A symmetrical expression in a triangle ABC.

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A symmetrical expression in a triangle ABC.

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#1 h Sep 22, 2010, 9:26 PM • 1 Y

Y by Adventure10

Prove that for any $\triangle ABC$ the expression $f(a,b,c)$ is symmetrically, where

$f(a,b,c)\equiv \frac {1}{a^2}\cdot \left[(b-c)^2(b+c-3a)^2+4r^2(b+c-2a)^2\right]$ ,

i.e. $f(a,b,c)=f(a,c,b)=f(b,a,c)=f(c,a,b)=f(b,c,a)=f(c,b,a)$ .

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#2 h Sep 29, 2010, 12:21 PM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula wrote:

Prove that for any $\triangle ABC$ the expression $f(a,b,c)$ is symmetrically, where

$f(a,b,c)\equiv \frac {1}{a^2}\cdot \left[(b-c)^2(b+c-3a)^2+4r^2(b+c-2a)^2\right]$ ,

i.e. $f(a,b,c)=f(a,c,b)=f(b,a,c)=f(c,a,b)=f(b,c,a)=f(c,b,a)$ .

just need to prove that $f(a,b,c) \equiv f(b,a,c)$ .

$a^2 b^2 \left( {f(a,b,c) - f(b,a,c)} \right)$

$=[ {b(b - c)(b + c - 3a) + a(a - c)(a + c - 3b)}][ {b(b - c)(b + c - 3a) - a(a - c)(a + c - 3b)}]+ 4r^2 [ {b(b + c - 2a) + a(a + c - 2b)} ][ {b(b + c - 2a) - a(a + c - 2b)}]$

$=[ ({a^2 + b^2 + c( {a + b} ) - 4ab})( {a + b - c})][ ({a - b})( {b + c - a})( {c + a - b})] + 4r^2 [a^2 + b^2 + c( {a + b} ) - 4ab][ (b - a)( {a + b + c} )]$

$=\left( {a - b} \right)\left( {a^2 + b^2 + c\left( {a + b} \right) - 4ab} \right)\frac{{16S^2 }}{{a + b + c}}$ $+ \frac{{16S^2 }}{{a + b + c}}\left( {a^2 + b^2 + c\left( {a + b} \right) - 4ab} \right)\left( {b - a} \right)$

$=0$

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