Pedoe's Two Triangle Inequality

In summary, the conversation discusses the need for help proving Pedoe's Inequality for two triangles, which states that the sum of the squared side lengths of one triangle multiplied by the squared side lengths of the other triangle is greater than or equal to 16 times their corresponding areas. The conversation explores different approaches to proving this inequality, including the use of the cosine rule and Hero's formula. Eventually, a user named Philip provides a proof using vectors and the condition for triangles to be similar.
  • #1
maverick280857
1,789
4
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).

Thanks...

Cheers
vivek
 
Last edited:
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  • #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
Vivek
 
  • #3
To the Moderator: Please shift this to the appropriate forum.
 
  • #4
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.
 
  • #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
vivek
 
  • #6
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
 
Last edited by a moderator:
  • #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
Vivek
 
  • #8
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.
 
  • #9
Hi Philip,

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

Cheers
Vivek
 

1. What is Pedoe's Two Triangle Inequality?

Pedoe's Two Triangle Inequality is a mathematical theorem that states that the sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

2. Who discovered Pedoe's Two Triangle Inequality?

Pedoe's Two Triangle Inequality was discovered by mathematician Edward W. Pedoe in 1953.

3. What is the significance of Pedoe's Two Triangle Inequality?

Pedoe's Two Triangle Inequality is an important theorem in geometry as it helps to determine whether a set of three given lengths can form a valid triangle. It also has applications in other mathematical fields, such as optimization problems.

4. Can Pedoe's Two Triangle Inequality be applied to all triangles?

Yes, Pedoe's Two Triangle Inequality applies to all triangles, regardless of their shape or size.

5. How can Pedoe's Two Triangle Inequality be proven?

Pedoe's Two Triangle Inequality can be proven using the triangle inequality theorem, which states that the sum of any two sides of a triangle is always greater than the third side. This can be used to show that the sum of two sides of a triangle must be greater than the third side, thus proving Pedoe's Two Triangle Inequality.

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