[math-fun] Tetrahedral altitudes
Richard's paper mentions mentions that the 4 altitude lines of a tetrahedron, through vertices and perpendicular to faces, do not necessarily concur in a common point. A long time ago I came across a charmingly antiquated tome: Nathan Altshiller-Court "Modern pure solid geometry" (1935) which I seem to remember stated the theorem that the altitudes always lie on a quadric; though his notation --- undefined --- for this property was something confusingly obscure. There were several other theorems of a similar nature. Can anybody confirm this? Has any further work in this direction taken place more recently? WFL On 10/29/11, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
... I've just submitted a paper, ``Five-point circles, the 76-point sphere, and the Pavillet tetrahedron'' to the Monthly. The theme is that triangle geometry is not dead. I would attach a copy if attachments were allowed; I will entertain individual requests. ...
On Sat, Oct 29, 2011 at 5:27 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
Richard's paper mentions mentions that the 4 altitude lines of a tetrahedron, through vertices and perpendicular to faces, do not necessarily concur in a common point.
I seem to recall that if one pair of altitudes is intersecting, then the other pair is as well. Does anyone know an easy proof or counterexample? --Joshua Zucker
This would follow almost immediately from the theorem I quoted: if two altitudes meet, the quadric must degenerate to a pair of planes; then either the other two altitudes lie on the other plane, or else three are coplanar and the tetrahedron degenerates to a prism. WFL On 10/30/11, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On Sat, Oct 29, 2011 at 5:27 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
Richard's paper mentions mentions that the 4 altitude lines of a tetrahedron, through vertices and perpendicular to faces, do not necessarily concur in a common point.
I seem to recall that if one pair of altitudes is intersecting, then the other pair is as well. Does anyone know an easy proof or counterexample?
--Joshua Zucker
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The argument below was incomplete: another possibility is that the quadric degenerates to a point, where all 4 altitudes meet: such tetrahedra are called "orthocentric" in Richard's paper. WFL On 10/30/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
This would follow almost immediately from the theorem I quoted: if two altitudes meet, the quadric must degenerate to a pair of planes; then either the other two altitudes lie on the other plane, or else three are coplanar and the tetrahedron degenerates to a prism. WFL
On 10/30/11, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On Sat, Oct 29, 2011 at 5:27 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
Richard's paper mentions mentions that the 4 altitude lines of a tetrahedron, through vertices and perpendicular to faces, do not necessarily concur in a common point.
I seem to recall that if one pair of altitudes is intersecting, then the other pair is as well. Does anyone know an easy proof or counterexample?
--Joshua Zucker
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 10/30/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
... A long time ago I came across a charmingly antiquated tome: Nathan Altshiller-Court "Modern pure solid geometry" (1935) which I seem to remember stated the theorem that the altitudes always lie on a quadric; though his notation --- undefined --- for this property was something confusingly obscure. There were several other theorems of a similar nature.
Can anybody confirm this? Has any further work in this direction taken place more recently?
WFL
Using Cl(3,0,1) geometric algebra (DCQ's), Court's theorem that the 4 altitudes lie on a quadric is reduced brutally to triviality. Let the equations of the tetrahedral face planes be represented by generic vectors E = [Eo,Ex,Ey,Ez], F, G, H; the vertices have covectors P = <F G H>_3, Q, R, S, where the product is Clifford; the altitudes have Pluecker bivectors K = <E P>_2, L, M, N, their components quadric polynomials in the faces' components. It is now a matter of computation to verify that the 4x6 matrix of the latter has rank 3, showing that the altitudes are linearly dependent, and therefore lie on a quadric. QED Fred Lunnon
A tetrahedron has freedom 12, since each its 4 vertex points has freedom 3; but 4 lines on a quadric regulus have freedom 4 + 9 = 13. Therefore there must be a single further constraint on the 4 altitudes, besides their lying on a quadric. What could this be? Regarding the altitudes as 4 points on the Grassman quadric (4 quadratic constraints) in projective 5-space, the quadric constraint is equivalent to their also lying on a plane (3 further linear constraints). The missing constraint must surely be (skew-)symmetric in the altitudes, and rational involving quadratic or quartic polynomial; it must also be projectively (semi-)invariant, at any rate with respect to 3-space. One (slightly asymmetric) possibility might have been the "cross-ratio" of lines K,L,M,N defined by (K^L)(M^N) / (K^M)(L^N), where "^" denotes the Grassman wedge inner product: however this turns out to be the squared ratio of two polynomials, each quartic in the faces' components. Anyone got any more suggestions? Fred Lunnon On 10/30/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 10/30/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
... A long time ago I came across a charmingly antiquated tome: Nathan Altshiller-Court "Modern pure solid geometry" (1935) which I seem to remember stated the theorem that the altitudes always lie on a quadric; though his notation --- undefined --- for this property was something confusingly obscure. There were several other theorems of a similar nature.
