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Question Number 202290 by MATHEMATICSAM last updated on 24/Dec/23

If ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) then show that ((a + b + c)/(x + y + z)) = ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) .

$$\mathrm{If}\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}}\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}}\:. \\ $$

Answered by Rasheed.Sindhi last updated on 24/Dec/23

If ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) then show that ((a + b + c)/(x + y + z)) = ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) ⇒((x+y)/(3a − b))=((y+z)/(3b − c))=((z+x)/(3c − a))=k (say) x+y=k(3a − b)...(i) y+z=k(3b − c)...(ii) z+x=k(3c − a)...(iii) (i)+(ii)+(iii): 2(x+y+z)=k{(3a − b)+(3b − c)+(3c − a)} x+y+z=k(a+b+c)...(iv) (iv)−(ii): x=k(a+b+c)−k(3b − c) =k(a−2b+2c) (iv)−(iii): y=k(a+b+c)−k(3c − a) =k(2a+b−2c) (iv)−(i): z=k(a+b+c)−k(3a − b) =k(−2a+2b+c) To prove: ((a + b + c)/(x + y + z)) = ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) lhs: ((a + b + c)/(x + y + z))=((a + b + c)/(k(a+b+c)))=(1/k) rhs: ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) = ((a^(2 ) + b^2 + c^2 )/(a(k(a−2b+2c)) + b(k(2a+b−2c)) + c(k(−2a+2b+c)))) = ((a^(2 ) + b^2 + c^2 )/(k{a(a−2b+2c)) + b(2a+b−2c)) + c(−2a+2b+c))})) = ((a^(2 ) + b^2 + c^2 )/(k{a^2 −2ab+2ca + 2ab+b^2 −2bc −2ca+2bc+c^2 })) = ((a^(2 ) + b^2 + c^2 )/(k{a^2 +b^2 +c^2 }))=(1/k) ∵ lhs=rhs=1/k ∴ Proved

$$\mathrm{If}\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}}\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}}\: \\ $$$$\: \\ $$$$\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}} \\ $$$$\Rightarrow\frac{{x}+{y}}{\mathrm{3}{a}\:−\:{b}}=\frac{{y}+{z}}{\mathrm{3}{b}\:−\:{c}}=\frac{{z}+{x}}{\mathrm{3}{c}\:−\:{a}}={k}\:\left({say}\right) \\ $$$${x}+{y}={k}\left(\mathrm{3}{a}\:−\:{b}\right)...\left({i}\right) \\ $$$${y}+{z}={k}\left(\mathrm{3}{b}\:−\:{c}\right)...\left({ii}\right) \\ $$$${z}+{x}={k}\left(\mathrm{3}{c}\:−\:{a}\right)...\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right): \\ $$$$\mathrm{2}\left({x}+{y}+{z}\right)={k}\left\{\left(\mathrm{3}{a}\:−\:{b}\right)+\left(\mathrm{3}{b}\:−\:{c}\right)+\left(\mathrm{3}{c}\:−\:{a}\right)\right\} \\ $$$${x}+{y}+{z}={k}\left({a}+{b}+{c}\right)...\left({iv}\right) \\ $$$$\: \\ $$$$\left({iv}\right)−\left({ii}\right):\: \\ $$$${x}={k}\left({a}+{b}+{c}\right)−{k}\left(\mathrm{3}{b}\:−\:{c}\right) \\ $$$$\:\:\:\:\:\:={k}\left({a}−\mathrm{2}{b}+\mathrm{2}{c}\right) \\ $$$$\left({iv}\right)−\left({iii}\right): \\ $$$${y}={k}\left({a}+{b}+{c}\right)−{k}\left(\mathrm{3}{c}\:−\:{a}\right) \\ $$$$\:\:\:={k}\left(\mathrm{2}{a}+{b}−\mathrm{2}{c}\right) \\ $$$$\left({iv}\right)−\left({i}\right): \\ $$$${z}={k}\left({a}+{b}+{c}\right)−{k}\left(\mathrm{3}{a}\:−\:{b}\right) \\ $$$$\:\:={k}\left(−\mathrm{2}{a}+\mathrm{2}{b}+{c}\right) \\ $$$$\:\mathcal{T}{o}\:{prove}: \\ $$$$\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}} \\ $$$${lhs}: \\ $$$$\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}=\frac{{a}\:+\:{b}\:+\:{c}}{{k}\left({a}+{b}+{c}\right)}=\frac{\mathrm{1}}{{k}} \\ $$$${rhs}: \\ $$$$\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}} \\ $$$$=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{a}\left({k}\left({a}−\mathrm{2}{b}+\mathrm{2}{c}\right)\right)\:+\:{b}\left({k}\left(\mathrm{2}{a}+{b}−\mathrm{2}{c}\right)\right)\:+\:{c}\left({k}\left(−\mathrm{2}{a}+\mathrm{2}{b}+{c}\right)\right)} \\ $$$$=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{\left.{k}\left.\left\{\left.{a}\left({a}−\mathrm{2}{b}+\mathrm{2}{c}\right)\right)\:+\:{b}\left(\mathrm{2}{a}+{b}−\mathrm{2}{c}\right)\right)\:+\:{c}\left(−\mathrm{2}{a}+\mathrm{2}{b}+{c}\right)\right)\right\}} \\ $$$$=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{k}\left\{{a}^{\mathrm{2}} −\cancel{\mathrm{2}{ab}}+\cancel{\mathrm{2}{ca}}\:+\:\cancel{\mathrm{2}{ab}}+{b}^{\mathrm{2}} −\cancel{\mathrm{2}{bc}}\:−\cancel{\mathrm{2}{ca}}+\cancel{\mathrm{2}{bc}}+{c}^{\mathrm{2}} \right\}} \\ $$$$=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{k}\left\{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right\}}=\frac{\mathrm{1}}{{k}} \\ $$$$\because\:{lhs}={rhs}=\mathrm{1}/{k} \\ $$$$\therefore\:\:\:\:\:\:\:\:\mathbb{P}\boldsymbol{\mathrm{roved}} \\ $$

