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Question Number 200471 by Mingma last updated on 19/Nov/23
Answered by AST last updated on 19/Nov/23
((CB_1 )/(B_1 A))×(1/2)×(1/4)=1⇒((CB_1 )/(B_1 A))=8 (([CAB])/([ADB]))=((CC_1 )/(DC_1 ))=((CD)/(DC_1 ))+1=((CB_1 )/(B_1 A))+((CA_1 )/(A_1 B))+1=13 ⇒[ADB]=((65)/(13))=5 (([CDA])/([CDB]))=(1/2)⇒[CDA]+[CDB]=3[CDA]=65−5 ⇒[CDA]=20;(([B_1 CD])/([B_1 AD]))=8 [B_1 CD]+[B_1 AD]=((9[B_1 CD])/8)=20⇒[B_1 CD]=((160)/9)
$$\frac{{CB}_{\mathrm{1}} }{{B}_{\mathrm{1}} {A}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{1}\Rightarrow\frac{{CB}_{\mathrm{1}} }{{B}_{\mathrm{1}} {A}}=\mathrm{8} \\ $$$$\frac{\left[{CAB}\right]}{\left[{ADB}\right]}=\frac{{CC}_{\mathrm{1}} }{{DC}_{\mathrm{1}} }=\frac{{CD}}{{DC}_{\mathrm{1}} }+\mathrm{1}=\frac{{CB}_{\mathrm{1}} }{{B}_{\mathrm{1}} {A}}+\frac{{CA}_{\mathrm{1}} }{{A}_{\mathrm{1}} {B}}+\mathrm{1}=\mathrm{13} \\ $$$$\Rightarrow\left[{ADB}\right]=\frac{\mathrm{65}}{\mathrm{13}}=\mathrm{5} \\ $$$$\frac{\left[{CDA}\right]}{\left[{CDB}\right]}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\left[{CDA}\right]+\left[{CDB}\right]=\mathrm{3}\left[{CDA}\right]=\mathrm{65}−\mathrm{5} \\ $$$$\Rightarrow\left[{CDA}\right]=\mathrm{20};\frac{\left[{B}_{\mathrm{1}} {CD}\right]}{\left[{B}_{\mathrm{1}} {AD}\right]}=\mathrm{8} \\ $$$$\left[{B}_{\mathrm{1}} {CD}\right]+\left[{B}_{\mathrm{1}} {AD}\right]=\frac{\mathrm{9}\left[{B}_{\mathrm{1}} {CD}\right]}{\mathrm{8}}=\mathrm{20}\Rightarrow\left[{B}_{\mathrm{1}} {CD}\right]=\frac{\mathrm{160}}{\mathrm{9}} \\ $$
Commented by Rupesh123 last updated on 19/Nov/23
Nice solution,sir!
Commented by klipto last updated on 28/Nov/23
160/9
Answered by mr W last updated on 19/Nov/23
AB^(→) =a BC^(→) =b CA^(→) =−a−b AA_1 ^(→) =a+(1/5)b CC_1 ^(→) =−(2/3)a−b say AD=sAA_1 =s(a+(1/5)b) CD=tCC_1 =t(−(2/3)a−b) AD=a+b+CD s(a+(1/5)b)=a+b+t(−(2/3)a−b) (1−s−((2t)/3))a+(1−(s/5)−t)b=0 1−s−((2t)/3)=0 ...(i) 1−(s/5)−t=0 ...(ii) ⇒s=(5/(13)), t=((12)/(13)) BD=−a+AD=−a+(5/(13))(a+(1/5)b)=−(8/(13))a+(1/(13))b say BD=uBB_1 , CB_1 =vCA v(−a−b)=−b+(1/u)(−(8/(13))a+(1/(13))b) ((8/(13u))−v)a+(1−(1/(13u))−v)b=0 (8/(13u))−v=0 ...(iii) 1−(1/(13u))−v=0 ...(iv) ⇒u=(9/(13)), v=(8/9) Δ_(B_1 CD) =(1−u)Δ_(B_1 CB) =(1−u)vΔ_(ABC) Δ_(B_1 CD) =(1−(9/(13)))×(8/9)Δ_(ABC) =((32)/(117))Δ_(ABC) =((32×65)/(117))=((160)/9)≈17.8
