Question Number 40857 by ajfour last updated on 28/Jul/18
Commented by ajfour last updated on 28/Jul/18
$${Find}\:{circumsphere}\:{radius}\:{R}\:{of} \\ $$$${a}\:{general}\:{triangular}\:{pyramid}. \\ $$
Commented by MJS last updated on 29/Jul/18
$$\mathrm{see}\:\mathrm{my}\:\mathrm{comment}\:\mathrm{to}\:\mathrm{qu}.\:\mathrm{40469} \\ $$$$\mathrm{the}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{coordinates}\:\mathrm{of}\:{S} \\ $$$$\mathrm{try}\:\mathrm{for}\:\mathrm{a}\:\mathrm{triangle},\:\mathrm{obviously}\:\mathrm{we}\:\mathrm{can}\:\mathrm{put} \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{{c}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{but}\:\mathrm{for}\:{C}\:\mathrm{we}\:\mathrm{need}\:\mathrm{Heron}'\mathrm{s}\:\mathrm{formula} \\ $$$${C}=\begin{pmatrix}{{x}_{{C}} }\\{{h}_{{c}} }\end{pmatrix}\:\mathrm{with}\:{h}_{{c}} =\frac{\mathrm{2}×\mathrm{area}\left(\bigtriangleup\right)}{{c}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{2}{c}} \\ $$$$\:\:\:\:\:\mathrm{and}\:{x}_{{C}} =\sqrt{{b}^{\mathrm{2}} −{h}_{{c}} ^{\mathrm{2}} } \\ $$$$\mathrm{then}\:\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{the}\:\mathrm{circumcircle} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{in}\:\mathbb{R}^{\mathrm{2}} \:\mathrm{but}\:\mathrm{in}\:\mathbb{R}^{\mathrm{3}} \:\mathrm{we}\:\mathrm{need}\:\mathrm{Euler}'\mathrm{s} \\ $$$$\mathrm{formula}\:\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{very}\:\mathrm{hard},\:\mathrm{maybe}\:\mathrm{impossible}, \\ $$$$\mathrm{to}\:\mathrm{handle}\:\mathrm{it}\:\mathrm{in}\:\mathrm{any}\:\mathrm{equation}\:\mathrm{system} \\ $$$${S}=\begin{pmatrix}{{x}_{{S}} }\\{{y}_{{S}} }\\{{h}}\end{pmatrix}\:\:\mathrm{with}\:{h}=\frac{\mathrm{3}×\mathrm{volume}\left(\mathrm{pyramid}\right)}{\mathrm{area}\left(\bigtriangleup{ABC}\right)} \\ $$$$\mathrm{then}\:\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{the}\:\mathrm{circumsphere}… \\ $$
Answered by ajfour last updated on 28/Jul/18
$${Let}\:\bigtriangleup{ABC}\:{be}\:{the}\:{base}\:{and}\:{S}\:{the} \\ $$$${vertex}.\:{Circumsphere}\:{centre}\:{G}. \\ $$$${Origin}\:{O}\:,{the}\:{centre}\:{of} \\ $$$$\:{circumcircle}\:{of}\:{base}\:\left({radius}\:\rho\right). \\ $$$${y}\:{axis}\:{towards}\:{right},\:{x}\:{axis} \\ $$$${outwards};\:{z}\:{axis}\:{upwards}. \\ $$$$\boldsymbol{\rho}={R}\mathrm{cos}\:\theta\:\:\:\:\:\:\:\:\:\:…..\left({i}\right) \\ $$$${a}=\mathrm{2}\rho\mathrm{cos}\:\alpha\:\:\:\:\:\:\:\:\:\:……\left({ii}\right) \\ $$$${b}=\mathrm{2}\rho\mathrm{cos}\:\left(\alpha+\beta\right)\:\:\:\:\:….\left({iii}\right) \\ $$$${c}=\mathrm{2}\rho\mathrm{cos}\:\beta\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({iv}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${A}\left(−\rho\mathrm{sin}\:\mathrm{2}\beta,\:\rho\mathrm{cos}\:\mathrm{2}\beta\:,\mathrm{0}\right) \\ $$$${B}\left(\mathrm{0},−\rho,\mathrm{0}\right) \\ $$$${C}\left(\rho\mathrm{sin}\:\mathrm{2}\alpha\:,\:\rho\mathrm{cos}\:\mathrm{2}\alpha\:,\:\mathrm{0}\right) \\ $$$${S}\left({R}\mathrm{cos}\:\gamma\mathrm{sin}\:\phi,\:{R}\mathrm{cos}\:\gamma\mathrm{cos}\:\phi,\:{R}\mathrm{sin}\:\theta+{R}\mathrm{sin}\:\gamma\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${SA}={p}\:\:\:\:\:\:\:\:\:\:…\left({v}\right) \\ $$$${SB}={q}\:\:\:\:\:\:\:\:\:….\left({vi}\right) \\ $$$${SC}={r}\:\:\:\:\:\:\:\:\:…..\left({vii}\right) \\ $$$${unknowns}:\:\:\:\boldsymbol{\alpha},\:\boldsymbol{\beta},\:\boldsymbol{\gamma},\:\boldsymbol{\theta},\:\boldsymbol{\phi},\:\boldsymbol{\rho},\:\boldsymbol{{R}} \\ $$$$\:{equations}\::\:\left({i}\right)\:\rightarrow\:\left({vii}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$
Commented by MrW3 last updated on 28/Jul/18
$${very}\:{nice}\:{try}\:{sir}!\: \\ $$$${It}\:{seems}\:{to}\:{be}\:{a}\:{hard}\:{task}. \\ $$