Question Number 173242 by AgniMath last updated on 08/Jul/22
Commented by som(math1967) last updated on 09/Jul/22
$$\mathrm{2}\:\:\:\:\:\:\:\:\:\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}\in\boldsymbol{{R}} \\ $$
Answered by AgniMath last updated on 09/Jul/22
![Another way (a + b + c) = (√3) or (a + b + c)^2 = 3 or a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 3 or a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 3ab + 3bc + 3ca [∵ ab + bc + ca = 1] or a^2 + b^2 + c^2 − ab − bc − ca = 0 or 2a^2 + 2b^2 + 2c^2 − 2ab − 2bc − 2ca = 0 or (a − b)^2 + (b − c)^2 + (c − a)^2 = 0 determinant ((((a − b)^2 = 0)),((or a − b = 0)),((or a = b))) determinant ((((b −c)^2 = 0)),((or b − c = 0)),((b = c))) determinant ((((c − a)^2 = 0)),((or c − a = 0)),((or c = a))) ∴ a = b = c ((a + b)/c) = ((a + a)/a) = ((2a)/a) = 2](Q173295.png)
$$\mathrm{Another}\:\mathrm{way} \\ $$$$ \\ $$$$\left({a}\:+\:{b}\:+\:{c}\right)\:=\:\sqrt{\mathrm{3}} \\ $$$${or}\:\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{2}} \:=\:\mathrm{3} \\ $$$${or}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:\mathrm{2}{ab}\:+\:\mathrm{2}{bc}\:+\:\mathrm{2}{ca}\:=\:\mathrm{3} \\ $$$${or}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:\mathrm{2}{ab}\:+\:\mathrm{2}{bc}\:+\:\mathrm{2}{ca}\:=\:\mathrm{3}{ab}\:+\:\mathrm{3}{bc}\:+\:\mathrm{3}{ca} \\ $$$$\left[\because\:{ab}\:+\:{bc}\:+\:{ca}\:=\:\mathrm{1}\right] \\ $$$${or}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:−\:{ab}\:−\:{bc}\:−\:{ca}\:=\:\mathrm{0} \\ $$$${or}\:\mathrm{2}{a}^{\mathrm{2}} \:+\:\mathrm{2}{b}^{\mathrm{2}} \:+\:\mathrm{2}{c}^{\mathrm{2}} \:−\:\mathrm{2}{ab}\:−\:\mathrm{2}{bc}\:−\:\mathrm{2}{ca}\:=\:\mathrm{0} \\ $$$${or}\:\left({a}\:−\:{b}\right)^{\mathrm{2}} \:+\:\left({b}\:−\:{c}\right)^{\mathrm{2}} \:+\:\left({c}\:−\:{a}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$ \\ $$$$\begin{vmatrix}{\left({a}\:−\:{b}\right)^{\mathrm{2}} \:=\:\mathrm{0}}\\{{or}\:{a}\:−\:{b}\:=\:\mathrm{0}}\\{{or}\:{a}\:=\:{b}}\end{vmatrix}\begin{vmatrix}{\left({b}\:−{c}\right)^{\mathrm{2}} \:=\:\mathrm{0}}\\{{or}\:{b}\:−\:{c}\:=\:\mathrm{0}}\\{{b}\:=\:{c}}\end{vmatrix}\begin{vmatrix}{\left({c}\:−\:{a}\right)^{\mathrm{2}} =\:\mathrm{0}}\\{{or}\:{c}\:−\:{a}\:=\:\mathrm{0}}\\{{or}\:{c}\:=\:{a}}\end{vmatrix} \\ $$$$\therefore\:{a}\:=\:{b}\:=\:{c} \\ $$$$ \\ $$$$\frac{{a}\:+\:{b}}{{c}}\:=\:\frac{{a}\:+\:{a}}{{a}}\:=\:\frac{\mathrm{2}\cancel{{a}}}{\cancel{{a}}}\:=\:\mathrm{2} \\ $$
Answered by a.lgnaoui last updated on 09/Jul/22
$${a}+{b}+{c}=\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{{a}+{b}}{{c}}+\mathrm{1}=\frac{\sqrt{\mathrm{3}}}{{c}} \\ $$$${ab}+{bc}+{ca}=\mathrm{1}\:\:\:\:\:\:\:\:\:\Rightarrow{ab}+{c}\left({a}+{b}\right)=\mathrm{1}\:\:\:\Rightarrow{ab}=\mathrm{1}−{c}\left({a}+{b}\right)\:\: \\ $$$$\frac{{ab}}{{c}}\:\:\:+\left({a}+{b}\right)=\frac{\mathrm{1}}{{c}}\:\:\:\:\:{ab}=\mathrm{1}−{c}\left({a}+{b}\right) \\ $$$$\:\:\frac{{a}+{b}}{{c}}+\mathrm{1}=\frac{\sqrt{\mathrm{3}}}{{c}}\:\:\:\:{a}+{b}=\sqrt{\mathrm{3}\:}−{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ab}=\mathrm{1}−\left(\sqrt{\mathrm{3}}−{c}\right){c}=\mathrm{1}−\sqrt{\mathrm{3}}{c}+{c}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}−{c}\right){x}+{c}^{\mathrm{2}} −\sqrt{\mathrm{3}}{c}+\mathrm{1} \\ $$$$\mathrm{3}+{c}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{c}−\mathrm{4}\left({c}^{\mathrm{2}} −\sqrt{\mathrm{3}}{c}+\mathrm{1}\right)=\mathrm{3}−\mathrm{4}−\mathrm{3}{c}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}{c} \\ $$$$\mathrm{3}{c}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{c}+\mathrm{1}\Rightarrow\left({c}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${c}=\frac{\mathrm{2}\sqrt{\mathrm{3}}−}{\mathrm{6}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:\:\:\:\:\Rightarrow{a}+{b}=\sqrt{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\frac{{a}+{b}}{{c}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}}=\mathrm{2}\:\:\:\:{is}\:{the}\:{possible}\:{solution} \\ $$
Commented by Tawa11 last updated on 11/Jul/22
$$\mathrm{Great}\:\mathrm{sirs} \\ $$