Question Number 226124 by ajfour last updated on 20/Nov/25
Commented by ajfour last updated on 20/Nov/25
$${If}\:{s}=\mathrm{1},\:{find}\:{R}. \\ $$$${or}\:{find}\:{R}/{s}. \\ $$
Answered by fantastic2 last updated on 22/Nov/25
$${a}+{b}={S} \\ $$$${R}\sqrt{\mathrm{2}}+{R}+{b}={S}\sqrt{\mathrm{2}} \\ $$$$\left({R}+{a}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\left({S}−{R}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow{a}=\mathrm{1}−{b} \\ $$$${putting}\:{in}\:{eq}\:\mathrm{3} \\ $$$$\left({R}+\mathrm{1}−{b}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\left(\mathrm{1}−{R}\right)^{\mathrm{2}} ..\mathrm{4} \\ $$$${We}\:{got}\: \\ $$$${R}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+{b}=\sqrt{\mathrm{2}}\Rightarrow{b}=\sqrt{\mathrm{2}}−{R}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$$\& \\ $$$${R}^{\mathrm{2}} +\mathrm{1}+{b}^{\mathrm{2}} +\mathrm{2}{R}−\mathrm{2}{b}−\mathrm{2}{Rb}={R}^{\mathrm{2}} +\mathrm{1}+{R}^{\mathrm{2}} −\mathrm{2}{R} \\ $$$$\Rightarrow{b}^{\mathrm{2}} +\mathrm{4}{R}−\mathrm{2}{b}\left(\mathrm{1}+{R}\right)={R}^{\mathrm{2}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} −\mathrm{2}{b}\left(\mathrm{1}+{R}\right)={R}^{\mathrm{2}} −\mathrm{4}{R} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}−{R}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\right)^{\mathrm{2}} −\mathrm{2}\left(\sqrt{\mathrm{2}}−{R}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\right)\left(\mathrm{1}+{R}\right)={R}\left({R}−\mathrm{4}\right) \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}−{R}\sqrt{\mathrm{2}}−{R}−{R}−\mathrm{1}\right)^{\mathrm{2}} ={R}\left({R}−\mathrm{4}\right)+\left(\mathrm{1}+{R}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−{R}\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}{R}+\mathrm{1}\right. \\ $$$${wait}\:{i}\:{think}\:{im}\:{making}\:{it}\:{complex}.. \\ $$$${Real}\:{solution} \\ $$$${R}=\frac{+\sqrt{\mathrm{2}}\sqrt{\mathrm{11}−\mathrm{2}\sqrt{\mathrm{2}}}−\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{11}−\mathrm{2}\sqrt{\mathrm{2}}}+\mathrm{3}}{\mathrm{4}} \\ $$$$\approx\mathrm{0}.\mathrm{3389105593} \\ $$
Commented by ajfour last updated on 20/Nov/25
$${yes},\:{and}\:{the}\:{answers}\:{are}\:{beautiful}\:{too}. \\ $$
Answered by ajfour last updated on 20/Nov/25
Answered by TonyCWX last updated on 22/Nov/25
![S=a+b b=S−a R(√2)+R+b=(a+b)(√2) R((√2)+1)=a(√2)+((√2)−1)b R=(2−(√2))a+(3−2(√2))(S−a) R=(2−(√2))a+(3−2(√2))S−(3−2(√2))a R=((√2)−1)a+(3−2(√2))S a=((√2)+1)R−((√2)−1)S ... E_1 R^2 +(S−R)^2 =(a+R)^2 a=(√(S^2 −2SR+2R^2 ))−R ... E_2 Equalise E_1 and E_2 (√(S^2 −2SR+2R^2 ))−R=((√2)+1)R−((√2)−1)S (√(S^2 −2SR+2R^2 ))=((√2)+2)R−((√2)−1)S S^2 −2SR+2R^2 =(4(√2)+6)R^2 −2(√2)SR+(3−2(√2))S^2 (4(√2)+4)R^2 +(2−2(√2))SR+(2−2(√2))S^2 =0 R=(((2(√2)−2)S±(√((12−8(√2))S^2 −(−32)S^2 )))/(8(√2)+8)) R=(((2(√2)−2)S±S(√((44−8(√2)))))/(8(√2)+8)) R=((S[(√2)−1+(√(11−2(√2)))])/(4(√2)+4)) S=1 ⇒ R=(((√2)−1+(√(11−2(√2))))/(4(√2)+4)) ≈ 0.3389105593...](https://www.tinkutara.com/question/Q226171.png)
