Since this has come up, here's a long time question I've had.... Take a random tetrahedron. For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron. For which triangle centers are the 4 lines concurrent? If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard.gibert... Ed Pegg Jr
A natural generalisation is to identify cases where the 4 lines lie on a quadric, rather than meeting in a point (which I imagine is probably unusual). WFL On 11/3/11, ed pegg <ed@mathpuzzle.com> wrote:
Since this has come up, here's a long time question I've had....
Take a random tetrahedron.
For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron.
For which triangle centers are the 4 lines concurrent?
If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard.gibert... Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
The 4-concurrent version seems a natural generalization of the corresponding plane geometry problem, which has a very pretty answer. Rich -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 1:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers A natural generalisation is to identify cases where the 4 lines lie on a quadric, rather than meeting in a point (which I imagine is probably unusual). WFL On 11/3/11, ed pegg <ed@mathpuzzle.com> wrote:
Since this has come up, here's a long time question I've had....
Take a random tetrahedron.
For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron.
For which triangle centers are the 4 lines concurrent?
If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard. gibert.pagesperso-orange.fr/index.htmlhttp://mathworld.wolfram.com/Kim berlingCenter.htmlhttp://en.wikipedia.org/wiki/Triangle_center Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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The answer being? WFL On 11/3/11, Schroeppel, Richard <rschroe@sandia.gov> wrote:
The 4-concurrent version seems a natural generalization of the corresponding plane geometry problem, which has a very pretty answer.
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 1:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
A natural generalisation is to identify cases where the 4 lines lie on a quadric, rather than meeting in a point (which I imagine is probably unusual). WFL
On 11/3/11, ed pegg <ed@mathpuzzle.com> wrote:
Since this has come up, here's a long time question I've had....
Take a random tetrahedron.
For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron.
For which triangle centers are the 4 lines concurrent?
If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard. gibert.pagesperso-orange.fr/index.htmlhttp://mathworld.wolfram.com/Kim berlingCenter.htmlhttp://en.wikipedia.org/wiki/Triangle_center Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Suppose lines are drawn from each vertex of a triangle to the opposite side. Each line divides its side into a clockwise piece and a counterclockwise piece. The three lines are concurrent iff the product of the clockwise pieces equals the product of the counterclockwise pieces. [Are we discussing the same thing? Surely WFL knows this thm.] Rich -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 5:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers The answer being? WFL On 11/3/11, Schroeppel, Richard <rschroe@sandia.gov> wrote:
The 4-concurrent version seems a natural generalization of the corresponding plane geometry problem, which has a very pretty answer.
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 1:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
A natural generalisation is to identify cases where the 4 lines lie on a quadric, rather than meeting in a point (which I imagine is probably unusual). WFL
On 11/3/11, ed pegg <ed@mathpuzzle.com> wrote:
Since this has come up, here's a long time question I've had....
Take a random tetrahedron.
For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron.
For which triangle centers are the 4 lines concurrent?
If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard. gibert.pagesperso-orange.fr/index.htmlhttp://mathworld.wolfram.com/Ki m berlingCenter.htmlhttp://en.wikipedia.org/wiki/Triangle_center Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Hi, all, Of course, this is Ceva's theorem, but what's the 3-D analog ? R. On Thu, 3 Nov 2011, Schroeppel, Richard wrote:
Suppose lines are drawn from each vertex of a triangle to the opposite side. Each line divides its side into a clockwise piece and a counterclockwise piece. The three lines are concurrent iff the product of the clockwise pieces equals the product of the counterclockwise pieces. [Are we discussing the same thing? Surely WFL knows this thm.]
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 5:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
The answer being? WFL
On 11/3/11, Schroeppel, Richard <rschroe@sandia.gov> wrote:
The 4-concurrent version seems a natural generalization of the corresponding plane geometry problem, which has a very pretty answer.
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 1:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
A natural generalisation is to identify cases where the 4 lines lie on a quadric, rather than meeting in a point (which I imagine is probably unusual). WFL
On 11/3/11, ed pegg <ed@mathpuzzle.com> wrote:
Since this has come up, here's a long time question I've had....
Take a random tetrahedron.
For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron.
