A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

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.

  • This Question is Closed
  1. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \((\pm (1-h^2),0,(1+h))\)

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How do you get that? I havent seen anything like that before for problems like these.

  3. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hahaha.. it is from https://en.wikipedia.org/wiki/Dodecahedron part "Catesian Coordinate" , read it.

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  5. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, hihihi.... I don't know how to prove.

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Lol, gotcha. No worries then. Yeah, I assumed there was some way to start incorporating triangles into it all.

  7. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Question: is it a regular one??

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 .

  9. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If it is a regular one, then its vertices are on the circle center (0,0,0) and radius 1, right?

  10. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the sphere, not circle.

  11. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (a,0,b) is the vertex on xz plan, we can find it by vector space, right?

  12. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh, forget it. I am crazy.

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah, thatd be too eays becuase yeah, itd be on the x-axis, (1,0,0). SO definitely dont think its regular

  14. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @zzr0ck3r any idea?? please.

  15. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Empty

  16. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  17. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Actually this doesn't seem right. There is no "equatorial plane at y=0" is there? Hahaha damn.

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Can we be certain its regular, though?

  19. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    They are supposed to be greater than 0. That's how its stated in the problem.

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And I suppose you're right, lol. I guess you could have mishapen thing you wished.

  22. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  23. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  25. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  26. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |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.

  27. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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!

  28. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |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.

  31. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  33. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.