RE: [math-fun] Ambiguously placed cities
Q2. Are there 10 road lengths that lead to distinct town configurations?
----------------------------
Q1. Draw a non-isosceles triangle ABC with rational sides. Let M be the midpoint of AB, and let DD' be a segment drawn through M that's perpendicular to CM, such that the distances d(D,M) and d(D',M) are equal and rational.
This implies distinct town configurations A,B,C,D and A,B,C,D' having the same sest of 6 distances.
The fact that DMD' is perpendicular to CM shows that the distances CD and CD' are equal. But I don't see why AD and AD' should be equal in this contstruction. Andy Latto andy.latto@pobox.com
On 11/26/06, Andy Latto <Andy.Latto@gensym.com> wrote:
The fact that DMD' is perpendicular to CM shows that the distances CD and CD' are equal. But I don't see why AD and AD' should be equal in this contstruction.
They aren't --- by congruent triangles, AD' = BD and AD = BD'. I now see the point of the original problem --- permuting the distances still produces a viable distance chart. So what the problem statement must _avoid_ specifying is the assignment of distances to specific edges! WFL
So why stop at ambiguity? In principle, it might be possible to choose the 6 (real) edge lengths to provide up to 5 essentially different distance charts. However ... Conjecture: if there are 3 or more assignments of 6 given lengths to edges, all consistent with planarity, then at least one pair of edges has equal lengths; the resulting symmetries render all charts indistinguishable. Now why this might be true? WFL
Good question. I don't know where you're getting the "5" below. There are 6! permutations of distances, 4! permutations of cities, leaving 30 possibilities to try that could potentially give different geometric arrangements. No doubt there are ways to further limit it, but it's not obvious to me what's obvious to whom about it. Dan's ambiguity involves a configuration where A, 1/2(B + C), D forms a right angle. The 6 distances can be thought of as edges of a tetrahedron; Dan's ambiguity interchanges two edges that have a vertex in common. I tried this out with Geometer's Sketchpad: draw 4 points, the 6 edges connecting them, and the 6 midpoints, and the 6 circles with the lines as diameters. Color opposite edges and their circles with like colors, e.g. R, G, B. You want to arrange so red circles pass through red midpoints, etc. You can readily arrange both red circles to pass through red midpoints, or one red circle to pass through a red midpoint and one green circle to pass through a green midpoint. This gives at least 3 configurations with the same set of distances. For one approximate solution, GSP measures with no duplicate distances. I don't see why you couldn't have more, but it's hard to verify this with this simple GSP experiment, since one would expect it to happen on a codimension 3 set, but you can only control 2 parameters at a time by moving a vertex with the mouse. Bill On Nov 26, 2006, at 10:01 PM, Fred lunnon wrote:
So why stop at ambiguity? In principle, it might be possible to choose the 6 (real) edge lengths to provide up to 5 essentially different distance charts. However ...
Conjecture: if there are 3 or more assignments of 6 given lengths to edges, all consistent with planarity, then at least one pair of edges has equal lengths; the resulting symmetries render all charts indistinguishable.
Now why this might be true? WFL
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Let the road lengths be arranged [AB, BC, CA, BD, AD, CD], where A,B,C,D denote cities. Then the follwoing two charts are distinct and planar, and both permute the same set of integers: [5, 8, 13, 11, 9, 17] [5, 17, 13, 11, 9, 8] [Given my capacity for wishful thinking, not to mention Maple's for wishful computation, I'd appreciate somebody verifying this independently!] Fred Lunnon
On 11/27/06, Bill Thurston <wpt4@cornell.edu> wrote:
Good question. I don't know where you're getting the "5" below. There are 6! permutations of distances, 4! permutations of cities, leaving 30 possibilities to try that could potentially give different geometric arrangements. No doubt there are ways to further limit it, but it's not obvious to me what's obvious to whom about it.
Yes indeed; but I also reasoned that we have only 6 homogeneous variables, and in general would expect one independent constraint to be imposed on them by each planarity equation, of the form tetrahedron volume = 0.
Dan's ambiguity involves a configuration where A, 1/2(B + C), D forms a right angle. The 6 distances can be thought of as edges of a tetrahedron; Dan's ambiguity interchanges two edges that have a vertex in common.
I tried this out with Geometer's Sketchpad: draw 4 points, the 6 edges connecting them, and the 6 midpoints, and the 6 circles with the lines as diameters. Color opposite edges and their circles with like colors, e.g. R, G, B. You want to arrange so red circles pass through red midpoints, etc. You can readily arrange both red circles to pass through red midpoints, or one red circle to pass through a red midpoint and one green circle to pass through a green midpoint. This gives at least 3 configurations with the same set of distances. For one approximate solution, GSP measures
Sadly, no data appears in my copy of the posting at this point ...
with no duplicate distances. I don't see why you couldn't have more, but it's hard to verify this with this simple GSP experiment, since one would expect it to happen on a codimension 3 set, but you can only control 2 parameters at a time by moving a vertex with the mouse.
