Question Number 54160 by ajfour last updated on 30/Jan/19
Commented by ajfour last updated on 30/Jan/19
$${Find}\:{maximum}\:{area}\:{of}\:\bigtriangleup{ABC}\:{in} \\ $$$${terms}\:{of}\:{R}\:{and}\:{r}. \\ $$
Answered by mr W last updated on 30/Jan/19
Commented by mr W last updated on 30/Jan/19
$$\alpha=\frac{\pi}{\mathrm{2}}−\frac{\beta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\alpha=\pi−\beta\:\:\:..\left({i}\right) \\ $$$$\beta=\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\beta=\pi−\alpha\:\:..\left({ii}\right) \\ $$$$\Rightarrow\alpha=\beta=\frac{\pi}{\mathrm{3}}=\mathrm{60}° \\ $$$${AC}=\mathrm{2}{R}\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}=\sqrt{\mathrm{3}}{R} \\ $$$${AB}=\mathrm{2}{r}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}}=\sqrt{\mathrm{3}}{r} \\ $$$$\angle{CAB}=\mathrm{180}°−\frac{\alpha}{\mathrm{2}}−\frac{\beta}{\mathrm{2}}=\mathrm{120}° \\ $$$$\Delta_{{max}} =\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{3}}{R}×\sqrt{\mathrm{3}}{r}×\mathrm{sin}\:\mathrm{120}° \\ $$$$\Rightarrow\Delta_{{max}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}{Rr}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 30/Jan/19
$${yes}\:{Sir},\:{straight}\:{and}\:{clear},\:{thanks} \\ $$$${again}\left({i}\:{am}\:{quite}\:{dumb}\:{at}\:{this}\right). \\ $$
Commented by mr W last updated on 30/Jan/19
$${i}\:{tried}\:{above}\:{to}\:{find}\:{the}\:{maximum}\:{triangle} \\ $$$${only}\:{using}\:{logic}.\:{but}\:{we}\:{can}\:{get}\:{the} \\ $$$${same}\:{result}\:{using}\:{calculus}: \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{R}\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}×\mathrm{2}{r}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}}×\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}{Rr}\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}\right) \\ $$$$\frac{\partial\left(\Delta\right)}{\partial\alpha}=\mathrm{2}{Rr}\mathrm{cos}\:\frac{\beta}{\mathrm{2}}\left\{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}\right)\right\}=\mathrm{0} \\ $$$$\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}\right)=\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}\right)=\mathrm{1}\:\:\:…\left({i}\right) \\ $$$$\frac{\partial\left(\Delta\right)}{\partial\beta}=\mathrm{2}{Rr}\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\left\{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\frac{\beta}{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\frac{\beta}{\mathrm{2}}\:\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}\right)\right\}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\beta}{\mathrm{2}}\:\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}\right)=\mathrm{1}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\mathrm{tan}\:\frac{\beta}{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\beta \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\alpha=\mathrm{1} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\frac{\alpha}{\mathrm{2}}=\mathrm{30}° \\ $$$$\Rightarrow\alpha=\beta=\mathrm{60}° \\ $$
Commented by ajfour last updated on 30/Jan/19
$${using}\:{calculus}\:{i}'{d}\:{got}\:{the}\:{same}, \\ $$$${but}\:{would}\:{you}\:{please}\:{explain}\:{your} \\ $$$${logic}\:{to}\:{me}\:{in}\:{some}\:{more}\:{detail},\:{Sir}? \\ $$
Commented by mr W last updated on 30/Jan/19
$${the}\:{logic}\:{is}: \\ $$$${with}\:{maximum}\:{triangle} \\ $$$${tangent}\:{at}\:{C}\:{is}\:{parallel}\:{to}\:{AB},\:{i}.{e}. \\ $$$${CP}\bot{AB}. \\ $$$${tangent}\:{at}\:{B}\:{is}\:{parallel}\:{to}\:{AC},\:{i}.{e}. \\ $$$${BQ}\bot{AC}. \\ $$
Commented by mr W last updated on 30/Jan/19
$${the}\:{logic}\:{is}\:{the}\:{same}\:{as}\:{in}\:{Q}\mathrm{53910}. \\ $$