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Thread: HSC 2017-2018 MX2 Marathon ADVANCED

  1. #26
    Supreme Member seanieg89's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    Find the maximal possible angle that a circle centred at the origin can meet the ellipse x^2/a^2 + y^2/b^2 = 1.

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    Supreme Member seanieg89's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    (The above Q isn't too hard, just something to get this thread rolling again .)

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    Taking a break! dan964's Avatar
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    Re: HSC 2017-2018 MX2 Marathon ADVANCED

    Update 2018
    This marathon will continue into 2018.
    2014 HSC
    2015 Gap Year
    2016-2018 UOW Bachelor of Maths/Maths Advanced.

    Links:
    View English Notes (no download)
    Maths Notes - Free Sample

    thsconline: https://thsconline.github.io/s/
    For 2017 papers, check the thread in the Mathematics forum.

    Dan will be back 14/10/17

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    -insert title here- Paradoxica's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    Quote Originally Posted by seanieg89 View Post
    Find the maximal possible angle that a circle centred at the origin can meet the ellipse x^2/a^2 + y^2/b^2 = 1.
    WLOG, assuming a=1, the maximum angle is 2cot⁻¹(b)

    If this is correct, I will post the working out later, I'm a bit all over the place atm.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Supreme Member seanieg89's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    Quote Originally Posted by Paradoxica View Post
    WLOG, assuming a=1, the maximum angle is 2cot⁻¹(b)

    If this is correct, I will post the working out later, I'm a bit all over the place atm.
    Honestly can't remember, but will check this afternoon. (It's just a simple calculation of course.)

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    Supreme Member seanieg89's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    Also, a new problem in the meantime of a different flavour:

    An ellipse is inscribed in the triangle ABC, with points of contact X,Y,Z on the triangle sides opposite A,B,C respectively.

    Prove that AX,BY,CZ are concurrent. (That is, prove that there exists a point that all three of these lines pass through.)

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    Supreme Member seanieg89's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    Quote Originally Posted by Paradoxica View Post
    WLOG, assuming a=1, the maximum angle is 2cot⁻¹(b)

    If this is correct, I will post the working out later, I'm a bit all over the place atm.
    I don't think that is quite correct, but it is close. (Perhaps an erroneous trig manipulation?)

    As a test, if , then we expect the maximal angle of contact to be close to , because the circle and the ellipse are basically identical. Also, the maximal angle should never exceed .

  8. #33
    -insert title here- Paradoxica's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    Quote Originally Posted by seanieg89 View Post
    I don't think that is quite correct, but it is close. (Perhaps an erroneous trig manipulation?)

    As a test, if , then we expect the maximal angle of contact to be close to , because the circle and the ellipse are basically identical. Also, the maximal angle should never exceed .
    what do you mean by that? is the circle not allowed to have arbitrary radius?

    also I know what my mistake was, inverse trig simplification is a pain :P
    Last edited by Paradoxica; 13 Oct 2017 at 3:34 PM.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  9. #34
    -insert title here- Paradoxica's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    Quote Originally Posted by seanieg89 View Post
    I don't think that is quite correct, but it is close. (Perhaps an erroneous trig manipulation?)

    As a test, if , then we expect the maximal angle of contact to be close to , because the circle and the ellipse are basically identical. Also, the maximal angle should never exceed .
    So I believe the answer is 2tan⁻¹(b) - π/2 or something like that.

    Probably should have an absolute around that since we're dealing with absolute angle measures.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  10. #35
    Supreme Member seanieg89's Avatar
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    Re: HSC 2017-2018 MX2 Marathon ADVANCED

    My meaning was that "angle of contact" means "acute angle of contact", otherwise the two circles that meet the ellipse tangentially give you the maximal angle (pi) and the problem is boring.

    Your new answer seems correct with absolute values.

    My answer (including the a-dependence) was the equivalent expression


  11. #36
    Supreme Member seanieg89's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    Quote Originally Posted by seanieg89 View Post
    Also, a new problem in the meantime of a different flavour:

    An ellipse is inscribed in the triangle ABC, with points of contact X,Y,Z on the triangle sides opposite A,B,C respectively.

    Prove that AX,BY,CZ are concurrent. (That is, prove that there exists a point that all three of these lines pass through.)
    Bump to move the unanswered question to the front. This one is a bit harder (in a certain sense) than the previous problem, unless there is a shortcut other than my intended method. Good to see some thread activity!

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    -insert title here- Paradoxica's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    Quote Originally Posted by seanieg89 View Post
    Bump to move the unanswered question to the front. This one is a bit harder (in a certain sense) than the previous problem, unless there is a shortcut other than my intended method. Good to see some thread activity!
    Affine Transformation Bash :P
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  13. #38
    Supreme Member seanieg89's Avatar
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    Re: HSC 2017 MX2 Marathon ADVANCED

    Quote Originally Posted by Paradoxica View Post
    Affine Transformation Bash :P
    Haha not even bashy, once you spot you can parallel project / affine transform the circle to an ellipse, there are no calculations to do, you just need to prove that the cevians meeting the incircle contact points are concurrent.

    This is within the scope of HSC geometry. Alternatively, at this stage the concurrence is trivial from Ceva's theorem, which isn't hard to prove itself.

  14. #39
    This too shall pass Sy123's Avatar
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    Re: HSC 2017-2018 MX2 Marathon ADVANCED


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