Bill
I tried solving equations numerically with Maple ---- there's a good deal of slack with 30 to choose from! For triple charts, all the combinations I did try produced solutions with equal pairs, even when I chased down more than one solution. Fred Lunnon
On Nov 27, 2006, at 2:51 AM, Fred lunnon wrote:
a green midpoint. This gives at least 3 configurations with the same set of distances. For one approximate solution, GSP measures
Sadly, no data appears in my copy of the posting at this point ...
Looking back at the message I sent, I see now that the text I copied and pasted from GSP was actually a TIFF graphic, and no doubt the math-fun-gremlins ate it up. I'll type it in: AB = 36.38 cm AC = 16.29 cm AD = 21.06 cm BC = 28.31 cm BD = 15.47 cm CD = 14.64 cm where C,.5(A+B),D and C,.5(A+D),B are (approximately) right angles. Bill
I typed Bill's hexad below into my program, which sure enough reported 3 small tetrahedra (out of the 30 possible) with absolute volumes 15.32733125 , 35.63804829 , 46.20211861 However, among the remainder were another 5 with volumes around 100, followed by a steady ramp up to the largest around 1000. Run-of-the mill integer hexads I've generated frequently show distinct volumes much closer together than these figures. So taken in the round, I have to say they're not terribly convincing. The search program has been souped up to generate integer hexads restricted to the region where all 20 possible faces are proper triangles. In short test runs it has so far turned up a handful of double proper tetrahedra (meaning two distinct shapes with the same volume and edge lengths), all in sets of threes with the same hexad; but not even a single planar hexad for edge-sum up to 288. It ain't looking good! Fred Lunnon On 11/27/06, Bill Thurston <wpt4@cornell.edu> wrote:
... Looking back at the message I sent, I see now that the text I copied and pasted from GSP was actually a TIFF graphic, and no doubt the math-fun-gremlins ate it up. I'll type it in: AB = 36.38 cm AC = 16.29 cm AD = 21.06 cm BC = 28.31 cm BD = 15.47 cm CD = 14.64 cm where C,.5(A+B),D and C,.5(A+D),B are (approximately) right angles.
On Nov 27, 2006, at 10:29 PM, Fred lunnon wrote:
I typed Bill's hexad below into my program, which sure enough reported 3 small tetrahedra (out of the 30 possible) with absolute volumes
15.32733125 , 35.63804829 , 46.20211861
However, among the remainder were another 5 with volumes around 100, followed by a steady ramp up to the largest around 1000.
Run-of-the mill integer hexads I've generated frequently show distinct volumes much closer together than these figures. So taken in the round, I have to say they're not terribly convincing.
What are you unconvinced of? These aren't purported to give integral solutions. The GSP demo is completely convincing, you can see there exists an exact solution because the two relevant intersection points of circles with edges independently move back and forth past the midpoints of the edges. I'm sure one could easily get numerically accurate solutions from these using Newton's method, but I don't see the point to doing so. If we could attach pictures, this would be much easier to talk about. But, here's a geometric construction that gives 3 configurations with the same edge lengths. Start with (almost) any triangle. Make a row of 3 congruent copies, by rotating the triangle around two of its edges. This gives a quadrilateral Q subdivided into 3 triangles, since the 3 angles meeting at a point add to pi. Let z be the intersection of the perpendicular bisectors of the two diagonals of Q. The cone of z to any of the three congruent triangles makes a 4-tuple having an identical set of edge lengths. As long as Q is not inscribed in a center (whose center would be z) these arrangements are not congruent. The condition for Q to be inscribed in a circle is that diagonally opposite angles sum to pi. If the angles of the original triangle are p, q, r, the angles of Q are p, pi-p, pi-r, r, so if the original triangle is not isosceles, Q is not so inscribed. Using sketchpad you can make this construction, which still has 2 degrees of freedom up to scaling. It appears you can get at least one more distinct configuration --- you can choose another flip of a vertex across an edge, with one distance-match to be made. You can draw a circle to see whether the two relevant distances are equal, and it's easy to arrange, in a way that the new 4-tuple (by visual inspection) is not equivalent to any of the others. I don't know if you can get 5---some attempts to impose distance equalities produce symmetry.
This looks like fun; wish I'd had the time to play with it. Starting with a 3-4-5 triangle, for example, and clearing denominators at the end, Bill's geometric construction yeilds the hexad 108, 144, 180, 5*sqrt(373), 13*sqrt(229), sqrt(44749) which I'm sure Fred's program will agree has three volume-zero tetrahedra and no degeneracy. I wonder whether this might be a universal construction for the case where there are three planar arrangements. That is, given (almost) any triangle ABC in the plane, Bill's method allows you to identify three points Z1, Z2, Z3 such that the three Z's distances to A,B,C are permutations of one another. (Bill says that sometimes he can produce a Z4 also, but that's beside the point here.) Is there only one such set of Z-distances for each ABC, or might there be more? How about for generic ABC? Anyway, now we're back to the problems Ed sent out two days ago:
Q1. Are there 6 integer road lengths that lead to distinct town configurations?
