Correlation between parallel lines and volumes of tetrahedrons.
$begingroup$
Given 4 parallel lines $d_1$, $d_2$, $d_3$, $d_4$ , no more than 2 of which can be on a same plane. Plane (P) intersects the 4 lines at 4 points A, B, C, D. Plane (Q) ( not identical to plane (P)) intersects the 4 lines at $A_1$, $B_1$, $C_1$, $D_1$. Proof that the volumes of the 2 tetrahedra ABC
$D_1$ and $A_1$$B_1$$C_1$$D$ are equal.
euclidean-geometry
asked Dec 9 '18 at 6:20
Nguyen Xuan MinhNguyen Xuan Minh
62
$endgroup$
add a comment |
$begingroup$
Given 4 parallel lines $d_1$, $d_2$, $d_3$, $d_4$ , no more than 2 of which can be on a same plane. Plane (P) intersects the 4 lines at 4 points A, B, C, D. Plane (Q) ( not identical to plane (P)) intersects the 4 lines at $A_1$, $B_1$, $C_1$, $D_1$. Proof that the volumes of the 2 tetrahedra ABC
$D_1$ and $A_1$$B_1$$C_1$$D$ are equal.
euclidean-geometry
asked Dec 9 '18 at 6:20
Nguyen Xuan MinhNguyen Xuan Minh
62
$endgroup$
$begingroup$
$ABCD$ and $A_1B_1C_1D_1$ are not tetrahedra, they are quadrilaterals.
$endgroup$
– Shubham Johri
Dec 9 '18 at 6:24
$begingroup$
Made a mistake, It's fixed now, thank you
$endgroup$
– Nguyen Xuan Minh
Dec 9 '18 at 6:32
$begingroup$
Counter-example, take $d_1, d_2, d_3, d_4$ be the lines joining origin and $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$, $D = (frac13,frac13,frac13)$ respectively. Let $P, Q$ be the planes $x+y+z = 1$ and $x+y+z = 2$. It is clear $A, B, C, D$ are the intersections of lines $d_i$ with $P$ and $A_1 = 2A$, $B_1 = 2B$, $C_1 = 2C$, $D_1 = 2D$ are the intersections with $Q$. However, $${rm Vol}(A_1B_1C_1D) = 4 {rm Vol}(ABCD_1) ne {rm Vol}(ABCD_1)$$
$endgroup$
– achille hui
Dec 12 '18 at 5:10
$begingroup$
I think you need conditions such as that $A$ and $A_1$ are on $d_1,$ $B$ and $B_1$ are on $d_2,$ etc. Inserting the word "respectively" in two places would do this. Otherwise it is still easy to come up with counterexamples.
$endgroup$
– David K
Dec 14 '18 at 14:20
add a comment |
$begingroup$
Given 4 parallel lines $d_1$, $d_2$, $d_3$, $d_4$ , no more than 2 of which can be on a same plane. Plane (P) intersects the 4 lines at 4 points A, B, C, D. Plane (Q) ( not identical to plane (P)) intersects the 4 lines at $A_1$, $B_1$, $C_1$, $D_1$. Proof that the volumes of the 2 tetrahedra ABC
$D_1$ and $A_1$$B_1$$C_1$$D$ are equal.
euclidean-geometry
asked Dec 9 '18 at 6:20
Nguyen Xuan MinhNguyen Xuan Minh
62
$endgroup$
Given 4 parallel lines $d_1$, $d_2$, $d_3$, $d_4$ , no more than 2 of which can be on a same plane. Plane (P) intersects the 4 lines at 4 points A, B, C, D. Plane (Q) ( not identical to plane (P)) intersects the 4 lines at $A_1$, $B_1$, $C_1$, $D_1$. Proof that the volumes of the 2 tetrahedra ABC
$D_1$ and $A_1$$B_1$$C_1$$D$ are equal.
euclidean-geometry
euclidean-geometry
asked Dec 9 '18 at 6:20
Nguyen Xuan MinhNguyen Xuan Minh
62
asked Dec 9 '18 at 6:20
Nguyen Xuan MinhNguyen Xuan Minh
62
edited Dec 14 '18 at 12:44
Nguyen Xuan Minh
asked Dec 9 '18 at 6:20
Nguyen Xuan MinhNguyen Xuan Minh
62
asked Dec 9 '18 at 6:20
Nguyen Xuan MinhNguyen Xuan Minh
62
asked Dec 9 '18 at 6:20
Nguyen Xuan MinhNguyen Xuan Minh
62
62
$begingroup$
$ABCD$ and $A_1B_1C_1D_1$ are not tetrahedra, they are quadrilaterals.
