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A problem with tetrahedron [closed]

Regular TetrahedronRelationship between angles in tetrahedronTetrahedron problemWhat determines the height of a tetrahedron?Tetrahedron with right dihedral angleSolve a tetrahedron with given lengths of three edges and a certain propertyAngle between two faces of a tetrahedronMeasure of the angleThe midpoints of an isosceles tetrahedronTetrahedron Centers

1

$begingroup$

Let ABCD be a tetrahedron with the property that opposite edges are equal. We know that the angle between the planes ABD and BCD is $90^circ$ and the angle between (BCD) and (CAD) is $60^circ$. Calculate the angle measure between (CAD) and (ABD).

geometry angle solid-geometry

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edited Mar 18 at 22:28

Aretino

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asked Mar 18 at 19:55

AdeleAdele

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closed as off-topic by Abcd, Xander Henderson, Eevee Trainer, Lee David Chung Lin, Alex Provost Mar 19 at 4:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Abcd, Xander Henderson, Eevee Trainer, Lee David Chung Lin, Alex Provost

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please help me!
  $endgroup$
  – Adele
  Mar 18 at 20:56
  

  
  
  
  

add a comment | 

1

$begingroup$

Let ABCD be a tetrahedron with the property that opposite edges are equal. We know that the angle between the planes ABD and BCD is $90^circ$ and the angle between (BCD) and (CAD) is $60^circ$. Calculate the angle measure between (CAD) and (ABD).

geometry angle solid-geometry

share|cite|improve this question

edited Mar 18 at 22:28

Aretino

25.8k31545

asked Mar 18 at 19:55

AdeleAdele

93

$endgroup$

closed as off-topic by Abcd, Xander Henderson, Eevee Trainer, Lee David Chung Lin, Alex Provost Mar 19 at 4:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Abcd, Xander Henderson, Eevee Trainer, Lee David Chung Lin, Alex Provost

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please help me!
  $endgroup$
  – Adele
  Mar 18 at 20:56
  

  
  
  
  

add a comment | 

$begingroup$

Let ABCD be a tetrahedron with the property that opposite edges are equal. We know that the angle between the planes ABD and BCD is $90^circ$ and the angle between (BCD) and (CAD) is $60^circ$. Calculate the angle measure between (CAD) and (ABD).

geometry angle solid-geometry

share|cite|improve this question

edited Mar 18 at 22:28

Aretino

25.8k31545

asked Mar 18 at 19:55

AdeleAdele

93

$endgroup$

share|cite|improve this question

edited Mar 18 at 22:28

Aretino

25.8k31545

asked Mar 18 at 19:55

AdeleAdele

93

share|cite|improve this question

edited Mar 18 at 22:28

Aretino

25.8k31545

asked Mar 18 at 19:55

AdeleAdele

93

edited Mar 18 at 22:28

Aretino

25.8k31545

edited Mar 18 at 22:28

Aretino

25.8k31545

asked Mar 18 at 19:55

AdeleAdele

93

asked Mar 18 at 19:55

AdeleAdele

93

2 Answers
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1

$begingroup$

The hypothesis that the tetrahedron has equal opposite edges means that opposite dihedral angles are also equal, and that all faces of the tetrahedron are congruent. A consequence of this is that for any edge $E$, the corresponding dihedral angle $Lambda$ satisfies $sin(Lambda)/|E| =$ constant, where the constant is independent of which edge we choose. (To prove this, note that the height of the tetrahedron is $hsin(Lambda)$, where $h$ is an altitude of a face $F$ and $Lambda$ is the dihedral angle between that face and the base of the tetrahedron, and then note that $h=A/|E|$, where $A$ is the area of any of the faces and $E$ is the edge joining face $F$ and the base.)

Choose coordinates so that $D$ is the origin and $B$ is at $(1,0,0)$. Then since $(ABD)$ and $(BCD)$ have dihedral angle $90^{circ}$, we can put $A$ somewhere in the $xy$ plane and $C$ somewhere in the $xz$ plane.

