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Question Number 150418 by ajfour last updated on 12/Aug/21
Commented by ajfour last updated on 12/Aug/21
For tetrahedron edge a, and sphere radius r, find fractional volume of sphere within the tetrahedron. f= { ((g_1 (a,r) ; (a/3)≤r≤..)),((g_2 (a,r) ; ..≤r≤..)),((g_3 (a,r) ; ..≤r≤..)) :} hope u understand..
$${For}\:{tetrahedron}\:{edge}\:{a}, \\ $$$${and}\:{sphere}\:{radius}\:{r},\:{find}\: \\ $$$${fractional}\:{volume}\:{of}\:{sphere}\: \\ $$$${within}\:{the}\:{tetrahedron}. \\ $$$${f}=\begin{cases}{{g}_{\mathrm{1}} \left({a},{r}\right)\:;\:\:\:\:\:\frac{{a}}{\mathrm{3}}\leqslant{r}\leqslant..}\\{{g}_{\mathrm{2}} \left({a},{r}\right)\:;\:\:\:..\leqslant{r}\leqslant..}\\{{g}_{\mathrm{3}} \left({a},{r}\right)\:;\:..\leqslant{r}\leqslant..}\end{cases} \\ $$$${hope}\:{u}\:{understand}.. \\ $$
Commented by mr W last updated on 12/Aug/21
touch on two faces and intersect on two otherfaces with intersection of equal size?
$${touch}\:{on}\:{two}\:{faces}\:{and}\:{intersect}\: \\ $$$${on}\:{two}\:{otherfaces}\:{with}\:{intersection}\: \\ $$$${of}\:{equal}\:{size}? \\ $$
Commented by ajfour last updated on 12/Aug/21
symmetrical for the three slant faces, just touching the bottom one..
$${symmetrical}\:{for}\:{the}\:{three} \\ $$$${slant}\:{faces},\:{just}\:{touching} \\ $$$${the}\:{bottom}\:{one}.. \\ $$
Answered by mr W last updated on 12/Aug/21
Commented by mr W last updated on 14/Aug/21
volume of tetrahedron V=(a^3 /(6(√2))) area of a face A=(((√3)a^2 )/4) mininum radius of sphere 4×(1/3)×(((√3)a^2 )/4)×r_(min) =V=(a^3 /(6(√2))) r_(min) =(((√6)a)/( 12))≈0.204a i.e. if r≤(((√6)a)/(12)), the sphere is completely inside the tetrahedron.
$${volume}\:{of}\:{tetrahedron} \\ $$$${V}=\frac{{a}^{\mathrm{3}} }{\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$${area}\:{of}\:{a}\:{face} \\ $$$${A}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${mininum}\:{radius}\:{of}\:{sphere} \\ $$$$\mathrm{4}×\frac{\mathrm{1}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}}×{r}_{{min}} ={V}=\frac{{a}^{\mathrm{3}} }{\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$${r}_{{min}} =\frac{\sqrt{\mathrm{6}}{a}}{\:\mathrm{12}}\approx\mathrm{0}.\mathrm{204}{a} \\ $$$${i}.{e}.\:{if}\:{r}\leqslant\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{12}},\:{the}\:{sphere}\:{is}\:{completely} \\ $$$${inside}\:{the}\:{tetrahedron}. \\ $$
Commented by mr W last updated on 14/Aug/21
Commented by mr W last updated on 14/Aug/21
case r>(((√6)a)/(12)) (1/3)×(((√3)a^2 )/4)(3h+r)=V=(a^3 /(6(√2))) ⇒h=(1/3)((((√6)a)/( 3))−r)=(((√6)a)/( 9))−(r/3) heigth of spherical cap: δ=r−h=((4r)/3)−(((√6)a)/( 9))=(1/3)(4r−(((√6)a)/3)) volume of a spherical cap: V_(cap) =πδ^2 (r−(δ/3)) V_(cap) =(π/(81))(4r−(((√6)a)/3))^2 (5r+(((√6)a)/( 3))) fractional volume of sphere inside of the tetrahedron: λ=1−((ΣV_(cap) )/V)=1−(3/(4πr^3 ))×((3π)/(81))(4r−(((√6)a)/3))^2 (5r+(((√6)a)/3)) λ=1−(1/(36))(4−(((√6)a)/(3r)))^2 (5+(((√6)a)/(3r))) this is valid till r=r_(max) with r_(max) all three spherical caps touch each other.