$endgroup$
– Shubham Johri
Dec 9 '18 at 6:24
$begingroup$
Made a mistake, It's fixed now, thank you
$endgroup$
– Nguyen Xuan Minh
Dec 9 '18 at 6:32
$begingroup$
Counter-example, take $d_1, d_2, d_3, d_4$ be the lines joining origin and $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$, $D = (frac13,frac13,frac13)$ respectively. Let $P, Q$ be the planes $x+y+z = 1$ and $x+y+z = 2$. It is clear $A, B, C, D$ are the intersections of lines $d_i$ with $P$ and $A_1 = 2A$, $B_1 = 2B$, $C_1 = 2C$, $D_1 = 2D$ are the intersections with $Q$. However, $${rm Vol}(A_1B_1C_1D) = 4 {rm Vol}(ABCD_1) ne {rm Vol}(ABCD_1)$$
$endgroup$
– achille hui
Dec 12 '18 at 5:10
$begingroup$
I think you need conditions such as that $A$ and $A_1$ are on $d_1,$ $B$ and $B_1$ are on $d_2,$ etc. Inserting the word "respectively" in two places would do this. Otherwise it is still easy to come up with counterexamples.
$endgroup$
– David K
Dec 14 '18 at 14:20
add a comment |
$begingroup$
$ABCD$ and $A_1B_1C_1D_1$ are not tetrahedra, they are quadrilaterals.
$endgroup$
– Shubham Johri
Dec 9 '18 at 6:24
$begingroup$
Made a mistake, It's fixed now, thank you
$endgroup$
– Nguyen Xuan Minh
Dec 9 '18 at 6:32
$begingroup$
Counter-example, take $d_1, d_2, d_3, d_4$ be the lines joining origin and $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$, $D = (frac13,frac13,frac13)$ respectively. Let $P, Q$ be the planes $x+y+z = 1$ and $x+y+z = 2$. It is clear $A, B, C, D$ are the intersections of lines $d_i$ with $P$ and $A_1 = 2A$, $B_1 = 2B$, $C_1 = 2C$, $D_1 = 2D$ are the intersections with $Q$. However, $${rm Vol}(A_1B_1C_1D) = 4 {rm Vol}(ABCD_1) ne {rm Vol}(ABCD_1)$$
$endgroup$
– achille hui
Dec 12 '18 at 5:10
$begingroup$
I think you need conditions such as that $A$ and $A_1$ are on $d_1,$ $B$ and $B_1$ are on $d_2,$ etc. Inserting the word "respectively" in two places would do this. Otherwise it is still easy to come up with counterexamples.
$endgroup$
– David K
Dec 14 '18 at 14:20
$begingroup$
$ABCD$ and $A_1B_1C_1D_1$ are not tetrahedra, they are quadrilaterals.
$endgroup$
– Shubham Johri
Dec 9 '18 at 6:24
$begingroup$
$ABCD$ and $A_1B_1C_1D_1$ are not tetrahedra, they are quadrilaterals.
$endgroup$
– Shubham Johri
Dec 9 '18 at 6:24
$begingroup$
Made a mistake, It's fixed now, thank you
$endgroup$
– Nguyen Xuan Minh
Dec 9 '18 at 6:32
$begingroup$
Made a mistake, It's fixed now, thank you
$endgroup$
– Nguyen Xuan Minh
Dec 9 '18 at 6:32
$begingroup$
Counter-example, take $d_1, d_2, d_3, d_4$ be the lines joining origin and $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$, $D = (frac13,frac13,frac13)$ respectively. Let $P, Q$ be the planes $x+y+z = 1$ and $x+y+z = 2$. It is clear $A, B, C, D$ are the intersections of lines $d_i$ with $P$ and $A_1 = 2A$, $B_1 = 2B$, $C_1 = 2C$, $D_1 = 2D$ are the intersections with $Q$. However, $${rm Vol}(A_1B_1C_1D) = 4 {rm Vol}(ABCD_1) ne {rm Vol}(ABCD_1)$$
$endgroup$
– achille hui
Dec 12 '18 at 5:10
$begingroup$
Counter-example, take $d_1, d_2, d_3, d_4$ be the lines joining origin and $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$, $D = (frac13,frac13,frac13)$ respectively. Let $P, Q$ be the planes $x+y+z = 1$ and $x+y+z = 2$. It is clear $A, B, C, D$ are the intersections of lines $d_i$ with $P$ and $A_1 = 2A$, $B_1 = 2B$, $C_1 = 2C$, $D_1 = 2D$ are the intersections with $Q$. However, $${rm Vol}(A_1B_1C_1D) = 4 {rm Vol}(ABCD_1) ne {rm Vol}(ABCD_1)$$
$endgroup$
– achille hui
Dec 12 '18 at 5:10
$begingroup$
I think you need conditions such as that $A$ and $A_1$ are on $d_1,$ $B$ and $B_1$ are on $d_2,$ etc. Inserting the word "respectively" in two places would do this. Otherwise it is still easy to come up with counterexamples.
