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