$$\boldsymbol{{case}}\:\:\boldsymbol{{r}}>\frac{\sqrt{\mathrm{6}}\boldsymbol{{a}}}{\mathrm{12}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{3}{h}+{r}\right)={V}=\frac{{a}^{\mathrm{3}} }{\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\sqrt{\mathrm{6}}{a}}{\:\mathrm{3}}−{r}\right)=\frac{\sqrt{\mathrm{6}}{a}}{\:\mathrm{9}}−\frac{{r}}{\mathrm{3}} \\ $$$${heigth}\:{of}\:{spherical}\:{cap}: \\ $$$$\delta={r}−{h}=\frac{\mathrm{4}{r}}{\mathrm{3}}−\frac{\sqrt{\mathrm{6}}{a}}{\:\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{4}{r}−\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}}\right) \\ $$$${volume}\:{of}\:{a}\:{spherical}\:{cap}: \\ $$$${V}_{{cap}} =\pi\delta^{\mathrm{2}} \left({r}−\frac{\delta}{\mathrm{3}}\right) \\ $$$${V}_{{cap}} =\frac{\pi}{\mathrm{81}}\left(\mathrm{4}{r}−\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}}\right)^{\mathrm{2}} \left(\mathrm{5}{r}+\frac{\sqrt{\mathrm{6}}{a}}{\:\mathrm{3}}\right) \\ $$$${fractional}\:{volume}\:{of}\:{sphere}\:{inside} \\ $$$${of}\:{the}\:{tetrahedron}: \\ $$$$\lambda=\mathrm{1}−\frac{\Sigma{V}_{{cap}} }{{V}}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}\pi{r}^{\mathrm{3}} }×\frac{\mathrm{3}\pi}{\mathrm{81}}\left(\mathrm{4}{r}−\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}}\right)^{\mathrm{2}} \left(\mathrm{5}{r}+\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}}\right) \\ $$$$\lambda=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{36}}\left(\mathrm{4}−\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}{r}}\right)^{\mathrm{2}} \left(\mathrm{5}+\frac{\sqrt{\mathrm{6}}{a}}{\mathrm{3}{r}}\right) \\ $$$${this}\:{is}\:{valid}\:{till}\:{r}={r}_{{max}} \\ $$$${with}\:{r}_{{max}} \:{all}\:{three}\:{spherical}\:{caps} \\ $$$${touch}\:{each}\:{other}. \\ $$
Commented by mr W last updated on 14/Aug/21
b^2 =r^2 −h^2 =r^2 −(1/9)((((√6)a)/( 3))−r)^2 =((8r^2 )/9)−((2a^2 )/( 27))+((2(√6)ar)/(27)) b^2 =(2/9)(4r^2 −(a^2 /( 3))+(((√6)ar)/3)) b=(1/3)(√(2(4r^2 −(a^2 /3)+(((√6)ar)/3)))) GD=(1/3)AD=(1/3)×(((√3)a)/2)=(((√3)a)/6) ED^2 +h^2 =GD^2 +r^2 ED^2 =(a^2 /(12))+r^2 −h^2 ED^2 =(a^2 /(12))+((8r^2 )/9)−((2a^2 )/( 27))+((2(√6)ar)/(27)) ED^2 =(1/9)((a^2 /(12))+8r^2 +((2(√6)ar)/( 3))) ED=(1/3)(√((a^2 /(12))+8r^2 +((2(√6)ar)/( 3))))
$${b}^{\mathrm{2}} ={r}^{\mathrm{2}} −{h}^{\mathrm{2}} ={r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\sqrt{\mathrm{6}}{a}}{\:\mathrm{3}}−{r}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{8}{r}^{\mathrm{2}} }{\mathrm{9}}−\frac{\mathrm{2}{a}^{\mathrm{2}} }{\:\mathrm{27}}+\frac{\mathrm{2}\sqrt{\mathrm{6}}{ar}}{\mathrm{27}} \\ $$$${b}^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{4}{r}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\:\mathrm{3}}+\frac{\sqrt{\mathrm{6}}{ar}}{\mathrm{3}}\right) \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{2}\left(\mathrm{4}{r}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{3}}+\frac{\sqrt{\mathrm{6}}{ar}}{\mathrm{3}}\right)} \\ $$$${GD}=\frac{\mathrm{1}}{\mathrm{3}}{AD}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{6}} \\ $$$$ \\ $$$${ED}^{\mathrm{2}} +{h}^{\mathrm{2}} ={GD}^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$${ED}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{12}}+{r}^{\mathrm{2}} −{h}^{\mathrm{2}} \\ $$$${ED}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{8}{r}^{\mathrm{2}} }{\mathrm{9}}−\frac{\mathrm{2}{a}^{\mathrm{2}} }{\:\mathrm{27}}+\frac{\mathrm{2}\sqrt{\mathrm{6}}{ar}}{\mathrm{27}} \\ $$$${ED}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{12}}+\mathrm{8}{r}^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{6}}{ar}}{\:\mathrm{3}}\right) \\ $$$${ED}=\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{12}}+\mathrm{8}{r}^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{6}}{ar}}{\:\mathrm{3}}} \\ $$
Commented by mr W last updated on 14/Aug/21
Commented by mr W last updated on 14/Aug/21
TD=(((√3)a)/2) TE=TD−ED TE=(((√3)a)/2)−(1/3)(√((a^2 /(12))+8r^2 +((2(√6)ar)/( 3)))) TE=2b=(2/3)(√(2(4r^2 −(a^2 /( 3))+(((√6)ar)/3)))) (((√3)a)/2)−(1/3)(√((a^2 /(12))+8r^2 +((2(√6)ar)/( 3))))=(2/3)(√(2(4r^2 −(a^2 /( 3))+(((√6)ar)/3)))) ((3(√3))/2)=(√((1/(12))+8(r^2 /a^2 )+((2(√6)r)/( 3a))))+2(√(2(4(r^2 /a^2 )−(1/( 3))+(((√6)r)/(3a))))) this is the equation to determine r_(max) . r_(max) ≈0.2989a
$${TD}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}} \\ $$$${TE}={TD}−{ED} \\ $$$${TE}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{12}}+\mathrm{8}{r}^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{6}}{ar}}{\:\mathrm{3}}} \\ $$$${TE}=\mathrm{2}{b}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{2}\left(\mathrm{4}{r}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\:\mathrm{3}}+\frac{\sqrt{\mathrm{6}}{ar}}{\mathrm{3}}\right)} \\ $$$$\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{12}}+\mathrm{8}{r}^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{6}}{ar}}{\:\mathrm{3}}}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{2}\left(\mathrm{4}{r}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\:\mathrm{3}}+\frac{\sqrt{\mathrm{6}}{ar}}{\mathrm{3}}\right)} \\ $$$$\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}}{\mathrm{12}}+\mathrm{8}\frac{{r}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{2}\sqrt{\mathrm{6}}{r}}{\:\mathrm{3}{a}}}+\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{4}\frac{{r}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{1}}{\:\mathrm{3}}+\frac{\sqrt{\mathrm{6}}{r}}{\mathrm{3}{a}}\right)} \\ $$$${this}\:{is}\:{the}\:{equation}\:{to}\:{determine}\:{r}_{{max}} . \\ $$$${r}_{{max}} \approx\mathrm{0}.\mathrm{2989}{a} \\ $$
Commented by mr W last updated on 14/Aug/21
Answered by ajfour last updated on 14/Aug/21
Commented by ajfour last updated on 14/Aug/21
h(altitude)=(((√2)a)/( (√3))) cos 2θ=(a/(2(√3)))÷((a(√3))/2)=(1/3) R_(min) =((a/2))tan 30°tan θ =(a/(2(√3)))(√((1−cos 2θ)/(1+cos 2θ))) R_(min) =(a/(2(√3)))(√((2/3)/(4/3)))=(a/(2(√6)))=((a(√6))/(12))
$${h}\left({altitude}\right)=\frac{\sqrt{\mathrm{2}}{a}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}}\boldsymbol{\div}\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${R}_{{min}} =\left(\frac{{a}}{\mathrm{2}}\right)\mathrm{tan}\:\mathrm{30}°\mathrm{tan}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}} \\ $$$${R}_{{min}} =\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\frac{\mathrm{2}/\mathrm{3}}{\mathrm{4}/\mathrm{3}}}=\frac{{a}}{\mathrm{2}\sqrt{\mathrm{6}}}=\frac{{a}\sqrt{\mathrm{6}}}{\mathrm{12}} \\ $$

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