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Question-158924




Question Number 158924 by ajfour last updated on 10/Nov/21
Commented by ajfour last updated on 10/Nov/21
a, b, c are slant edges of outer  tetrahedron. Find minimum  volume of inner tetrahedron  whose vertices lie on faces of  outer one when the outer one      has a maximum volume.
$${a},\:{b},\:{c}\:{are}\:{slant}\:{edges}\:{of}\:{outer} \\ $$$${tetrahedron}.\:{Find}\:{minimum} \\ $$$${volume}\:{of}\:{inner}\:{tetrahedron} \\ $$$${whose}\:{vertices}\:{lie}\:{on}\:{faces}\:{of} \\ $$$${outer}\:{one}\:{when}\:{the}\:{outer}\:{one} \\ $$$$\:\:\:\:{has}\:{a}\:{maximum}\:{volume}. \\ $$
Commented by ajfour last updated on 10/Nov/21
thanks,Sir; i did blunder!
$${thanks},{Sir};\:{i}\:{did}\:{blunder}! \\ $$
Commented by mr W last updated on 10/Nov/21
center of sphere (r,r,r)   plane ABC: (x/a)+(y/b)+(z/c)=1  r=±(((r/a)+(r/b)+(r/c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 ))))) (we take −)  r=(1/((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))))
$${center}\:{of}\:{sphere}\:\left({r},{r},{r}\right)\: \\ $$$${plane}\:{ABC}:\:\frac{{x}}{{a}}+\frac{{y}}{{b}}+\frac{{z}}{{c}}=\mathrm{1} \\ $$$${r}=\pm\frac{\frac{{r}}{{a}}+\frac{{r}}{{b}}+\frac{{r}}{{c}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}\:\left({we}\:{take}\:−\right) \\ $$$${r}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$
Answered by mr W last updated on 10/Nov/21
Part I  say θ=angle between OA and OB  say ϕ=angle betw. OC and plane OAB  then the volume of the outer  tetrahedron is  V_O =(1/3)×((ab sin θ)/2)×c sin ϕ=((abc sin θ sin ϕ)/6)  it′s clear V_O  is maximum when  θ=ϕ=90°, i.e. a,b,c are perpendicular  to each other.
$${Part}\:{I} \\ $$$${say}\:\theta={angle}\:{between}\:{OA}\:{and}\:{OB} \\ $$$${say}\:\varphi={angle}\:{betw}.\:{OC}\:{and}\:{plane}\:{OAB} \\ $$$${then}\:{the}\:{volume}\:{of}\:{the}\:{outer} \\ $$$${tetrahedron}\:{is} \\ $$$${V}_{{O}} =\frac{\mathrm{1}}{\mathrm{3}}×\frac{{ab}\:\mathrm{sin}\:\theta}{\mathrm{2}}×{c}\:\mathrm{sin}\:\varphi=\frac{{abc}\:\mathrm{sin}\:\theta\:\mathrm{sin}\:\varphi}{\mathrm{6}} \\ $$$${it}'{s}\:{clear}\:{V}_{{O}} \:{is}\:{maximum}\:{when} \\ $$$$\theta=\varphi=\mathrm{90}°,\:{i}.{e}.\:{a},{b},{c}\:{are}\:{perpendicular} \\ $$$${to}\:{each}\:{other}. \\ $$

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