Q2. Are there 10 road lengths that lead to distinct [five-]town configurations?
--Michael Kleber On 11/27/06, Bill Thurston <wpthurston@mac.com> wrote:
On Nov 27, 2006, at 10:29 PM, Fred lunnon wrote:
I typed Bill's hexad below into my program, which sure enough reported 3 small tetrahedra (out of the 30 possible) with absolute volumes
15.32733125 , 35.63804829 , 46.20211861
However, among the remainder were another 5 with volumes around 100, followed by a steady ramp up to the largest around 1000.
Run-of-the mill integer hexads I've generated frequently show distinct volumes much closer together than these figures. So taken in the round, I have to say they're not terribly convincing.
What are you unconvinced of? These aren't purported to give integral solutions. The GSP demo is completely convincing, you can see there exists an exact solution because the two relevant intersection points of circles with edges independently move back and forth past the midpoints of the edges. I'm sure one could easily get numerically accurate solutions from these using Newton's method, but I don't see the point to doing so. If we could attach pictures, this would be much easier to talk about.
But, here's a geometric construction that gives 3 configurations with the same edge lengths.
Start with (almost) any triangle. Make a row of 3 congruent copies, by rotating the triangle around two of its edges. This gives a quadrilateral Q subdivided into 3 triangles, since the 3 angles meeting at a point add to pi.
Let z be the intersection of the perpendicular bisectors of the two diagonals of Q.
The cone of z to any of the three congruent triangles makes a 4-tuple having an identical set of edge lengths. As long as Q is not inscribed in a center (whose center would be z) these arrangements are not congruent.
The condition for Q to be inscribed in a circle is that diagonally opposite angles sum to pi. If the angles of the original triangle are p, q, r, the angles of Q are p, pi-p, pi-r, r, so if the original triangle is not isosceles, Q is not so inscribed.
Using sketchpad you can make this construction, which still has 2 degrees of freedom up to scaling. It appears you can get at least one more distinct configuration --- you can choose another flip of a vertex across an edge, with one distance-match to be made. You can draw a circle to see whether the two relevant distances are equal, and it's easy to arrange, in a way that the new 4-tuple (by visual inspection) is not equivalent to any of the others.
I don't know if you can get 5---some attempts to impose distance equalities produce symmetry.
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On Nov 28, 2006, at 10:07 AM, Michael Kleber wrote:
I wonder whether this might be a universal construction for the case where there are three planar arrangements. That is, given (almost) any triangle ABC in the plane, Bill's method allows you to identify three points Z1, Z2, Z3 such that the three Z's distances to A,B,C are permutations of one another.
The construction I described actually gives 3 triples of Z's, since it depends on choosing two sides of the triangle, to do 180 degree rotations around. Michael's rephrasing brings it back closer to Dan's construction: given a triangle, draw the 3 medians, and draw their perpendicular lines at their feet. These three lines form another triangle, similar to the first. Any vertex of this larger triangle is the Z1 of a triple, where Z2 and Z3 are obtained by reflecting it across two of the medians. You can't do any more reflections unless Z2 say is another apex of the second triangle, but that only happens when the original triangle is isosceles.
(Bill says that sometimes he can produce a Z4 also, but that's beside the point here.) Is there only one such set of Z-distances for each ABC, or might there be more? How about for generic ABC?
No, the 4th configuration I was talking about didn't come from a Z4: in this 4th configuration, there would be no triangle congruent to ABC, these lengths would be rearranged linearly.
Anyway, now we're back to the problems Ed sent out two days ago:
Q1. Are there 6 integer road lengths that lead to distinct town
configurations?
Q2. Are there 10 road lengths that lead to distinct [five-]town
configurations?
Here's how you can get arbitrarily many towns with the same set of distances, generalizing Dan's construction: Let rho: E^2 -> E^2 be reflection through the origin, and sigma be reflection in the x-axis. Let X be any collection of points such that sigma(X) = X Y be any collection such that rho(Y) = Y, such that X U Y has no symmetry. Claim: for any subset Z of the y-axis, the distances among X U Y U Z and X U Y U rho(Z) are equal. That's because rho(Z) = sigma(Z). There are clearly some more variations of this, e.g. using other symmetries, as well as other geomtries and other dimensions --- it would be intriguing to get the full picture --- but I should stop for now! Bill
Here's a nice mental image for ghost symmetries giving multiple arrangements with all the same pairwise distance data. Imagine a figure in the plane --- say a pine tree --- that is symmetric by reflection in a vertical line, another figure --- say a log --- that is symmetric by reflection in a horizontal line, and a third figure --- say a letter S --- that is symmetric by 180 degree rotation about the intersection of these two lines. There is a `ghost' symmetry group of order 4 suggested by the picture, but each of the three figures is invariant only by a subgroup of order 2, and there is no actual symmetry of the actual figure. Each of the three things has one other image under the ghost symmetry group. Now if you replace any one of the three items by its other image under the ghost group --- say, you replace the S by a mirror S --- you get a different picture, yet any two of the three new items is isomorphic to the original. It's visually kind of intriguing. Bill On Nov 28, 2006, at 1:22 PM, Bill Thurston wrote:
On Nov 28, 2006, at 10:07 AM, Michael Kleber wrote:
I wonder whether this might be a universal construction for the case where there are three planar arrangements. That is, given (almost) any triangle ABC in the plane, Bill's method allows you to identify three points Z1, Z2, Z3 such that the three Z's distances to A,B,C are permutations of one another.
