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Pedoe's Two Triangle Inequality

  1. May 17, 2005 #1
    Hi everyone

    I need some help proving Pedoe's Inequality for two triangles, which states that

    [tex]a_{1}^2(b_{2}^2+c_{2}^2-a_{2}^2) + b_{1}^2(c_{2}^2 + a_{2}^2 - b_{2}^2) + c_{1}^2(a_{2}^2 + b_{2}^2 - c_{2}^2) \geq 16F_{1}F_{2}[/tex]

    where [itex](a_{1},b_{1},c_{1})[/itex] and [itex](a_{2},b_{2},c_{2})[/itex] are the sides of triangles [itex]A_{1}B_{1}C_{1}[/itex] and [itex]A_{2}B_{2}C_{2}[/itex] respectively and [itex]F_{1}[/itex], [itex]F_{2}[/itex] are their areas.

    The expressions in the brackets suggest usage of the cosine rule, which gives [itex]b_{2}^2 + c_{2}^2 - a_{2}^2 = 2b_{2}c_{2}\cos A_{2}[/itex]. Using this the left hand side transforms to three terms of the type [itex]2a_{1}^2b_{2}c_{2}\cos A_{2}[/itex] but this doesn't seem to help. The right hand side can be transformed using Hero's formula for the area of either triangle. This also gets rid of 16. But I don't know how to proceed further.

    I would be grateful if someone could suggest a way out. In case there is a proof available on the internet, please let me know...I am searching for it myself on google right now....so far I have found several pages just listing the theorem's statement (mostly copied from wiki).


    Last edited: May 17, 2005
  2. jcsd
  3. May 19, 2005 #2
    Hi again

    I posted this sometime back...apparently it was read by very few people. With this reminder, I am hoping someone who has some idea about the inequality can help me out. Its urgent (but not homework)....

    Thanks and Cheers
  4. May 20, 2005 #3
    To the Moderator: Please shift this to the appropriate forum.
  5. May 20, 2005 #4


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    Do you know when equality is supposed to happen? If so, you could use this type of proof technique:

    Theorem: 2^x > x^2 for all x > 4.


    Consider f(x) = 2^x - x^2. Then f(4) = 0.
    f'(x) = 2^x (log 2) - 2x
    f'(4) > 0, so f is increasing at x = 4.
    f''(x) = 2^x (log 2)^2 - 2
    f''(x) > 0 for all x >= 4, so f'(x) is increasing for x >= 4.
    Therefore, f(x) is increasing on x >= 4, so 0 is its minimum, which occurs at x = 4.
    Therefore, 2^x > x^2 for all x > 4.
  6. May 20, 2005 #5
    Hi Hurkyl

    Thanks for your reply. Equality occurs iff the two triangles are similar. Could you please elaborate on your idea a bit?

    Thanks and cheers
  7. May 20, 2005 #6


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    Dearly Missed

    Last edited: May 21, 2005
  8. May 22, 2005 #7
    Hi arildno

    Thanks so much for the pdf file. I am sorry for this delayed reply (I was busy with tests). I've seen some of it and will go through it in detail.

    Thanks and cheers
  9. Jun 8, 2005 #8
    Let a and b be the vectot , let v be the angle between a and b, then
    (axb)*(axb)= a^2*b^2*sin(v)^2=a^2*b^2*(1-cos(v)^2)=a^2*b^2-(a.b)^2
    let c=a-b, then the area of triangle a,b,c is
    F^2 = (1/4)*a^2*b^2*sin(v)^2 = (1/4)*(axb)*(axb)=(1/4)(a^2*b^2-(a.b)^2)

    Similiarly, define a' and b' as the vector,let v' be the angle between a' and b', let c'=a'-c'.
    The area of triangle a', b', c' is
    F1^2 = (1/4)*(a'^2*b'^2-(a'.b')^2)

    (16*F1*F)^2 = 16*(a^2*b^2-(a.b)^2)(a'^2*b'^2-(a'.b')^2)
    = 16*(a^2*b^2*a'^2*b'^2 - a^2*b^2*(a'.b')^2 - a'^2*b'^2*(a.b)^2 + (a.b)^2*(a'.b')^2) ------(1)

    Let A = a'^2*(-a^2+b^2+c^2) + b'^2*(a^2-b^2+c^2) +c'^2*(a^2+b^2-c^2)
    because c^2= a^2+b^2-2*a*b*cos(v) = a^2+b^2-2*(a.b)
    c'^2 =a'^2+b'^2-2*(a'.b')
    subsitute c, c' in A

    A = 2*(a'^2*(b^2-(a.b)) + 2*(b'^2*(a^2-(a.b)) + (a'^2+b'^2-2*(a'.b'))*2*(a.b)
    = 2*(a'^2*b^2 + b'^2*a^2 -2*(a.b)*(a'.b'))
    = 2*((a'*b-a*b')^2 + 2*a*b*a'*b' - 2*(a.b)*(a'.b'))

    A^2 = 4*((a'*b - a*b')^4 + 4*(a'*b -a*b')^2*a*b*a'*b' + 4*(a*b*a'*b')^2 - 4*(a'*b-a*b')^2*(a.b)*(a'.b')
    -8*a*b*a'*b'*(a.b)*(a'.b') + 4(a.b)^2(a'.b')^2) -------------(2)

    Substract (1) from (2)

    A^2 - (16*F1*F)^2 = 4*((a'*b - a*b')^4 +4*(a'*b-a*b')^2*(a*b*a'*b' - (a.b)*(a'.b')) +
    4*a^2*b^2(a'.b')^2 + 4*a'^2*b'^2*(a.b)^2 - 8*a*b*a'*b'*(a.b)*(a'.b'))
    = 4*(a'*b-a*b')^4 + 16*(a'*b-a*b')^2*(a*b*a'*b'-(a.b)*(a'.b')) +
    It's easy to see that on the right hand side, the first and the third term are large than zero, the second term only need
    concern a*b*a'*b'-(a.b)*(a'.b'), because
    so a*b*a'*b'-(a.b)*(a'.b') = a*b*a'*b'*(1-cos(v)*cos(v'))
    since cos(v)<=1 cos(v')<=1 so cos(v)*cos(v')<=1, so a*b*a'*b'-(a.b)*(a'*b') >=0
    We draw conclusion that A^2-(16*F*F1)^2 >= 0 therefore A>=16*F*F1

    To make the equal sign hold, we must have
    a'*b-a*b'=0 ======> a'/b' = a/b
    a*b*(a'.b')-a'*b'*(a.b)=0 ===> a*b*a'*b'*(cos(v')-cos(v))=0 =====> cos(v')=cos(v) ====> v'=v
    This is equivalent to the codition that the triangles abc and a'b'c' are similar.
  10. Nov 28, 2005 #9
    Hi Philip,

    I'm sorry I didn't see your post earlier. Interesting. Thanks.

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