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Question Number 202630 by ajfour last updated on 30/Dec/23

Commented by Frix last updated on 31/Dec/23

$$[\mathrm{I}\mathrm{forgot}\mathrm{to}\mathrm{check}\Rightarrow \mathrm{the}{2}^{\mathrm{nd}}\mathrm{positive}\mathrm{root}$$$$\mathrm{is}\mathrm{false}]$$

Commented by Frix last updated on 31/Dec/23

$$\mathrm{I}\mathrm{think}$$$$R=(2k^2+1+2k\sqrt{k^2+1})r$$$$\mathrm{with}k>0\mathrm{and}{k}^6+\frac{9k^4}{16}-\frac{5k^2}{8}+\frac{1}{16}=0$$$$[k=\tan \frac{\alpha }{2}\mathrm{with}\alpha =\mathrm{angle}\mathrm{at}\mathrm{top}]$$$$\mathrm{I}\mathrm{get}$$$$k\approx .337774062352$$$$\Rightarrow$$$$r\approx .228182634396\wedge R\approx .442954341401$$$$[\alpha \approx 37.327°;\mathrm{height}\approx 1.3114]$$

Answered by mr W last updated on 31/Dec/23

Commented by Frix last updated on 31/Dec/23

$$\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{consider}k=\sqrt{1-2R}$$$$\Rightarrow$$$$k^6+\frac{9k^4}{16}-\frac{5k^2}{8}+\frac{1}{16}=0$$$$\mathrm{is}\mathrm{equal}\mathrm{to}$$$$R^3-\frac{57R^2}{32}+\frac{7R}{8}-\frac{1}{8}=0$$

Commented by mr W last updated on 31/Dec/23

Commented by ajfour last updated on 31/Dec/23

$$Iget(r,R)\equiv$$$$(0.2282,0.44295)$$$$(0.4435,0.3893)$$$$(0.0546,0.4864)$$

Commented by mr W last updated on 31/Dec/23

$${(R+r)}^2-{(R-r)}^2+{(1-r)}^2=1^2$$$$\Rightarrow r=2-4R$$$$\sin \theta =\frac{R-r}{R+r}=\frac{5R-2}{2-3R}\to 0.4<R<0.5$$$$\frac{R}{1-R}=\cos 2\theta =1-2{\left(\frac{5R-2}{2-3R}\right)}^2$$$$32R^3-57R^2+28R-4=0$$$$\Rightarrow R\approx 0.44295433$$$$\Rightarrow r\approx 0.22818264$$

Commented by mr W last updated on 31/Dec/23

$$butonlythefirstoneisvalid.$$

Commented by ajfour last updated on 31/Dec/23

$$Thankyousir,yourwayisvery$$$$charming.Ihavntcheckedtoreject.$$

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