The construction I described actually gives 3 triples of Z's, since it depends on choosing two sides of the triangle, to do 180 degree rotations around. Michael's rephrasing brings it back closer to Dan's construction: given a triangle, draw the 3 medians, and draw their perpendicular lines at their feet. These three lines form another triangle, similar to the first. Any vertex of this larger triangle is the Z1 of a triple, where Z2 and Z3 are obtained by reflecting it across two of the medians. You can't do any more reflections unless Z2 say is another apex of the second triangle, but that only happens when the original triangle is isosceles.
(Bill says that sometimes he can produce a Z4 also, but that's beside the point here.) Is there only one such set of Z-distances for each ABC, or might there be more? How about for generic ABC?
No, the 4th configuration I was talking about didn't come from a Z4: in this 4th configuration, there would be no triangle congruent to ABC, these lengths would be rearranged linearly.
Anyway, now we're back to the problems Ed sent out two days ago:
Q1. Are there 6 integer road lengths that lead to distinct town
configurations?
Q2. Are there 10 road lengths that lead to distinct [five-]town
configurations?
Here's how you can get arbitrarily many towns with the same set of distances, generalizing Dan's construction:
Let rho: E^2 -> E^2 be reflection through the origin, and sigma be reflection in the x-axis.
Let X be any collection of points such that sigma(X) = X Y be any collection such that rho(Y) = Y, such that X U Y has no symmetry.
Claim: for any subset Z of the y-axis, the distances among X U Y U Z and X U Y U rho(Z) are equal.
That's because rho(Z) = sigma(Z).
There are clearly some more variations of this, e.g. using other symmetries, as well as other geomtries and other dimensions --- it would be intriguing to get the full picture --- but I should stop for now!
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Okay, I had another drip of time... WPT:
The construction I described actually gives 3 triples of Z's, since it depends on choosing two sides of the triangle, to do 180 degree rotations around. Michael's rephrasing brings it back closer to Dan's construction: given a triangle, draw the 3 medians, and draw their perpendicular lines at their feet. These three lines form another triangle, similar to the first. Any vertex of this larger triangle is the Z1 of a triple, where Z2 and Z3 are obtained by reflecting it across two of the medians.
Oof, of course; very nice. What's more, starting with a 3-4-5 and rotating about the midpoints of the length 3 and 5 sides gets you one extra rational (integer, even!) value; you end up with {3, 4, 5, 13, 5*sqrt(145), sqrt(7081)}. Jim Buddenhagen likewise mentioned four integer sides. Jim, is yours any smaller than this one? Going back further, I just reread Fred Lunnon's mail from two days ago:
Let the road lengths be arranged [AB, BC, CA, BD, AD, CD], where A,B,C,D denote cities. Then the follwoing two charts are distinct and planar, and both permute the same set of integers:
[5, 8, 13, 11, 9, 17] [5, 17, 13, 11, 9, 8]
I'm not sure why you decided it was somehow inferior, Fred! Yes, the first hexad has three cities collinear, but when handed a problem about distances along roads, nothing seems more likely than that some city is along the route between two others. Did anyone verify this, per Fred's request? It certainly seems like the prettiest configuration seen so far. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
On 11/29/06, Michael Kleber <michael.kleber@gmail.com> wrote:
Okay, I had another drip of time...
Oof, of course; very nice. What's more, starting with a 3-4-5 and rotating about the midpoints of the length 3 and 5 sides gets you one extra rational (integer, even!) value; you end up with {3, 4, 5, 13, 5*sqrt(145), sqrt(7081)}.
Sorry, MK --- this one is a dud, with no planar perms at all! If you're only going for double distinct planar charts (flat tetrahedra) with unequal lengths, you might just as well choose arbitrary integers for the first four --- for example [3, 4, 5, 6, 7.885487244, 3.133542872] in the same order as usual; also with the two quartic irrationals reversed. Inter alia, we can generate triple charts along the lines of the Thurston-Kleber example given earlier, from three arbitrary integer lengths u,v,w and three rational square roots x,y,z, as follows: Let a^2 = 9*(u+v+w)*(u+v-w)*(v+w-u)*(w+u-v) = (12*ABC)^2; then [u,v,w,x,y,z], [u,v,w,z,y,x], [u,v,w,y,x,z] are all planar, where (a*x)^2 = v^2*u^4+9*w^6-u^6-u^2*v^2*w^2-v^6+5*u^4*w^2 -9*w^4*u^2+5*w^2*v^4+v^4*u^2-9*w^4*v^2, (a*y)^2 = (2*v^4+2*u^4-4*u^2*v^2-6*v^2*w^2-3*u^2*w^2) *(v^2-2*w^2-2*u^2), (a*z)^2 = (2*v^4+2*u^4-4*u^2*v^2-6*u^2*w^2-3*v^2*w^2) *(u^2-2*w^2-2*v^2). Besides the triple planar chart, this hexad also yields a triple proper tetrahedron and a two double tetrahedra; though not all these are necessarily real. Of course, if we could find u,v,w such that all four polynomials were simultaneously rational squares, we'd have a triply-planar integer chart. Over to the number-theorists!