$endgroup$
– David K
Dec 14 '18 at 14:20
$begingroup$
I think you need conditions such as that $A$ and $A_1$ are on $d_1,$ $B$ and $B_1$ are on $d_2,$ etc. Inserting the word "respectively" in two places would do this. Otherwise it is still easy to come up with counterexamples.
$endgroup$
– David K
Dec 14 '18 at 14:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $(P)||(Q).$
Thus, we need to prove that areas of $Delta ABC$ and $Delta A_1B_1C_1$ they are equal,
which is obviously wrong: move $(Q)$ such that $(Q)||(P).$
answered Dec 9 '18 at 7:16
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
1
$begingroup$
Can someone, which down-voted, explain me why did you do it?
$endgroup$
– Michael Rozenberg
Dec 11 '18 at 5:32
add a comment |
$begingroup$
It is not explicitly stated that one of the given four parallel lines passes through both $D$ and $D_1,$ but I will assume this is given since otherwise
the two tetrahedra might have different volumes.
If you know that a shear transformation of three-dimensional Euclidean space preserves volume, you can construct a plane $P'$ perpendicular to $d_1$
(hence also perpendicular to the other three given lines)
and apply a shear transformation parallel to $d_1$ that maps $P$ to $P'.$
This transformation maps $A,$ $B,$ and $C$ to points $A',$ $B',$ and $C'$
in plane $P'$, and the tetrahedron $ABCD_1$ has volume equal to $A'B'C'D_1,$
which has base $triangle A'B'C'$ and height $DD_1.$
Likewise you can construct a plane $Q'$ perpendicular to $d_1$ and apply a shear transformation parallel to $d_1$ that maps $Q$ to $Q',$
and in particular maps $A_1,$ $B_1,$ and $C_1$ to $A_1',$ $B_1',$ and $C_1'.$
Then the tetrahedron $A_1B_1C_1D$ has volume equal to $A_1'B_1'C_1'D,$
which has base $triangle A_1'B_1'C_1'$ and height $DD_1.$
But since $triangle A_1'B_1'C_1'$ is congruent to $triangle A'B'C',$
the tetrahedron $triangle A_1'B_1'C_1'D$ has the same volume as $A'B'C'D_1,$
and hence all four tetrahedra have the same volume.
answered Dec 15 '18 at 4:00
David KDavid K
54.1k342116
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $(P)||(Q).$
Thus, we need to prove that areas of $Delta ABC$ and $Delta A_1B_1C_1$ they are equal,
which is obviously wrong: move $(Q)$ such that $(Q)||(P).$
answered Dec 9 '18 at 7:16
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
1
$begingroup$
Can someone, which down-voted, explain me why did you do it?
$endgroup$
– Michael Rozenberg
Dec 11 '18 at 5:32
add a comment |
$begingroup$
Let $(P)||(Q).$
Thus, we need to prove that areas of $Delta ABC$ and $Delta A_1B_1C_1$ they are equal,
which is obviously wrong: move $(Q)$ such that $(Q)||(P).$
answered Dec 9 '18 at 7:16
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
1
$begingroup$
Can someone, which down-voted, explain me why did you do it?
$endgroup$
– Michael Rozenberg
Dec 11 '18 at 5:32
add a comment |
$begingroup$
Let $(P)||(Q).$
Thus, we need to prove that areas of $Delta ABC$ and $Delta A_1B_1C_1$ they are equal,
which is obviously wrong: move $(Q)$ such that $(Q)||(P).$
answered Dec 9 '18 at 7:16
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
Let $(P)||(Q).$
Thus, we need to prove that areas of $Delta ABC$ and $Delta A_1B_1C_1$ they are equal,
which is obviously wrong: move $(Q)$ such that $(Q)||(P).$
answered Dec 9 '18 at 7:16
Michael RozenbergMichael Rozenberg
103k1891195
answered Dec 9 '18 at 7:16
Michael RozenbergMichael Rozenberg
103k1891195
answered Dec 9 '18 at 7:16
Michael RozenbergMichael Rozenberg
103k1891195
answered Dec 9 '18 at 7:16
Michael RozenbergMichael Rozenberg
103k1891195
103k1891195
1
$begingroup$
Can someone, which down-voted, explain me why did you do it?
