Question-202630

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

Commented by Frix last updated on 31/Dec/23

[I forgot to check ⇒ the 2^(nd) positive root is false]

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

Commented by Frix last updated on 31/Dec/23

I think R=(2k^2 +1+2k(√(k^2 +1)))r with k>0 and k^6 +((9k^4 )/(16))−((5k^2 )/8)+(1/(16))=0 [k=tan (α/2) with α=angle at top] I get k≈.337774062352 ⇒ r≈.228182634396∧R≈.442954341401 [α≈37.327°; height ≈1.3114]

$$\mathrm{I}\:\mathrm{think} \\ $$$${R}=\left(\mathrm{2}{k}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{k}\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}\right){r} \\ $$$$\mathrm{with}\:{k}>\mathrm{0}\:\mathrm{and}\:{k}^{\mathrm{6}} +\frac{\mathrm{9}{k}^{\mathrm{4}} }{\mathrm{16}}−\frac{\mathrm{5}{k}^{\mathrm{2}} }{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{16}}=\mathrm{0} \\ $$$$\left[{k}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{with}\:\alpha=\mathrm{angle}\:\mathrm{at}\:\mathrm{top}\right] \\ $$$$\mathrm{I}\:\mathrm{get} \\ $$$${k}\approx.\mathrm{337774062352} \\ $$$$\Rightarrow \\ $$$${r}\approx.\mathrm{228182634396}\wedge{R}\approx.\mathrm{442954341401} \\ $$$$\left[\alpha\approx\mathrm{37}.\mathrm{327}°;\:\mathrm{height}\:\approx\mathrm{1}.\mathrm{3114}\right] \\ $$

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

Commented by Frix last updated on 31/Dec/23

I didn′t consider k=(√(1−2R)) ⇒ k^6 +((9k^4 )/(16))−((5k^2 )/( 8))+(1/(16))=0 is equal to R^3 −((57R^2 )/(32))+((7R)/8)−(1/8)=0

$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{consider}\:{k}=\sqrt{\mathrm{1}−\mathrm{2}{R}} \\ $$$$\Rightarrow \\ $$$${k}^{\mathrm{6}} +\frac{\mathrm{9}{k}^{\mathrm{4}} }{\mathrm{16}}−\frac{\mathrm{5}{k}^{\mathrm{2}} }{\:\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{16}}=\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$${R}^{\mathrm{3}} −\frac{\mathrm{57}{R}^{\mathrm{2}} }{\mathrm{32}}+\frac{\mathrm{7}{R}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$

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

Commented by ajfour last updated on 31/Dec/23

I get (r,R)≡ (0.2282 , 0.44295) (0.4435 , 0.3893) (0.0546 , 0.4864)

$${I}\:{get}\:\:\left({r},{R}\right)\equiv \\ $$$$\left(\mathrm{0}.\mathrm{2282}\:,\:\mathrm{0}.\mathrm{44295}\right) \\ $$$$\left(\mathrm{0}.\mathrm{4435}\:,\:\mathrm{0}.\mathrm{3893}\right) \\ $$$$\left(\mathrm{0}.\mathrm{0546}\:,\:\mathrm{0}.\mathrm{4864}\right) \\ $$

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

(R+r)^2 −(R−r)^2 +(1−r)^2 =1^2 ⇒r=2−4R sin θ=((R−r)/(R+r))=((5R−2)/(2−3R)) → 0.4<R<0.5 (R/(1−R))=cos 2θ=1−2(((5R−2)/(2−3R)))^2 32R^3 −57R^2 +28R−4=0 ⇒R≈0.44295433 ⇒r≈0.22818264

$$\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} +\left(\mathrm{1}−{r}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\mathrm{2}−\mathrm{4}{R} \\ $$$$\mathrm{sin}\:\theta=\frac{{R}−{r}}{{R}+{r}}=\frac{\mathrm{5}{R}−\mathrm{2}}{\mathrm{2}−\mathrm{3}{R}}\:\:\rightarrow\:\mathrm{0}.\mathrm{4}<{R}<\mathrm{0}.\mathrm{5} \\ $$$$\frac{{R}}{\mathrm{1}−{R}}=\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{5}{R}−\mathrm{2}}{\mathrm{2}−\mathrm{3}{R}}\right)^{\mathrm{2}} \\ $$$$\mathrm{32}{R}^{\mathrm{3}} −\mathrm{57}{R}^{\mathrm{2}} +\mathrm{28}{R}−\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{R}\approx\mathrm{0}.\mathrm{44295433} \\ $$$$\Rightarrow{r}\approx\mathrm{0}.\mathrm{22818264} \\ $$

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

$${but}\:{only}\:{the}\:{first}\:{one}\:{is}\:{valid}. \\ $$

Commented by ajfour last updated on 31/Dec/23

$${Thank}\:{you}\:{sir},\:{your}\:{way}\:{is}\:{very} \\ $$$${charming}.\:{I}\:{havnt}\:{checked}\:{to}\:{reject}. \\ $$

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