Question Number 154910 by mr W last updated on 23/Sep/21
Commented by mr W last updated on 23/Sep/21
$${tetrahedron}\:{T}−{ABC}\:{has}\:{edge}\:{lengthes} \\ $$$${a},{b},{c},{u},{v},{w}. \\ $$$${the}\:{faces}\:{of}\:{tetrahedron}\:{T}'−{A}'{B}'{C}' \\ $$$${have}\:{the}\:{distances}\:{d}_{\mathrm{1}} ,{d}_{\mathrm{2}} ,{d}_{\mathrm{3}} ,{d}_{\mathrm{4}} \:{to}\:{the} \\ $$$${corresponding}\:{faces}\:{of}\:{T}−{ABC}. \\ $$$${find}\:{the}\:{volume}\:{of}\:{new}\:{tetrahedron}. \\ $$
Commented by Rasheed.Sindhi last updated on 23/Sep/21
$${Guess} \\ $$$${V}\:'=\left(\mathrm{1}+\frac{\left(\overset{{area}} {{face}\mathrm{1}}\right){d}_{\mathrm{1}} +\left(\overset{{area}} {{face}\mathrm{2}}\right){d}_{\mathrm{2}} +\left(\overset{{area}} {{face}\mathrm{3}}\right){d}_{\mathrm{3}} +\left(\overset{{area}} {{face}\mathrm{4}}\right){d}_{\mathrm{4}} }{\mathrm{3}{V}}\right)^{\mathrm{3}} .{V} \\ $$
Answered by mr W last updated on 24/Sep/21
![tetrahedron T−ABC: edges: a,b,c,u,v,w volume: V in terms of a,b,c,u,v,w^(∗)) face 1: ΔTBC, area A_1 in terms of a,v,w ^(∗∗)) face 2: ΔTCA, area A_2 in terms of b,w,u face 3: ΔTAB, area A_3 in terms of c,u,v face 4: ΔABC, area A_4 in terms of a,b,c radius of insphere of T−ABC: r ((r(A_1 +A_2 +A_3 +A_4 ))/3)=V tetrahedrons T−ABC and T′−A′B′C′ are similar. say the edges of T′−A′B′C′ are a′,b′,c′,u′,v′,w′ and a′=ka, b′=kb, c′=kc, u′=ku, v′=kv, w′=kw with k=factor of magnification. we have area of face 1 of T′−A′B′C′ A_1 ′=k^2 A_1 area of face 2 of T′−A′B′C′ A_2 ′=k^2 A_2 area of face 3 of T′−A′B′C′ A_3 ′=k^2 A_3 area of face 4 of T′−A′B′C′ A_4 ′=k^2 A_4 volume of T′−A′B′C′ V′=k^3 V on the other side, V′=((A_1 ′(r+d_1 ))/3)+((A_2 ′(r+d_2 ))/3)+((A_3 ′(r+d_3 ))/3)+((A_4 ′(r+d_4 ))/3) V′=((k^2 A_1 (r+d_1 ))/3)+((k^2 A_2 (r+d_2 ))/3)+((k^2 A_3 (r+d_3 ))/3)+((k^2 A_4 (r+d_4 ))/3) V′=((k^2 [(A_1 +A_2 +A_3 +A_4 )r+A_1 d_1 +A_2 d_2 +A_3 d_3 +A_4 d_4 ])/3) V′=((k^2 (3V+A_1 d_1 +A_2 d_2 +A_3 d_3 +A_4 d_4 ))/3) ((k^2 (3V+A_1 d_1 +A_2 d_2 +A_3 d_3 +A_4 d_4 ))/3)=k^3 V ⇒k=1+((A_1 d_1 +A_2 d_2 +A_3 d_3 +A_4 d_4 )/(3V)) volume of T′−A′B′C′ is V′=(1+((A_1 d_1 +A_2 d_2 +A_3 d_3 +A_4 d_4 )/(3V)))^3 V ___________________________ ^(∗)) using Euler′ formula, see Q40469 ^(∗∗)) using Heron′s formula](Q155015.png)
$${tetrahedron}\:{T}−{ABC}: \\ $$$${edges}:\:{a},{b},{c},{u},{v},{w} \\ $$$${volume}:\:{V}\:{in}\:{terms}\:{of}\:{a},{b},{c},{u},{v},{w}\:^{\left.\ast\right)} \\ $$$${face}\:\mathrm{1}:\:\Delta{TBC},\:{area}\:{A}_{\mathrm{1}} \:{in}\:{terms}\:{of}\:{a},{v},{w}\:\:^{\left.\ast\ast\right)} \\ $$$${face}\:\mathrm{2}:\:\Delta{TCA},\:{area}\:{A}_{\mathrm{2}} \:{in}\:{terms}\:{of}\:{b},{w},{u} \\ $$$${face}\:\mathrm{3}:\:\Delta{TAB},\:{area}\:{A}_{\mathrm{3}} \:{in}\:{terms}\:{of}\:{c},{u},{v} \\ $$$${face}\:\mathrm{4}:\:\Delta{ABC},\:{area}\:{A}_{\mathrm{4}} \:{in}\:{terms}\:{of}\:{a},{b},{c} \\ $$$${radius}\:{of}\:{insphere}\:{of}\:{T}−{ABC}:\:{r} \\ $$$$\frac{{r}\left({A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}_{\mathrm{3}} +{A}_{\mathrm{4}} \right)}{\mathrm{3}}={V} \\ $$$${tetrahedrons}\:{T}−{ABC}\:{and}\:{T}'−{A}'{B}'{C}' \\ $$$${are}\:{similar}. \\ $$$${say}\:{the}\:{edges}\:{of}\:{T}'−{A}'{B}'{C}' \\ $$$${are}\:{a}',{b}',{c}',{u}',{v}',{w}'\:{and} \\ $$$${a}'={ka},\:{b}'={kb},\:{c}'={kc},\:{u}'={ku},\:{v}'={kv},\:{w}'={kw} \\ $$$${with}\:{k}={factor}\:{of}\:{magnification}. \\ $$$$ \\ $$$${we}\:{have} \\ $$$${area}\:{of}\:{face}\:\mathrm{1}\:{of}\:\:{T}'−{A}'{B}'{C}'\:{A}_{\mathrm{1}} '={k}^{\mathrm{2}} {A}_{\mathrm{1}} \\ $$$${area}\:{of}\:{face}\:\mathrm{2}\:{of}\:\:{T}'−{A}'{B}'{C}'\:{A}_{\mathrm{2}} '={k}^{\mathrm{2}} {A}_{\mathrm{2}} \\ $$$${area}\:{of}\:{face}\:\mathrm{3}\:{of}\:\:{T}'−{A}'{B}'{C}'\:{A}_{\mathrm{3}} '={k}^{\mathrm{2}} {A}_{\mathrm{3}} \\ $$$${area}\:{of}\:{face}\:\mathrm{4}\:{of}\:\:{T}'−{A}'{B}'{C}'\:{A}_{\mathrm{4}} '={k}^{\mathrm{2}} {A}_{\mathrm{4}} \\ $$$${volume}\:{of}\:{T}'−{A}'{B}'{C}'\:{V}'={k}^{\mathrm{3}} {V} \\ $$$$ \\ $$$${on}\:{the}\:{other}\:{side}, \\ $$$${V}'=\frac{{A}_{\mathrm{1}} '\left({r}+{d}_{\mathrm{1}} \right)}{\mathrm{3}}+\frac{{A}_{\mathrm{2}} '\left({r}+{d}_{\mathrm{2}} \right)}{\mathrm{3}}+\frac{{A}_{\mathrm{3}} '\left({r}+{d}_{\mathrm{3}} \right)}{\mathrm{3}}+\frac{{A}_{\mathrm{4}} '\left({r}+{d}_{\mathrm{4}} \right)}{\mathrm{3}} \\ $$$${V}'=\frac{{k}^{\mathrm{2}} {A}_{\mathrm{1}} \left({r}+{d}_{\mathrm{1}} \right)}{\mathrm{3}}+\frac{{k}^{\mathrm{2}} {A}_{\mathrm{2}} \left({r}+{d}_{\mathrm{2}} \right)}{\mathrm{3}}+\frac{{k}^{\mathrm{2}} {A}_{\mathrm{3}} \left({r}+{d}_{\mathrm{3}} \right)}{\mathrm{3}}+\frac{{k}^{\mathrm{2}} {A}_{\mathrm{4}} \left({r}+{d}_{\mathrm{4}} \right)}{\mathrm{3}} \\ $$$${V}'=\frac{{k}^{\mathrm{2}} \left[\left({A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}_{\mathrm{3}} +{A}_{\mathrm{4}} \right){r}+{A}_{\mathrm{1}} {d}_{\mathrm{1}} +{A}_{\mathrm{2}} {d}_{\mathrm{2}} +{A}_{\mathrm{3}} {d}_{\mathrm{3}} +{A}_{\mathrm{4}} {d}_{\mathrm{4}} \right]}{\mathrm{3}} \\ $$$${V}'=\frac{{k}^{\mathrm{2}} \left(\mathrm{3}{V}+{A}_{\mathrm{1}} {d}_{\mathrm{1}} +{A}_{\mathrm{2}} {d}_{\mathrm{2}} +{A}_{\mathrm{3}} {d}_{\mathrm{3}} +{A}_{\mathrm{4}} {d}_{\mathrm{4}} \right)}{\mathrm{3}} \\ $$$$\frac{{k}^{\mathrm{2}} \left(\mathrm{3}{V}+{A}_{\mathrm{1}} {d}_{\mathrm{1}} +{A}_{\mathrm{2}} {d}_{\mathrm{2}} +{A}_{\mathrm{3}} {d}_{\mathrm{3}} +{A}_{\mathrm{4}} {d}_{\mathrm{4}} \right)}{\mathrm{3}}={k}^{\mathrm{3}} {V} \\ $$$$\Rightarrow{k}=\mathrm{1}+\frac{{A}_{\mathrm{1}} {d}_{\mathrm{1}} +{A}_{\mathrm{2}} {d}_{\mathrm{2}} +{A}_{\mathrm{3}} {d}_{\mathrm{3}} +{A}_{\mathrm{4}} {d}_{\mathrm{4}} }{\mathrm{3}{V}} \\ $$$${volume}\:{of}\:{T}'−{A}'{B}'{C}'\:{is} \\ $$$${V}'=\left(\mathrm{1}+\frac{{A}_{\mathrm{1}} {d}_{\mathrm{1}} +{A}_{\mathrm{2}} {d}_{\mathrm{2}} +{A}_{\mathrm{3}} {d}_{\mathrm{3}} +{A}_{\mathrm{4}} {d}_{\mathrm{4}} }{\mathrm{3}{V}}\right)^{\mathrm{3}} {V} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:^{\left.\ast\right)} \:{using}\:{Euler}'\:{formula},\:{see}\:{Q}\mathrm{40469} \\ $$$$\:^{\left.\ast\ast\right)} \:{using}\:{Heron}'{s}\:{formula} \\ $$