Jim Buddenhagen likewise mentioned four integer sides. Jim, is yours any smaller than this one?
Going back further, I just reread Fred Lunnon's mail from two days ago:
I'm not sure why you decided it was somehow inferior, Fred! Yes, the first hexad has three cities collinear, but when handed a problem about distances along roads, nothing seems more likely than that some city is along the route between two others.
I suppose it's a matter of aesthetic opinion --- like Ed, I'd prefer to see a solution with unequal lengths and proper triangles.
Did anyone verify this, per Fred's request? It certainly seems like the prettiest configuration seen so far.
The program had been (badly) designed to omit such cases: I could have generated many similar but smaller examples. An efficient search needs to be refined to exclude boring cases --- at this stage I just haven't got around to trying. ------------------------------------------------------------------------------------------ I have now conducted a search for integer planar charts with unequal sides and proper triangles, and while there are some, they're pretty sparse --- for edge sum up to 215, there are just 15 (single) examples --- I'll post a list if requested. Since at a rough (and probably optimistic) guess based on program statistics, I expect on average one doubly planar example among every 360 singly planar, it's clear we'll have to wait a good deal longer than the 17 hours this run cost, for a half-decent chance of striking lucky. The two smallest such (singly planar) hexads found were [15, 19, 20, 18, 9, 13]; [15, 19, 20, 13, 26, 18]; Notice the curious quasi-symmetry relating them, which also appears amongst other examples: ABC remains the same, BD swaps with CD, only AD is new. Fred Lunnon
Before people go too far looking for such charts, I have a question: is it possible to have 4 points in a plane, no 3 colinear, so that all 6 distances are integers? And with the additional constraint that no 2 distances are equal? Franklin T. Adams-Watters ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.
On Wed, 29 Nov 2006, franktaw@netscape.net wrote:
Before people go too far looking for such charts, I have a question: is it possible to have 4 points in a plane, no 3 colinear, so that all 6 distances are integers? And with the additional constraint that no 2 distances are equal?
Mr. Pegg seems to have found one: http://mathworld.wolfram.com/TriangleDissection.html -J
Sadly nobody has answered the question that was actually asked below, so I'll take up the challenge. As remarked in my last posting, I've found a handful, with the added constraint that all 20 possible triangles formed from the distances are proper. The examples I gave there were [AB, BC, CA, BD, AD, CD] = [15, 19, 20, 18, 9, 13]; [15, 19, 20, 13, 26, 18]. Fred Lunnon On 11/30/06, franktaw@netscape.net <franktaw@netscape.net> wrote:
Before people go too far looking for such charts, I have a question: is it possible to have 4 points in a plane, no 3 colinear, so that all 6 distances are integers? And with the additional constraint that no 2 distances are equal?
Franklin T. Adams-Watters
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On 11/29/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
Sadly nobody has answered the question that was actually asked below, so I'll take up the challenge. As remarked in my last posting, I've found a handful, with the added constraint that all 20 possible triangles formed from the distances are proper. The examples I gave there were [AB, BC, CA, BD, AD, CD] = [15, 19, 20, 18, 9, 13]; [15, 19, 20, 13, 26, 18].
Fred Lunnon
On 11/30/06, franktaw@netscape.net <franktaw@netscape.net> wrote:
Before people go too far looking for such charts, I have a question: is it possible to have 4 points in a plane, no 3 colinear, so that all 6 distances are integers? And with the additional constraint that no 2 distances are equal?
Franklin T. Adams-Watters
Your arrangements are planar (so solve Franklin T. Adams-Watters question). I didn't check carefully, but it appears that the (6 take 3) triangles formed by the distances are non-degenerate, but so far as I can tell neither arrangement can be rearranged in a planar way to give an integer solution to the ambiguous towns problem. I didn't check carefully, so maybe I am missing the point of these arrangements or something. JB
I have now found may solutions to the ambiguous towns problem with 5 integers distances and 1 non-integer distance. These were found by first finding a parameterized family of solutions with 4 integer distances and 2 non-integer distances and then searching on this family. They all belong to the class of solutions illustrated in Eg Pegg's gif file picture. They are pretty ugly, here is one of the simpler ones: [AB,BC,CA,DA,DB,DC]= [70200, 219024, 2808*sqrt(6709), 70425, 5625, 224649] is planar and can be re-arranged to [AB,BC,CA,DA,DB,DC]= [70200, 219024, 2808*sqrt(6709), 5625, 70425, 224649] which is also planar. Jim Buddenhagen
On 11/30/06, James Buddenhagen <jbuddenh@gmail.com> wrote:
I have now found may solutions to the ambiguous towns problem with 5 integers distances and 1 non-integer distance. These were found by first finding a parameterized family of solutions with 4 integer distances and 2 non-integer distances and then searching on this family. They all belong to the class of solutions illustrated in Eg Pegg's gif file picture.
