Question Number 211058 by mr W last updated on 27/Aug/24
Commented by mr W last updated on 27/Aug/24
$${find}\:{the}\:{radius}\:{R}\:{of}\:{the}\:{largest}\:{sphere} \\ $$$${which}\:{can}\:{be}\:{placed}\:{between}\:{the}\:{red} \\ $$$${and}\:{blue}\:{planes}\:{in}\:{the}\:{first}\:{octant}. \\ $$$${assume}\:\mathrm{0}<{p}\leqslant{a},\:\mathrm{0}<{q}\leqslant{b},\:\mathrm{0}<{r}\leqslant{c}. \\ $$
Answered by mr W last updated on 29/Aug/24
Commented by mr W last updated on 28/Aug/24
$${the}\:{largest}\:{sphere}\:{above}\:{and}\:{touching} \\ $$$${the}\:{lower}\:{plane}\:\frac{\boldsymbol{{x}}}{\boldsymbol{{p}}}+\frac{\boldsymbol{{y}}}{\boldsymbol{{q}}}+\frac{\boldsymbol{{z}}}{\boldsymbol{{r}}}=\mathrm{1}: \\ $$$${say}\:{its}\:{radius}\:{is}\:{R}_{{o}} ,\:{then}\:{its}\:{center} \\ $$$${is}\:{at}\:\left({R}_{{o}} ,{R}_{{o}} ,{R}_{{o}} \right). \\ $$$${we}\:{have} \\ $$$$\frac{\frac{{R}_{{o}} }{{p}}+\frac{{R}_{{o}} }{{q}}+\frac{{R}_{{o}} }{{r}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}}={R}_{{o}} \\ $$$$\Rightarrow{R}_{{o}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}} \\ $$
Commented by mr W last updated on 29/Aug/24
Commented by mr W last updated on 28/Aug/24
$${the}\:{largest}\:{sphere}\:{under}\:{the} \\ $$$${upper}\:{plane}\:\frac{\boldsymbol{{x}}}{\boldsymbol{{a}}}+\frac{\boldsymbol{{y}}}{\boldsymbol{{b}}}+\frac{\boldsymbol{{z}}}{\boldsymbol{{c}}}=\mathrm{1}: \\ $$$${say}\:{its}\:{radius}\:{is}\:{R}_{{i}} ,\:{then}\:{its}\:{center} \\ $$$${is}\:{at}\:\left({R}_{{i}} ,{R}_{{i}} ,{R}_{{i}} \right). \\ $$$${we}\:{have} \\ $$$$\frac{\frac{{R}_{{i}} }{{a}}+\frac{{R}_{{i}} }{{b}}+\frac{{R}_{{i}} }{{c}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}=−{R}_{{i}} \\ $$$$\Rightarrow{R}_{{i}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$
Commented by mr W last updated on 28/Aug/24
Commented by mr W last updated on 29/Aug/24
$${let}'{s}\:{ignore}\:{the}\:{lower}\:{plane}\:{at}\:{first}. \\ $$$${as}\:{shown}\:{previously},\:{the}\:{largest}\: \\ $$$${sphere}\:{under}\:{the}\:{upper}\:{plane}\:{has}\: \\ $$$${the}\:{radius}\:{R}_{{i}} \:{and}\:{and}\:{its}\:{center}\:{is}\: \\ $$$${at}\:\left({R}_{{i}} ,{R}_{{i}} ,{R}_{{i}} \right). \\ $$$${R}_{{i}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$$${if}\:{the}\:{distance}\:{from}\:\left({R}_{{i}} ,{R}_{{i}} ,{R}_{{i}} \right)\:{to} \\ $$$${the}\:{lower}\:{plane}\:{is}\:{equal}\:{to}\:{or}\:{larger} \\ $$$${than}\:{R}_{{i}} ,\:{i}.{e}.\:{if} \\ $$$$\frac{\frac{{R}_{{i}} }{{p}}+\frac{{R}_{{i}} }{{q}}+\frac{{R}_{{i}} }{{r}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}}\geqslant{R}_{{i}} ,\:{i}.{e}.\:{if} \\ $$$$\frac{\mathrm{1}}{\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}}\leqslant{R}_{{i}} \\ $$$${i}.