For which triangle centers are the 4 lines concurrent?
If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard. gibert.pagesperso-orange.fr/index.htmlhttp://mathworld.wolfram.com/Ki m berlingCenter.htmlhttp://en.wikipedia.org/wiki/Triangle_center Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I believe I have a generalised theorem for all simplices: "Consider an arbitrary edge, and mark a point on that edge. Draw the hyperplane through this point and the n-1 simplex vertices not on this edge. Then, the n(n+1)/2 hyperplanes formed in this manner concur at a point iff there exists a quadric tangent to each edge at the marked point on that edge." The 'only if' part is easy by performing a projective transformation to replace the configuration with a regular simplex and its centre. I leave the converse as an exercise to the reader. Counting all the degrees of freedom suggests that the theorem is probably true. It certainly works in two dimensions, anyway, by a degenerate case of either Brianchon's or Chasles' theorem. Sincerely, Adam P. Goucher
Hi, all, Of course, this is Ceva's theorem, but what's the 3-D analog ? R.
On Thu, 3 Nov 2011, Schroeppel, Richard wrote:
Suppose lines are drawn from each vertex of a triangle to the opposite side. Each line divides its side into a clockwise piece and a counterclockwise piece. The three lines are concurrent iff the product of the clockwise pieces equals the product of the counterclockwise pieces. [Are we discussing the same thing? Surely WFL knows this thm.]
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 5:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
The answer being? WFL
On 11/3/11, Schroeppel, Richard <rschroe@sandia.gov> wrote:
The 4-concurrent version seems a natural generalization of the corresponding plane geometry problem, which has a very pretty answer.
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 1:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
A natural generalisation is to identify cases where the 4 lines lie on a quadric, rather than meeting in a point (which I imagine is probably unusual). WFL
On 11/3/11, ed pegg <ed@mathpuzzle.com> wrote:
Since this has come up, here's a long time question I've had....
Take a random tetrahedron.
For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron.
For which triangle centers are the 4 lines concurrent?
If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard. gibert.pagesperso-orange.fr/index.htmlhttp://mathworld.wolfram.com/Ki m berlingCenter.htmlhttp://en.wikipedia.org/wiki/Triangle_center Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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For "n-1" earlier read "n" = spatial dimension. Converse in 3-space, synthetically: Given a tetrahedron (with vertices) PQRS, and some quadric tangent to 6 points along its edges with 6 planes as specified earlier, project the quadric to the unit sphere. [The tetrahedron now has a mid-sphere: opposite pairs of edges have equal total lengths, so there are 2 extra constraints on its shape. See http://www.ac-noumea.nc/maths/polyhedr/tetra_.htm .] The quotient group fixing the sphere is just the Moebius group in 2-space, of which the general "loxodromic" element fixes two points, and transports any given point to an arbitrary point (all on the sphere). By means of such, project the tangencies on edges PQ, PR (fixing that on PQ), PS (fixing those on PQ, PR) to 3 vertices around one face of an inscribed regular octahedron. This fixes point P, lines PQ,PR,PS, planes PQR,PRS,PSQ, coincident with those of a regular tetrahedron. Vertex Q must lie on line PQ, and fixing it fixes lines QR tangent to the sphere and meeting PR, and fixes line QS similarly; the entire tetrahedron is then determined. It is now easy to see that the distance of line RS from the origin is a decreasing function of length PQ; so for only one location of Q can RS be tangent to the sphere, and that occurs when PQRS is regular. As before, by symmetry the 6 planes meet in a point. QED Converse in n-space, "analytically": The above argument fails in higher dimensions, where projective n-space and Moebius (n-1)-space groups have freedom n^2-1 and n(n+1)/2 resp. The algebraic constraints on n(n+1)/2 planes concurrent in a point are that their projective vectors have rank only n-1 instead of n; furthermore, these determinants are linear in each point along the edges, so in general determine them uniquely by vanishing. By the forward argument, the constraints on n(n+1)/2 points being tangencies of a quadric must be the same expressions; therefore one condition is equivalent to the other. QED Fred Lunnon On 11/6/11, Adam P. Goucher <apgoucher@gmx.com> wrote:
I believe I have a generalised theorem for all simplices:
"Consider an arbitrary edge, and mark a point on that edge. Draw the hyperplane through this point and the n-1 simplex vertices not on this edge. Then, the n(n+1)/2 hyperplanes formed in this manner concur at a point iff there exists a quadric tangent to each edge at the marked point on that edge."
