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Question Number 194464 by Mingma last updated on 08/Jul/23

Answered by mr W last updated on 08/Jul/23

Commented by mr W last updated on 08/Jul/23

$$R=radiusofbigcircle$$$$GF=\sqrt{{\left(2R\right)}^2-x^2}=\sqrt{4R^2-x^2}$$$$\frac{AF}{OF}=\frac{HF}{GF}$$$$\Rightarrow \frac{\sqrt{4R^2-x^2}-x}{R}=\frac{2R}{\sqrt{4R^2-x^2}}$$$$4R^2-x^2-x\sqrt{4R^2-x^2}=2R^2$$$$2R^2-x^2=x\sqrt{4R^2-x^2}$$$$4R^4-4R^2{x}^2+x^4=4R^2{x}^2-x^4$$$$x^4-4R^2{x}^2+2R^4=0$$$$\Rightarrow x^2=(2-\sqrt{2})R^2$$$$\Rightarrow x=\sqrt{2-\sqrt{2}}R$$$$OE=\sqrt{{(R-r)}^2-r^2}=\sqrt{R(R-2r)}$$$$\frac{AD}{AB}=\frac{GF}{HF}$$$$\Rightarrow \frac{AD}{x}=\frac{\sqrt{4R^2-x^2}-x}{2R}$$$$\Rightarrow AD=\frac{x\sqrt{4R^2-x^2}-x^2}{2R}=(\sqrt{2}-1)R$$$$\frac{BD}{AB}=\frac{GH}{HF}$$$$\Rightarrow \frac{BD}{x}=\frac{x}{2R}$$$$\Rightarrow BD=\frac{x^2}{2R}=\frac{(2-\sqrt{2})R}{2}$$$$\frac{AO}{OF}=\frac{GH}{GF}$$$$\Rightarrow \frac{AO}{x}=\frac{x}{\sqrt{4R^2-x^2}}$$$$\Rightarrow AO=\frac{x^2}{\sqrt{4R^2-x^2}}=(\sqrt{2}-1)\sqrt{2-\sqrt{2}}R$$$$DE=AO+OE-AD$$$$=(\sqrt{2}-1)\sqrt{2-\sqrt{2}}R+\sqrt{R(R-2r)}-(\sqrt{2}-1)R$$$$=(\sqrt{2}-1)(\sqrt{2-\sqrt{2}}-1)R+\sqrt{R(R-2r)}$$$$CE-BD=r-\frac{(2-\sqrt{2})R}{2}$$$${[r-\frac{(2-\sqrt{2})R}{2}]}^2+{[(\sqrt{2}-1)(\sqrt{2-\sqrt{2}}-1)R+\sqrt{R(R-2r)}]}^2=r^2$$$$(exactsolutionomited...)$$$$\Rightarrow r\approx 0.387978R$$$$\Rightarrow \frac{x}{R}\approx \frac{\sqrt{2-\sqrt{2}}}{0.387978}\approx 1.9727$$

Commented by Mingma last updated on 08/Jul/23

Very great solution, sir!

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