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Question-155657




Question Number 155657 by mr W last updated on 03/Oct/21
Commented by mr W last updated on 03/Oct/21
Q154594
$${Q}\mathrm{154594} \\ $$
Answered by mr W last updated on 03/Oct/21
Commented by mr W last updated on 03/Oct/21
it′s obvious:  AE=diameter of circle  ΔABG∼ΔFDE∼ΔFCG  α+β=45°    CI=BI=DI=radius=R  BK=BI×cos 2α=R cos 2α  (((BK)/(CI)))^2 =cos^2  2α  similarly  (((DH)/(CI)))^2 =cos^2  2β=cos^2  (90°−2α)=sin^2  2α  ⇒(((BK)/(CI)))^2 +(((DH)/(CI)))^2 =cos^2  2α+sin^2  2α=1  (Δ_(ABG) /Δ_(GCF) )=(((BK)/(CI)))^2   (Δ_(EDF) /Δ_(GCF) )=(((DH)/(CI)))^2   (Δ_(ABG) /Δ_(GCF) )+(Δ_(EDF) /Δ_(GCF) )=(((BK)/(CI)))^2 +(((DH)/(CI)))^2 =1  ⇒Δ_(ABG) +Δ_(EDF) =Δ_(GCF)   ⇒yellow area=half of circle  ⇒green area=half of circle  ⇒((yellow)/(green))=1
$${it}'{s}\:{obvious}: \\ $$$${AE}={diameter}\:{of}\:{circle} \\ $$$$\Delta{ABG}\sim\Delta{FDE}\sim\Delta{FCG} \\ $$$$\alpha+\beta=\mathrm{45}° \\ $$$$ \\ $$$${CI}={BI}={DI}={radius}={R} \\ $$$${BK}={BI}×\mathrm{cos}\:\mathrm{2}\alpha={R}\:\mathrm{cos}\:\mathrm{2}\alpha \\ $$$$\left(\frac{{BK}}{{CI}}\right)^{\mathrm{2}} =\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\alpha \\ $$$${similarly} \\ $$$$\left(\frac{{DH}}{{CI}}\right)^{\mathrm{2}} =\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\beta=\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{90}°−\mathrm{2}\alpha\right)=\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\alpha \\ $$$$\Rightarrow\left(\frac{{BK}}{{CI}}\right)^{\mathrm{2}} +\left(\frac{{DH}}{{CI}}\right)^{\mathrm{2}} =\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\alpha+\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\alpha=\mathrm{1} \\ $$$$\frac{\Delta_{{ABG}} }{\Delta_{{GCF}} }=\left(\frac{{BK}}{{CI}}\right)^{\mathrm{2}} \\ $$$$\frac{\Delta_{{EDF}} }{\Delta_{{GCF}} }=\left(\frac{{DH}}{{CI}}\right)^{\mathrm{2}} \\ $$$$\frac{\Delta_{{ABG}} }{\Delta_{{GCF}} }+\frac{\Delta_{{EDF}} }{\Delta_{{GCF}} }=\left(\frac{{BK}}{{CI}}\right)^{\mathrm{2}} +\left(\frac{{DH}}{{CI}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\Delta_{{ABG}} +\Delta_{{EDF}} =\Delta_{{GCF}} \\ $$$$\Rightarrow{yellow}\:{area}={half}\:{of}\:{circle} \\ $$$$\Rightarrow{green}\:{area}={half}\:{of}\:{circle} \\ $$$$\Rightarrow\frac{{yellow}}{{green}}=\mathrm{1} \\ $$
Commented by mr W last updated on 03/Oct/21
Commented by amin96 last updated on 03/Oct/21
Nice sir really greate
$${Nice}\:{sir}\:{really}\:{greate} \\ $$
Commented by puissant last updated on 03/Oct/21
WOW Mr W i appreciate..
$${WOW}\:{Mr}\:{W}\:{i}\:{appreciate}.. \\ $$
Commented by Tawa11 last updated on 03/Oct/21
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by ajfour last updated on 03/Oct/21
Commented by ajfour last updated on 03/Oct/21
(A+D)+E+F = (A+E)+(D+F)  =(r^2 /2)(sin 2α+sin 2β)  =r^2 sin θcos (α−β)  =△_1   (B+C)+E+F=(B+E)+(C+F)  =(r^2 /2)(sin (90°+2α)+sin (90°+2β)  =(r^2 /2)(cos 2α+cos 2β)  =r^2 cos θcos (α−β)=△_2   hence for  θ=45°  ,  △_1 =△_2   ⇒  A+D=B+C  we thus exchange green A+D  with saffron  B+C  hence     saffron area=semi-circle area    saffron area=green area.
$$\left({A}+{D}\right)+{E}+{F}\:=\:\left({A}+{E}\right)+\left({D}+{F}\right) \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta\right) \\ $$$$={r}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\left(\alpha−\beta\right)\:\:=\bigtriangleup_{\mathrm{1}} \\ $$$$\left({B}+{C}\right)+{E}+{F}=\left({B}+{E}\right)+\left({C}+{F}\right) \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{sin}\:\left(\mathrm{90}°+\mathrm{2}\alpha\right)+\mathrm{sin}\:\left(\mathrm{90}°+\mathrm{2}\beta\right)\right. \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{cos}\:\mathrm{2}\alpha+\mathrm{cos}\:\mathrm{2}\beta\right) \\ $$$$={r}^{\mathrm{2}} \mathrm{cos}\:\theta\mathrm{cos}\:\left(\alpha−\beta\right)=\bigtriangleup_{\mathrm{2}} \\ $$$${hence}\:{for}\:\:\theta=\mathrm{45}°\:\:,\:\:\bigtriangleup_{\mathrm{1}} =\bigtriangleup_{\mathrm{2}} \\ $$$$\Rightarrow\:\:{A}+{D}={B}+{C} \\ $$$${we}\:{thus}\:{exchange}\:{green}\:{A}+{D} \\ $$$${with}\:{saffron}\:\:{B}+{C} \\ $$$${hence}\: \\ $$$$\:\:{saffron}\:{area}={semi}-{circle}\:{area} \\ $$$$\:\:{saffron}\:{area}={green}\:{area}. \\ $$
Commented by mr W last updated on 05/Oct/21
nice solution!
$${nice}\:{solution}! \\ $$

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