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Thread: HSC 2018 MX2 Marathon

  1. #151
    I love trials pikachu975's Avatar
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by CapitalSwine View Post
    Cant seem to figure out how to do this question:
    https://imgur.com/a/jRaWrmq
    BSQ = PQB (alternate angle theorem)
    PQB = 180-RPQ (co-interior angles in a parallelogram are supplementary)
    BSQ + RPQ = 180

    Hence opposite angles are supplementary so RSQP is a cyclic quad

    Hope this reasoning is legit, forgot maths

    EDIT: Nvm too slow
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  2. #152
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    Re: HSC 2018 MX2 Marathon

    It is easy to prove RSQP is cyclic; so that this can be easily shown in many ways.

    e.g.

    Let angle AQS = @; .: angle QBS = @; .: angle PRS = @ (PR // QB)

    .: angle AQS = angle PRS

    .: external angle AQS of quadrilateral RSQP = int opp angle PRS

    .: RSQP must be cyclic

    QED
    Last edited by Drongoski; 29 May 2018 at 9:29 AM.
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  3. #153
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    Re: HSC 2018 MX2 Marathon











    HSC 2018

    English Adv - MX1 - MX2 - Physics - Economics

  4. #154
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    Re: HSC 2018 MX2 Marathon

    a)


    b)



    c)

    For , .

    If , then clearly .

    Therefore, .

    From b) . Since ,



    Replacing with , we get with and so



    Giving, as required,



    d)

    From b),

    .







    From c),

    altSwift likes this.

  5. #155
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    Re: HSC 2018 MX2 Marathon






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