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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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