$$\overset{\rightarrow} {{AB}}=\boldsymbol{{a}} \\ $$$$\overset{\rightarrow} {{BC}}=\boldsymbol{{b}} \\ $$$$\overset{\rightarrow} {{CA}}=−\boldsymbol{{a}}−\boldsymbol{{b}} \\ $$$$\overset{\rightarrow} {{AA}_{\mathrm{1}} }=\boldsymbol{{a}}+\frac{\mathrm{1}}{\mathrm{5}}\boldsymbol{{b}} \\ $$$$\overset{\rightarrow} {{CC}_{\mathrm{1}} }=−\frac{\mathrm{2}}{\mathrm{3}}\boldsymbol{{a}}−\boldsymbol{{b}} \\ $$$${say}\:{AD}={sAA}_{\mathrm{1}} ={s}\left(\boldsymbol{{a}}+\frac{\mathrm{1}}{\mathrm{5}}\boldsymbol{{b}}\right) \\ $$$${CD}={tCC}_{\mathrm{1}} ={t}\left(−\frac{\mathrm{2}}{\mathrm{3}}\boldsymbol{{a}}−\boldsymbol{{b}}\right) \\ $$$${AD}=\boldsymbol{{a}}+\boldsymbol{{b}}+{CD} \\ $$$${s}\left(\boldsymbol{{a}}+\frac{\mathrm{1}}{\mathrm{5}}\boldsymbol{{b}}\right)=\boldsymbol{{a}}+\boldsymbol{{b}}+{t}\left(−\frac{\mathrm{2}}{\mathrm{3}}\boldsymbol{{a}}−\boldsymbol{{b}}\right) \\ $$$$\left(\mathrm{1}−{s}−\frac{\mathrm{2}{t}}{\mathrm{3}}\right)\boldsymbol{{a}}+\left(\mathrm{1}−\frac{{s}}{\mathrm{5}}−{t}\right)\boldsymbol{{b}}=\mathrm{0} \\ $$$$\mathrm{1}−{s}−\frac{\mathrm{2}{t}}{\mathrm{3}}=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\mathrm{1}−\frac{{s}}{\mathrm{5}}−{t}=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow{s}=\frac{\mathrm{5}}{\mathrm{13}},\:{t}=\frac{\mathrm{12}}{\mathrm{13}} \\ $$$${BD}=−\boldsymbol{{a}}+{AD}=−\boldsymbol{{a}}+\frac{\mathrm{5}}{\mathrm{13}}\left(\boldsymbol{{a}}+\frac{\mathrm{1}}{\mathrm{5}}\boldsymbol{{b}}\right)=−\frac{\mathrm{8}}{\mathrm{13}}\boldsymbol{{a}}+\frac{\mathrm{1}}{\mathrm{13}}\boldsymbol{{b}} \\ $$$${say}\:{BD}={uBB}_{\mathrm{1}} ,\:{CB}_{\mathrm{1}} ={vCA} \\ $$$${v}\left(−\boldsymbol{{a}}−\boldsymbol{{b}}\right)=−\boldsymbol{{b}}+\frac{\mathrm{1}}{{u}}\left(−\frac{\mathrm{8}}{\mathrm{13}}\boldsymbol{{a}}+\frac{\mathrm{1}}{\mathrm{13}}\boldsymbol{{b}}\right) \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{13}{u}}−{v}\right)\boldsymbol{{a}}+\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{13}{u}}−{v}\right)\boldsymbol{{b}}=\mathrm{0} \\ $$$$\frac{\mathrm{8}}{\mathrm{13}{u}}−{v}=\mathrm{0}\:\:\:…\left({iii}\right) \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{13}{u}}−{v}=\mathrm{0}\:\:\:…\left({iv}\right) \\ $$$$\Rightarrow{u}=\frac{\mathrm{9}}{\mathrm{13}},\:{v}=\frac{\mathrm{8}}{\mathrm{9}} \\ $$$$ \\ $$$$\Delta_{{B}_{\mathrm{1}} {CD}} =\left(\mathrm{1}−{u}\right)\Delta_{{B}_{\mathrm{1}} {CB}} =\left(\mathrm{1}−{u}\right){v}\Delta_{{ABC}} \\ $$$$\Delta_{{B}_{\mathrm{1}} {CD}} =\left(\mathrm{1}−\frac{\mathrm{9}}{\mathrm{13}}\right)×\frac{\mathrm{8}}{\mathrm{9}}\Delta_{{ABC}} =\frac{\mathrm{32}}{\mathrm{117}}\Delta_{{ABC}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{32}×\mathrm{65}}{\mathrm{117}}=\frac{\mathrm{160}}{\mathrm{9}}\approx\mathrm{17}.\mathrm{8} \\ $$
Commented by Rupesh123 last updated on 19/Nov/23
Nice solution,sir!
Commented by Anonim_X last updated on 19/Nov/23
great solution
$${great}\:{solution} \\ $$
Commented by mr W last updated on 20/Nov/23
certainly we can directly apply the Ceva′s theorem, see below. but here i applied the vector method.
