Question Number 46073 by ajfour last updated on 20/Oct/18
Commented by ajfour last updated on 20/Oct/18
$${Determine}\:{coordinates}\:{of}\:{A}; \\ $$$$\:\:\:{b}\:>\:{c}\:.\:{P}\:{is}\:{contact}\:{point}\:{of}\:{side} \\ $$$${AB}\:{of}\:\bigtriangleup{ABC}\:{with}\:{upper}\:{rim}\:{of} \\ $$$${cylinder}\:\left({height}\:{H},\:{radius}\:{R}\right). \\ $$
Answered by MrW3 last updated on 23/Oct/18
$${let}\:{d}=\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${x}_{{B}} =\frac{{a}}{\mathrm{2}} \\ $$$${y}_{{B}} =−{d} \\ $$$${x}_{{C}} =−\frac{{a}}{\mathrm{2}} \\ $$$${y}_{{C}} =−{d} \\ $$$${z}_{{B}} ={z}_{{C}} =\mathrm{0} \\ $$$${x}_{{P}} ={R}\:\mathrm{cos}\:\phi \\ $$$${y}_{{P}} ={R}\:\mathrm{sin}\:\phi \\ $$$${z}_{{P}} ={H} \\ $$$${x}_{{A}} =\frac{{a}}{\mathrm{2}}+\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)\lambda \\ $$$$\Rightarrow{x}_{{A}} −{x}_{{B}} =\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)\lambda \\ $$$$\Rightarrow{x}_{{A}} −{x}_{{C}} ={a}+\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)\lambda \\ $$$${y}_{{A}} =−{d}+\left({R}\:\mathrm{sin}\:\phi+{d}\right)\lambda \\ $$$$\Rightarrow{y}_{{A}} −{y}_{{B}} =\left({R}\:\mathrm{sin}\:\phi+{d}\right)\lambda \\ $$$$\Rightarrow{y}_{{A}} −{y}_{{C}} =\left({R}\:\mathrm{sin}\:\phi+{d}\right)\lambda \\ $$$${z}_{{A}} ={H}\lambda \\ $$$$\Rightarrow{z}_{{A}} −{z}_{{B}} ={H}\lambda \\ $$$$\Rightarrow{z}_{{A}} −{z}_{{C}} ={H}\lambda \\ $$$${AB}=\sqrt{\left({x}_{{A}} −{x}_{{B}} \right)^{\mathrm{2}} +\left({y}_{{A}} −{y}_{{B}} \right)^{\mathrm{2}} +\left({z}_{{A}} −{z}_{{B}} \right)^{\mathrm{2}} }={c} \\ $$$$\Rightarrow\lambda\sqrt{\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({R}\:\mathrm{sin}\:\phi+{d}\right)^{\mathrm{2}} +{H}^{\mathrm{2}} }={c}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${AC}=\sqrt{\left({x}_{{A}} −{x}_{{C}} \right)^{\mathrm{2}} +\left({y}_{{A}} −{y}_{{C}} \right)^{\mathrm{2}} +\left({z}_{{A}} −{z}_{{C}} \right)^{\mathrm{2}} }={b} \\ $$$$\Rightarrow\lambda\sqrt{\left(\frac{{a}}{\lambda}+{R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({R}\:\mathrm{sin}\:\phi+{d}\right)^{\mathrm{2}} +{H}^{\mathrm{2}} }={b}\:\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\lambda\sqrt{\left(\frac{{a}}{\lambda}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{a}}{\lambda}\right)\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)+\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({R}\:\mathrm{sin}\:\phi+{d}\right)^{\mathrm{2}} +{H}^{\mathrm{2}} }={b} \\ $$$$\Rightarrow\lambda\sqrt{\left(\frac{{a}}{\lambda}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{a}}{\lambda}\right)\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)+\frac{{c}^{\mathrm{2}} }{\lambda^{\mathrm{2}} }}={b} \\ $$$$\Rightarrow\sqrt{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{a}\lambda\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)}={b} \\ $$$$\Rightarrow\mathrm{2}{a}\lambda\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)={b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{cos}\:\phi=\frac{\mathrm{1}}{\mathrm{2}{R}}\left[{a}−\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}\lambda}\right]\:\:\:…\left({iii}\right) \\ $$$$\left({i}\right): \\ $$$$\lambda\sqrt{{R}^{\mathrm{2}} −{aR}\:\mathrm{cos}\:\phi+\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{2}{dR}\:\mathrm{sin}\:\phi+{d}^{\mathrm{2}} +{H}^{\mathrm{2}} }={c} \\ $$$$\lambda\sqrt{\mathrm{2}{R}^{\mathrm{2}} +{H}^{\mathrm{2}} −{R}\left({a}\:\mathrm{cos}\:\phi−\mathrm{2}{d}\:\mathrm{sin}\:\phi\right)}={c} \\ $$$${H}^{\mathrm{2}} +\mathrm{2}{R}^{\mathrm{2}} \left[\mathrm{1}−\mathrm{sin}\:\left(\alpha−\phi\right)\right]=\frac{{c}^{\mathrm{2}} }{\lambda^{\mathrm{2}} } \\ $$$$\mathrm{sin}\:\left(\alpha−\phi\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{R}^{\mathrm{2}} }\left(\frac{{c}^{\mathrm{2}} }{\lambda^{\mathrm{2}} }\:−{H}^{\mathrm{2}} \right)\:\:…\left({iv}\right) \\ $$$${with}\:\alpha=\mathrm{sin}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}} \\ $$$$ \\ $$$${put}\:\left({iii}\right)\:{into}\:\left({iv}\right)\:{we}\:{get}\:{an}\:{eqn}.\:{for}\:\lambda. \\ $$$${with}\:\lambda\:{and}\:\phi\:{we}\:{can}\:{get}\:{the}\:{coordinates} \\ $$$${of}\:{point}\:{A}. \\ $$$$ \\ $$$$\Rightarrow\mathrm{cos}^{−\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{2}{aR}}\left({a}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\lambda}\right)\right]\:+\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{R}^{\mathrm{2}} }\left(\frac{{c}^{\mathrm{2}} }{\lambda^{\mathrm{2}} }\:−{H}^{\mathrm{2}} \right)\right]=\mathrm{sin}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}} \\ $$
Commented by ajfour last updated on 21/Oct/18
$${Thanks}\:{a}\:{lot}\:{Sir},\:{but}\:{can}\:{you} \\ $$$${please}\:{check}\:{my}\:{equation}\:{for}\:\boldsymbol{{z}}_{{A}} \:? \\ $$$${If}\:\:{c}={b},\:\:{and}\:{even}\:{a}=\mathrm{2}{R},\:{i}\:{think} \\ $$$${i}\:{get}\:{from}\:{my}\:{answer}, \\ $$$$\:\:\:\:\:{z}\:=\:\sqrt{\frac{{R}^{\:\mathrm{4}} }{{H}^{\mathrm{2}} }+{b}^{\mathrm{2}} }\:−\frac{{R}^{\:\mathrm{2}} }{{H}}\:. \\ $$$$\:{Does}\:{this}\:{seem}\:{true},\:{Sir}\:? \\ $$
Commented by MrW3 last updated on 21/Oct/18
$${a}=\mathrm{2}{R}\Rightarrow{d}=\mathrm{0} \\ $$$${c}={b} \\ $$$$\Rightarrow\mathrm{cos}\:\phi=\mathrm{1}−\frac{\mathrm{1}}{\lambda} \\ $$$$\lambda\sqrt{{H}^{\mathrm{2}} +\frac{\mathrm{2}{R}^{\mathrm{2}} }{\lambda}}={b} \\ $$$$\lambda^{\mathrm{2}} \left({H}^{\mathrm{2}} +\frac{\mathrm{2}{R}^{\mathrm{2}} }{\lambda}\right)={b}^{\mathrm{2}} \\ $$$${H}^{\mathrm{2}} \lambda^{\mathrm{2}} +\mathrm{2}{R}^{\mathrm{2}} \lambda−{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\lambda=\frac{−{R}^{\mathrm{2}} +\sqrt{{R}^{\mathrm{4}} +{H}^{\mathrm{2}} {b}^{\mathrm{2}} }}{{H}^{\mathrm{2}} }=\sqrt{\left(\frac{{R}}{{H}}\right)^{\mathrm{4}} +\left(\frac{{b}}{{H}}\right)^{\mathrm{2}} }−\left(\frac{{R}}{{H}}\right)^{\mathrm{2}} \\ $$$${z}_{{A}} =\lambda{H}=\sqrt{\frac{{R}^{\mathrm{4}} }{{H}^{\mathrm{2}} }+{b}^{\mathrm{2}} }−\frac{{R}^{\mathrm{2}} }{{H}} \\ $$$$\Rightarrow{your}\:{answer}\:{is}\:{correct}. \\ $$