They are pretty ugly, here is one of the simpler ones:
[AB,BC,CA,DA,DB,DC]= [70200, 219024, 2808*sqrt(6709), 70425, 5625, 224649]
is planar and can be re-arranged to
[AB,BC,CA,DA,DB,DC]= [70200, 219024, 2808*sqrt(6709), 5625, 70425, 224649]
which is also planar.
Jim Buddenhagen
Oops, I forgot that 3 colinear points was a design element in my program, so the above is not all that noteworthy. Here is a much simpler example in the same vein: [AB,BC,CA,DA,DB,DC]= [360, 400, 40*sqrt(181), 81, 369, 481] is planar and can be re-arranged to [AB,BC,CA,DA,DB,DC]= [360, 400, 40*sqrt(181), 369, 81,481] JB
On 11/30/06, James Buddenhagen <jbuddenh@gmail.com> wrote:
... Your arrangements are planar (so solve Franklin T. Adams-Watters question). I didn't check carefully, but it appears that the (6 take 3) triangles formed by the distances are non-degenerate, but so far as I can tell neither arrangement can be rearranged in a planar way to give an integer solution to the ambiguous towns problem.
Correct.
I didn't check carefully, so maybe I am missing the point of these arrangements or something.
My only point was to answer FTAW's actual question, which previous attempts at response had failed to achieve. On 11/30/06, James Buddenhagen <jbuddenh@gmail.com> wrote:
... Oops, I forgot that 3 colinear points was a design element in my program, so the above is not all that noteworthy.
Still, nice to see a family. Hang in there! ------------------------------------------------------------------------------------------ I've now got a search program running which copes with equal lengths and possible imaginary faces. I'm not sure it's hoovering up every single proper planar chart, but there are a lot more than previously --- so far approx 120 up to edge- sum 85, but none more than singly planar. Fred Lunnon
On 11/30/06, James Buddenhagen <jbuddenh@gmail.com> wrote:
... Your arrangements are planar (so solve Franklin T. Adams-Watters question). I didn't check carefully, but it appears that the (6 take 3) triangles formed by the distances are non-degenerate, but so far as I can tell neither arrangement can be rearranged in a planar way to give an integer solution to the ambiguous towns problem.
Correct.
I didn't check carefully, so maybe I am missing the point of these arrangements or something.
My only point was to answer FTAW's actual question, which previous attempts at response had failed to achieve. On 11/30/06, James Buddenhagen <jbuddenh@gmail.com> wrote:
... Oops, I forgot that 3 colinear points was a design element in my program, so the above is not all that noteworthy.
Still, nice to see a family. Hang in there! ------------------------------------------------------------------------------------------ I've now got a search program running which copes with equal lengths and possible imaginary faces. I'm not sure it's hoovering up every proper planar integer chart, but there are a lot more than previously --- so far approx 120 up to edge- sum 85, but none more than singly planar. Fred Lunnon
On 11/30/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
I've now got a search program running which copes with equal lengths and possible imaginary faces. I'm not sure it's hoovering up every proper planar integer chart, but there are a lot more than previously --- so far approx 120 up to edge- sum 85, but none more than singly planar.
Fred Lunnon
This search program has trawled every (I hope) planar chart with integer edges and 4 proper triangles up to edge-sum 111, at which point I'm calling it a day (or two) for now. It found 250 planar hexads, quite a few of which show also a second "flawed" permutation also planar, but with one linear "face"; however _no_ genuine double charts were detected. In a remarkably large proportion of cases the vertices ("cities") are also concyclic. There are also many cases of multiples of smaller hexads; rhombuses and rectangles (Pythagorean); parallelograms; trapeziums with diagonals equal to one side; all of which can probably be organised into orderly families without excessive effort. The list is available on request. Since I can't meet Ed's challenge, I'll substitute a (gentler) problem of my own ---- Four cities, no three lying on a straight line, are linked by straight roads, each stretching an integer number of miles. The first road is 2 miles long; how long are the other five? Fred Lunnon
On 12/2/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
Since I can't meet Ed's challenge, I'll substitute a (gentler) problem of my own ----
Four cities, no three lying on a straight line, are linked by straight roads, each stretching an integer number of miles. The first road is 2 miles long; how long are the other five?
I think this problem, or maybe just a similar one, is solved in Ian Stewart's column entitled "A six-pack for the tree gods", collected in his book _Another Vine Math You've Got Me Into_ if memory serves me, which it may not ... --Joshua Zucker
Huh --- scooped! I haven't chased up the article, but if they look that similar, the chances are they're the same problem. OK, try this one for size instead (quite literally) --- Four cities, lying on a circle, are linked by straight roads, each stretching an integer number of miles. If the circle has radius 147, what lengths do the roads have? On 12/3/06, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On 12/2/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
Since I can't meet Ed's challenge, I'll substitute a (gentler) problem of my own ----
Four cities, no three lying on a straight line, are linked by straight roads, each stretching an integer number of miles. The first road is 2 miles long; how long are the other five?
