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If-you-know-b-2-c-2-a-2-2bc-2-c-2-a-2-b-2-2ca-2-a-2-b-2-c-2-2ab-2-3-then-what-s-the-value-of-b-2-c-2-a-2-2bc-c-2-a-2-b-2-2ac-a-2-b-2-c-2-2

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Question Number 84871 by tw000001 last updated on 17/Mar/20
If you know (((b^2 +c^2 −a^2 )/(2bc)))^2 +(((c^2 +a^2 −b^2 )/(2ca)))^2 +(((a^2 +b^2 −c^2 )/(2ab)))^2 =3, then what′s the value of ((b^2 +c^2 −a^2 )/(2bc))+((c^2 +a^2 −b^2 )/(2ac))+((a^2 +b^2 −c^2 )/(2ab))?
$$\mathrm{If}\:\mathrm{you}\:\mathrm{know} \\ $$$$\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\right)^{\mathrm{2}} +\left(\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ca}}\right)^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}\right)^{\mathrm{2}} =\mathrm{3}, \\ $$$$\mathrm{then}\:\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}+\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}? \\ $$
Commented by ajfour last updated on 17/Mar/20
1 ?
$$\mathrm{1}\:? \\ $$
Answered by MJS last updated on 17/Mar/20
(((b^2 +c^2 −a^2 )/(2bc)))^2 +(((c^2 +a^2 −b^2 )/(2ca)))^2 +(((a^2 +b^2 −c^2 )/(2ab)))^2 =3 ⇔ (a^2 +b^2 +c^2 )(a+b+c)(−a+b+c)(a−b+c)(a+b−c)=0 ⇔ c=−a−b∨c=−a+b∨c=a−b∨c=a+b ⇒ ((b^2 +c^2 −a^2 )/(2bc))+((c^2 +a^2 −b^2 )/(2ca))+((a^2 +b^2 −c^2 )/(2ab))=x x=−3∨x=1 the red c gives −3, the others give 1
$$\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\right)^{\mathrm{2}} +\left(\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ca}}\right)^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}\right)^{\mathrm{2}} =\mathrm{3} \\ $$$$\Leftrightarrow \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${c}=−{a}−{b}\vee{c}=−{a}+{b}\vee{c}={a}−{b}\vee{c}={a}+{b} \\ $$$$\Rightarrow \\ $$$$\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}+\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ca}}+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}={x} \\ $$$${x}=−\mathrm{3}\vee{x}=\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{red}\:{c}\:\mathrm{gives}\:−\mathrm{3},\:\mathrm{the}\:\mathrm{others}\:\mathrm{give}\:\mathrm{1} \\ $$
Commented by MJS last updated on 17/Mar/20
you′re welcome
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$
Commented by tw000001 last updated on 17/Mar/20
Thank you for your solution.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{your}\:\mathrm{solution}. \\ $$

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