Can anybody confirm this? Has any further work in this direction taken place more recently?
WFL
Using Cl(3,0,1) geometric algebra (DCQ's), Court's theorem that the 4 altitudes lie on a quadric is reduced brutally to triviality.
Let the equations of the tetrahedral face planes be represented by generic vectors E = [Eo,Ex,Ey,Ez], F, G, H; the vertices have covectors P = <F G H>_3, Q, R, S, where the product is Clifford; the altitudes have Pluecker bivectors K = <E P>_2, L, M, N, their components quadric polynomials in the faces' components.
It is now a matter of computation to verify that the 4x6 matrix of the latter has rank 3, showing that the altitudes are linearly dependent, and therefore lie on a quadric. QED
Fred Lunnon
A tetrahedron has freedom 12, since each its 4 vertex points has freedom 3; but 4 lines on a quadric regulus have freedom 4 + 9 = 13. Therefore there must be a single further constraint on the 4 altitudes, besides their lying on a quadric. What could this be? Regarding the altitudes as 4 points on the Grassman quadric (4 quadratic constraints) in projective 5-space, the quadric constraint is equivalent to their also lying on a plane (3 further linear constraints). The missing constraint must surely be (skew-)symmetric in the altitudes, and rational involving quadratic or quartic polynomial; it must also be projectively (semi-)invariant, at any rate with respect to 3-space. One (slightly asymmetric) possibility might have been the "cross-ratio" of lines K,L,M,N defined by (K^L)(M^N) / (K^M)(L^N), where "^" denotes the Grassman wedge inner product: however this turns out to be the squared ratio of two polynomials, each quartic in the faces' components. Anyone got any more suggestions? Fred Lunnon On 10/30/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 10/30/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
... A long time ago I came across a charmingly antiquated tome: Nathan Altshiller-Court "Modern pure solid geometry" (1935) which I seem to remember stated the theorem that the altitudes always lie on a quadric; though his notation --- undefined --- for this property was something confusingly obscure. There were several other theorems of a similar nature.
Can anybody confirm this? Has any further work in this direction taken place more recently?
WFL
Using Cl(3,0,1) geometric algebra (DCQ's), Court's theorem that the 4 altitudes lie on a quadric is reduced brutally to triviality.
Let the equations of the tetrahedral face planes be represented by generic vectors E = [Eo,Ex,Ey,Ez], F, G, H; the vertices have covectors P = <F G H>_3, Q, R, S, where the product is Clifford; the altitudes have Pluecker bivectors K = <E P>_2, L, M, N, their components quadric polynomials in the faces' components.
It is now a matter of computation to verify that the 4x6 matrix of the latter has rank 3, showing that the altitudes are linearly dependent, and therefore lie on a quadric. QED
Fred Lunnon
On 10/31/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
A tetrahedron has freedom 12, since each its 4 vertex points has freedom 3; but 4 lines on a quadric regulus have freedom 4 + 9 = 13.
Therefore there must be a single further constraint on the 4 altitudes, besides their lying on a quadric. What could this be?
Regarding the altitudes as 4 points on the Grassman quadric (4 quadratic constraints) in projective 5-space, the quadric constraint is equivalent to their also lying on a plane (3 further linear constraints).
The missing constraint must surely be (skew-)symmetric in the altitudes, and rational involving quadratic or quartic polynomial; it must also be projectively (semi-)invariant, at any rate with respect to 3-space.
One (slightly asymmetric) possibility might have been the "cross-ratio" of lines K,L,M,N defined by (K^L)(M^N) / (K^M)(L^N), where "^" denotes the Grassman wedge inner product: however this turns out to be the squared ratio of two polynomials, each quartic in the faces' components.
Anyone got any more suggestions?
Fred Lunnon
On 10/30/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 10/30/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
... A long time ago I came across a charmingly antiquated tome: Nathan Altshiller-Court "Modern pure solid geometry" (1935) which I seem to remember stated the theorem that the altitudes always lie on a quadric; though his notation --- undefined --- for this property was something confusingly obscure. There were several other theorems of a similar nature.
Can anybody confirm this? Has any further work in this direction taken place more recently?