{e}.\:{if}\:{R}_{{o}} \leqslant{R}_{{i}} , \\ $$$${that}\:{means}\:{if}\: \\ $$$$\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }} \\ $$$$\geqslant\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$${then}\:{the}\:{lower}\:{plane}\:{is}\:{beneath} \\ $$$${the}\:{sphere}\:{with}\:{radius}\:{R}_{{i}} \:{and}\:{this}\: \\ $$$${sphere}\:{is}\:{also}\:{the}\:{largest}\:{sphere} \\ $$$${between}\:{both}\:{planes}.\:{that}\:{means} \\ $$$${in}\:{this}\:{case} \\ $$$${R}={R}_{{i}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$$$ \\ $$$${if}\:{R}_{{o}} >{R}_{{i}} ,\:{i}.{e}.\:{if} \\ $$$$\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }} \\ $$$$<\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$${then}\:{the}\:{largest}\:{sphere}\:{between} \\ $$$${both}\:{planes}\:{is}\:{the}\:{maximum}\:{sphere} \\ $$$${from}\:{following}\:{three}\:{cases}: \\ $$$$ \\ $$$${case}\:\mathrm{1}:\: \\ $$$${center}\:{at}\:\left({u},\:{R}_{\mathrm{1}} ,{R}_{\mathrm{1}} \right),\:{radius}\:{R}_{\mathrm{1}} \\ $$$$\frac{\frac{{u}}{{a}}+\frac{{R}_{\mathrm{1}} }{{b}}+\frac{{R}_{\mathrm{1}} }{{c}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}=−{R}_{\mathrm{1}} \\ $$$$\Rightarrow{u}={a}−{aR}_{\mathrm{1}} \left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{u}={a}−{aR}_{\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}\right)+{R}_{\mathrm{1}} \\ $$$$\Rightarrow{u}={a}−\frac{{aR}_{\mathrm{1}} }{{R}_{{i}} }+{R}_{\mathrm{1}} \\ $$$$\frac{\frac{{u}}{{p}}+\frac{{R}_{\mathrm{1}} }{{q}}+\frac{{R}_{\mathrm{1}} }{{r}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}}={R}_{\mathrm{1}} \\ $$$$\Rightarrow{u}={p}−{pR}_{\mathrm{1}} \left(\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{u}={p}−{pR}_{\mathrm{1}} \left(\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}\right)+{R}_{\mathrm{1}} \\ $$$$\Rightarrow{u}={p}−\frac{{pR}_{\mathrm{1}} }{{R}_{{o}} }+{R}_{\mathrm{1}} \\ $$$${p}−\frac{{pR}_{\mathrm{1}} }{{R}_{{o}} }+{R}_{\mathrm{1}} ={a}−\frac{{aR}_{\mathrm{1}} }{{R}_{{i}} }+{R}_{\mathrm{1}} \\ $$$$\frac{{R}_{\mathrm{1}} }{{R}_{{i}} }\left(\mathrm{1}−\frac{{p}}{{a}}×\frac{{R}_{{i}} }{{R}_{{o}} }\right)=\mathrm{1}−\frac{{p}}{{a}} \\ $$$$\Rightarrow{R}_{\mathrm{1}} =\frac{\mathrm{1}−\frac{{p}}{{a}}}{\mathrm{1}−\frac{{p}}{{a}}×\frac{{R}_{{i}} }{{R}_{{o}} }}×{R}_{{i}} \\ $$$${u}={a}\left(\mathrm{1}−\frac{{R}_{\mathrm{1}} }{{R}_{{i}} }\right)+{R}_{\mathrm{1}} \\ $$$$\Rightarrow{u}={a}\left(\mathrm{1}−\frac{\mathrm{1}−\frac{{p}}{{a}}}{\mathrm{1}−\frac{{p}}{{a}}×\frac{{R}_{{i}} }{{R}_{{o}} }}\right)+{R}_{\mathrm{1}} \:>\:{R}_{\mathrm{1}} \\ $$$$ \\ $$$${similarly} \\ $$$${case}\:\mathrm{2}:\: \\ $$$${center}\:{at}\:\left({R}_{\mathrm{2}} ,{v},{R}_{\mathrm{2}} \right),\:{radius}\:{R}_{\mathrm{2}} \\ $$$$\Rightarrow{R}_{\mathrm{2}} =\frac{\mathrm{1}−\frac{{q}}{{b}}}{\mathrm{1}−\frac{{q}}{{b}}×\frac{{R}_{{i}} }{{R}_{{o}} }}×{R}_{{i}} \\ $$$$\Rightarrow{v}={b}\left(\mathrm{1}−\frac{\mathrm{1}−\frac{{q}}{{b}}}{\mathrm{1}−\frac{{q}}{{b}}×\frac{{R}_{{i}} }{{R}_{{o}} }}\right)+{R}_{\mathrm{2}} \:>\:{R}_{\mathrm{2}} \\ $$$$ \\ $$$${case}\:\mathrm{3}:\: \\ $$$${center}\:{at}\:\left({R}_{\mathrm{3}} ,{R}_{\mathrm{3}} ,{w}\right),\:{radius}\:{R}_{\mathrm{3}} \\ $$$$\Rightarrow{R}_{\mathrm{3}} =\frac{\mathrm{1}−\frac{{r}}{{c}}}{\mathrm{1}−\frac{{r}}{{c}}×\frac{{R}_{{i}} }{{R}_{{o}} }}×{R}_{{i}} \\ $$$$\Rightarrow{w}={c}\left(\mathrm{1}−\frac{\mathrm{1}−\frac{{r}}{{c}}}{\mathrm{1}−\frac{{r}}{{c}}×\frac{{R}_{{i}} }{{R}_{{o}} }}\right)+{R}_{\mathrm{3}} \:>\:{R}_{\mathrm{3}} \\ $$$$ \\ $$$${since}\:\frac{{R}_{{i}} }{{R}_{{o}} }<\mathrm{1}\:{and}\:\mathrm{0}<\frac{{p}}{{a}},\:\frac{{q}}{{b}},\:\frac{{r}}{{c}}\leqslant\mathrm{1} \\ $$$${and}\:{f}\left({x}\right)=\frac{\mathrm{1}−{x}}{\mathrm{1}−{kx}}\:{is}\:{a}\:{strctly}\: \\ $$$${decreasing}\:{function}\:{for}\:\mathrm{0}<{k}<\mathrm{1}\:{and} \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{1},\:{we}\:{get} \\ $$$${R}={max}\left({R}_{\mathrm{1}} ,{R}_{\mathrm{2}} ,{R}_{\mathrm{3}} \right) \\ $$$$\:\:\:=\frac{\mathrm{1}−{min}\left(\frac{{p}}{{a}},\frac{{q}}{{b}},\frac{{r}}{{c}}\right)}{\mathrm{1}−{min}\left(\frac{{p}}{{a}},\frac{{q}}{{b}},\frac{{r}}{{c}}\right)×\frac{{R}_{{i}} }{{R}_{{o}} }}×{R}_{{i}} \\ $$
Commented by mr W last updated on 27/Aug/24
$${summary}: \\ $$$${with}\: \\ $$$${R}_{{i}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$$${R}_{{o}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}} \\ $$$${if}\:{R}_{{o}} \leqslant{R}_{{i}} ,\:{then} \\ $$$${R}={R}_{{i}} \\ $$$${if}\:{R}_{{o}} >{R}_{{i}} ,\:{then} \\ $$$${R}=\frac{\mathrm{1}−{min}\left(\frac{{p}}{{a}},\frac{{q}}{{b}},\frac{{r}}{{c}}\right)}{\mathrm{1}−{min}\left(\frac{{p}}{{a}},\frac{{q}}{{b}},\frac{{r}}{{c}}\right)\centerdot\frac{{R}_{{i}} }{{R}_{{o}} }}\centerdot{R}_{{i}} \\ $$
Commented by mr W last updated on 27/Aug/24
$${formula}\:{used}: \\ $$$${distance}\:{from}\:{point}\:{to}\:{plane}\:{in}\:{space} \\ $$
Commented by mr W last updated on 27/Aug/24
Commented by ajfour last updated on 27/Aug/24
$${immense}\:!\:{i}\:{will}\:{look}\:{carefully}.\:{get} \\ $$$${the}\:{idea}\:{very}\:{well}.\:{great}\:{thing}\:{for} \\ $$$${question}\:{treasure}. \\ $$
Commented by mr W last updated on 27/Aug/24
$${thanks}\:{for}\:{checking}\:{sir}! \\ $$
Commented by mr W last updated on 27/Aug/24
