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Question Number 88820 by ajfour last updated on 13/Apr/20

Commented by ajfour last updated on 13/Apr/20

If both coloured regions have equal area, find B(x_B , y_B ) in terms of c.

$${If}\:{both}\:{coloured}\:{regions}\:{have} \\ $$$${equal}\:{area},\:{find}\:{B}\left({x}_{{B}} \:,\:{y}_{{B}} \right)\:{in} \\ $$$${terms}\:{of}\:{c}. \\ $$

Commented by john santu last updated on 13/Apr/20

total area = (4/3)c(√c) line BQ : y=((√b)−(√c))x +(√(bc)) blue area = (((b+c+2(√(bc)))^(3/2) )/6) = (2/3)(c)^(3/2)

$${total}\:{area}\:=\:\frac{\mathrm{4}}{\mathrm{3}}{c}\sqrt{{c}} \\ $$$${line}\:{BQ}\::\:{y}=\left(\sqrt{{b}}−\sqrt{{c}}\right){x}\:+\sqrt{{bc}} \\ $$$${blue}\:{area}\:=\:\frac{\left({b}+{c}+\mathrm{2}\sqrt{{bc}}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{6}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\left({c}\right)^{\mathrm{3}/\mathrm{2}} \\ $$$$ \\ $$

Commented by jagoll last updated on 13/Apr/20

((√b)+(√c))^3 = 4((√c))^3 (√b) + (√( c)) = (4)^(1/(3 )) (√c) (√b) = (√c) [ (4)^(1/(3 )) −1 ] (√(b/c)) = (4)^(1/(3 )) −1

$$\left(\sqrt{{b}}+\sqrt{{c}}\right)^{\mathrm{3}} \:=\:\mathrm{4}\left(\sqrt{{c}}\right)^{\mathrm{3}} \\ $$$$\sqrt{{b}}\:+\:\sqrt{\:{c}}\:=\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{4}}\:\sqrt{{c}} \\ $$$$\sqrt{{b}}\:=\:\sqrt{{c}}\:\left[\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{4}}\:−\mathrm{1}\:\right] \\ $$$$\sqrt{\frac{{b}}{{c}}}\:=\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{4}}\:−\mathrm{1} \\ $$

Commented by jagoll last updated on 13/Apr/20

i continue to work mr john

$${i}\:{continue}\:{to}\:{work}\:{mr}\:{john} \\ $$

Commented by john santu last updated on 13/Apr/20

good mr jagoll. thank you

$${good}\:{mr}\:{jagoll}.\:{thank}\:{you} \\ $$

Answered by ajfour last updated on 13/Apr/20

A_(total) =2×(c(√c)−((c(√c))/3))=((4c(√c))/3) A_(blue) =(1/2)(x_B ^2 +c)(x_B +(√c)) −((x_B ^3 /3)+((c(√c))/3)) = ((2c(√c))/3) say x_B = t (t^3 /6)+((√c)/2)t^2 +(c/2)t−((c(√c))/2)=0 ⇒ t^3 +3(√c)t^2 +3ct+c(√c)=4c(√c) ⇒ (t+(√c))^3 =4c(√c) ⇒ x_B = t = (√c)( (4)^(1/3) −1) .

