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Question Number 191458 by Mingma last updated on 24/Apr/23

Answered by a.lgnaoui last updated on 26/Apr/23

$$△\mathrm{ACD}\mathrm{et}△\mathrm{BCD}\mathrm{semblablables}$$$$\mathrm{AB}\mathrm{tangdnte}\mathrm{au}\mathrm{quart}\mathrm{cercle}\mathrm{Rouge}\mathrm{en}\mathrm{D}$$$$\mathrm{CD}=\mathrm{R}2\mathrm{CD}\mathrm{\perp }\mathrm{AB};\mathrm{BC}=\mathrm{R}1+\mathrm{R}2$$$$(\mathrm{R}1\mathrm{rayon}\mathrm{du}\mathrm{quart}\mathrm{cercle}\mathrm{vert}.$$$$\mathrm{ABC}\mathrm{triangle}\mathrm{recrangle}\mathrm{en}\mathrm{C}$$$${\mathrm{AB}}^2={\mathrm{AC}}^2+{\mathrm{BC}}^2(\mathrm{AB}=\mathrm{diametre}\mathrm{cercle}\mathrm{jaune}=2\mathrm{R})$$$${4\mathrm{R}}^2={\mathrm{AC}}^2+{(\mathrm{R}1+\mathrm{R}2)}^2\left(1\right)$$$$△\mathrm{ACD}{\mathrm{AC}}^2={(\mathrm{AB}-\mathrm{BCcos}\theta )}^2+\left({\mathrm{R}2}^2\right)$$$$={[2\mathrm{R}-(\mathrm{R}1+\mathrm{R}2)\cos \theta )}^2]+{\left(\mathrm{R}2\right)}^2$$$$\theta =\measuredangle \mathrm{CBA}$$$$\left(1\right)\Rightarrow 4{\mathbf{R}}^2=[{(2\mathbf{R}-(\mathbf{R}1+\mathbf{R}2)\cos \mathit{\theta })}^2+\left({\mathrm{R}2}^2\right)]+{(\mathrm{R}1+\mathrm{R}2)}^2$$$$\Rightarrow {4\mathrm{R}}^2={4\mathrm{R}}^2+{(\mathrm{R}1+\mathrm{R}2)}^2{\cos }^2\theta -4\mathrm{R}(\mathrm{R}1+\mathrm{R}2)\cos \theta )+{\left(\mathrm{R}2\right)}^2$$$${(\mathrm{R}1+\mathrm{R}2)}^2$$$$$$$$0={(\mathbf{R}1+\mathbf{R}2)}^2(1+\cos {}^2\mathit{\theta })-4\mathbf{R}(\mathbf{R}1+\mathbf{R}2)\cos \theta +{\left(\mathbf{R}2\right)}^2$$$$$$$$0=({\mathrm{R}1}^2+{\mathrm{R}2}^2+2\mathrm{R}1\times \mathrm{R}2)(1+{\cos }^2\theta )$$$$-4\mathrm{R}(\mathrm{R}1+\mathrm{R}2)\cos \theta +\left({\mathrm{R}2}^2\right)\left(2\right)$$$$$$$$△\mathrm{BCF}{\mathrm{CF}}^2=\left({\mathrm{R}1}^2\right)+{\left(\mathrm{R}2\right)}^2={2\mathrm{R}}^2$$$$$$$$\Rightarrow {\left(\mathbf{R}1\right)}^2+\mathbf{R}{2}^2=2{\mathbf{R}}^2\left(3\right)$$$$$$$$\Rightarrow 0=2[({\mathrm{R}}^2+\left(\mathrm{R}1\times \mathrm{R}2\right)](1+\cos {}^2\theta )$$$$-4\mathrm{R}(\mathrm{R}1+\mathrm{R}2)\cos \theta +{\left(\mathrm{R}2\right)}^2$$$$$$$$2({\mathrm{R}}^2+\mathrm{R}1\times \mathrm{R}2)-4\mathrm{R}(\mathrm{R}1+\mathrm{R}2)\cos \theta +{\mathrm{R}2}^2$$$$$$$$+2({\mathrm{R}}^2+\mathrm{R}1\times \mathrm{R}2)\cos {}^2\theta =0$$$$$$$$2{\mathbf{R}}^2(1+{\cos }^2\mathit{\theta })+2\left(\mathbf{R}1\times \mathbf{R}2\right)(1+\cos {}^2\mathit{\theta })$$$$+\mathbf{R}{2}^2-4\mathbf{R}(\mathbf{R}1+\mathbf{R}2)\cos \mathit{\theta }=0$$$$$$$$△\mathrm{BCD}\mathrm{et}△\mathrm{EHB}\mathrm{semblables}$$$$\sin \theta =\frac{\mathrm{R}2}{\mathrm{R}1+\mathrm{R}2}=\frac{\mathrm{EH}}{\mathrm{R}1}\Rightarrow \mathrm{EH}=\mathrm{R}1\sin \theta$$$$\Rightarrow \mathbf{R}2=\mathbf{R}1\sin \mathit{\theta }+\mathbf{R}2\sin \mathit{\theta }$$$$[\mathbf{R}1=\frac{\mathbf{R}2(1-\sin \mathit{\theta })}{\sin \mathit{\theta }}]\mathrm{ou}$$$$[\sin \mathit{\theta }=\frac{\mathbf{R}2}{\mathbf{R}1+\mathbf{R}2}$$$$△\mathrm{BIC}$$$$\mathrm{BC}=2\mathrm{Rcos}\theta \Rightarrow (\mathrm{R}1+\mathrm{R}2)=2\mathrm{Rcos}\theta$$$$[\cos \mathit{\theta }=\frac{\mathbf{R}1+\mathbf{R}2}{2\mathbf{R}}]$$$$$$$$\sin \theta \times \cos \theta =\frac{\mathbf{R}2}{2\mathbf{R}}$$$$\Rightarrow [\sin 2\mathit{\theta }=\frac{\mathrm{R}2}{\mathrm{R}}]\left(4\right)$$$$\mathrm{Surface}\mathrm{quart}\mathrm{de}\mathrm{cercle}\mathrm{veet}=\frac{{\left(\pi \mathrm{R}1\right)}^2}{4}$$$$\mathrm{Celle}\mathrm{du}\mathrm{quart}\mathrm{cercle}\mathrm{Rouge}=\frac{\pi {\left(\mathrm{R}2\right)}^2}{4}$$$$\left({\mathrm{R}1}^2\right)+\left({\mathrm{R}2}^2\right)={2\mathrm{R}}^2\Rightarrow$$$$\mathbf{S}1+\mathbf{S}2=\frac{\pi \left(2{\mathbf{R}}^2\right)}{4}=\frac{\pi {\mathbf{R}}^2}{2}$$$$$$$$\mathrm{Surface}\mathbf{de}\mathbf{qusrt}\mathbf{ceecle}\mathbf{vert}\mathbf{et}\mathbf{rouge}$$$$=\mathbf{moitie}\mathbf{de}\mathbf{surface}\mathbf{du}\mathbf{cercle}$$$$\left(4\right)\Rightarrow$$$$\frac{\mathbf{Surfsce}\mathbf{Vert}}{\mathbf{Surface}\mathbf{Rouge}}=\frac{{\left(\mathbf{R}1\right)}^2}{{\left(\mathbf{R}2\right)}^2}=\frac{{\mathbf{R}}^2-{\left(\mathbf{R}2\right)}^2}{{\left(\mathbf{R}\sin 2\mathit{\theta }\right)}^2}$$$$$$$$=\frac{1}{\sin {}^{2}2\theta }-{\left(\frac{\mathrm{R}2}{\mathrm{Rsin}2\theta }\right)}^2=\frac{1}{\sin {}^{2}2\theta }-1$$$$=\frac{1}{\tan {}^{2}2\mathit{\theta }}$$$$.\mathrm{to}\mathrm{continoius}....$$$$$$