Since $(DBA)$ and $(DBC)$ are congruent triangles (modulo inversion), $A$ and $C$ have coordinates $(u,v,0)$ and $(1-u,0,v)$, respectively, for some $u$ and $v$. Since the length of $AC$ equals the length of $DB$, which is 1, we have $(1-2u)^2 + 2v^2 = 1$.

By the first remark, we have $sin(60^circ) /sqrt{(1-u)^2 + v^2} = sin(90^circ) / 1 = 1$ from the edges $DC$ and $DB$, and we similarly have $sin(Lambda) /sqrt{u^2+v^2} = 1$ from edge $DA$, where $Lambda$ is the angle we want to know. The first equation implies $((1-u)^2 + v^2) = frac{3}{4}$. Combining with the above equation $(1-2u)^2 + 2v^2 = 1$ we find $1-2u^2 = frac{1}{2}$, so that $u=frac{1}{2}$ and thus $v=frac{1}{sqrt{2}}$. Then $sin(Lambda) = sqrt{frac{3}{4}}$, so $$Lambda = sin^{-1}left(sqrt{frac{3}{4}}right) = 60^circ$$

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answered Mar 18 at 22:39

YlyYly

6,85921539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, very much!
  $endgroup$
  – Adele
  Mar 24 at 15:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Adele If you found this answer helpful, you can upvote it and/or accept it.
  $endgroup$
  – Yly
  Mar 24 at 17:44
  

  
  
  
  

add a comment | 

0

$begingroup$

You can set up a coordinate system such that:
$$
D=(0,0,0),quad B=(t,0,0),quad A=(t-a,0,b),quad C=(a,b,0),
$$
where $t$ must be chosen such that $AC=BD$.
Then it is not difficult to set up normal vectors to faces $BDC$, $ADC$ and $DAB$:
$$
N_{BDC}=(0,0,1),quad N_{DAB}=(0,1,0),quad N_{ADC}=Atimes C=(-b^2,ab,b(t-a)).
$$
From these one can compute the cosine of the angle between planes $ADC$ and $BCD$: equating it to $cos 60°$ will give the ratio $a^2/b^2$. Finally, one can compute the cosine of the angle between planes $ADC$ and $DAB$.

share|cite|improve this answer

answered Mar 18 at 21:33

AretinoAretino

25.8k31545

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add a comment | 

1

$begingroup$

The hypothesis that the tetrahedron has equal opposite edges means that opposite dihedral angles are also equal, and that all faces of the tetrahedron are congruent. A consequence of this is that for any edge $E$, the corresponding dihedral angle $Lambda$ satisfies $sin(Lambda)/|E| =$ constant, where the constant is independent of which edge we choose. (To prove this, note that the height of the tetrahedron is $hsin(Lambda)$, where $h$ is an altitude of a face $F$ and $Lambda$ is the dihedral angle between that face and the base of the tetrahedron, and then note that $h=A/|E|$, where $A$ is the area of any of the faces and $E$ is the edge joining face $F$ and the base.)

Choose coordinates so that $D$ is the origin and $B$ is at $(1,0,0)$. Then since $(ABD)$ and $(BCD)$ have dihedral angle $90^{circ}$, we can put $A$ somewhere in the $xy$ plane and $C$ somewhere in the $xz$ plane.

Since $(DBA)$ and $(DBC)$ are congruent triangles (modulo inversion), $A$ and $C$ have coordinates $(u,v,0)$ and $(1-u,0,v)$, respectively, for some $u$ and $v$. Since the length of $AC$ equals the length of $DB$, which is 1, we have $(1-2u)^2 + 2v^2 = 1$.

By the first remark, we have $sin(60^circ) /sqrt{(1-u)^2 + v^2} = sin(90^circ) / 1 = 1$ from the edges $DC$ and $DB$, and we similarly have $sin(Lambda) /sqrt{u^2+v^2} = 1$ from edge $DA$, where $Lambda$ is the angle we want to know. The first equation implies $((1-u)^2 + v^2) = frac{3}{4}$. Combining with the above equation $(1-2u)^2 + 2v^2 = 1$ we find $1-2u^2 = frac{1}{2}$, so that $u=frac{1}{2}$ and thus $v=frac{1}{sqrt{2}}$. Then $sin(Lambda) = sqrt{frac{3}{4}}$, so $$Lambda = sin^{-1}left(sqrt{frac{3}{4}}right) = 60^circ$$

share|cite|improve this answer

answered Mar 18 at 22:39

YlyYly

6,85921539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, very much!
  $endgroup$
  – Adele
  Mar 24 at 15:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Adele If you found this answer helpful, you can upvote it and/or accept it.
  $endgroup$
  – Yly
  Mar 24 at 17:44
  

