Question Number 214297 by ajfour last updated on 04/Dec/24
Commented by ajfour last updated on 04/Dec/24
$${the}\:{tangents}\:{are}\:{at}\:{right}\:{angles}\:{to} \\ $$$${one}\:{another}\:{and}\:{corner}\:{maynot}\:{be} \\ $$$${at}\:{semicircle}\:{centre}. \\ $$
Commented by mr W last updated on 05/Dec/24
Commented by ajfour last updated on 05/Dec/24
https://youtu.be/ys6xN29xNUM?si=9rk6nXD0GcmHTp9Z
Answered by mr W last updated on 04/Dec/24
Commented by mr W last updated on 05/Dec/24
$${a}=\mathrm{1} \\ $$$${R}=\mathrm{3} \\ $$$${b}={r} \\ $$$$\frac{{b}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}={x}+\sqrt{\left({R}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }={x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\beta}{\mathrm{2}}=\frac{{b}}{{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}} \\ $$$${similarly} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}} \\ $$$$\frac{\beta}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\frac{\beta}{\mathrm{2}}=\frac{\mathrm{1}−\mathrm{tan}\:\frac{{a}}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:\frac{{a}}{\mathrm{2}}} \\ $$$$\frac{{b}}{{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}}=\frac{\mathrm{1}−\frac{{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}}}{\mathrm{1}+\frac{{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}}} \\ $$$$\frac{{b}}{{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}}=\frac{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}−{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}+{a}} \\ $$$${let}\:\alpha=\frac{{a}}{{R}},\:\beta=\frac{{b}}{{R}},\:\xi=\frac{{x}}{{R}} \\ $$$$\frac{\beta}{\xi+\sqrt{\mathrm{1}−\mathrm{2}\beta}}=\frac{−\xi+\sqrt{\mathrm{1}−\mathrm{2}\alpha}−\alpha}{−\xi+\sqrt{\mathrm{1}−\mathrm{2}\alpha}+\alpha} \\ $$$${let}\:\lambda=\sqrt{\mathrm{1}−\mathrm{2}\beta},\:{p}=\sqrt{\mathrm{1}−\mathrm{2}\alpha}+\alpha,\:{q}=\sqrt{\mathrm{1}−\mathrm{2}\alpha}−\alpha \\ $$$$\frac{\mathrm{1}−\lambda^{\mathrm{2}} }{\mathrm{2}\left(\xi+\lambda\right)}=\frac{{q}−\xi}{{p}−\xi} \\ $$$$\left({p}−\xi\right)\lambda^{\mathrm{2}} +\mathrm{2}\left({q}−\xi\right)\lambda+\left(\mathrm{2}{q}+\mathrm{1}\right)\xi−\left[\mathrm{2}\xi^{\mathrm{2}} −\left(\mathrm{2}{q}+\mathrm{1}\right)\xi+{p}\right]=\mathrm{0} \\ $$$$\lambda=\frac{\xi−{q}+\sqrt{\left(\xi−{q}\right)^{\mathrm{2}} +\left({p}−\xi\right)\left[\mathrm{2}\xi^{\mathrm{2}} −\left(\mathrm{2}{q}+\mathrm{1}\right)\xi+{p}\right]}}{{p}−\xi} \\ $$$${b}=\left\{\mathrm{1}−\left\{\frac{\xi−{q}+\sqrt{\left(\xi−{q}\right)^{\mathrm{2}} +\left({p}−\xi\right)\left[\mathrm{2}\xi^{\mathrm{2}} −\left(\mathrm{2}{q}+\mathrm{1}\right)\xi+{p}\right]}}{{p}−\xi}\right\}^{\mathrm{2}} \right\}\frac{{R}}{\mathrm{2}} \\ $$$${b}_{{max}} \approx\mathrm{0}.\mathrm{6659}\:{at}\:\xi=\frac{{x}}{{R}}\approx−\mathrm{0}.\mathrm{1399} \\ $$$$================= \\ $$
Commented by mr W last updated on 05/Dec/24
Commented by mr W last updated on 05/Dec/24