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Math Help - deductive geometry questions

  1. #1
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    deductive geometry questions

    hi all, first time poster here

    i've been scratching my head for a couple of hours now regarding three questions on an exercise sheet i have to do. i feel like as if there's some rule or theorem that i don't know to get the answers. please help me. thanks in advance.

    Imageshack - dsc00104hr.jpg

    they would be questions 18, 20 and 21
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  2. #2
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    Re: deductive geometry questions

    Question 18. Maybe this solution is a little long, but still... I'll write, e.g., just ABC instead of "the area of ABC".

    Lemma. AB / BP = 2.
    Proof. Draw a line through R parallel to CB and let S be the intersection with AB. AR = RC => AS = SB; RQ = QP => SB = BP.
    End proof.

    Let PAC = 1. Then PRC = PRA = 1/2; CQR = CQP = 1/4. By lemma, CBA = 2 CBP = 2/3 => ARQB = 2/3 - 1/4 = 5/12. PQB = PRA - ARQB = 1/2 - 5/12 = 1/12. Finally, BQ / QC = BQP / PQC = (1/12) / (1/4) = 1/3.
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  3. #3
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    Re: deductive geometry questions

    A shorter solution to question 18. Draw a line parallel to AB through R and let T be the intersection with CB. Then AR = RC implies that BQ = QC and PQ = QR implies that BQ = QT.
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  4. #4
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    Re: deductive geometry questions

    Hi emakarov,

    after your construction

    triangles are similar as follows

    RTP -QBP ART - ACB
    RT = 2 QB =4
    BC=-2RT = 8
    QC= 8 - 2 = 6
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  5. #5
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    Re: deductive geometry questions

    Quote Originally Posted by bjhopper View Post
    triangles are similar as follows

    RTP -QBP ART - ACB
    RT = 2 QB =4
    Hmm, the vertices in your picture must be named differently from mine because the problem statement says that QB = 2. I also don't see why triangles RTP and QBP are similar...
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  6. #6
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    Re: deductive geometry questions

    Quote Originally Posted by emakarov View Post
    Hmm, the vertices in your picture must be named differently from mine because the problem statement says that QB = 2. I also don't see why triangles RTP and QBP are similar...

    my vertices are as shown by OP

    RT ppl CB as constructed
    angle RTB = QBP ca ppl lines
    angle QPB is common therefore triangle QBP is similar to RPT
    BP =TB since Q is midpoint of RP

    QB is 2 and it follows that RT is 4 BC= 8 QC the answer is 8-2=6
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  7. #7
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    Re: deductive geometry questions

    OK, it's just that in post #3 I drew the line RT parallel to AB, not BC, and T was its intersection with BC.
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