  
  
  
  

add a comment | 

1

$begingroup$

The hypothesis that the tetrahedron has equal opposite edges means that opposite dihedral angles are also equal, and that all faces of the tetrahedron are congruent. A consequence of this is that for any edge $E$, the corresponding dihedral angle $Lambda$ satisfies $sin(Lambda)/|E| =$ constant, where the constant is independent of which edge we choose. (To prove this, note that the height of the tetrahedron is $hsin(Lambda)$, where $h$ is an altitude of a face $F$ and $Lambda$ is the dihedral angle between that face and the base of the tetrahedron, and then note that $h=A/|E|$, where $A$ is the area of any of the faces and $E$ is the edge joining face $F$ and the base.)

Choose coordinates so that $D$ is the origin and $B$ is at $(1,0,0)$. Then since $(ABD)$ and $(BCD)$ have dihedral angle $90^{circ}$, we can put $A$ somewhere in the $xy$ plane and $C$ somewhere in the $xz$ plane.

Since $(DBA)$ and $(DBC)$ are congruent triangles (modulo inversion), $A$ and $C$ have coordinates $(u,v,0)$ and $(1-u,0,v)$, respectively, for some $u$ and $v$. Since the length of $AC$ equals the length of $DB$, which is 1, we have $(1-2u)^2 + 2v^2 = 1$.

By the first remark, we have $sin(60^circ) /sqrt{(1-u)^2 + v^2} = sin(90^circ) / 1 = 1$ from the edges $DC$ and $DB$, and we similarly have $sin(Lambda) /sqrt{u^2+v^2} = 1$ from edge $DA$, where $Lambda$ is the angle we want to know. The first equation implies $((1-u)^2 + v^2) = frac{3}{4}$. Combining with the above equation $(1-2u)^2 + 2v^2 = 1$ we find $1-2u^2 = frac{1}{2}$, so that $u=frac{1}{2}$ and thus $v=frac{1}{sqrt{2}}$. Then $sin(Lambda) = sqrt{frac{3}{4}}$, so $$Lambda = sin^{-1}left(sqrt{frac{3}{4}}right) = 60^circ$$

share|cite|improve this answer

answered Mar 18 at 22:39

YlyYly

6,85921539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, very much!
  $endgroup$
  – Adele
  Mar 24 at 15:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Adele If you found this answer helpful, you can upvote it and/or accept it.
  $endgroup$
  – Yly
  Mar 24 at 17:44
  

  
  
  
  

add a comment | 

$begingroup$

The hypothesis that the tetrahedron has equal opposite edges means that opposite dihedral angles are also equal, and that all faces of the tetrahedron are congruent. A consequence of this is that for any edge $E$, the corresponding dihedral angle $Lambda$ satisfies $sin(Lambda)/|E| =$ constant, where the constant is independent of which edge we choose. (To prove this, note that the height of the tetrahedron is $hsin(Lambda)$, where $h$ is an altitude of a face $F$ and $Lambda$ is the dihedral angle between that face and the base of the tetrahedron, and then note that $h=A/|E|$, where $A$ is the area of any of the faces and $E$ is the edge joining face $F$ and the base.)

Choose coordinates so that $D$ is the origin and $B$ is at $(1,0,0)$. Then since $(ABD)$ and $(BCD)$ have dihedral angle $90^{circ}$, we can put $A$ somewhere in the $xy$ plane and $C$ somewhere in the $xz$ plane.

Since $(DBA)$ and $(DBC)$ are congruent triangles (modulo inversion), $A$ and $C$ have coordinates $(u,v,0)$ and $(1-u,0,v)$, respectively, for some $u$ and $v$. Since the length of $AC$ equals the length of $DB$, which is 1, we have $(1-2u)^2 + 2v^2 = 1$.