$${example}\:\mathrm{1}: \\ $$$${upper}\:{plane}\:\frac{{x}}{\mathrm{6}}+\frac{{y}}{\mathrm{9}}+\frac{{z}}{\mathrm{10}}=\mathrm{1} \\ $$$${lower}\:{plane}\:\frac{{x}}{\mathrm{5}}+\frac{{y}}{\mathrm{3}}+\frac{{z}}{\mathrm{4}}=\mathrm{1} \\ $$$${a}=\mathrm{6},\:{b}=\mathrm{9},\:{c}=\mathrm{10} \\ $$$${p}=\mathrm{5},\:{q}=\mathrm{3},\:{r}=\mathrm{4} \\ $$$$\frac{{p}}{{a}}=\frac{\mathrm{5}}{\mathrm{6}},\:\frac{{q}}{{b}}=\frac{\mathrm{3}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{3}},\:\frac{{r}}{{c}}=\frac{\mathrm{4}}{\mathrm{10}}=\frac{\mathrm{2}}{\mathrm{5}}\: \\ $$$$\Rightarrow{min}=\frac{{q}}{{b}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${R}_{{i}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{10}}+\sqrt{\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{2}} }}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{102}−\mathrm{3}\sqrt{\mathrm{406}}}{\mathrm{25}}\approx\mathrm{1}.\mathrm{662} \\ $$$${R}_{{o}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\sqrt{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{47}+\sqrt{\mathrm{769}}}{\mathrm{24}}\approx\mathrm{3}.\mathrm{114} \\ $$$${R}_{{o}} >{R}_{{i}} \: \\ $$$${R}=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{24}×\left(\mathrm{102}−\mathrm{3}\sqrt{\mathrm{406}}\right)}{\mathrm{25}×\left(\mathrm{47}+\sqrt{\mathrm{769}}\right)}}×\frac{\mathrm{102}−\mathrm{3}\sqrt{\mathrm{406}}}{\mathrm{25}} \\ $$$$\:\:\:\:\approx\mathrm{1}.\mathrm{348} \\ $$
Commented by behi834171 last updated on 27/Aug/24
$${mathmatic}\:{is}\:{very}\:{beautiful}. \\ $$$${your}\:{Q}'{s}\:\&\:{A}'{s}\:,{are}\:{more}\:{than}\: \\ $$$${beautiful}! \\ $$$${They}\:{are}\:{amazing}\:{sir}\:{mrW}. \\ $$$${Thanks}\:{god}\:{for}\:{being}\:{of}\:{you}\:{in}\:{this}\:{forum}. \\ $$
Commented by mr W last updated on 27/Aug/24
$${thanks}\:{back}\:{to}\:{you},\:{father}\:{from}\:{Behi},\: \\ $$$${for}\:{your}\:{appreciation}\:{and}\:{support}! \\ $$
Commented by necx122 last updated on 27/Aug/24
$${Mr}.\:{Behi}\:{its}\:{been}\:{a}\:{really}\:{long}\:{time}.\:{Feels} \\ $$$${good}\:{to}\:{see}\:{your}\:{post}\:{here}\:{again}. \\ $$
Commented by ajfour last updated on 27/Aug/24
$${yes}\:{so}\:{glad}\:{for}\:{your}\:{remark}\:{bsck}\:{on}\:{this}\:{forum}. \\ $$
Commented by behi834171 last updated on 28/Aug/24
$${Red}\:{heart}\:{to}\:{you}\:{dear}\:{mr}.{W} \\ $$
Commented by behi834171 last updated on 28/Aug/24
$${Thank}\:{you}\:{so}\:{much}\:{sir}\:{with}\:{red}\:{herat}! \\ $$
Commented by behi834171 last updated on 28/Aug/24
$${so}\:{happy}\:{for}\:{see}\:{you}\:\:{again}\:{sir}\:{ajfour} \\ $$$$\:{and}\:{red}\:{heart}\:{to}\:{you}! \\ $$
Commented by BHOOPENDRA last updated on 01/Sep/24
$${great}\:{Mr}.{W} \\ $$