WFL
Using Cl(3,0,1) geometric algebra (DCQ's), Court's theorem that the 4 altitudes lie on a quadric is reduced brutally to triviality.
Let the equations of the tetrahedral face planes be represented by generic vectors E = [Eo,Ex,Ey,Ez], F, G, H; the vertices have covectors P = <F G H>_3, Q, R, S, where the product is Clifford; the altitudes have Pluecker bivectors K = <E P>_2, L, M, N, their components quadric polynomials in the faces' components.
It is now a matter of computation to verify that the 4x6 matrix of the latter has rank 3, showing that the altitudes are linearly dependent, and therefore lie on a quadric. QED
Fred Lunnon
Apologies for earlier multiple postings, resulting from Gmail server hiccoughs! I've just noticed the following recent observation by Richard Guy --- "In 1827 Jacob Steiner noted that the altitudes ofa tetrahedron are the generators of an equilateral hyperboloid." I haven't yet discovered exactly what "equilateral" means in this context --- possibly a hyperbolic paraboloid? --- but it looks as though Steiner knew the answer to the problem below a while back. WFL On 10/31/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
A tetrahedron has freedom 12, since each its 4 vertex points has freedom 3; but 4 lines on a quadric regulus have freedom 4 + 9 = 13.
Therefore there must be a single further constraint on the 4 altitudes, besides their lying on a quadric. What could this be?
Regarding the altitudes as 4 points on the Grassman quadric (4 quadratic constraints) in projective 5-space, the quadric constraint is equivalent to their also lying on a plane (3 further linear constraints).
The missing constraint must surely be (skew-)symmetric in the altitudes, and rational involving quadratic or quartic polynomial; it must also be projectively (semi-)invariant, at any rate with respect to 3-space.
One (slightly asymmetric) possibility might have been the "cross-ratio" of lines K,L,M,N defined by (K^L)(M^N) / (K^M)(L^N), where "^" denotes the Grassman wedge inner product: however this turns out to be the squared ratio of two polynomials, each quartic in the faces' components.
Anyone got any more suggestions?
Fred Lunnon
Fred, One source said it's got by rotating a rectangular hyperbola about its asymptotes, but they meant axes? Definitely not a hyperbolic paraboloid. Rotate a rect hyp about its `minor' (imaginary?) axis, giving a hyperboloid of one sheet. Asymptotic cone would have a vertical semi-angle pi/4. Does that use up all your degrees of freedom? I believe that the centre of the quadric is the Monge point of the tetrahedron. R. On Mon, 31 Oct 2011, Fred lunnon wrote:
Apologies for earlier multiple postings, resulting from Gmail server hiccoughs!
I've just noticed the following recent observation by Richard Guy --- "In 1827 Jacob Steiner noted that the altitudes ofa tetrahedron are the generators of an equilateral hyperboloid."
I haven't yet discovered exactly what "equilateral" means in this context --- possibly a hyperbolic paraboloid? --- but it looks as though Steiner knew the answer to the problem below a while back. WFL
On 10/31/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
A tetrahedron has freedom 12, since each its 4 vertex points has freedom 3; but 4 lines on a quadric regulus have freedom 4 + 9 = 13.
Therefore there must be a single further constraint on the 4 altitudes, besides their lying on a quadric. What could this be?
Regarding the altitudes as 4 points on the Grassman quadric (4 quadratic constraints) in projective 5-space, the quadric constraint is equivalent to their also lying on a plane (3 further linear constraints).
The missing constraint must surely be (skew-)symmetric in the altitudes, and rational involving quadratic or quartic polynomial; it must also be projectively (semi-)invariant, at any rate with respect to 3-space.
One (slightly asymmetric) possibility might have been the "cross-ratio" of lines K,L,M,N defined by (K^L)(M^N) / (K^M)(L^N), where "^" denotes the Grassman wedge inner product: however this turns out to be the squared ratio of two polynomials, each quartic in the faces' components.
Anyone got any more suggestions?
Fred Lunnon
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Richard, I had confused myself again: of course the constraint is not a projective invariant, since altitudes are invariant only under similarity! Then "hyperboloid of rotation" becomes the natural interpretation, and the additional fixed conical angle pi/4 also looks plausible. However: freedom of hyperboloid axis line 4, centre point 1, radius 1, (cone angle 1); of 4 altitude lines along regulus 4. So total freedom of altitudes would then be 4+1+1+4 = 10, less by 2 than the tetrahedron's! Anyway, I shall go away to think up a lazy brute-force way to decide the matter algebraically. Fred On 10/31/11, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Fred, One source said it's got by rotating a rectangular hyperbola about its asymptotes, but they meant axes? Definitely not a hyperbolic paraboloid. Rotate a rect hyp about its `minor' (imaginary?) axis, giving a hyperboloid of one sheet. Asymptotic cone would have a vertical semi-angle pi/4. Does that use up all your degrees of freedom? I believe that the centre of the quadric is the Monge point of the tetrahedron. R.
