Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Hi, maverick.
    I googled a bit, and found the following article.
    Have fun!
    http://jipam.vu.edu.au/images/106_03_JIPAM/106_03.pdf [Broken]
    Your inequality is 3.13
    Last edited by a moderator: May 2, 2017
  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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook