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Can someone explain this proof to me?

  1. Oct 17, 2009 #1
    I posted this in the geometry subforum, but I think that might have been the wrong place to post this.

    Given two triangles with vertices A1, B1, C1 and A2, B2, C2 respectively. A1A2, B1B2, C1C2 are extended to meet at a point V say. Now, B1C1 and B2C2 are extended to meet at L, A1B1 and A2B2 meet at N and A1C1 and A2C2 meet at M. Prove that L, M and N are concurrent.

    Proof (as given in text):

    Let A1B1C1 be the reference triangle and V be the unit point (1,1,1). A2 is on the join of A1(1,0,0) and V(1,1,1), so it can be taken as (1+p,1,1). Similarly, the point B2 is given by (1,1+q,1) and C2 by (1,1,1+r).

    Now, the line B2C2 is

    [tex]\left|\stackrel{\stackrel{x}{1}}{1}\stackrel{\stackrel{y}{1+q}}{1}\stackrel{\stackrel{z}{1}}{1+r} \right|[/tex] = 0.


    The point L is given by x = 0, y{1-(1+r)} + z{1 - (1+q)} = 0

    i.e. x=0, [tex]\frac{y}{q} + \frac{z}{r}[/tex] = 0

    Therefore, L lies on the line [tex]\frac{x}{p}+ \frac{y}{q}+ \frac{z}{r}[/tex] = 0. By symmetry, so do M and N.

    Hence proved


    From start to finish, I can't get it. Can someone please explain to me what all this means?

    For the diagram that goes with the proof, please refer to my topic in the geometry section

    https://www.physicsforums.com/showthread.php?t=345248
     
  2. jcsd
  3. Oct 17, 2009 #2

    tiny-tim

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    Hi anirudh215! :smile:
    Sorry, I don't understand this at all :redface:

    how are A B and C (1,0,0) etc? and why is V (1,1,1)? and what's a "reference triangle"? :confused:
     
    Last edited by a moderator: Apr 24, 2017
  4. Oct 18, 2009 #3
    Exactly what I want to know! I saw this in a geometry book entitled "Teaching of Higher Geometry in schools - A report for the Mathematical Association" Published by Bell and Sons. I've been blinking over many proofs in this text. What gets me down is it says it's for high school students!
     
  5. Oct 18, 2009 #4

    tiny-tim

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    mathworld and archive.org

    hmm … that's at http://openlibrary.org/b/OL6159907M/teaching_of_higher_geometry_in_schools", but with no preview … why are you looking at a book published in 1953? :confused:

    This proof seems to be using trilinear coordinates … see Eric W. Weisstein's http://mathworld.wolfram.com/ReferenceTriangle.html" [Broken] :smile:

    You can find downloadable books on "trilinear coordinates" (or any subject!) by doing a search at archive.org … see http://www.archive.org/search.php?query=trilinear AND mediatype:texts"
     
    Last edited by a moderator: May 4, 2017
  6. Oct 18, 2009 #5
    Re: mathworld and archive.org

    Heh. I found it on my prof's shelf, and while he made me wait in his office (as all professors do), I picked up this one and began reading it. It had a few very cool proofs. This one was so intriguing, I just HAD TO find out about it.

    So I did find out about it. It's not trilinear co ordinates but is something that is called barycentric co-ordinates. If you are also interested, you can find out more about it from H.M. Coxeter, Introduction to Geometry, page 216. This is damn interesting!!
     
    Last edited by a moderator: Apr 24, 2017
  7. Oct 18, 2009 #6

    tiny-tim

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    Last edited by a moderator: Apr 24, 2017
  8. Oct 18, 2009 #7
    Oops. Well, I don't know what trilinear coordinates are either, so I thought you'd made a mistake. Sorry. But that's basically it. Now how does one explain the problem?
     
  9. Oct 18, 2009 #8

    tiny-tim

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    erm :redface: … one (ie you) looks at mathworld and wikipedia and the archive.org free books until one has it! :wink:

    and then one explains it to everyone here! :smile:
     
  10. Oct 18, 2009 #9
    Will do! :)
     
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