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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 △ABC   ((OA)/(OC))=((AM)/(BM))   ⇒((AC)/(OC))=((AB)/(MB))⇒((AC)/(AB))=((OC)/(MB))  In △LHC (large hadron Collider !:)and △OLX_4   Since △LHC∼△OLX_4   ((LH)/(CH))=((OX_4 )/(X_4 L)) ⇒(5/(HC))=((13−5)/5)⇒HC=((25)/8)  Similarly i △GJB and MX_4 G ,((MX_3 )/(GX_3 ))=((GJ)/(JB))⇒((13−12)/(12))=((12)/(BJ))  BJ=144  OC=OL+LC=(√((13−5)^2 +5^2 ))+(√(((25^2 )/(64))+25))=((13(√(89)))/8)  BM=GM+BM=(√(12^2 +1^2 ))+(√(12^2 +12^4 ))=13(√(145))  ((AC)/(AB))=((OC)/(MB))=((√(89))/(8(√(145))))

$${My}\:{try} \\ $$$${In}\:\bigtriangleup{ABC}\:\:\:\frac{{OA}}{{OC}}=\frac{{AM}}{{BM}}\:\:\:\Rightarrow\frac{{AC}}{{OC}}=\frac{{AB}}{{MB}}\Rightarrow\frac{{AC}}{{AB}}=\frac{{OC}}{{MB}} \\ $$$${In}\:\bigtriangleup{LHC}\:\left({large}\:{hadron}\:{Collider}\:!:\right){and}\:\bigtriangleup{OLX}_{\mathrm{4}} \\ $$$${Since}\:\bigtriangleup{LHC}\sim\bigtriangleup{OLX}_{\mathrm{4}} \\ $$$$\frac{{LH}}{{CH}}=\frac{{OX}_{\mathrm{4}} }{{X}_{\mathrm{4}} {L}}\:\Rightarrow\frac{\mathrm{5}}{{HC}}=\frac{\mathrm{13}−\mathrm{5}}{\mathrm{5}}\Rightarrow{HC}=\frac{\mathrm{25}}{\mathrm{8}} \\ $$$${Similarly}\:{i}\:\bigtriangleup{GJB}\:{and}\:{MX}_{\mathrm{4}} {G}\:,\frac{{MX}_{\mathrm{3}} }{{GX}_{\mathrm{3}} }=\frac{{GJ}}{{JB}}\Rightarrow\frac{\mathrm{13}−\mathrm{12}}{\mathrm{12}}=\frac{\mathrm{12}}{{BJ}} \\ $$$${BJ}=\mathrm{144} \\ $$$${OC}={OL}+{LC}=\sqrt{\left(\mathrm{13}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }+\sqrt{\frac{\mathrm{25}^{\mathrm{2}} }{\mathrm{64}}+\mathrm{25}}=\frac{\mathrm{13}\sqrt{\mathrm{89}}}{\mathrm{8}} \\ $$$${BM}={GM}+{BM}=\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }+\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{12}^{\mathrm{4}} }=\mathrm{13}\sqrt{\mathrm{145}} \\ $$$$\frac{{AC}}{{AB}}=\frac{{OC}}{{MB}}=\frac{\sqrt{\mathrm{89}}}{\mathrm{8}\sqrt{\mathrm{145}}} \\ $$$$ \\ $$$$ \\ $$

Answered by mr W last updated on 13/Mar/21

Commented by mr W last updated on 13/Mar/21

(x/8)×5+(x/1)×12=13  x=((8×13)/(5+96))=((104)/(101))  AC=((13+x)/8)×(√(8^2 +5^2 ))=((1417(√(89)))/(808))  AB=((13+x)/1)×(√(1^2 +12^2 ))=((1417(√(145)))/(101))  ((AC)/(AB))=((1417(√(89)))/(808))×((101)/(1417(√(145))))=((√(89))/(8(√(145))))  ⇒a=89, b=145  ⇒a+b=234

$$\frac{{x}}{\mathrm{8}}×\mathrm{5}+\frac{{x}}{\mathrm{1}}×\mathrm{12}=\mathrm{13} \\ $$$${x}=\frac{\mathrm{8}×\mathrm{13}}{\mathrm{5}+\mathrm{96}}=\frac{\mathrm{104}}{\mathrm{101}} \\ $$$${AC}=\frac{\mathrm{13}+{x}}{\mathrm{8}}×\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }=\frac{\mathrm{1417}\sqrt{\mathrm{89}}}{\mathrm{808}} \\ $$$${AB}=\frac{\mathrm{13}+{x}}{\mathrm{1}}×\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\frac{\mathrm{1417}\sqrt{\mathrm{145}}}{\mathrm{101}} \\ $$$$\frac{{AC}}{{AB}}=\frac{\mathrm{1417}\sqrt{\mathrm{89}}}{\mathrm{808}}×\frac{\mathrm{101}}{\mathrm{1417}\sqrt{\mathrm{145}}}=\frac{\sqrt{\mathrm{89}}}{\mathrm{8}\sqrt{\mathrm{145}}} \\ $$$$\Rightarrow{a}=\mathrm{89},\:{b}=\mathrm{145} \\ $$$$\Rightarrow{a}+{b}=\mathrm{234} \\ $$

Commented by Dwaipayan Shikari last updated on 13/Mar/21

Thanks sir ! Even more shorter and Nice

$${Thanks}\:{sir}\:!\:{Even}\:{more}\:{shorter}\:{and}\:{Nice} \\ $$

Commented by mr W last updated on 13/Mar/21

actually we even don′t need to know  the value of x:  AC=((13+x)/8)×(√(8^2 +5^2 ))  AB=((13+x)/1)×(√(1^2 +12^2 ))  ((AC)/(AB))=((√(8^2 +5^2 ))/(8(√(1^2 +12^2 ))))=((√(89))/(8(√(145))))

$${actually}\:{we}\:{even}\:{don}'{t}\:{need}\:{to}\:{know} \\ $$$${the}\:{value}\:{of}\:{x}: \\ $$$${AC}=\frac{\mathrm{13}+{x}}{\mathrm{8}}×\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} } \\ $$$${AB}=\frac{\mathrm{13}+{x}}{\mathrm{1}}×\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} } \\ $$$$\frac{{AC}}{{AB}}=\frac{\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }}{\mathrm{8}\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{89}}}{\mathrm{8}\sqrt{\mathrm{145}}} \\ $$

Commented by Dwaipayan Shikari last updated on 13/Mar/21

Sir this works for every Pythagorean Triplets.Isn′t it ?

$${Sir}\:{this}\:{works}\:{for}\:{every}\:{Pythagorean}\:{Triplets}.{Isn}'{t}\:{it}\:? \\ $$

Commented by mr W last updated on 13/Mar/21

i think yes.

$${i}\:{think}\:{yes}. \\ $$

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