Question Number 202290 by MATHEMATICSAM last updated on 24/Dec/23
$$\mathrm{If}\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}}\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}}\:. \\ $$
Answered by Rasheed.Sindhi last updated on 24/Dec/23
$$\mathrm{If}\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}}\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}}\: \\ $$$$\: \\ $$$$\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}} \\ $$$$\Rightarrow\frac{{x}+{y}}{\mathrm{3}{a}\:−\:{b}}=\frac{{y}+{z}}{\mathrm{3}{b}\:−\:{c}}=\frac{{z}+{x}}{\mathrm{3}{c}\:−\:{a}}={k}\:\left({say}\right) \\ $$$${x}+{y}={k}\left(\mathrm{3}{a}\:−\:{b}\right)…\left({i}\right) \\ $$$${y}+{z}={k}\left(\mathrm{3}{b}\:−\:{c}\right)…\left({ii}\right) \\ $$$${z}+{x}={k}\left(\mathrm{3}{c}\:−\:{a}\right)…\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right): \\ $$$$\mathrm{2}\left({x}+{y}+{z}\right)={k}\left\{\left(\mathrm{3}{a}\:−\:{b}\right)+\left(\mathrm{3}{b}\:−\:{c}\right)+\left(\mathrm{3}{c}\:−\:{a}\right)\right\} \\ $$$${x}+{y}+{z}={k}\left({a}+{b}+{c}\right)…\left({iv}\right) \\ $$$$\: \\ $$$$\left({iv}\right)−\left({ii}\right):\: \\ $$$${x}={k}\left({a}+{b}+{c}\right)−{k}\left(\mathrm{3}{b}\:−\:{c}\right) \\ $$$$\:\:\:\:\:\:={k}\left({a}−\mathrm{2}{b}+\mathrm{2}{c}\right) \\ $$$$\left({iv}\right)−\left({iii}\right): \\ $$$${y}={k}\left({a}+{b}+{c}\right)−{k}\left(\mathrm{3}{c}\:−\:{a}\right) \\ $$$$\:\:\:={k}\left(\mathrm{2}{a}+{b}−\mathrm{2}{c}\right) \\ $$$$\left({iv}\right)−\left({i}\right): \\ $$$${z}={k}\left({a}+{b}+{c}\right)−{k}\left(\mathrm{3}{a}\:−\:{b}\right) \\ $$$$\:\:={k}\left(−\mathrm{2}{a}+\mathrm{2}{b}+{c}\right) \\ $$$$\:\mathcal{T}{o}\:{prove}: \\ $$$$\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}} \\ $$$${lhs}: \\ $$$$\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}=\frac{{a}\:+\:{b}\:+\:{c}}{{k}\left({a}+{b}+{c}\right)}=\frac{\mathrm{1}}{{k}} \\ $$$${rhs}: \\ $$$$\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}} \\ $$$$=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{a}\left({k}\left({a}−\mathrm{2}{b}+\mathrm{2}{c}\right)\right)\:+\:{b}\left({k}\left(\mathrm{2}{a}+{b}−\mathrm{2}{c}\right)\right)\:+\:{c}\left({k}\left(−\mathrm{2}{a}+\mathrm{2}{b}+{c}\right)\right)} \\ $$$$=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{\left.{k}\left.\left\{\left.{a}\left({a}−\mathrm{2}{b}+\mathrm{2}{c}\right)\right)\:+\:{b}\left(\mathrm{2}{a}+{b}−\mathrm{2}{c}\right)\right)\:+\:{c}\left(−\mathrm{2}{a}+\mathrm{2}{b}+{c}\right)\right)\right\}} \\ $$$$=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{k}\left\{{a}^{\mathrm{2}} −\cancel{\mathrm{2}{ab}}+\cancel{\mathrm{2}{ca}}\:+\:\cancel{\mathrm{2}{ab}}+{b}^{\mathrm{2}} −\cancel{\mathrm{2}{bc}}\:−\cancel{\mathrm{2}{ca}}+\cancel{\mathrm{2}{bc}}+{c}^{\mathrm{2}} \right\}} \\ $$$$=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{k}\left\{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right\}}=\frac{\mathrm{1}}{{k}} \\ $$$$\because\:{lhs}={rhs}=\mathrm{1}/{k} \\ $$$$\therefore\:\:\:\:\:\:\:\:\mathbb{P}\boldsymbol{\mathrm{roved}} \\ $$
Answered by som(math1967) last updated on 24/Dec/23
$${Each}\:{ratio} \\ $$$$=\frac{\mathrm{3}{a}−{b}+\mathrm{3}{b}−{c}+\mathrm{3}{c}−{a}}{{x}+{y}+{y}+{z}+{z}+{x}} \\ $$$$=\frac{\mathrm{2}\left({a}+{b}+{c}\right)}{\mathrm{2}\left({x}+{y}+{z}\right)}=\frac{{a}+{b}+{c}}{{x}+{y}+{z}} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{{a}+{b}+{c}−\mathrm{3}{a}+{b}}{{x}+{y}+{z}−{x}−{y}} \\ $$$$=\frac{{a}+{b}+{c}−\mathrm{3}{b}+{c}}{{x}+{y}+{z}−{y}−{z}}=\frac{{a}+{b}+{c}−\mathrm{3}{c}+{a}}{{x}+{y}+{z}−{z}−{x}} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{\mathrm{2}{b}−\mathrm{2}{a}+{c}}{{z}}=\frac{{a}−\mathrm{2}{b}+\mathrm{2}{c}}{{x}} \\ $$$$=\frac{\mathrm{2}{a}−\mathrm{2}{c}+{b}}{{y}} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{\mathrm{2}{bc}−\mathrm{2}{ac}+{c}^{\mathrm{2}} }{{cz}}=\frac{{a}^{\mathrm{2}} −\mathrm{2}{ab}+\mathrm{2}{ac}}{{ax}} \\ $$$$=\frac{\mathrm{2}{ab}−\mathrm{2}{bc}+{b}^{\mathrm{2}} }{{by}} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{x}+{y}+{z}} \\ $$$$=\frac{\mathrm{2}{bc}−\mathrm{2}{ac}+{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ab}+\mathrm{2}{ac}+\mathrm{2}{ab}−\mathrm{2}{bc}+{b}^{\mathrm{2}} }{{cz}+{ax}+{by}}\: \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{ax}+{by}+{cz}} \\ $$