Answered by som(math1967) last updated on 24/Dec/23

Each ratio =((3a−b+3b−c+3c−a)/(x+y+y+z+z+x)) =((2(a+b+c))/(2(x+y+z)))=((a+b+c)/(x+y+z)) ⇒((a+b+c)/(x+y+z))=((a+b+c−3a+b)/(x+y+z−x−y)) =((a+b+c−3b+c)/(x+y+z−y−z))=((a+b+c−3c+a)/(x+y+z−z−x)) ⇒((a+b+c)/(x+y+z))=((2b−2a+c)/z)=((a−2b+2c)/x) =((2a−2c+b)/y) ⇒((a+b+c)/(x+y+z))=((2bc−2ac+c^2 )/(cz))=((a^2 −2ab+2ac)/(ax)) =((2ab−2bc+b^2 )/(by)) ⇒((a+b+c)/(x+y+z)) =((2bc−2ac+c^2 +a^2 −2ab+2ac+2ab−2bc+b^2 )/(cz+ax+by)) ⇒((a+b+c)/(x+y+z))=((a^2 +b^2 +c^2 )/(ax+by+cz))

$${Each}\:{ratio} \\ $$$$=\frac{\mathrm{3}{a}−{b}+\mathrm{3}{b}−{c}+\mathrm{3}{c}−{a}}{{x}+{y}+{y}+{z}+{z}+{x}} \\ $$$$=\frac{\mathrm{2}\left({a}+{b}+{c}\right)}{\mathrm{2}\left({x}+{y}+{z}\right)}=\frac{{a}+{b}+{c}}{{x}+{y}+{z}} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{{a}+{b}+{c}−\mathrm{3}{a}+{b}}{{x}+{y}+{z}−{x}−{y}} \\ $$$$=\frac{{a}+{b}+{c}−\mathrm{3}{b}+{c}}{{x}+{y}+{z}−{y}−{z}}=\frac{{a}+{b}+{c}−\mathrm{3}{c}+{a}}{{x}+{y}+{z}−{z}−{x}} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{\mathrm{2}{b}−\mathrm{2}{a}+{c}}{{z}}=\frac{{a}−\mathrm{2}{b}+\mathrm{2}{c}}{{x}} \\ $$$$=\frac{\mathrm{2}{a}−\mathrm{2}{c}+{b}}{{y}} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{\mathrm{2}{bc}−\mathrm{2}{ac}+{c}^{\mathrm{2}} }{{cz}}=\frac{{a}^{\mathrm{2}} −\mathrm{2}{ab}+\mathrm{2}{ac}}{{ax}} \\ $$$$=\frac{\mathrm{2}{ab}−\mathrm{2}{bc}+{b}^{\mathrm{2}} }{{by}} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{x}+{y}+{z}} \\ $$$$=\frac{\mathrm{2}{bc}−\mathrm{2}{ac}+{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ab}+\mathrm{2}{ac}+\mathrm{2}{ab}−\mathrm{2}{bc}+{b}^{\mathrm{2}} }{{cz}+{ax}+{by}}\: \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{ax}+{by}+{cz}} \\ $$

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