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Exercise in Geometry

  1. May 10, 2007 #1
    It is a nice question.
    It has a few different kinds of proofes.


    So who is smart here? :wink:
    Last edited: May 10, 2007
  2. jcsd
  3. May 11, 2007 #2


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    While I don't claim to be 'smart', I am at least knowledgable enough to recognize that this has nothing to do with "Tensor Analysis and Differential Geometry". I am moving this thread to "General Math".
  4. May 11, 2007 #3
    Clearly, neither propositions is true. For example, you can lengthen EO by moving A and B up the circle a bit without changing DC. There's no relationship between them at all.
  5. May 11, 2007 #4


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    How do you do that, without changing angles BTC and DTC? (From the figure, I'm assuming ATC and BTD are straight lines, though it doesn't explicitly say that).

    Proof: join A, B, C, D to O, then use the facts that a chord subtends equal angles at any point on the circumference, and the angle at the centre is twice the angle at the circumference, to get lots of similar triangles.
  6. May 12, 2007 #5
    I just couldn't find which triangles are similar..
    Last edited: May 12, 2007
  7. May 12, 2007 #6
    Haha, I read that ATC=BTD < 90 for some reason. I think all those angle symbols got me dyslexic. The picture didn't help correct this misreading.
  8. May 12, 2007 #7


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    See attachement.
    1. The angle at the bottom right is 45 by symmetry (the angles at point T are 90).
    2. The angle at the centre is twice 45 = 90.
    3. The two angles [itex]\alpha[/itex] are equal
    4 The two lengths are both [itex]R \sin \alpha[/itex] where R is the radius of the circle.

    Attached Files:

  9. May 13, 2007 #8


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    I haven't seen alephzero's solution yet because it's still "pending approval".

    I looked at this problem and realized that if you can prove that the mid point of OT is also the mid-point of EF then the rest was trivial. But then I couldn't seem to see an easy geometric proof for that mid-point result. I did however find a very easy analytic proof of this.
  10. May 14, 2007 #9


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    What happened to that Aleph's attachment, it seems to have spent a very long time (several days) "pending approval.

    Anyway I may as well post my non-geometric proof.

    Lets assume we have already done the basics and shown that angle ATE equals angle BTE equals 45 degrees. (Radius bisects chord therefore triangles ATE and BTE are congruent etc).

    1. Take a coordinate reference at "O", with x axis parallel to AB.

    2. Denote the distance OT as a. The midpoint of OT is clearly (0,a/2).

    3. The system is similar to one with circle having equation x^2 + y^2 = 1 and line DE having equation y = x + a.

    4. The coordinates of "D" and "E" are therefore given by the simultaneous solutions of equation set :

    x^2 + y^2 = 1 and y = x + a

    5. Solving the above for "y" gives :

    y^2 - ay - (1- a^2)/2 = 0

    y1, y2 = a/2 +/- sqrt(...)

    6. Clearly (y1+y1)/2 = a/2, so the midpoint of OT is coincident with the midpoint of EF.

    That's pretty much it. From there it's fairly trivial to show that lenght OF equals length ET equals one half of length AB.
    Last edited: May 14, 2007
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