Commented by ajfour last updated on 23/Oct/18
$${Quite}\:{compact}\:{Sir}! \\ $$
Answered by ajfour last updated on 21/Oct/18
$$\left[\mathrm{4}{R}^{\mathrm{2}} +\mathrm{4}\left({c}^{\mathrm{2}} −{z}^{\mathrm{2}} \right)\left(\frac{{z}−{H}}{{z}}\right)−\left(\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}}\right)^{\mathrm{2}} \right] \\ $$$$×\left({R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$=\left[\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}\right)−\mathrm{2}{R}^{\mathrm{2}} −{zH}−{c}^{\mathrm{2}} \left(\frac{{z}−{H}}{{z}}\right)\right]^{\mathrm{2}} \\ $$$$ \\ $$
Answered by ajfour last updated on 21/Oct/18
$${A}\left({r}\mathrm{sin}\:\theta,\:{r}\mathrm{cos}\:\theta,\:{z}\right) \\ $$$$\:\theta\:{being}\:{angle}\:{of}\:{radius}\:{from} \\ $$$${y}-{axes}. \\ $$$${R}\:\mathrm{sin}\:\alpha\:=\:\frac{{a}}{\mathrm{2}}\:\:,\:{R}\mathrm{cos}\:\alpha\:=\:\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${B}\left(\frac{{a}}{\mathrm{2}},\:−{R}\mathrm{cos}\:\alpha,\:\mathrm{0}\right) \\ $$$${C}\left(−\frac{{a}}{\mathrm{2}},\:−{R}\mathrm{cos}\:\alpha,\:\mathrm{0}\right) \\ $$$$\left({r}\mathrm{sin}\:\theta−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({r}\mathrm{cos}\:\theta+{R}\mathrm{cos}\:\alpha\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{z}^{\mathrm{2}} \:=\:{c}^{\mathrm{2}} \:\:\:\:\:\:…\left({i}\right) \\ $$$$\left({r}\mathrm{sin}\:\theta+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({r}\mathrm{cos}\:\theta+{R}\mathrm{cos}\:\alpha\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{z}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \:\:\:\:\:\:….\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right)\:{gives} \\ $$$$\:\:\:\:\:\:\mathrm{2}\boldsymbol{{ar}}\mathrm{sin}\:\boldsymbol{\theta}\:=\:\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \:\:\:\:\:….\left({I}\right) \\ $$$$\:\mathrm{2}{r}\mathrm{cos}\:\theta\:=\:\sqrt{\mathrm{4}{r}^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}}\right)^{\mathrm{2}} }\:\:..\left({II}\right) \\ $$$${APB}\:{is}\:{straight},\:{hence} \\ $$$$\frac{{r}\mathrm{sin}\:\theta−{R}\mathrm{sin}\:\phi}{{r}\mathrm{sin}\:\theta−\frac{{a}}{\mathrm{2}}}\:=\:\frac{{r}\mathrm{cos}\:\theta−{R}\mathrm{cos}\:\phi}{{r}\mathrm{cos}\:\theta+{R}\mathrm{cos}\:\alpha} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{z}−{H}}{{z}}\:\:\:\:\:\:\:….\left({iii}\right) \\ $$$$\:\:\:\:\left({R}\mathrm{sin}\:\phi\right)^{\mathrm{2}} +\left({R}\mathrm{cos}\:\phi\right)^{\mathrm{2}} =\:{R}^{\mathrm{2}} \:\:\:, \\ $$$${so} \\ $$$$\left[{r}\mathrm{sin}\:\theta−\left(\frac{{z}−{H}}{{z}}\right)\left({r}\mathrm{sin}\:\theta−\frac{{a}}{\mathrm{2}}\right)\right]^{\mathrm{2}} \\ $$$$\:+\left[{r}\mathrm{cos}\:\theta−\left(\frac{{z}−{H}}{{z}}\right)\left({r}\mathrm{cos}\:\theta+{R}\mathrm{cos}\:\alpha\right)\right]^{\mathrm{2}} =\:{R}^{\mathrm{2}} \\ $$$$….. \\ $$$$….. \\ $$