axis and left and right are the x-axis. So now where are (a,0,b)? Well it'll be one of the outer most points, since that's going to be along the equator at the plane where y=0. So there are 10 points equally spaced at y=0 so the angle between them will be \(\frac{2 \pi}{10} = \frac{\pi}{5}\). Rotating down from the top point we can see we'll have for the points pretty clearly: \[(a_1,0,b_1) = ( \cos ( \frac{\pi}{2} - \frac{\pi}{5}), 0, \sin ( \frac{\pi}{2} - \frac{\pi}{5}) )\] \[(a_2,0,b_2) = ( \cos ( \frac{\pi}{2} - 2 \frac{\pi}{5}), 0, \sin ( \frac{\pi}{2} - 2\frac{\pi}{5}) )\] You can simplify it but I am writing it this way to make it clear how I'm looking at it geometrically.
Empty
  • Empty
Actually this doesn't seem right. There is no "equatorial plane at y=0" is there? Hahaha damn.
anonymous
  • anonymous
Can we be certain its regular, though?
Empty
  • Empty
I was under the false assumption that both a and b were greater than zero, there's actually no way what I describe can be done. Also if it's not regular then doesn't this entire question become arbitrary? What's stopping us from placing multiple points anywhere we like?
anonymous
  • anonymous
They are supposed to be greater than 0. That's how its stated in the problem.
anonymous
  • anonymous
And I suppose you're right, lol. I guess you could have mishapen thing you wished.
Empty
  • Empty
Hmmm I don't know if it's possible to have a regular dodecahedron with two vertices at (0,0,1) and (0,0,-1) while having 2 other points in the first quadrant of the (x,z) plane formed at y=0.
Empty
  • Empty
Just looking at the orthogonal projection you can see where all the points will lie, there are multiple ones but none of them can be along the same meridian from +z to -z while tracing a line from the north to south pole without zig-zagging.
anonymous
  • anonymous
I wasnt sure either. I couldnt find a good image to help me see it. I was supposing you could have it tilted oddly such that it might work, but I had no idea. Not the easiest thing to imagine.
Empty
  • Empty
Well they say it's centered at the origin with points at (0,0,1) and (0,0,-1) so we are restricted to just rotating it along this fixed axis and any way we interchange the two vertices here while keeping it centered will be symmerical and unnoticeable so seems kinda funny to me. Hmmm
Empty
  • Empty
|dw:1439256007225:dw| This is the part of the picture I'm focusing on rotating. We can't rotate around the z-axis to put more than one in the y=0 plane at a time.
Empty
  • Empty
Oh wait no, what I said originally is right. I was just assuming both values of a and b had to be fulfilled simultaneously but they don't!
Empty
  • Empty
Still something seems fishy about this orthographic projection. Maybe it's not truly othographic because the dimensions seem off on it to me. Maybe I'm wrong idk.
anonymous
  • anonymous
Yeah, just two different possibilities position wise, not simultaneous :P But yeah, just making sense of what you posted before. Theres another graph there on the wiki that definitely makes it look lined up differently. But I would also think that theres more than one way to have a regular dodecahedron
anonymous
  • anonymous
|dw:1439256529549:dw| Not sure really how to draw it, but I think theres definitely a way to get the points lines up. THe bottom point is supposed to be disappearing under the solid, but Im not an artist, lol. But it looks like something like this is feasible. And, if that orthogonal projection is correct, that seems to work as well. But it does look weird because that seems to satisfy both points at once where the image im thinking of (based off another picture), only has one possibility at a time.
Empty
  • Empty
Yeah exactly what I am looking at too. Sorry my internet died for the past hour while I fumbled around with my router so sorry for the delay lol.
anonymous
  • anonymous
No worries. But yeah, thats the only thing I could imagine, honestly. If you think the orthogonal picture is trustworthy, then its pretty basic from there. If youre not sure and dont think that both of those points can exist at the same time, then maybe theres one we can eliminate or maybe theres a way to get an expression based on what I drew up, assuming it works as an idea.

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