I think this problem, or maybe just a similar one, is solved in Ian Stewart's column entitled "A six-pack for the tree gods", collected in his book _Another Vine Math You've Got Me Into_ if memory serves me, which it may not ...
--Joshua Zucker
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 12/7/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
Huh --- scooped! I haven't chased up the article, but if they look that similar, the chances are they're the same problem.
OK, try this one for size instead (quite literally) --- Four cities, lying on a circle, are linked by straight roads, each stretching an integer number of miles. If the circle has radius 147, what lengths do the roads have?
A square root sign had gone AWOL in the above, as they have a habit of doing every so often. That should have read --- Four cities, lying on a circle, are linked by straight roads, each stretching an integer number of miles. If the circle has radius 50, what possible lengths might the roads have? Apologies to everybody. WFL
On 12/3/06, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On 12/2/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
Since I can't meet Ed's challenge, I'll substitute a (gentler) problem of my own ----
Four cities, no three lying on a straight line, are linked by straight roads, each stretching an integer number of miles. The first road is 2 miles long; how long are the other five?
I think this problem, or maybe just a similar one, is solved in Ian Stewart's column entitled "A six-pack for the tree gods", collected in his book _Another Vine Math You've Got Me Into_ if memory serves me, which it may not ...
--Joshua Zucker
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
And that one's dead in the water as well --- [30,40,50,50,40,30] cooks it straight away! This is harder than I thought ... Still, it's nice to know I'm not the only one to have difficulties with square roots --- when asked to find the square root of 625/4, Maple came up today with the ingenious answer 1/168*sqrt(4410000). WFL On 12/7/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 12/7/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
Huh --- scooped! I haven't chased up the article, but if they look that similar, the chances are they're the same problem.
OK, try this one for size instead (quite literally) --- Four cities, lying on a circle, are linked by straight roads, each stretching an integer number of miles. If the circle has radius 147, what lengths do the roads have?
A square root sign had gone AWOL in the above, as they have a habit of doing every so often. That should have read ---
Four cities, lying on a circle, are linked by straight roads, each stretching an integer number of miles. If the circle has radius 50, what possible lengths might the roads have?
Apologies to everybody. WFL
On 12/3/06, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On 12/2/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
Since I can't meet Ed's challenge, I'll substitute a (gentler) problem of my own ----
Four cities, no three lying on a straight line, are linked by straight roads, each stretching an integer number of miles. The first road is 2 miles long; how long are the other five?
I think this problem, or maybe just a similar one, is solved in Ian Stewart's column entitled "A six-pack for the tree gods", collected in his book _Another Vine Math You've Got Me Into_ if memory serves me, which it may not ...
--Joshua Zucker
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
The following 4-parameter expression generates a planar chart of 4 concyclic vertices for any choice of integer parameters a,b,c,d, excepting just one odd; furthermore it generates precisely _all_ those for which two (or three) pairs of opposite edges are equal, modulo permutations of the vertices A,B,C,D: [AB, BC, CA, BD, AD, CD] = [a*b, (a*c+b*d)/2, (a*c-b*d)/2, (a*c-b*d)/2, (a*c+b*d)/2, c*d] I don't know whether a 5-parameter rational parameterisation exists for all concyclic charts. Fred Lunnon
On 12/8/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
The following 4-parameter expression generates a planar chart of 4 concyclic vertices for any choice of integer parameters a,b,c,d, excepting just one odd; furthermore it generates precisely _all_ those for which two (or three) pairs of opposite edges are equal, modulo permutations of the vertices A,B,C,D:
[AB, BC, CA, BD, AD, CD] = [a*b, (a*c+b*d)/2, (a*c-b*d)/2, (a*c-b*d)/2, (a*c+b*d)/2, c*d]
I don't know whether a 5-parameter rational parameterisation exists for all concyclic charts.
[I'll have to quote myself, since nobody else seems to be going to --- just in case anybody's still listening out there ...] The above seems to be the best that can be expected in the way of a rational parameterisation: it might just be worth using to search for an ambiguous example. I reason as follows [I can't actually quote a theorem here, and I'd be grateful for any relevant references.] If there exists a rational parameterisation for the general planar chart, i.e. flat tetrahedron in terms of its edge lengths, then the variety in projective 5-space defined by the degree-6 equation in 6 homogeneous variables volume(u,v,w,x,y,z) = 0 has a high-order singularity at some point Q. The only candidates for Q turn out to be a folded umbrella [0,0,0,1,1,1], a folded square [0,1,1,1,1,0], and their symmetries. The parameterisation is then constructed by intersecting the variety with the line joining Q to an arbitrary point P: if the singularity was sufficiently fierce, there will only be one other possible point R of intersection with the surface remaining. Unfortunately, for the umbrella there tirn out to be 2 remaining, and for the square 4. So, no rational solution. Turning to the concyclic case, the construction above has to be modified to intersect a plane joining two singularities Q,Q' and arbitrary P with the 4-variety defined by volume(u,v,w,x,y,z) = 0; u z + v y + w x = 0 (Ptolemy relation). Once again, there turn out to be two resp four points of intersection with the variety remaining. In practice, a quadratically irrational parameterisation might still be utilised to restrict a search for charts --- it's just rather more complicated and less restrictive than a rational one! Fred Lunnon
Oof, of course; very nice. What's more, starting with a 3-4-5 and rotating about the midpoints of the length 3 and 5 sides gets you one extra rational (integer, even!) value; you end up with {3, 4, 5, 13, 5*sqrt(145), sqrt(7081)}.