Ho-ho! Suppose we are given (Pluecker coordinates of) altitudes K,L,M,N of a tetrahedron, and want to solve for faces (projective) E,F,G,H. The 3-vector directions of paired altitudes and faces are proportional, so we can immediately set them equal (freedom 8). The 3-vector moments of the altitudes now give 12 linear equations for the "constant" components Eo,Fo,Go,Ho of the faces, with coefficients cubic in their direction components; as polynomials these can be computed to have rank 4, so can in general be solved uniquely [also consistently, since the tetrahedron exists]. Therefore the altitudes must also have freedom 12, the quadric freedom 8, and it cannot be constrained to be only a hyperboloid of rotation (freedom 7), let alone one with conical angle pi/4 (freedom 6). The mystery deepens ... what single Euclidean invariant could possibly apply to the quadric? Perhaps Steiner was completely mistaken, and the constraint is after all concerned with (say) the cross-ratio of the spacing of the lines along the regulus? The Pluecker vector for N in terms of faces' components is shown below: moment first, direction second, each with components ordered x,y,z. N = [ + Gy Fx Hy Eo – Gy Fo Hy Ex – Fy Eo Hy Gx + Ey Hy Fo Gx + Fy Ex Go Hy – Ey Hy Fx Go + Gz Fx Hz Eo – Gz Fo Hz Ex – Fz Eo Hz Gx + Ez Hz Fo Gx + Fz Ex Go Hz – Ez Hz Fx Go, – Fx Eo Hx Gy + Ex Hx Fo Gy + Gx Fy Hx Eo – Gx Fo Hx Ey – Ex Hx Fy Go + Fx Ey Go Hx + Gz Fy Hz Eo – Gz Fo Hz Ey – Fz Eo Hz Gy + Ez Hz Fo Gy + Fz Ey Go Hz – Ez Hz Fy Go, – Fx Eo Hx Gz + Ex Hx Fo Gz + Gx Fz Hx Eo – Gx Fo Hx Ez – Ex Hx Fz Go + Fx Ez Go Hx – Fy Eo Hy Gz + Ey Hy Fo Gz + Gy Fz Hy Eo – Gy Fo Hy Ez – Ey Hy Fz Go + Fy Ez Go Hy, – Hx (Gz Ex Fy – Gz Ey Fx – Gy Ex Fz + Gy Ez Fx + Gx Ey Fz – Gx Ez Fy), – Hy (Gz Ex Fy – Gz Ey Fx – Gy Ex Fz + Gy Ez Fx + Gx Ey Fz – Gx Ez Fy), – Hz (Gz Ex Fy – Gz Ey Fx – Gy Ex Fz + Gy Ez Fx + Gx Ey Fz – Gx Ez Fy) ] Fred Lunnon On 11/1/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Richard,
I had confused myself again: of course the constraint is not a projective invariant, since altitudes are invariant only under similarity! Then "hyperboloid of rotation" becomes the natural interpretation, and the additional fixed conical angle pi/4 also looks plausible.
However: freedom of hyperboloid axis line 4, centre point 1, radius 1, (cone angle 1); of 4 altitude lines along regulus 4. So total freedom of altitudes would then be 4+1+1+4 = 10, less by 2 than the tetrahedron's!
Anyway, I shall go away to think up a lazy brute-force way to decide the matter algebraically.
Fred
On 10/31/11, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Fred, One source said it's got by rotating a rectangular hyperbola about its asymptotes, but they meant axes? Definitely not a hyperbolic paraboloid. Rotate a rect hyp about its `minor' (imaginary?) axis, giving a hyperboloid of one sheet. Asymptotic cone would have a vertical semi-angle pi/4. Does that use up all your degrees of freedom? I believe that the centre of the quadric is the Monge point of the tetrahedron. R.
