0
$\begingroup$

I was trying to prove the Heron's Formula myself. I came to the expression $ 2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4) $ ). In the next step, I have to find $ (a+b+c)(-a+b+c)(a-b+c)(a+b-c) $ from it. But I can't find a way to proceed. How do I do it?

6 Answers 6

5

$ 2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)$

$=4a^2b^2-(a^4+b^4+c^4-2c^2a^2-2b^2c^2+2a^2b^2)$

$=(2ab)^2-(a^2+b^2-c^2)^2$

$=(2ab-a^2-b^2+c^2)(2ab+a^2+b^2-c^2)$

$=\{c^2-(a-b)^2\}\{(a+b)^2-c^2\}$

$=(c-a+b)(c+a-b)(a+b-c)(a+b+c)$

$=(-a+b+c)(a-b+c)(a+b-c)(a+b+c)$

2

Starting from the LHS $2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)$ if you wished to factorize this expression, you could start considering it as a polynomial in $a$ and looking for roots.

For $a$ to be a root of this polynomial it will have to be some polynomial combination of $b$ and $c$, the simplest possibly such one is just $a+b$ so let's try that (using $(b+c)^2 = b^2 + 2bc + c^2$ and $(b+c)^4 = b^4 + 4b^3c + 6 b^2c^2 + 4bc^3 + c^4$) we multiply out to get get $2(b^2b^2+2bcb^2+c^2b^2+b^2c^2+2bcc^2+c^2c^2+b^2c^2)-(4b^3c + 6 b^2c^2 + 4bc^3+2b^4+2c^4)$ now collecting up like terms to simplify we have $(2b^4+4b^3c+6b^2c^2+4bc^3+2c^4)-(4b^3c + 6 b^2c^2 + 4bc^3+2b^4+2c^4)$ which is just $0$!

This shows that $a = b+c$ is a root of the polynomial, so $a - b - c | 2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)$ we could then do the long division to get another polynomial of lower degree and try the same sort of thing, but we don't have to do so much hard work. The polynomial is symmetric so a symmetric argument proves that $b - a - c$ and $c - a - b$ also divide it, hence the polynomial is a multiple of $(a-b-c)(b-a-c)(c-a-b)$. Since it's a degree 4 polynomial it can only have one more linear factor. Furthermore that linear factor must be invariant under symmetry so clearly it must be $a+b+c$ up to a constant factor, but by inspection of the polynomial that constant must be 1.

This proves that $2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4) = (a+b+c)(a-b-c)(b-a-c)(c-a-b).$

  • 0
    This answer is using symmetry arguments to cut down the amount of work done. It also uses the fact that $P(a)=0$ implies $(x-a)|P(x)$.2012-12-10
1

To prove $A=B$ it is enough to prove $A-B=0$, this can be done in a matter of nanoseconds in pari/gp:

? 2*(a^2*b^2+a^2*c^2+b^2*c^2)-(a^4+b^4+c^4) - (a+b+c)*(-a+b+c)*(a-b+c)*(a+b-c) %1 = 0 
  • 1
    this isn't LaTeX code and it's not supposed to be please don't edit.2012-12-10
1

First note that $A=2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)=[a^2 b^2+a^2c^2-a^4+2a^2bc]+[-2a^2bc+a^2b^2+a^2c^2]+[2b^2c^2-b^4-c^4] $ Now $a^2 b^2+a^2c^2-a^4+2a^2bc=a^2[b^2+2bc+c^2-a^2]=a^2[(b+c)^2-a^2]$ and $-2a^2bc+a^2b^2+a^2c^2=a^2[-2bc+b^2+c^2]=a^2(b-c)^2$ and $2b^2c^2-b^4-c^4=-(b^4-2b^2c^2+c^4)=-(b^2-c^2)^2=-(b-c)^2(b+c)^2$ Therefore, $A=a^2[(b+c)^2-a^2]+a^2(b-c)^2-(b-c)^2(b+c)^2=a^2[(b+c)^2-a^2]-(b-c)^2[(b+c)^2-a^2]=[(b+c)^2-a^2][a^2-(b-c)^2]$ I think the OP can continue this

1

\begin{align*} 2a^2b^2 +2b^2c^2+2c^2a^2-a^4-b^4-c^4 &=4a^2b^2-(2a^2b^2-2b^2c^2-2c^2a^2+a^4+b^4+c^4)\\ &=(2ab)^2-(b^2+a^2-c^2)\\ &=(2ab+b^2+a^2-c^2)(2ab-b^2-a^2+c^2)\\ &=\bigl((a+b)^2-c^2\bigr)\bigl(c^2-(a-b)^2\bigr)\\&=(a+b+c)(a+b-c)(c+a-b)(c-a+b)\end{align*}

1

You can also work backwards, and use $(x+y)(x-y)=x^2-y^2$ twice, but you have to shuffle the terms a bit to make the pattern match:

$ (a+b+c)(-a + b+c)(a-b+c)(a+b-c) = $ (rearrange the terms) $ ((b+c) + a) ((b+c) -a)(a - (b-c)) (a + (b-c)) = $ (apply $(x+y)(x-y)=x^2-y^2$) $ ((b+c)^2 - a^2) (a^2 - (b-c)^2) = $ (expand using $(x\pm y)^2 = x^2 \pm 2xy + y^2$) $ (b^2+c^2+2bc-a^2) (a^2-b^2-c^2 + 2bc) = $ (rearrange the terms again, bringing $2bc$ to the front) $ (2bc + (b^2+c^2-a^2)) (2bc - (b^2+c^2-a^2)) = $ (apply $(x+y)(x-y)=x^2-y^2$) $ 4b^2c^2 - (b^2+c^2-a^2)^2 $

After this, it's a straightforward expansion.