$${A}_{{total}} =\mathrm{2}×\left({c}\sqrt{{c}}−\frac{{c}\sqrt{{c}}}{\mathrm{3}}\right)=\frac{\mathrm{4}{c}\sqrt{{c}}}{\mathrm{3}}\: \\ $$$${A}_{{blue}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}_{{B}} ^{\mathrm{2}} +{c}\right)\left({x}_{{B}} +\sqrt{{c}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\left(\frac{{x}_{{B}} ^{\mathrm{3}} }{\mathrm{3}}+\frac{{c}\sqrt{{c}}}{\mathrm{3}}\right)\:\:=\:\frac{\mathrm{2}{c}\sqrt{{c}}}{\mathrm{3}} \\ $$$$\:\:\:\:{say}\:\:{x}_{{B}} =\:{t} \\ $$$$\:\:\:\:\:\frac{{t}^{\mathrm{3}} }{\mathrm{6}}+\frac{\sqrt{{c}}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{{c}}{\mathrm{2}}{t}−\frac{{c}\sqrt{{c}}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{3}} +\mathrm{3}\sqrt{{c}}{t}^{\mathrm{2}} +\mathrm{3}{ct}+{c}\sqrt{{c}}=\mathrm{4}{c}\sqrt{{c}} \\ $$$$\Rightarrow\:\:\:\left({t}+\sqrt{{c}}\right)^{\mathrm{3}} =\mathrm{4}{c}\sqrt{{c}} \\ $$$$\Rightarrow\:\:\:{x}_{{B}} =\:{t}\:=\:\sqrt{{c}}\left(\:\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\:. \\ $$

Answered by mr W last updated on 13/Apr/20

(2/3)(x_B +(√c))[((c+x_B ^2 )/2)−(((x_B −(√c))/2))^2 ]=(1/2)×(2/3)×2c(√c) let C=(√c) (x_B +C)[((C^2 +x_B ^2 +2Cx_B )/4)]=C^3 ((x_B /C))^3 +3((x_B /C))^2 +3((x_B /C))−3=0 ((x_B /C)+1)^3 =4 (x_B /C)=(4)^(1/3) −1≈0.5874 ⇒x_B ≈0.5874 (√c)

$$\frac{\mathrm{2}}{\mathrm{3}}\left({x}_{{B}} +\sqrt{{c}}\right)\left[\frac{{c}+{x}_{{B}} ^{\mathrm{2}} }{\mathrm{2}}−\left(\frac{{x}_{{B}} −\sqrt{{c}}}{\mathrm{2}}\right)^{\mathrm{2}} \right]=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{2}{c}\sqrt{{c}} \\ $$$${let}\:{C}=\sqrt{{c}} \\ $$$$\left({x}_{{B}} +{C}\right)\left[\frac{{C}^{\mathrm{2}} +{x}_{{B}} ^{\mathrm{2}} +\mathrm{2}{Cx}_{{B}} }{\mathrm{4}}\right]={C}^{\mathrm{3}} \\ $$$$\left(\frac{{x}_{{B}} }{{C}}\right)^{\mathrm{3}} +\mathrm{3}\left(\frac{{x}_{{B}} }{{C}}\right)^{\mathrm{2}} +\mathrm{3}\left(\frac{{x}_{{B}} }{{C}}\right)−\mathrm{3}=\mathrm{0} \\ $$$$\left(\frac{{x}_{{B}} }{{C}}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{4} \\ $$$$\frac{{x}_{{B}} }{{C}}=\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\approx\mathrm{0}.\mathrm{5874} \\ $$$$\Rightarrow{x}_{{B}} \approx\mathrm{0}.\mathrm{5874}\:\sqrt{{c}} \\ $$

Commented by ajfour last updated on 13/Apr/20

(x_B /C)+1 = (4)^(1/3) x_B =(√c) ((4)^(1/3) −1) . Isn′t it Sir.

$$\frac{{x}_{{B}} }{{C}}+\mathrm{1}\:=\:\:\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$$${x}_{{B}} =\sqrt{{c}}\:\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\:.\:\:{Isn}'{t}\:{it}\:{Sir}. \\ $$

Commented by mr W last updated on 13/Apr/20

Commented by mr W last updated on 13/Apr/20

yes. the exact value is x_B =((4)^(1/3) −1)(√c).

$${yes}.\:{the}\:{exact}\:{value}\:{is}\:{x}_{{B}} =\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\sqrt{{c}}. \\ $$

Commented by ajfour last updated on 13/Apr/20

sir i will need some time understanding your method thoroughly, this on and circle one..

$${sir}\:{i}\:{will}\:{need}\:{some}\:{time}\: \\ $$$${understanding}\:{your}\:{method} \\ $$$${thoroughly},\:{this}\:{on}\:{and}\:{circle}\:{one}.. \\ $$

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