Question Number 135458 by Dwaipayan Shikari last updated on 13/Mar/21
Commented by Dwaipayan Shikari last updated on 13/Mar/21
Commented by Dwaipayan Shikari last updated on 13/Mar/21
$${My}\:{try} \\ $$$${In}\:\bigtriangleup{ABC}\:\:\:\frac{{OA}}{{OC}}=\frac{{AM}}{{BM}}\:\:\:\Rightarrow\frac{{AC}}{{OC}}=\frac{{AB}}{{MB}}\Rightarrow\frac{{AC}}{{AB}}=\frac{{OC}}{{MB}} \\ $$$${In}\:\bigtriangleup{LHC}\:\left({large}\:{hadron}\:{Collider}\:!:\right){and}\:\bigtriangleup{OLX}_{\mathrm{4}} \\ $$$${Since}\:\bigtriangleup{LHC}\sim\bigtriangleup{OLX}_{\mathrm{4}} \\ $$$$\frac{{LH}}{{CH}}=\frac{{OX}_{\mathrm{4}} }{{X}_{\mathrm{4}} {L}}\:\Rightarrow\frac{\mathrm{5}}{{HC}}=\frac{\mathrm{13}−\mathrm{5}}{\mathrm{5}}\Rightarrow{HC}=\frac{\mathrm{25}}{\mathrm{8}} \\ $$$${Similarly}\:{i}\:\bigtriangleup{GJB}\:{and}\:{MX}_{\mathrm{4}} {G}\:,\frac{{MX}_{\mathrm{3}} }{{GX}_{\mathrm{3}} }=\frac{{GJ}}{{JB}}\Rightarrow\frac{\mathrm{13}−\mathrm{12}}{\mathrm{12}}=\frac{\mathrm{12}}{{BJ}} \\ $$$${BJ}=\mathrm{144} \\ $$$${OC}={OL}+{LC}=\sqrt{\left(\mathrm{13}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }+\sqrt{\frac{\mathrm{25}^{\mathrm{2}} }{\mathrm{64}}+\mathrm{25}}=\frac{\mathrm{13}\sqrt{\mathrm{89}}}{\mathrm{8}} \\ $$$${BM}={GM}+{BM}=\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }+\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{12}^{\mathrm{4}} }=\mathrm{13}\sqrt{\mathrm{145}} \\ $$$$\frac{{AC}}{{AB}}=\frac{{OC}}{{MB}}=\frac{\sqrt{\mathrm{89}}}{\mathrm{8}\sqrt{\mathrm{145}}} \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 13/Mar/21
Commented by mr W last updated on 13/Mar/21
$$\frac{{x}}{\mathrm{8}}×\mathrm{5}+\frac{{x}}{\mathrm{1}}×\mathrm{12}=\mathrm{13} \\ $$$${x}=\frac{\mathrm{8}×\mathrm{13}}{\mathrm{5}+\mathrm{96}}=\frac{\mathrm{104}}{\mathrm{101}} \\ $$$${AC}=\frac{\mathrm{13}+{x}}{\mathrm{8}}×\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }=\frac{\mathrm{1417}\sqrt{\mathrm{89}}}{\mathrm{808}} \\ $$$${AB}=\frac{\mathrm{13}+{x}}{\mathrm{1}}×\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\frac{\mathrm{1417}\sqrt{\mathrm{145}}}{\mathrm{101}} \\ $$$$\frac{{AC}}{{AB}}=\frac{\mathrm{1417}\sqrt{\mathrm{89}}}{\mathrm{808}}×\frac{\mathrm{101}}{\mathrm{1417}\sqrt{\mathrm{145}}}=\frac{\sqrt{\mathrm{89}}}{\mathrm{8}\sqrt{\mathrm{145}}} \\ $$$$\Rightarrow{a}=\mathrm{89},\:{b}=\mathrm{145} \\ $$$$\Rightarrow{a}+{b}=\mathrm{234} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Mar/21
$${Thanks}\:{sir}\:!\:{Even}\:{more}\:{shorter}\:{and}\:{Nice} \\ $$
Commented by mr W last updated on 13/Mar/21
$${actually}\:{we}\:{even}\:{don}'{t}\:{need}\:{to}\:{know} \\ $$$${the}\:{value}\:{of}\:{x}: \\ $$$${AC}=\frac{\mathrm{13}+{x}}{\mathrm{8}}×\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} } \\ $$$${AB}=\frac{\mathrm{13}+{x}}{\mathrm{1}}×\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} } \\ $$$$\frac{{AC}}{{AB}}=\frac{\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }}{\mathrm{8}\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{89}}}{\mathrm{8}\sqrt{\mathrm{145}}} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Mar/21
$${Sir}\:{this}\:{works}\:{for}\:{every}\:{Pythagorean}\:{Triplets}.{Isn}'{t}\:{it}\:? \\ $$
Commented by mr W last updated on 13/Mar/21
$${i}\:{think}\:{yes}. \\ $$