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Question Number 164018 by cortano1 last updated on 13/Jan/22

Answered by mr W last updated on 13/Jan/22

Commented by cortano1 last updated on 13/Jan/22

$$superbsolution$$

Commented by mr W last updated on 13/Jan/22

$$b=2R$$$$a=\sqrt{{(R+r)}^2-{(R-r)}^2}+\sqrt{{(R+r)}^2-{(2R-r)}^2}$$$$a=2\sqrt{Rr}+\sqrt{3R(2r-R)}$$$$BC=2\sqrt{{(R+r)}^2-r^2}=2\sqrt{R(R+2r)}$$$$a=\sqrt{{BC}^2-R^2}=\sqrt{4R(R+2r)-R^2}=\sqrt{R(3R+8r)}$$$$2\sqrt{Rr}+\sqrt{3R(2r-R)}=\sqrt{R(3R+8r)}$$$$let\lambda =\frac{R}{r}$$$$2+\sqrt{3(2-\lambda )}=\sqrt{3\lambda +8}$$$$3(2-\lambda )=3\lambda +8+4-4\sqrt{3\lambda +8}$$$$3(\lambda +1)=2\sqrt{3\lambda +8}$$$$9({\lambda }^2+2\lambda +1)=4(3\lambda +8)$$$$9{\lambda }^2+6\lambda -23=0$$$$\lambda =\frac{-1+2\sqrt{6}}{3}$$$$\Rightarrow \frac{R}{r}=\frac{2\sqrt{6}-1}{3}✓$$

Commented by Tawa11 last updated on 13/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

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