That's quite nice --- though SBG did inadvertantly hint at the answer by asking for an explanation! Here's my stab at the impossibility: given some arrangement of rays R1,...,Rn meeting at mutually obtuse angles in the origin, let F1,...,Fn denote their perpendicular planes through the orgin. Evidently R2,...,Rn lie on the far side of F1 from R1; similarly R3,...,Rn lie in the acute slice between F1 and F2; and R4,...,Rn lie in the acute trihedron formed by F1,F2,F3. Therefore n <= 4. Fred Lunnon On 8/30/10, rcs@xmission.com <rcs@xmission.com> wrote:
Forwarded from SBG ...
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Date: Mon, 30 Aug 2010 12:08:08 -0700 Subject: Re: [math-fun] left vs. right From: "Stephen B. Gray" <stevebg@roadrunner.com>
Here's an exercise in 3D visualization. Given a point P, is it possible to construct FIVE rays coming out from P such that every ray makes an obtuse angle (>90 degrees) with every other one? (That's 10 angles that must be obtuse.) Explain your answer.
Steve Gray
Here's an obvious followup question: In n dimensions, is the maximum number of mutually-obtuse rays with a common origin always n+1? Here's a suggested construction that I believe generates canonical, symmetrical solutions: For n dimensions, start with the n-1 dimension solution that has n rays. Add the new dimension perpendicular to the previous n-1 dimensions. Then "bend" the n-1 rays all in the same direction in the new dimension, and add a new ray in the opposite direction along the new dimension. If the previous rays are all bent by the correct amount, then I believe all rays, including the new one, will be isomorphic to each other (and all angles will be the same). Starting with one dimension, the solution involves 2 rays at 180 degrees to each other. To extend that to 2 dimensions, bend the two rays to an angle of 120 degrees, then add the new ray in the opposite direction. For 3 dimensions, bend those three rays, then add a fourth in the opposite direction, resulting in tetrahedral symmetry. And so on. Tom On 8/30/10, rcs@xmission.com <rcs@xmission.com> wrote:
Forwarded from SBG ...
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Date: Mon, 30 Aug 2010 12:08:08 -0700 Subject: Re: [math-fun] left vs. right From: "Stephen B. Gray" <stevebg@roadrunner.com>
Here's an exercise in 3D visualization. Given a point P, is it possible to construct FIVE rays coming out from P such that every ray makes an obtuse angle (>90 degrees) with every other one? (That's 10 angles that must be obtuse.) Explain your answer.
Steve Gray
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Sorry, I should have said: Then "bend" the n rays ... (as opposed to n-1 rays). Tom
Here's an obvious followup question: In n dimensions, is the maximum number of mutually-obtuse rays with a common origin always n+1?
Here's a suggested construction that I believe generates canonical, symmetrical solutions: For n dimensions, start with the n-1 dimension solution that has n rays. Add the new dimension perpendicular to the previous n-1 dimensions. Then "bend" the n-1 rays all in the same direction in the new dimension, and add a new ray in the opposite direction along the new dimension. If the previous rays are all bent by the correct amount, then I believe all rays, including the new one, will be isomorphic to each other (and all angles will be the same).
Starting with one dimension, the solution involves 2 rays at 180 degrees to each other. To extend that to 2 dimensions, bend the two rays to an angle of 120 degrees, then add the new ray in the opposite direction. For 3 dimensions, bend those three rays, then add a fourth in the opposite direction, resulting in tetrahedral symmetry. And so on.
Tom
On 8/30/10, rcs@xmission.com <rcs@xmission.com> wrote:
Forwarded from SBG ...
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Date: Mon, 30 Aug 2010 12:08:08 -0700 Subject: Re: [math-fun] left vs. right From: "Stephen B. Gray" <stevebg@roadrunner.com>
Here's an exercise in 3D visualization. Given a point P, is it possible to construct FIVE rays coming out from P such that every ray makes an obtuse angle (>90 degrees) with every other one? (That's 10 angles that must be obtuse.) Explain your answer.
Steve Gray
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On Mon, Aug 30, 2010 at 10:34 PM, Tom Karzes <karzes@sonic.net> wrote:
Here's an obvious followup question: In n dimensions, is the maximum number of mutually-obtuse rays with a common origin always n+1?
Yes, my proof generalizes. If one ray passes through (0, 0, 0, ... 1), then the last coordinate of all the other rays is negative. Project them all onto the plane perpendicular to the first ray. By the inductive hypothesis, the angle between two of the projections is non-obtuse. The angle between the rays before being projected is smaller (because the final coordinate makes a positive contribution to the dot product), so is not obtuse.
Here's a suggested construction that I believe generates canonical, symmetrical solutions:
Just place a simplex with its center at the origin, and use the rays through its vertices. The symmetrical way to express this is by embedding the n-dimensional space as a hyperplane in n+1 dimensional space, specifically the hyperplane where the coordinates sum to 1. Put the center point at the point whose coordinates are all 1/(n+1), and use the rays from there to the n+1 points that have one coordinate 1 and the rest 0. Andy
Thanks. The simplex is exactly what I was aiming for. Tom Andy Latto writes:
On Mon, Aug 30, 2010 at 10:34 PM, Tom Karzes <karzes@sonic.net> wrote:
Here's an obvious followup question: In n dimensions, is the maximum number of mutually-obtuse rays with a common origin always n+1?
Yes, my proof generalizes. If one ray passes through (0, 0, 0, ... 1), then the last coordinate of all the other rays is negative. Project them all onto the plane perpendicular to the first ray. By the inductive hypothesis, the angle between two of the projections is non-obtuse. The angle between the rays before being projected is smaller (because the final coordinate makes a positive contribution to the dot product), so is not obtuse.
Here's a suggested construction that I believe generates canonical, symmetrical solutions:
Just place a simplex with its center at the origin, and use the rays through its vertices. The symmetrical way to express this is by embedding the n-dimensional space as a hyperplane in n+1 dimensional space, specifically the hyperplane where the coordinates sum to 1. Put the center point at the point whose coordinates are all 1/(n+1), and use the rays from there to the n+1 points that have one coordinate 1 and the rest 0.
Andy
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participants (3)
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Andy Latto -
Fred lunnon -
Tom Karzes