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Euclidian Proof

  1. Jan 22, 2009 #1
    No, this isn't a homework problem.

    Given any triangle with a fixed length for one edge, and a fixed internal angle for the opposite vertex, the vertex will lay on a circle, where the other edges are allowed to vary in length. But I can't seem to construct a geometrical proof. Any takers? It's deceptively simple.
     

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  3. Jan 23, 2009 #2

    tiny-tim

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    Hi Phrak! :smile:

    The angle to the circumcentre is fixed (it's double the given angle, and you can prove it using isoceles triangles), and the circumcentre lies on a fixed line. :wink:
     
  4. Jan 23, 2009 #3
    And I was just bemoaning the fact that no one responded. :smile: But wait. What's a circumcentre, and what angle from whence and to where?
     
    Last edited: Jan 23, 2009
  5. Jan 23, 2009 #4

    tiny-tim

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    I think most PF members don't like "real" geometry! :biggrin:
    he he :rofl:

    The circumcentre is the centre of the circumcircle, which is the circle which goes through all three vertices of a triangle, and the angle is the angle subtended by the original line at the circumcentre. :wink:

    :cool: join CAMREG … the CAMpaign for REal Geometry!
     
  6. Jan 23, 2009 #5
    So true. Parallel lines that meet a a point. Hubris! :surprised

    I can't imagine how you know all this. Way classical education? I had one Euclidean geometry class in High School, sans circumcentres.

    If I understand you correctly, I draw the pependicular bisectors through the remaining two sides. These will meet at an angle alpha=180-beta at the circumcentre I should expect alpha to be constant as much as beta is expected to be constant. But I can prove alpha is constant using isosoles triangles, right?
     
    Last edited: Jan 23, 2009
  7. Jan 23, 2009 #6

    tiny-tim

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    ooh, i'm sorry, Phrak :redface:

    i hadn't realised so little classical geometry is taught nowadays …

    ok, it's best to introduce this idea the other away round, by starting with a circle, and seeing what happens inside it …

    one of the most useful circle theorems is that a fixed chord AB subtends the same angle ACB at any point C on the circumference on the smaller arc, and 180º minus that angle on the larger arc, and that the angle AOB subtended at the centre is twice the latter angle …

    to prove that, let O be the centre of the circle, and let the line CO (with C on the larger arc) poke a little beyond O to Q …

    then OAC and OBC are isoceles triangles ('cos it's a circle! :wink:)

    and so you can prove that AOQ = 2ACO and QOB = 2OCB …

    and then for your problem, simply do the same proof "backwards" :smile:
     
  8. Jan 23, 2009 #7
    No problem at all. Though it's been a while since high school for me. The classical studies are more common in the UK than the US, that makes the difference.

    And thanks for the proof. :smile:
     
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