$endgroup$
– Michael Rozenberg
Dec 11 '18 at 5:32
add a comment |
1
$begingroup$
Can someone, which down-voted, explain me why did you do it?
$endgroup$
– Michael Rozenberg
Dec 11 '18 at 5:32
1
1
$begingroup$
Can someone, which down-voted, explain me why did you do it?
$endgroup$
– Michael Rozenberg
Dec 11 '18 at 5:32
$begingroup$
Can someone, which down-voted, explain me why did you do it?
$endgroup$
– Michael Rozenberg
Dec 11 '18 at 5:32
add a comment |
$begingroup$
It is not explicitly stated that one of the given four parallel lines passes through both $D$ and $D_1,$ but I will assume this is given since otherwise
the two tetrahedra might have different volumes.
If you know that a shear transformation of three-dimensional Euclidean space preserves volume, you can construct a plane $P'$ perpendicular to $d_1$
(hence also perpendicular to the other three given lines)
and apply a shear transformation parallel to $d_1$ that maps $P$ to $P'.$
This transformation maps $A,$ $B,$ and $C$ to points $A',$ $B',$ and $C'$
in plane $P'$, and the tetrahedron $ABCD_1$ has volume equal to $A'B'C'D_1,$
which has base $triangle A'B'C'$ and height $DD_1.$
Likewise you can construct a plane $Q'$ perpendicular to $d_1$ and apply a shear transformation parallel to $d_1$ that maps $Q$ to $Q',$
and in particular maps $A_1,$ $B_1,$ and $C_1$ to $A_1',$ $B_1',$ and $C_1'.$
Then the tetrahedron $A_1B_1C_1D$ has volume equal to $A_1'B_1'C_1'D,$
which has base $triangle A_1'B_1'C_1'$ and height $DD_1.$
But since $triangle A_1'B_1'C_1'$ is congruent to $triangle A'B'C',$
the tetrahedron $triangle A_1'B_1'C_1'D$ has the same volume as $A'B'C'D_1,$
and hence all four tetrahedra have the same volume.
answered Dec 15 '18 at 4:00
David KDavid K
54.1k342116
$endgroup$
add a comment |
$begingroup$
It is not explicitly stated that one of the given four parallel lines passes through both $D$ and $D_1,$ but I will assume this is given since otherwise
the two tetrahedra might have different volumes.
If you know that a shear transformation of three-dimensional Euclidean space preserves volume, you can construct a plane $P'$ perpendicular to $d_1$
(hence also perpendicular to the other three given lines)
and apply a shear transformation parallel to $d_1$ that maps $P$ to $P'.$
This transformation maps $A,$ $B,$ and $C$ to points $A',$ $B',$ and $C'$
in plane $P'$, and the tetrahedron $ABCD_1$ has volume equal to $A'B'C'D_1,$
which has base $triangle A'B'C'$ and height $DD_1.$
Likewise you can construct a plane $Q'$ perpendicular to $d_1$ and apply a shear transformation parallel to $d_1$ that maps $Q$ to $Q',$
and in particular maps $A_1,$ $B_1,$ and $C_1$ to $A_1',$ $B_1',$ and $C_1'.$
Then the tetrahedron $A_1B_1C_1D$ has volume equal to $A_1'B_1'C_1'D,$
which has base $triangle A_1'B_1'C_1'$ and height $DD_1.$
But since $triangle A_1'B_1'C_1'$ is congruent to $triangle A'B'C',$
the tetrahedron $triangle A_1'B_1'C_1'D$ has the same volume as $A'B'C'D_1,$
and hence all four tetrahedra have the same volume.
answered Dec 15 '18 at 4:00
David KDavid K
54.1k342116
$endgroup$
add a comment |
$begingroup$
It is not explicitly stated that one of the given four parallel lines passes through both $D$ and $D_1,$ but I will assume this is given since otherwise
the two tetrahedra might have different volumes.
