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Question Number 214402 by golsendro last updated on 07/Dec/24
If ((a+b)/c) = ((b+c)/a) = ((a+c)/b) then find ((a+b)/c) .
$$\:\:\:\mathrm{If}\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}\:=\:\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}\:\mathrm{then}\:\mathrm{find}\: \\ $$$$\:\:\:\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:. \\ $$
Answered by efronzo1 last updated on 07/Dec/24
Let ((a+b)/c) = ((b+c)/a) = ((a+c)/b) = m then a+b = mc ; b+c = ma ; a+c = mb ⇒ a+b+b+c+a+c = mc+ma+mb 2a+2b+2c = m(a+b+c) 2(a+b+c) = m(a+b+c) m = 2 ⇒((a+b)/c) = 2
$$\:\:\:\:\mathrm{Let}\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}\:=\:\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}\:=\:\mathrm{m}\: \\ $$$$\:\:\:\:\:\mathrm{then}\:\mathrm{a}+\mathrm{b}\:=\:\mathrm{mc}\:;\:\mathrm{b}+\mathrm{c}\:=\:\mathrm{ma}\:;\:\mathrm{a}+\mathrm{c}\:=\:\mathrm{mb} \\ $$$$\:\:\Rightarrow\:\mathrm{a}+\mathrm{b}+\mathrm{b}+\mathrm{c}+\mathrm{a}+\mathrm{c}\:=\:\mathrm{mc}+\mathrm{ma}+\mathrm{mb} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2a}+\mathrm{2b}+\mathrm{2c}\:=\:\mathrm{m}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\:=\:\mathrm{m}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{m}\:=\:\mathrm{2}\:\Rightarrow\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:=\:\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 07/Dec/24
((a+b)/c) = ((b+c)/a) = ((a+c)/b) ((a+b)/c)+1 = ((b+c)/a)+1 = ((a+c)/b)+1 ((a+b+c)/c)=((a+b+c)/a)=((a+b+c)/b) a+b+c=0 ∣ (1/c)=(1/a)=(1/b) { ((a+b+c=0⇒a+b=−c⇒((a+b)/c)=((−c)/c)=−1)),((a=b=c⇒((a+b)/c)=((a+a)/a)=((2a)/a)=2)) :} ((a+b)/c)=−1 or 2
$$\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}\:=\:\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}} \\ $$$$\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}+\mathrm{1}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}+\mathrm{1}\:=\:\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}+\mathrm{1} \\ $$$$\frac{{a}+{b}+{c}}{{c}}=\frac{{a}+{b}+{c}}{{a}}=\frac{{a}+{b}+{c}}{{b}} \\ $$$${a}+{b}+{c}=\mathrm{0}\:\mid\:\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{{b}}\:\:\:\:\: \\ $$$$\begin{cases}{{a}+{b}+{c}=\mathrm{0}\Rightarrow{a}+{b}=−{c}\Rightarrow\frac{{a}+{b}}{{c}}=\frac{−{c}}{{c}}=−\mathrm{1}}\\{{a}={b}={c}\Rightarrow\frac{{a}+{b}}{{c}}=\frac{{a}+{a}}{{a}}=\frac{\mathrm{2}{a}}{{a}}=\mathrm{2}}\end{cases} \\ $$$$\frac{{a}+{b}}{{c}}=−\mathrm{1}\:{or}\:\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 07/Dec/24
((a+b)/c)=((b+c)/a)=((c+a)/b)=(((a+b)+(b+c)+c+a))/(c+a+b)) =((2a+2b+2c)/(a+b+c))=((2(a+b+c))/(a+b+c))=2 ((a+b)/c)=2
$$\frac{{a}+{b}}{{c}}=\frac{{b}+{c}}{{a}}=\frac{{c}+{a}}{{b}}=\frac{\left.\left({a}+{b}\right)+\left({b}+{c}\right)+{c}+{a}\right)}{{c}+{a}+{b}} \\ $$$$=\frac{\mathrm{2}{a}+\mathrm{2}{b}+\mathrm{2}{c}}{{a}+{b}+{c}}=\frac{\mathrm{2}\left({a}+{b}+{c}\right)}{{a}+{b}+{c}}=\mathrm{2} \\ $$$$\frac{{a}+{b}}{{c}}=\mathrm{2} \\ $$

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