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Thread: [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 $\displaystyle \vec{a}, \vec{b}, ...,$ then

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

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

    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 $\displaystyle \vec{f}$.

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

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

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

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

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

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

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

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

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

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

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

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

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