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Math Help - Help with two proofs involving transversals

  1. #1
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    Help with two proofs involving transversals

    Let a triangle with the sides a, b, c be given with three vertex transversals that pass through a point. Show that if for each of the three vertex transversals, one forms the quotient of their upper section {t_{a}}^0 and the whole segment t_a, then the sum of the three quotients has the constant value 2:
    \frac {{t_{a}}^0} {t_b} +\frac {{t_{b}}^0} {t_b}+\frac {{t_{c}}^0} {t_c} = 2

    I am stuck on what is meant by the "upper section." Is it referring to the segment past the point on the side of the triangle that the transversal passes through?

    I appreciate the help.
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  2. #2
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    Hello ordinalhigh
    Quote Originally Posted by ordinalhigh View Post
    Let a triangle with the sides a, b, c be given with three vertex transversals that pass through a point. Show that if for each of the three vertex transversals, one forms the quotient of their upper section {t_{a}}^0 and the whole segment t_a, then the sum of the three quotients has the constant value 2:
    \frac {{t_{a}}^0} {t_b} +\frac {{t_{b}}^0} {t_b}+\frac {{t_{c}}^0} {t_c} = 2

    I am stuck on what is meant by the "upper section." Is it referring to the segment past the point on the side of the triangle that the transversal passes through?

    I appreciate the help.
    I think that the question can be re-written like this:
    P is any point in the interior of a triangle ABC. Transversals AP, BP and CP are produced to meet the opposite sides at points L, M and N respectively. Prove that:
    \frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}= 2
    The reason I think this is correct is that when P is the centroid of the triangle, each of the ratios is equal to \tfrac23, and the sum is therefore \tfrac23+\tfrac23+\tfrac23=2.

    I think that you might be able to prove it using Ceva's Theorem, but so far I'm afraid I've drawn a blank.

    Grandad
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  3. #3
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    Hello ordinalhigh

    I have a proof that does indeed use Ceva's Theorem. Here's the outline.

    Look at the attached diagram. We have to prove that:
    \frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}= 2
    Let L,\; M,\; N divide BC,\; CA,\; AB in the ratios p:1,\; q:1,\; r:1. Then, as shown in the diagram:
    AN = \frac{rc}{1+r}

    NB = \frac{c}{1+r}

    ... etc
    Ceva's Theorem states that:
    \frac{AN}{NB}\cdot\frac{BL}{LC}\cdot\frac{CM}{MA}=  1

    \Rightarrow pqr=1
    ...(1)
    Next, we construct the line LQ \parallel BA, meeting CN at Q and, using similar triangles, we prove that:
    LQ = \frac{BN.LC}{BC}=\frac{AN.PL}{AP}
    and hence that
    \frac{PL}{AP}= \frac{BN.LC}{AN.BC}

    \Rightarrow 1+\frac{PL}{AP}= 1+\frac{BN.LC}{AN.BC}

    \Rightarrow \frac{AL}{AP}=\frac{AN.BC+BN.LC}{AN.BC}
    = ... etc

    = \frac{1+r+rp}{r+rp}
    \Rightarrow \frac{AP}{AL}=\frac{r+rp}{1+r+rp}
    Similarly:
    \frac{BP}{BM}=\frac{p+pq}{1+p+q}
    and:
    \frac{CP}{CN}=\frac{q+qr}{1+q+qr}
    Adding these results together:
    \frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}=\frac{r+  rp}{1+r+rp}+\frac{p+pq}{1+p+pq}+\frac{q+qr}{1+q+qr  }
    From (1):
    r = \frac{1}{pq}

    \Rightarrow \frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}=\frac{\d  frac{1}{pq}+\dfrac{1}{q}}{1+\frac{1}{pq}+\frac{1}{  p}}+\frac{p+pq}{1+p+pq}+\frac{q+\dfrac{1}{p}}{1+q+  \dfrac{1}{p}}
    = ... etc
    =2
    I'm sure you can fill in the gaps.

    Grandad
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