Question Number 81972 by ajfour last updated on 17/Feb/20
Commented by ajfour last updated on 17/Feb/20
$$\bigtriangleup{PQR}\:{is}\:{equilateral},\:{find} \\ $$$${s}_{{min},} \:{s}_{{max}} . \\ $$
Answered by mr W last updated on 17/Feb/20
Commented by mr W last updated on 17/Feb/20
$$\Delta_{{OAB}} =\sqrt{\frac{\mathrm{11}}{\mathrm{2}}\left(\frac{\mathrm{11}}{\mathrm{2}}−\mathrm{2}\right)\left(\frac{\mathrm{11}}{\mathrm{2}}−\mathrm{4}\right)\left(\frac{\mathrm{11}}{\mathrm{2}}−\mathrm{5}\right)}=\frac{\sqrt{\mathrm{231}}}{\mathrm{4}} \\ $$$$\Delta_{{OAB}} =\frac{\mathrm{5}\delta}{\mathrm{2}}=\frac{\sqrt{\mathrm{231}}}{\mathrm{4}} \\ $$$$\Rightarrow\delta=\frac{\sqrt{\mathrm{231}}}{\mathrm{10}} \\ $$$$\sigma=\sqrt{\mathrm{2}^{\mathrm{2}} −\delta^{\mathrm{2}} }=\sqrt{\mathrm{4}−\frac{\mathrm{231}}{\mathrm{100}}}=\frac{\mathrm{13}}{\mathrm{10}} \\ $$$${O}\left(\frac{\mathrm{13}}{\mathrm{10}},−\frac{\sqrt{\mathrm{231}}}{\mathrm{10}}\right)=\left(\sigma,−\delta\right) \\ $$$${eqn}.\:{of}\:{circle}: \\ $$$$\left({x}−\sigma\right)^{\mathrm{2}} +\left({y}+\delta\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$$ \\ $$$${say}\:{P}\left({p},\mathrm{0}\right),\:{Q}\left(\mathrm{0},{q}\right) \\ $$$${q}=\sqrt{{s}^{\mathrm{2}} −{p}^{\mathrm{2}} } \\ $$$$ \\ $$$${x}^{\mathrm{2}} +\left({y}−{q}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={s}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({x}−\sigma\right)^{\mathrm{2}} +\left({y}+\delta\right)^{\mathrm{2}} =\mathrm{16}\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{2}{px}−{p}^{\mathrm{2}} −\mathrm{2}{qy}+{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{{q}}{{p}}{y}+\frac{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{2}{p}} \\ $$$$ \\ $$$$\left(\frac{{q}}{{p}}{y}+\frac{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{2}{p}}−{p}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{qy}−{p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{p}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{4}{p}^{\mathrm{2}} {s}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −{qy}+\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{\mathrm{4}}−\frac{{p}^{\mathrm{2}} {s}^{\mathrm{2}} }{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }=\mathrm{0} \\ $$$${y}^{\mathrm{2}} −{qy}−\frac{\mathrm{3}{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{\sqrt{\mathrm{3}}{p}+{q}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{p}+\sqrt{{s}^{\mathrm{2}} −{p}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{{p}+\sqrt{\mathrm{3}}{q}}{\mathrm{2}}=\frac{{p}+\sqrt{\mathrm{3}\left({s}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$$\left(\frac{{p}+\sqrt{\mathrm{3}\left({s}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)}}{\mathrm{2}}−\sigma\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}{p}+\sqrt{{s}^{\mathrm{2}} −{p}^{\mathrm{2}} }}{\mathrm{2}}+\delta\right)^{\mathrm{2}} =\mathrm{16} \\ $$$$\Rightarrow\left({p}+\sqrt{\mathrm{3}\left({s}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)}−\mathrm{2}\sigma\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}{p}+\sqrt{{s}^{\mathrm{2}} −{p}^{\mathrm{2}} }+\mathrm{2}\delta\right)^{\mathrm{2}} =\mathrm{64} \\ $$$${for}\:{minimum}\:{and}\:{maximum}\:{s}:\:\frac{{ds}}{{dp}}=\mathrm{0} \\ $$$${s}_{{min}} \approx\mathrm{2}.\mathrm{3935}\:{at}\:{p}\approx\mathrm{1}.\mathrm{9567} \\ $$
Commented by mr W last updated on 17/Feb/20
Commented by ajfour last updated on 17/Feb/20
$${superb}!\:{Sir};\:{requires}\:{enough} \\ $$$${determination}.. \\ $$
Commented by mr W last updated on 17/Feb/20
Commented by mr W last updated on 17/Feb/20
$${s}_{{max}} \:{is}\:{possible}\:{when} \\ $$$$\left({q}+\frac{\sqrt{\mathrm{231}}}{\mathrm{10}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{13}}{\mathrm{10}}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$$\Rightarrow{q}=\frac{\mathrm{3}\sqrt{\mathrm{159}}−\sqrt{\mathrm{231}}}{\mathrm{10}}\approx\mathrm{2}.\mathrm{2630} \\ $$$$\Rightarrow{p}\approx\mathrm{1}.\mathrm{3022} \\ $$$$\Rightarrow{s}_{{max}} \approx\mathrm{2}.\mathrm{6109} \\ $$$${or} \\ $$$${q}=\mathrm{0} \\ $$$$\Rightarrow{p}\approx\mathrm{2}.\mathrm{8613} \\ $$$$\Rightarrow{s}_{{max}} \approx\mathrm{2}.\mathrm{8613} \\ $$
Commented by mr W last updated on 17/Feb/20
Commented by mr W last updated on 17/Feb/20
Commented by ajfour last updated on 17/Feb/20
$${Too}\:{Good}\:{Sir},\:{thanks}\:{plentiful}. \\ $$$${you}\:{have}\:{dealt}\:{all}\:{cases}. \\ $$