The 'only if' part is easy by performing a projective transformation to replace the configuration with a regular simplex and its centre. I leave the converse as an exercise to the reader. Counting all the degrees of freedom suggests that the theorem is probably true.
It certainly works in two dimensions, anyway, by a degenerate case of either Brianchon's or Chasles' theorem.
Sincerely,
Adam P. Goucher
Hi, all, Of course, this is Ceva's theorem, but what's the 3-D analog ? R.
On Thu, 3 Nov 2011, Schroeppel, Richard wrote:
Suppose lines are drawn from each vertex of a triangle to the opposite side. Each line divides its side into a clockwise piece and a counterclockwise piece. The three lines are concurrent iff the product of the clockwise pieces equals the product of the counterclockwise pieces. [Are we discussing the same thing? Surely WFL knows this thm.]
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 5:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
The answer being? WFL
On 11/3/11, Schroeppel, Richard <rschroe@sandia.gov> wrote:
The 4-concurrent version seems a natural generalization of the corresponding plane geometry problem, which has a very pretty answer.
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 1:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
A natural generalisation is to identify cases where the 4 lines lie on a quadric, rather than meeting in a point (which I imagine is probably unusual). WFL
On 11/3/11, ed pegg <ed@mathpuzzle.com> wrote:
Since this has come up, here's a long time question I've had....
Take a random tetrahedron.
For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron.
For which triangle centers are the 4 lines concurrent?
If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard. gibert.pagesperso-orange.fr/index.htmlhttp://mathworld.wolfram.com/Ki m berlingCenter.htmlhttp://en.wikipedia.org/wiki/Triangle_center Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Oh dear, "analytical" argument was hopelessly holey (not holy, ho-ho): one would need at least to verify that the quadric constraint was also linear in the edge-points (specified eg. via the ratio in which their edges are divided). More generally, such constraints generate polynomial ideals, and it would be necessary to show that both ideals have the same root ideal: a tall order in the absence of the polynomials in question; and often even even in their presence! WFL On 11/7/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
For "n-1" earlier read "n" = spatial dimension.
Converse in 3-space, synthetically:
Given a tetrahedron (with vertices) PQRS, and some quadric tangent to 6 points along its edges with 6 planes as specified earlier, project the quadric to the unit sphere. [The tetrahedron now has a mid-sphere: opposite pairs of edges have equal total lengths, so there are 2 extra constraints on its shape. See http://www.ac-noumea.nc/maths/polyhedr/tetra_.htm .]
The quotient group fixing the sphere is just the Moebius group in 2-space, of which the general "loxodromic" element fixes two points, and transports any given point to an arbitrary point (all on the sphere). By means of such, project the tangencies on edges PQ, PR (fixing that on PQ), PS (fixing those on PQ, PR) to 3 vertices around one face of an inscribed regular octahedron. This fixes point P, lines PQ,PR,PS, planes PQR,PRS,PSQ, coincident with those of a regular tetrahedron.
Vertex Q must lie on line PQ, and fixing it fixes lines QR tangent to the sphere and meeting PR, and fixes line QS similarly; the entire tetrahedron is then determined. It is now easy to see that the distance of line RS from the origin is a decreasing function of length PQ; so for only one location of Q can RS be tangent to the sphere, and that occurs when PQRS is regular. As before, by symmetry the 6 planes meet in a point. QED
Converse in n-space, "analytically":
The above argument fails in higher dimensions, where projective n-space and Moebius (n-1)-space groups have freedom n^2-1 and n(n+1)/2 resp.
The algebraic constraints on n(n+1)/2 planes concurrent in a point are that their projective vectors have rank only n-1 instead of n; furthermore, these determinants are linear in each point along the edges, so in general determine them uniquely by vanishing.
