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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

$$Q154594$$

Answered by mr W last updated on 03/Oct/21

Commented by mr W last updated on 03/Oct/21

$${it}^{\prime }sobvious:$$$$AE=diameterofcircle$$$$\mathrm{\Delta }ABG\sim \mathrm{\Delta }FDE\sim \mathrm{\Delta }FCG$$$$\alpha +\beta =45°$$$$$$$$CI=BI=DI=radius=R$$$$BK=BI\times \cos 2\alpha =R\cos 2\alpha$$$${\left(\frac{BK}{CI}\right)}^2={\cos }^{2}2\alpha$$$$similarly$$$${\left(\frac{DH}{CI}\right)}^2={\cos }^{2}2\beta ={\cos }^2(90°-2\alpha )={\sin }^{2}2\alpha$$$$\Rightarrow {\left(\frac{BK}{CI}\right)}^2+{\left(\frac{DH}{CI}\right)}^2={\cos }^{2}2\alpha +{\sin }^{2}2\alpha =1$$$$\frac{{\mathrm{\Delta }}_{ABG}}{{\mathrm{\Delta }}_{GCF}}={\left(\frac{BK}{CI}\right)}^2$$$$\frac{{\mathrm{\Delta }}_{EDF}}{{\mathrm{\Delta }}_{GCF}}={\left(\frac{DH}{CI}\right)}^2$$$$\frac{{\mathrm{\Delta }}_{ABG}}{{\mathrm{\Delta }}_{GCF}}+\frac{{\mathrm{\Delta }}_{EDF}}{{\mathrm{\Delta }}_{GCF}}={\left(\frac{BK}{CI}\right)}^2+{\left(\frac{DH}{CI}\right)}^2=1$$$$\Rightarrow {\mathrm{\Delta }}_{ABG}+{\mathrm{\Delta }}_{EDF}={\mathrm{\Delta }}_{GCF}$$$$\Rightarrow yellowarea=halfofcircle$$$$\Rightarrow greenarea=halfofcircle$$$$\Rightarrow \frac{yellow}{green}=1$$

Commented by mr W last updated on 03/Oct/21

Commented by amin96 last updated on 03/Oct/21

$$Nicesirreallygreate$$

Commented by puissant last updated on 03/Oct/21

$$WOWMrWiappreciate..$$

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

$$(A+D)+E+F=(A+E)+(D+F)$$$$=\frac{r^2}{2}(\sin 2\alpha +\sin 2\beta )$$$$=r^2\sin \theta \cos (\alpha -\beta )={△}_1$$$$(B+C)+E+F=(B+E)+(C+F)$$$$=\frac{r^2}{2}(\sin (90°+2\alpha )+\sin (90°+2\beta )$$$$=\frac{r^2}{2}(\cos 2\alpha +\cos 2\beta )$$$$=r^2\cos \theta \cos (\alpha -\beta )={△}_2$$$$hencefor\theta =45°,{△}_1={△}_2$$$$\Rightarrow A+D=B+C$$$$wethusexchangegreenA+D$$$$withsaffronB+C$$$$hence$$$$saffronarea=semi-circlearea$$$$saffronarea=greenarea.$$

Commented by mr W last updated on 05/Oct/21

$$nicesolution!$$

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