Question Number 214297 by ajfour last updated on 04/Dec/24
Commented by ajfour last updated on 04/Dec/24
$${the}\:{tangents}\:{are}\:{at}\:{right}\:{angles}\:{to} \\ $$$${one}\:{another}\:{and}\:{corner}\:{maynot}\:{be} \\ $$$${at}\:{semicircle}\:{centre}. \\ $$
Answered by mr W last updated on 04/Dec/24
Commented by mr W last updated on 04/Dec/24
$${a}=\mathrm{1} \\ $$$${R}=\mathrm{3} \\ $$$${b}={r} \\ $$$$\frac{{b}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}={x}+\sqrt{\left({R}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }={x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\beta}{\mathrm{2}}=\frac{{b}}{{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}} \\ $$$${similarly} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}} \\ $$$$\frac{\beta}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\frac{\beta}{\mathrm{2}}=\frac{\mathrm{1}−\mathrm{tan}\:\frac{{a}}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:\frac{{a}}{\mathrm{2}}} \\ $$$$\frac{{b}}{{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}}=\frac{\mathrm{1}−\frac{{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}}}{\mathrm{1}+\frac{{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}}} \\ $$$$\frac{{b}}{{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}}=\frac{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}−{a}}{−{x}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}+{a}} \\ $$$$\frac{{b}}{{x}+\sqrt{\mathrm{3}\left(\mathrm{3}−\mathrm{2}{b}\right)}}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}−{x}}{\:\sqrt{\mathrm{3}}+\mathrm{1}−{x}} \\ $$$$\Rightarrow{b}_{{max}} \approx\mathrm{0}.\mathrm{66472} \\ $$
Commented by mr W last updated on 04/Dec/24
Commented by ajfour last updated on 04/Dec/24
$$\frac{\alpha}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}+\theta \\ $$$$\frac{\beta}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}−\theta \\ $$$$\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)=\frac{{a}}{\:{p}+\sqrt{\left({R}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)=\frac{{b}}{\:−{p}+\sqrt{\left({R}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$${As}\:\frac{{a}}{\:{p}+\sqrt{\left({R}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }}×\frac{{b}}{\:−{p}+\sqrt{\left({R}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }}=\mathrm{1} \\ $$$${p}^{\mathrm{2}} +{p}\left\{\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}−\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}\right\} \\ $$$$+{R}\sqrt{{R}−\mathrm{2}{a}}\sqrt{{R}−\mathrm{2}{b}}={ab} \\ $$$$… \\ $$