See http://www.geometrie.tuwien.ac.at/havlicek/pub/hoehen.pdf Hans Havlicek, Gunter Weiß "Altitudes of a Tetrahedron and Traceless Quadratic Forms" Tech. Univ. Vienna (no date?) The altitudes of the tetrahedron lie on a quadric regulus, and by counting freedoms they must satisfy one further constraint: that the hyperboloid is "equilateral". This turns out to mean simply that it is _traceless_, that is reducible to the form a^2 x^2 + b^2 y^2 - c^2 z^2, where c^2 = a^2 + b^2. Geometrically, the asymptotic cone apparently comprises a regulus of orthogonal triplets of generator lines. Also the hyperboloid is swept out by a second "conjugate" regulus; the perpendiculars at the faces' orthocentres apparently belong to this other regulus, so lying on the same hyperboloid and meeting all the altitudes. Pity I had to slog through figuring most of this out for myself, before being in a position to do a sufficiently specific search to discover it was already well-known! Fred Lunnon On 11/1/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Ho-ho! Suppose we are given (Pluecker coordinates of) altitudes K,L,M,N of a tetrahedron, and want to solve for faces (projective) E,F,G,H.
The 3-vector directions of paired altitudes and faces are proportional, so we can immediately set them equal (freedom 8). The 3-vector moments of the altitudes now give 12 linear equations for the "constant" components Eo,Fo,Go,Ho of the faces, with coefficients cubic in their direction components; as polynomials these can be computed to have rank 4, so can in general be solved uniquely [also consistently, since the tetrahedron exists].
Therefore the altitudes must also have freedom 12, the quadric freedom 8, and it cannot be constrained to be only a hyperboloid of rotation (freedom 7), let alone one with conical angle pi/4 (freedom 6).
The mystery deepens ... what single Euclidean invariant could possibly apply to the quadric? Perhaps Steiner was completely mistaken, and the constraint is after all concerned with (say) the cross-ratio of the spacing of the lines along the regulus?
The Pluecker vector for N in terms of faces' components is shown below: moment first, direction second, each with components ordered x,y,z.
N = [
+ Gy Fx Hy Eo – Gy Fo Hy Ex – Fy Eo Hy Gx + Ey Hy Fo Gx + Fy Ex Go Hy – Ey Hy Fx Go + Gz Fx Hz Eo – Gz Fo Hz Ex – Fz Eo Hz Gx + Ez Hz Fo Gx + Fz Ex Go Hz – Ez Hz Fx Go,
– Fx Eo Hx Gy + Ex Hx Fo Gy + Gx Fy Hx Eo – Gx Fo Hx Ey – Ex Hx Fy Go + Fx Ey Go Hx + Gz Fy Hz Eo – Gz Fo Hz Ey – Fz Eo Hz Gy + Ez Hz Fo Gy + Fz Ey Go Hz – Ez Hz Fy Go,
– Fx Eo Hx Gz + Ex Hx Fo Gz + Gx Fz Hx Eo – Gx Fo Hx Ez – Ex Hx Fz Go + Fx Ez Go Hx – Fy Eo Hy Gz + Ey Hy Fo Gz + Gy Fz Hy Eo – Gy Fo Hy Ez – Ey Hy Fz Go + Fy Ez Go Hy,
– Hx (Gz Ex Fy – Gz Ey Fx – Gy Ex Fz + Gy Ez Fx + Gx Ey Fz – Gx Ez Fy),
– Hy (Gz Ex Fy – Gz Ey Fx – Gy Ex Fz + Gy Ez Fx + Gx Ey Fz – Gx Ez Fy),
– Hz (Gz Ex Fy – Gz Ey Fx – Gy Ex Fz + Gy Ez Fx + Gx Ey Fz – Gx Ez Fy)
]
Fred Lunnon
On 11/1/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Richard,
I had confused myself again: of course the constraint is not a projective invariant, since altitudes are invariant only under similarity! Then "hyperboloid of rotation" becomes the natural interpretation, and the additional fixed conical angle pi/4 also looks plausible.
However: freedom of hyperboloid axis line 4, centre point 1, radius 1, (cone angle 1); of 4 altitude lines along regulus 4. So total freedom of altitudes would then be 4+1+1+4 = 10, less by 2 than the tetrahedron's!
Anyway, I shall go away to think up a lazy brute-force way to decide the matter algebraically.
Fred
On 10/31/11, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Fred, One source said it's got by rotating a rectangular hyperbola about its asymptotes, but they meant axes? Definitely not a hyperbolic paraboloid. Rotate a rect hyp about its `minor' (imaginary?) axis, giving a hyperboloid of one sheet. Asymptotic cone would have a vertical semi-angle pi/4. Does that use up all your degrees of freedom? I believe that the centre of the quadric is the Monge point of the tetrahedron. R.
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