Question Number 213323 by Spillover last updated on 02/Nov/24
Commented by Frix last updated on 03/Nov/24
$$\mathrm{No}\:\mathrm{nice}\:\mathrm{numbers} \\ $$$$\mathrm{Let}\:{r}=\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\rho_{{q}.{c}.} \approx\mathrm{4}.\mathrm{87164313433}\:\left[\:_{\mathrm{polynome}\:\mathrm{of}\:\mathrm{8}^{\mathrm{th}} \:\mathrm{degree}\:} ^{\mathrm{Solution}\:\mathrm{of}\:\mathrm{a}} \right] \\ $$$$\rho_{{s}.{c}.} \approx\mathrm{3}.\mathrm{49740513991} \\ $$$${R}\approx\mathrm{1}.\mathrm{66629777247} \\ $$
Answered by mr W last updated on 03/Nov/24
Commented by mr W last updated on 03/Nov/24
$${a}={radius}\:{of}\:{quarter}\:{circle} \\ $$$${b}={radius}\:{of}\:{semicircle} \\ $$$$\left({a}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\left({b}+{r}\right)^{\mathrm{2}} −\left({b}−{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} −\mathrm{2}{ar}=\mathrm{4}{br}\:\:\:\:…\left({i}\right) \\ $$$$\sqrt{\left({a}−{R}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} }+\sqrt{\left({b}−{R}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} }={b} \\ $$$$\Rightarrow\:\sqrt{{a}^{\mathrm{2}} −\mathrm{2}{aR}}+\sqrt{{b}^{\mathrm{2}} −\mathrm{2}{bR}}={b}\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\sqrt{\left({a}+{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }={b}+\sqrt{\left({b}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{ar}}={b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{2}{br}}\:\:\:\:\:…\left({iii}\right) \\ $$$${let}\:\alpha=\frac{{a}}{{r}},\:\beta=\frac{{b}}{{r}},\:\lambda=\frac{{R}}{{r}} \\ $$$$\alpha^{\mathrm{2}} −\mathrm{2}\alpha=\mathrm{4}\beta\:\:\:\:…\left({i}\right) \\ $$$$\:\sqrt{\alpha^{\mathrm{2}} −\mathrm{2}\alpha\lambda}+\sqrt{\beta^{\mathrm{2}} −\mathrm{2}\beta\lambda}=\beta\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\sqrt{\alpha^{\mathrm{2}} +\mathrm{2}\alpha}=\beta+\sqrt{\beta^{\mathrm{2}} −\mathrm{2}\beta}\:\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow\lambda\approx\mathrm{1}.\mathrm{666297772} \\ $$
Commented by Spillover last updated on 03/Nov/24
$${perfect} \\ $$