Sorry, MK --- this one is a dud, with no planar perms at all!
Er, well, yes. One does need to clear *all* one's denominators. How do you like {3*18, 4*18, 5*18, 13, 5*sqrt(145), sqrt(7081)} ? Triply planar, I hope. --Michael*18 Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
On 11/29/06, Michael Kleber <michael.kleber@gmail.com> wrote:
Oof, of course; very nice. What's more, starting with a 3-4-5 and rotating about the midpoints of the length 3 and 5 sides gets you one extra rational (integer, even!) value; you end up with {3, 4, 5, 13, 5*sqrt(145), sqrt(7081)}.
Sorry, MK --- this one is a dud, with no planar perms at all!
Er, well, yes. One does need to clear *all* one's denominators. How do you like {3*18, 4*18, 5*18, 13, 5*sqrt(145), sqrt(7081)} ? Triply planar, I hope.
--Michael*18 Kleber
Looks OK to me. My program only checks for two configurations not three, however. Here is one with no right triangles: [ 272, 2*sqrt(6401), 130, 290, 50, 2*sqrt(6890)] Jim Buddenhagen
It only just occurred to me that the geometric constructions we've seen so far for doubly planar charts (no rationality requirement) all build families that preserve a single triangle, and provide alternate placements for the fourth point. Fred's various examples -- integers with the collinearity blemish and generic quartics -- both do likewise. Is this necessary? That is, given the six distances, can you at least identify three of them that must form a triangle, or might there be two planar charts with no triangles in common? (It's possible that someone has posted an example I've missed which already settles this question; I haven't tried to do an exhaustive literature search :-). --Michael -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
In practice, it just seems to happen that if you do not fix a triangle, the only solutions you find have pairs of equal length. Perhaps light might be cast on the subject via an exhaustive analysis of the formal solutions of --- say --- the (essentially) 29_C_2 = 406 possible subsets of equations for a triple planar hexad. Uuurghhh! WFL On 11/30/06, Michael Kleber <michael.kleber@gmail.com> wrote:
It only just occurred to me that the geometric constructions we've seen so far for doubly planar charts (no rationality requirement) all build families that preserve a single triangle, and provide alternate placements for the fourth point.
Fred's various examples -- integers with the collinearity blemish and generic quartics -- both do likewise.
Is this necessary? That is, given the six distances, can you at least identify three of them that must form a triangle, or might there be two planar charts with no triangles in common? (It's possible that someone has posted an example I've missed which already settles this question; I haven't tried to do an exhaustive literature search :-).
On Nov 29, 2006, at 3:52 PM, Michael Kleber wrote:
Let the road lengths be arranged [AB, BC, CA, BD, AD, CD], where A,B,C,D denote cities. Then the follwoing two charts are distinct and planar, and both permute the same set of integers:
[5, 8, 13, 11, 9, 17] [5, 17, 13, 11, 9, 8]
I'm not sure why you decided it was somehow inferior, Fred! Yes, the first hexad has three cities collinear, but when handed a problem about distances along roads, nothing seems more likely than that some city is along the route between two others.
Did anyone verify this, per Fred's request? It certainly seems like the prettiest configuration seen so far. I drew this out, and it checks. It's an instance of the construction Dan first mentioned, triangle ABD with sides 5,11,9 is common to both, and vertex C is on the perpendicular at the foot of the median of this triangle from A to BD. Bill
On 11/28/06, Michael Kleber <michael.kleber@gmail.com> wrote:
This looks like fun; wish I'd had the time to play with it.
Starting with a 3-4-5 triangle, for example, and clearing denominators at the end, Bill's geometric construction yeilds the hexad
108, 144, 180, 5*sqrt(373), 13*sqrt(229), sqrt(44749)
which I'm sure Fred's program will agree has three volume-zero tetrahedra and no degeneracy.
Yup, it checks out [blowing me out of the water once more] --- three distinct planar charts, besides one triple and three double proper tetrahedra [distinct shapes, equal volumes]. My dismal prognosis for finding an integer hexad turns out to result from an ingeniously gross program bug. BTTDB again, WFL
participants (10)
-
Andy Latto -
Bill Thurston -
Bill Thurston -
franktaw＠netscape.net -
Fred lunnon -
James Buddenhagen -
Jason Holt -
Joshua Zucker -
Marc LeBrun -
Michael Kleber