$${S}={a}+{b} \\ $$$${b}={S}−{a} \\ $$$$ \\ $$$${R}\sqrt{\mathrm{2}}+{R}+{b}=\left({a}+{b}\right)\sqrt{\mathrm{2}} \\ $$$${R}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)={a}\sqrt{\mathrm{2}}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){b} \\ $$$${R}=\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){a}+\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\left({S}−{a}\right) \\ $$$${R}=\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){a}+\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right){S}−\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right){a} \\ $$$${R}=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){a}+\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right){S} \\ $$$${a}=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){R}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){S}\:…\:{E}_{\mathrm{1}} \\ $$$$ \\ $$$${R}^{\mathrm{2}} +\left({S}−{R}\right)^{\mathrm{2}} =\left({a}+{R}\right)^{\mathrm{2}} \\ $$$${a}=\sqrt{{S}^{\mathrm{2}} −\mathrm{2}{SR}+\mathrm{2}{R}^{\mathrm{2}} }−{R}\:…\:{E}_{\mathrm{2}} \\ $$$$ \\ $$$${Equalise}\:{E}_{\mathrm{1}} \:{and}\:{E}_{\mathrm{2}} \\ $$$$\sqrt{{S}^{\mathrm{2}} −\mathrm{2}{SR}+\mathrm{2}{R}^{\mathrm{2}} }−{R}=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){R}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){S} \\ $$$$\sqrt{{S}^{\mathrm{2}} −\mathrm{2}{SR}+\mathrm{2}{R}^{\mathrm{2}} }=\left(\sqrt{\mathrm{2}}+\mathrm{2}\right){R}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){S} \\ $$$${S}^{\mathrm{2}} −\mathrm{2}{SR}+\mathrm{2}{R}^{\mathrm{2}} =\left(\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{6}\right){R}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{SR}+\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right){S}^{\mathrm{2}} \\ $$$$\left(\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{4}\right){R}^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\right){SR}+\left(\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\right){S}^{\mathrm{2}} =\mathrm{0} \\ $$$${R}=\frac{\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}\right){S}\pm\sqrt{\left(\mathrm{12}−\mathrm{8}\sqrt{\mathrm{2}}\right){S}^{\mathrm{2}} −\left(−\mathrm{32}\right){S}^{\mathrm{2}} }}{\mathrm{8}\sqrt{\mathrm{2}}+\mathrm{8}} \\ $$$${R}=\frac{\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}\right){S}\pm{S}\sqrt{\left(\mathrm{44}−\mathrm{8}\sqrt{\mathrm{2}}\right)}}{\mathrm{8}\sqrt{\mathrm{2}}+\mathrm{8}} \\ $$$${R}=\frac{{S}\left[\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{11}−\mathrm{2}\sqrt{\mathrm{2}}}\right]}{\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{4}} \\ $$$${S}=\mathrm{1}\:\Rightarrow\:{R}=\frac{\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{11}−\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{4}}\:\approx\:\mathrm{0}.\mathrm{3389105593}… \\ $$
Commented by ajfour last updated on 22/Nov/25
$${yes}\:{R}\approx\mathrm{0}.\mathrm{33891} \\ $$$$\:\:{R}=\:\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\sqrt{\frac{\mathrm{41}}{\mathrm{16}}−\frac{\mathrm{7}\sqrt{\mathrm{2}}}{\mathrm{4}}}\: \\ $$