By the forward argument, the constraints on n(n+1)/2 points being tangencies of a quadric must be the same expressions; therefore one condition is equivalent to the other. QED
Fred Lunnon
On 11/6/11, Adam P. Goucher <apgoucher@gmx.com> wrote:
I believe I have a generalised theorem for all simplices:
"Consider an arbitrary edge, and mark a point on that edge. Draw the hyperplane through this point and the n-1 simplex vertices not on this edge. Then, the n(n+1)/2 hyperplanes formed in this manner concur at a point iff there exists a quadric tangent to each edge at the marked point on that edge."
The 'only if' part is easy by performing a projective transformation to replace the configuration with a regular simplex and its centre. I leave the converse as an exercise to the reader. Counting all the degrees of freedom suggests that the theorem is probably true.
It certainly works in two dimensions, anyway, by a degenerate case of either Brianchon's or Chasles' theorem.
Sincerely,
Adam P. Goucher
Hi, all, Of course, this is Ceva's theorem, but what's the 3-D analog ? R.
On Thu, 3 Nov 2011, Schroeppel, Richard wrote:
Suppose lines are drawn from each vertex of a triangle to the opposite side. Each line divides its side into a clockwise piece and a counterclockwise piece. The three lines are concurrent iff the product of the clockwise pieces equals the product of the counterclockwise pieces. [Are we discussing the same thing? Surely WFL knows this thm.]
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 5:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
The answer being? WFL
On 11/3/11, Schroeppel, Richard <rschroe@sandia.gov> wrote:
The 4-concurrent version seems a natural generalization of the corresponding plane geometry problem, which has a very pretty answer.
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 1:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
A natural generalisation is to identify cases where the 4 lines lie on a quadric, rather than meeting in a point (which I imagine is probably unusual). WFL
On 11/3/11, ed pegg <ed@mathpuzzle.com> wrote:
Since this has come up, here's a long time question I've had....
Take a random tetrahedron.
For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron.
For which triangle centers are the 4 lines concurrent?
If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard. gibert.pagesperso-orange.fr/index.htmlhttp://mathworld.wolfram.com/Ki m berlingCenter.htmlhttp://en.wikipedia.org/wiki/Triangle_center Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Sure --- but do such points constitute "centres" of the edges --- in Kimberling's sense reduced appropriately to 1-D, anyway? WFL On 11/3/11, Schroeppel, Richard <rschroe@sandia.gov> wrote:
Suppose lines are drawn from each vertex of a triangle to the opposite side. Each line divides its side into a clockwise piece and a counterclockwise piece. The three lines are concurrent iff the product of the clockwise pieces equals the product of the counterclockwise pieces. [Are we discussing the same thing? Surely WFL knows this thm.]
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 5:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
The answer being? WFL
On 11/3/11, Schroeppel, Richard <rschroe@sandia.gov> wrote:
The 4-concurrent version seems a natural generalization of the corresponding plane geometry problem, which has a very pretty answer.
Rich
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, November 03, 2011 1:14 PM To: math-fun Subject: Re: [math-fun] Tetrahedral Centers
A natural generalisation is to identify cases where the 4 lines lie on a quadric, rather than meeting in a point (which I imagine is probably unusual). WFL
On 11/3/11, ed pegg <ed@mathpuzzle.com> wrote:
Since this has come up, here's a long time question I've had....
Take a random tetrahedron.
For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron.
For which triangle centers are the 4 lines concurrent?
If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard. gibert.pagesperso-orange.fr/index.htmlhttp://mathworld.wolfram.com/Ki m berlingCenter.htmlhttp://en.wikipedia.org/wiki/Triangle_center Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Why should X333 be of particular interest in this context, I wonder? [Note that the long line below had 3 URL's run together!] WFL On 11/3/11, ed pegg <ed@mathpuzzle.com> wrote:
Since this has come up, here's a long time question I've had....
Take a random tetrahedron.
For a given triangle center, connect it to the opposing corner of the tetrahedron. Repeat for each face of the tetrahedron.
For which triangle centers are the 4 lines concurrent?
If X333 had that property, I suppose Y333 would be a fine name for that Tetrahedral center. Links: http://faculty.evansville.edu/ck6/encyclopedia/ETC.htmlhttp://bernard.gibert... Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (5)
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Adam P. Goucher -
ed pegg -
Fred lunnon -
Richard Guy -
Schroeppel, Richard