By the first remark, we have $sin(60^circ) /sqrt{(1-u)^2 + v^2} = sin(90^circ) / 1 = 1$ from the edges $DC$ and $DB$, and we similarly have $sin(Lambda) /sqrt{u^2+v^2} = 1$ from edge $DA$, where $Lambda$ is the angle we want to know. The first equation implies $((1-u)^2 + v^2) = frac{3}{4}$. Combining with the above equation $(1-2u)^2 + 2v^2 = 1$ we find $1-2u^2 = frac{1}{2}$, so that $u=frac{1}{2}$ and thus $v=frac{1}{sqrt{2}}$. Then $sin(Lambda) = sqrt{frac{3}{4}}$, so $$Lambda = sin^{-1}left(sqrt{frac{3}{4}}right) = 60^circ$$

share|cite|improve this answer

answered Mar 18 at 22:39

YlyYly

6,85921539

$endgroup$

share|cite|improve this answer

answered Mar 18 at 22:39

YlyYly

6,85921539

answered Mar 18 at 22:39

YlyYly

6,85921539

answered Mar 18 at 22:39

YlyYly

6,85921539

0

$begingroup$

You can set up a coordinate system such that:
$$
D=(0,0,0),quad B=(t,0,0),quad A=(t-a,0,b),quad C=(a,b,0),
$$
where $t$ must be chosen such that $AC=BD$.
Then it is not difficult to set up normal vectors to faces $BDC$, $ADC$ and $DAB$:
$$
N_{BDC}=(0,0,1),quad N_{DAB}=(0,1,0),quad N_{ADC}=Atimes C=(-b^2,ab,b(t-a)).
$$
From these one can compute the cosine of the angle between planes $ADC$ and $BCD$: equating it to $cos 60°$ will give the ratio $a^2/b^2$. Finally, one can compute the cosine of the angle between planes $ADC$ and $DAB$.

share|cite|improve this answer

answered Mar 18 at 21:33

AretinoAretino

25.8k31545

$endgroup$

add a comment | 

0

$begingroup$

You can set up a coordinate system such that:
$$
D=(0,0,0),quad B=(t,0,0),quad A=(t-a,0,b),quad C=(a,b,0),
$$
where $t$ must be chosen such that $AC=BD$.
Then it is not difficult to set up normal vectors to faces $BDC$, $ADC$ and $DAB$:
$$
N_{BDC}=(0,0,1),quad N_{DAB}=(0,1,0),quad N_{ADC}=Atimes C=(-b^2,ab,b(t-a)).
$$
From these one can compute the cosine of the angle between planes $ADC$ and $BCD$: equating it to $cos 60°$ will give the ratio $a^2/b^2$. Finally, one can compute the cosine of the angle between planes $ADC$ and $DAB$.

share|cite|improve this answer

answered Mar 18 at 21:33

AretinoAretino

25.8k31545

$endgroup$

add a comment | 

$begingroup$

You can set up a coordinate system such that:
$$
D=(0,0,0),quad B=(t,0,0),quad A=(t-a,0,b),quad C=(a,b,0),
$$
where $t$ must be chosen such that $AC=BD$.
Then it is not difficult to set up normal vectors to faces $BDC$, $ADC$ and $DAB$:
$$
N_{BDC}=(0,0,1),quad N_{DAB}=(0,1,0),quad N_{ADC}=Atimes C=(-b^2,ab,b(t-a)).
$$
From these one can compute the cosine of the angle between planes $ADC$ and $BCD$: equating it to $cos 60°$ will give the ratio $a^2/b^2$. Finally, one can compute the cosine of the angle between planes $ADC$ and $DAB$.

share|cite|improve this answer

answered Mar 18 at 21:33

AretinoAretino

25.8k31545

$endgroup$

share|cite|improve this answer

answered Mar 18 at 21:33

AretinoAretino

25.8k31545

answered Mar 18 at 21:33

AretinoAretino

25.8k31545

answered Mar 18 at 21:33

AretinoAretino

25.8k31545