$${certainly}\:{we}\:{can}\:{directly}\:{apply}\:{the} \\ $$$${Ceva}'{s}\:{theorem},\:{see}\:{below}.\:{but}\:{here} \\ $$$${i}\:{applied}\:{the}\:{vector}\:{method}. \\ $$
Commented by mr W last updated on 20/Nov/23
Answered by ajfour last updated on 19/Nov/23
△_0 =((5×3)/2)sin B=6.5 ⇒ sin B=((13)/(15)) let D(0,0); A(−a,0); C(0,b) y_(B1) =−(b/4) sin B=(((b+(b/4)))/5)=((13)/(15)) ⇒ b=((52)/(15)) cos B=(((9+25−AC^2 )/(2×3×5)))=(√(1−(((13)/(15)))^2 )) AC^2 =34−2(√(56)) =a^2 +b^2 x_(C1) =0=((−2a+x_B )/3) ⇒ x_B =2a eq of AC y=(b/a)x+b eq of BB_1 y=((y_B /x_B ))x =−(b/(8a))x intersection (x_(B1) , y_(B1) ) (b/a)x+b=−(b/(8a))x ⇒ x_(B1) =−((8a)/9) ⇒ y_(B1) =(b/9) △_(B_1 CD) =((b(−x_B_1 ))/2)=((4ab)/9) =((4b(√((a^2 +b^2 )−b^2 )))/9) =(4/9)(((52)/(15)))(√(34−2(√(56))−(((52)/(15)))^2 )) ≈ 4.0809
$$\bigtriangleup_{\mathrm{0}} =\frac{\mathrm{5}×\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:{B}=\mathrm{6}.\mathrm{5}\:\:\Rightarrow\:\mathrm{sin}\:{B}=\frac{\mathrm{13}}{\mathrm{15}} \\ $$$$ \\ $$$${let}\:\:{D}\left(\mathrm{0},\mathrm{0}\right);\:\:{A}\left(−{a},\mathrm{0}\right);\:\:{C}\left(\mathrm{0},{b}\right) \\ $$$$\:\:{y}_{{B}\mathrm{1}} =−\frac{{b}}{\mathrm{4}}\:\:\: \\ $$$$\mathrm{sin}\:{B}=\frac{\left({b}+\frac{{b}}{\mathrm{4}}\right)}{\mathrm{5}}=\frac{\mathrm{13}}{\mathrm{15}}\:\:\: \\ $$$$\Rightarrow\:\:{b}=\frac{\mathrm{52}}{\mathrm{15}} \\ $$$$\mathrm{cos}\:{B}=\left(\frac{\mathrm{9}+\mathrm{25}−{AC}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\mathrm{5}}\right)=\sqrt{\mathrm{1}−\left(\frac{\mathrm{13}}{\mathrm{15}}\right)^{\mathrm{2}} } \\ $$$${AC}^{\mathrm{2}} =\mathrm{34}−\mathrm{2}\sqrt{\mathrm{56}}\:={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${x}_{{C}\mathrm{1}} =\mathrm{0}=\frac{−\mathrm{2}{a}+{x}_{{B}} }{\mathrm{3}}\:\:\Rightarrow\:\:{x}_{{B}} =\mathrm{2}{a} \\ $$$${eq}\:{of}\:{AC}\:\:\:\:\:\:{y}=\frac{{b}}{{a}}{x}+{b} \\ $$$${eq}\:{of}\:{BB}_{\mathrm{1}} \:\:\:\:\:{y}=\left(\frac{{y}_{{B}} }{{x}_{{B}} }\right){x}\:\:\:=−\frac{{b}}{\mathrm{8}{a}}{x} \\ $$$${intersection}\:\:\left({x}_{{B}\mathrm{1}} ,\:{y}_{{B}\mathrm{1}} \right) \\ $$$$\frac{{b}}{{a}}{x}+{b}=−\frac{{b}}{\mathrm{8}{a}}{x} \\ $$$$\Rightarrow\:\:{x}_{{B}\mathrm{1}} =−\frac{\mathrm{8}{a}}{\mathrm{9}}\:\Rightarrow\:\:{y}_{{B}\mathrm{1}} =\frac{{b}}{\mathrm{9}} \\ $$$$\bigtriangleup_{{B}_{\mathrm{1}} {CD}} =\frac{{b}\left(−{x}_{{B}_{\mathrm{1}} } \right)}{\mathrm{2}}=\frac{\mathrm{4}{ab}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}{b}\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} }}{\mathrm{9}} \\ $$$$\:\:\:=\frac{\mathrm{4}}{\mathrm{9}}\left(\frac{\mathrm{52}}{\mathrm{15}}\right)\sqrt{\mathrm{34}−\mathrm{2}\sqrt{\mathrm{56}}−\left(\frac{\mathrm{52}}{\mathrm{15}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\approx\:\mathrm{4}.\mathrm{0809} \\ $$
Commented by mr W last updated on 19/Nov/23
i think in the diagram 1, 2 and 1, 4 just mean the ratios are 1:2 and 1:4.
$${i}\:{think}\:{in}\:{the}\:{diagram}\:\mathrm{1},\:\mathrm{2}\:{and}\:\mathrm{1},\:\mathrm{4} \\ $$$${just}\:{mean}\:{the}\:{ratios}\:{are}\:\mathrm{1}:\mathrm{2}\:{and}\:\mathrm{1}:\mathrm{4}. \\ $$
Commented by ajfour last updated on 19/Nov/23
oh! thanks sir. dint dawn on me so..
$${oh}!\:{thanks}\:{sir}.\:{dint}\:{dawn}\:{on}\:{me}\:{so}.. \\ $$

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