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Math Help - [SOLVED] Geometry

  1. #1
    kgk
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    [SOLVED] Geometry

    In triangleABC
    D is a point on BA such that BDA = 1:2
    E is a point on CB such that CE:EB = 1:4
    DC and AE intersect at F
    I am suppose to find CF:FD(express the ratio in terms of two positive whole numbers) Help!
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  2. #2
    MHF Contributor
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    Geometry

    Hello kgk
    Quote Originally Posted by kgk View Post
    In triangleABC
    D is a point on BA such that BDA = 1:2
    E is a point on CB such that CE:EB = 1:4
    DC and AE intersect at F
    I am suppose to find CF:FD(express the ratio in terms of two positive whole numbers) Help!
    You can solve this using ratios and position vectors. See my post at http://www.mathhelpforum.com/math-he...tml#post262231 for the basic formula.

    In your problem, if we denote the position vectors of A, B, ... by \vec{a}, \vec{b}, ..., then

    \vec{d} = \frac{1}{3}\vec{a} +\frac{2}{3}\vec{b}

    \vec{e} = \frac{4}{5}\vec{c} + \frac{1}{5}\vec{b}

    If we suppose that F divides CD in the ratio 1:n, and EA in the ratio 1:m, then have two ways of expressing \vec{f}.

    Using CF:CD = 1:n, \vec{f} = \frac{n}{n+1}\vec{c} + \frac{1}{n+1}\vec{d}

    =\frac{1}{3(n+1)}\vec{a}+ \frac{2}{3(n+1)}\vec{b} +\frac{n}{n+1}\vec{c}

    Using EF:FA = 1:m, \vec{f} = \frac{m}{m+1}\vec{e}+\frac{1}{m+1}\vec{a}

    = \frac{1}{m+1}\vec{a}+\frac{m}{5(m+1)}\vec{b}+\frac  {4m}{5(m+1)}\vec{c}

    If we equate these two expressions for \vec{f} and compare coefficients of \vec{a}, \vec{b} and \vec{c}, we get:

    \frac{1}{3(n+1)}=\frac{1}{m+1} (1)

    \frac{2}{3(n+1)}=\frac{m}{5(m+1)} (2)

    \frac{n}{n+1} =\frac{4m}{5(m+1)} (3)

    From (2) and (3): \frac{n}{n+1} = \frac{8}{3(n+1)}

    \Rightarrow n = \frac{8}{3}

    Check: \Rightarrow m = 10. Subst into (1): \frac{1}{3\times \frac{11}{3}} = \frac{1}{11}

    So the ratio CF:FD = 1:\frac{8}{3} = 3:8.

    Grandad
    Last edited by Grandad; February 9th 2009 at 01:30 AM. Reason: Typo
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