Pedoe's Two Triangle Inequality

**maverick280857** · May17-05, 05:06 AM

Hi everyone

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

$LaTeX Code: a_{1}^2(b_{2}^2+c_{2}^2-a_{2}^2) + b_{1}^2(c_{2}^2 + a_{2}^2 - b_{2}^2) + c_{1}^2(a_{2}^2 + b_{2}^2 - c_{2}^2) \\geq 16F_{1}F_{2}$

where $LaTeX Code: (a_{1},b_{1},c_{1})$ and $LaTeX Code: (a_{2},b_{2},c_{2})$ are the sides of triangles $LaTeX Code: A_{1}B_{1}C_{1}$ and $LaTeX Code: A_{2}B_{2}C_{2}$ respectively and $LaTeX Code: F_{1}$ , $LaTeX Code: F_{2}$ are their areas.

The expressions in the brackets suggest usage of the cosine rule, which gives $LaTeX Code: b_{2}^2 + c_{2}^2 - a_{2}^2 = 2b_{2}c_{2}\\cos A_{2}$ . Using this the left hand side transforms to three terms of the type $LaTeX Code: 2a_{1}^2b_{2}c_{2}\\cos A_{2}$ 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).

Thanks...

Cheers
vivek

**maverick280857** · May19-05, 11:13 AM

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
Vivek

**maverick280857** · May20-05, 05:55 AM

To the Moderator: Please shift this to the appropriate forum.

**Hurkyl** · May20-05, 07:05 AM

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.

Proof:

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.

**maverick280857** · May20-05, 09:05 AM

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
vivek

**arildno** · May20-05, 09:45 AM

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
Your inequality is 3.13

**maverick280857** · May22-05, 12:32 PM

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
Vivek

**Philip Lam** · Jun8-05, 05:13 PM

Let a and b be the vectot , let v be the angle between a and b, then
aXb=a*b*sin(v)
a.b=a*b*cos(v)
(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=(1/2)*a*b*sin(v)
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=(1/2)*a'*b'*sin(v')
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')) +
16*(a*b*(a'.b')-a'*b'*(a.b))^2
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
(a.b)=a*b*cos(v)
(a'.b')=a'*b'*cos(v')
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
AND
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.

**maverick280857** · Nov28-05, 04:50 AM

Hi Philip,

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

Cheers
Vivek