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Question-211277




Question Number 211277 by mr W last updated on 04/Sep/24
Commented by mr W last updated on 04/Sep/24
find the volume of the tetrahedron  in terms of a,b,c, θ with 0<θ≤120°.
$${find}\:{the}\:{volume}\:{of}\:{the}\:{tetrahedron} \\ $$$${in}\:{terms}\:{of}\:{a},{b},{c},\:\theta\:{with}\:\mathrm{0}<\theta\leqslant\mathrm{120}°. \\ $$
Commented by BHOOPENDRA last updated on 06/Sep/24
Mr.W the condition  you have given  is it  talking about fermat point  then i think i can try further  like if we see bottom view of this then  the point O should be fermat point ?
$${Mr}.{W}\:{the}\:{condition}\:\:{you}\:{have}\:{given} \\ $$$${is}\:{it}\:\:{talking}\:{about}\:{fermat}\:{point} \\ $$$${then}\:{i}\:{think}\:{i}\:{can}\:{try}\:{further} \\ $$$${like}\:{if}\:{we}\:{see}\:{bottom}\:{view}\:{of}\:{this}\:{then} \\ $$$${the}\:{point}\:{O}\:{should}\:{be}\:{fermat}\:{point}\:? \\ $$
Commented by BHOOPENDRA last updated on 06/Sep/24
Commented by mr W last updated on 07/Sep/24
for θ≠120° i don′t think point O is  the fermat point in bottom view.   if you think it is, please prove that  and show your working.  i don′t have the solution.
$${for}\:\theta\neq\mathrm{120}°\:{i}\:{don}'{t}\:{think}\:{point}\:{O}\:{is} \\ $$$${the}\:{fermat}\:{point}\:{in}\:{bottom}\:{view}.\: \\ $$$${if}\:{you}\:{think}\:{it}\:{is},\:{please}\:{prove}\:{that} \\ $$$${and}\:{show}\:{your}\:{working}. \\ $$$${i}\:{don}'{t}\:{have}\:{the}\:{solution}. \\ $$
Commented by Ghisom last updated on 21/Jun/25
I don′t think we can reach an explicit  general solution. I tried for a=3, b=4,  c=5, α=60°. we get 2 solutions, but we  have to solve a quartic with no “nice”  solution.  I′m interested in the ellipse determined  by A, B, C and the cone wrapping the  tetrahedron. will post my result tomorrow,  it′s 23:40 where I live...
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{reach}\:\mathrm{an}\:\mathrm{explicit} \\ $$$$\mathrm{general}\:\mathrm{solution}.\:\mathrm{I}\:\mathrm{tried}\:\mathrm{for}\:{a}=\mathrm{3},\:{b}=\mathrm{4}, \\ $$$${c}=\mathrm{5},\:\alpha=\mathrm{60}°.\:\mathrm{we}\:\mathrm{get}\:\mathrm{2}\:\mathrm{solutions},\:\mathrm{but}\:\mathrm{we} \\ $$$$\mathrm{have}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{a}\:\mathrm{quartic}\:\mathrm{with}\:\mathrm{no}\:“\mathrm{nice}'' \\ $$$$\mathrm{solution}. \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{interested}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ellipse}\:\mathrm{determined} \\ $$$$\mathrm{by}\:{A},\:{B},\:{C}\:\mathrm{and}\:\mathrm{the}\:\mathrm{cone}\:\mathrm{wrapping}\:\mathrm{the} \\ $$$$\mathrm{tetrahedron}.\:\mathrm{will}\:\mathrm{post}\:\mathrm{my}\:\mathrm{result}\:\mathrm{tomorrow}, \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{23}:\mathrm{40}\:\mathrm{where}\:\mathrm{I}\:\mathrm{live}… \\ $$
Commented by Ghisom last updated on 22/Jun/25
...it′s even more complicated.  the conic is a byperbola in this case  ⇒  we must determine either  • for a given triangle which angles α are  leading to a closed tetrahedron or  • for a given α which triangles are possible
$$…\mathrm{it}'\mathrm{s}\:\mathrm{even}\:\mathrm{more}\:\mathrm{complicated}. \\ $$$$\mathrm{the}\:\mathrm{conic}\:\mathrm{is}\:\mathrm{a}\:\mathrm{byperbola}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$$\Rightarrow \\ $$$$\mathrm{we}\:\mathrm{must}\:\mathrm{determine}\:\mathrm{either} \\ $$$$\bullet\:\mathrm{for}\:\mathrm{a}\:\mathrm{given}\:\mathrm{triangle}\:\mathrm{which}\:\mathrm{angles}\:\alpha\:\mathrm{are} \\ $$$$\mathrm{leading}\:\mathrm{to}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{tetrahedron}\:\mathrm{or} \\ $$$$\bullet\:\mathrm{for}\:\mathrm{a}\:\mathrm{given}\:\alpha\:\mathrm{which}\:\mathrm{triangles}\:\mathrm{are}\:\mathrm{possible} \\ $$

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