Commented by a.lgnaoui last updated on 26/Apr/23

Answered by mr W last updated on 26/Apr/23

Commented by mr W last updated on 27/Apr/23

$$a=radiusofbluequatercircle$$$$b=radiusofredquatercircle$$$$R=radiusofbigcircle$$$$$$$$\frac{a+b}{b}=\frac{\sqrt{a^2+x^2}}{x}$$$$\Rightarrow x^2=\frac{{ab}^2}{a+2b}$$$$$$$$2R=\frac{\sqrt{a^2+b^2}}{\sin 45°}=\sqrt{2(a^2+b^2)}$$$$\Rightarrow 2R^2=a^2+b^2$$$$\Rightarrow \frac{\pi R^2}{2}=\frac{\pi a^2}{4}+\frac{\pi b^2}{4}$$$$\Rightarrow 50\mathrm{\%}circle=A+B✓$$$$$$$$\frac{2R}{a+b}=\frac{\sqrt{a^2+x^2}}{a}$$$$\sqrt{2(a^2+b^2)}=\frac{a+b}{a}\times \sqrt{a^2+\frac{{ab}^2}{a+2b}}$$$$2(a^2+b^2)=\frac{{(a+b)}^4}{a(a+2b)}$$$$a^4-4a^2{b}^2-b^4=0$$$${(a^2-2b^2)}^2=5b^4$$$$a^2-2b^2=\sqrt{5}{b}^2$$$$\Rightarrow \frac{a^2}{b^2}=2+\sqrt{5}={\phi }^3=\frac{A}{B}✓$$

Commented by Mingma last updated on 27/Apr/23

Excellent, sir!

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