If you know that a shear transformation of three-dimensional Euclidean space preserves volume, you can construct a plane $P'$ perpendicular to $d_1$
(hence also perpendicular to the other three given lines)
and apply a shear transformation parallel to $d_1$ that maps $P$ to $P'.$
This transformation maps $A,$ $B,$ and $C$ to points $A',$ $B',$ and $C'$
in plane $P'$, and the tetrahedron $ABCD_1$ has volume equal to $A'B'C'D_1,$
which has base $triangle A'B'C'$ and height $DD_1.$
Likewise you can construct a plane $Q'$ perpendicular to $d_1$ and apply a shear transformation parallel to $d_1$ that maps $Q$ to $Q',$
and in particular maps $A_1,$ $B_1,$ and $C_1$ to $A_1',$ $B_1',$ and $C_1'.$
Then the tetrahedron $A_1B_1C_1D$ has volume equal to $A_1'B_1'C_1'D,$
which has base $triangle A_1'B_1'C_1'$ and height $DD_1.$
But since $triangle A_1'B_1'C_1'$ is congruent to $triangle A'B'C',$
the tetrahedron $triangle A_1'B_1'C_1'D$ has the same volume as $A'B'C'D_1,$
and hence all four tetrahedra have the same volume.
answered Dec 15 '18 at 4:00
David KDavid K
54.1k342116
$endgroup$
It is not explicitly stated that one of the given four parallel lines passes through both $D$ and $D_1,$ but I will assume this is given since otherwise
the two tetrahedra might have different volumes.
If you know that a shear transformation of three-dimensional Euclidean space preserves volume, you can construct a plane $P'$ perpendicular to $d_1$
(hence also perpendicular to the other three given lines)
and apply a shear transformation parallel to $d_1$ that maps $P$ to $P'.$
This transformation maps $A,$ $B,$ and $C$ to points $A',$ $B',$ and $C'$
in plane $P'$, and the tetrahedron $ABCD_1$ has volume equal to $A'B'C'D_1,$
which has base $triangle A'B'C'$ and height $DD_1.$
Likewise you can construct a plane $Q'$ perpendicular to $d_1$ and apply a shear transformation parallel to $d_1$ that maps $Q$ to $Q',$
and in particular maps $A_1,$ $B_1,$ and $C_1$ to $A_1',$ $B_1',$ and $C_1'.$
Then the tetrahedron $A_1B_1C_1D$ has volume equal to $A_1'B_1'C_1'D,$
which has base $triangle A_1'B_1'C_1'$ and height $DD_1.$
But since $triangle A_1'B_1'C_1'$ is congruent to $triangle A'B'C',$
the tetrahedron $triangle A_1'B_1'C_1'D$ has the same volume as $A'B'C'D_1,$
and hence all four tetrahedra have the same volume.
answered Dec 15 '18 at 4:00
David KDavid K
54.1k342116
answered Dec 15 '18 at 4:00
David KDavid K
54.1k342116
answered Dec 15 '18 at 4:00
David KDavid K
54.1k342116
answered Dec 15 '18 at 4:00
David KDavid K
54.1k342116
54.1k342116
add a comment |
add a comment |
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$begingroup$
$ABCD$ and $A_1B_1C_1D_1$ are not tetrahedra, they are quadrilaterals.
$endgroup$
– Shubham Johri
Dec 9 '18 at 6:24
$begingroup$
Made a mistake, It's fixed now, thank you
$endgroup$
– Nguyen Xuan Minh
Dec 9 '18 at 6:32
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Counter-example, take $d_1, d_2, d_3, d_4$ be the lines joining origin and $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$, $D = (frac13,frac13,frac13)$ respectively. Let $P, Q$ be the planes $x+y+z = 1$ and $x+y+z = 2$. It is clear $A, B, C, D$ are the intersections of lines $d_i$ with $P$ and $A_1 = 2A$, $B_1 = 2B$, $C_1 = 2C$, $D_1 = 2D$ are the intersections with $Q$. However, $${rm Vol}(A_1B_1C_1D) = 4 {rm Vol}(ABCD_1) ne {rm Vol}(ABCD_1)$$
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– achille hui
Dec 12 '18 at 5:10
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I think you need conditions such as that $A$ and $A_1$ are on $d_1,$ $B$ and $B_1$ are on $d_2,$ etc. Inserting the word "respectively" in two places would do this. Otherwise it is still easy to come up with counterexamples.
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– David K
Dec 14 '18 at 14:20