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Maj. Pestich
guest

Nov2905, 10:44 PM (EST) 

"4 Travellers problem"

Hello Alex, Ca fait longtemps... I read the solution to 4 travellers by S. Anderson and the conclusion that they are colliner at all times does not seem right. While it is true that when two travellers move, the line connecting them does move parallel to itself if they move at constant speeds, but it does not require that the 3rd one stay on the same line. Lines connecting each T to others must stay parallel to themselves. Thus if we have 4 general position lines and let's assume that rendezvous happen in the natural order: 1 meets 2 first, then 3 etc. We have 'fan' of lines 1,2,3,4. Choose random position of T1 on line 1. Connect it by a horizontal segment to a point on line 2. This position will determine relative speeds of T1 and T2. Do the same for positions of T3 and T4. The segments from T1 to T3 and from T1 to T4 will be rotated clockwise wrt T1T2. Connect T2 to T3 and T2 to T4. Now complete the quadrilateral with T3T4. We have flexibility for all except the last segment. The speeds are now determined by the positions of the 4 Ts. They will move with the quadrilateral T1T2T3T4 proportionally shrinking. Double ratio of 4 points, 'perehodia na jargon'. We can project any quadrilateral into any other one (into one made by lines 1,2,3,4). When Ts are collinear the quadrilateral is 'degenerate'. Now I wonder if the lines 1,2,3,4 could be arranged in such a way that it will be impossible to arrange all 4 on one line initially because of irrationality of the rendezvous distances. Will need irrational speeds, I guess. But this is kind of hazy for my military mind. I hope it is correct and makes at least some sense... No referrences to this note are necessary. "Po prochtenii szhech' !" Regards, Maj. Pestich. 

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mr_homm
Member since May2205

Dec0105, 09:16 AM (EST) 

1. "RE: 4 Travellers problem"
In response to message #0

Hello Maj. Pestich, What you say here is correct, and it proves that the 4 travelers must all be on a single line.
Choose random position of T1 on line 1. Connect it by a horizontal segment to a point on line 2. This position will determine relative speeds of T1 and T2. Do the same for positions of T3 and T4. The segments from T1 to T3 and from T1 to T4 will be rotated clockwise wrt T1T2. Connect T2 to T3 and T2 to T4. Now complete the quadrilateral with T3T4. We have flexibility for all except the last segment. The speeds are now determined by the positions of the 4 Ts. They will move with the quadrilateral T1T2T3T4 proportionally shrinking. Double ratio of 4 points, 'perehodia na jargon'. We can project any quadrilateral into any other one (into one made by lines 1,2,3,4).
 The important idea here is that the quadrilateral is proportionally shrinking. Because of this, if one side shrinks to zero, all the other sides must also shrink to zero at the same time. This means all 4 travelers meet at the same time, which shows that all 4 roads must meet at a single point. This contradicts the assumption that the roads are in "general position," which means that they must meet each other separately, not all at once. Therefore it is impossible for the quadrilateral to shrink proportionally. If the quadrilateral is nondegenerate, it MUST shrink proportionally. Therefore, it cannont be nondegenerate. Thus the quadrilateral is a degenerate one, and is really a straight line. > When Ts are collinear the quadrilateral is 'degenerate'. > Correct! There is further discussion of this problem in this thread, where I originally posted the material that Alex kindly made into the page you saw. There, we discussed the cases where the roads were not in general position. It turns out that either the travelers are collinear or the roads are concurrent, or a mixture of the two cases. Since the original problem assumes that the roads are not all concurrent at the same point, only the collinear travelers case is possible. 

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Maj. Pestich
guest

Dec0205, 12:29 PM (EST) 

2. "RE: 4 Travellers problem"
In response to message #1

Hello Stuart, It'seems natural for me to align my people for marching, and they will stay aligned while I look at them. The moment I turn away, the straight lines vanish by real life magic, they start wander in all directions.
Thus it is possible that some of our Ts will travel in opposite directions and the segment connecting two of them will rotate around a fixed point. Also, taking in consideration the kind of world we live in, some lines connecting two Ts (moving at constant speeds!) will envelope parabolas (BTW, we in military use parabolas extensively to shoot around corners by rotating cannons in horizontal position). Thus it gets hazier every moment when we venture beyond 'staying parallel course'. What about my question with irrationalities? Is it possible to arrange paths so that it is quite impossible to stay aligned to maintain proper ratios? And if they differ even well to the right from the decimal pont two Ts may not meet. That would be a failed mission. Maj. Pestich


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Lt. Pestich
guest

Dec1505, 04:26 PM (EST) 

3. "RE: 4 Travellers problem"
In response to message #2

Recently I was demoted for revealing some geometry secrets, but this will not stop me from closing in on the truth about 4 Ts problem. It just seems inconceivable that a problem that old (surely it was created before automobile and even bicycle transportation) would pass such time 'filters' as the great geometers of the past with a 'degenerate' solution of a straight line starting points. It would have a chance to sneak by a blind Euler, but young Pascal would have rejected it outright well before. Yet both of them (as well as countless others) allowed it to exist and reach our time in the form it appeared on CTK pages. Like many, if not all, puzzles on these pages, the 4 Ts problem is a 'play' on one or more of the FUNDAMENTAL relatioships in math and/or physics. I consulted with my superiors and my subordinates and we came up with just 2 theorems 'sponsored' by number 4 :  the double ratio of 4 points  the Abel theorem of solving polinomials of powers less then 5 in radicals There must be more theorems around number 4, but we just do not use them (unlike the two above) in our dry army life on a daily basis. So, is 4 Ts based on one of them, or both of them ? Or neither one, but something else? My last question is: What role, if any, the Heisenberg's uncertainty principle plays in the 4 Ts puzzle? Think about it: if they are aligned exactly on the line, then we can not assure their constant speeds, and under these conditions they will miss each other, granted not by a mile, but still... Yet neither Euler, nor Pascal, not even Poncelet heard about Werner Heisenberg... Salute, Lt. Pestich 

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mr_homm
Member since May2205

Dec1505, 06:29 PM (EST) 

5. "RE: 4 Travellers problem"
In response to message #3

Hello again Lt Pestich! > There must be more theorems around number 4, but we just >do not use > them (unlike the two above) in our dry army life on a >daily basis. Surely you use the famous 4 color map theorem, since maps are a great help in military life.... > So, is 4 Ts based on one of them, or both of them ? > Or neither one, but something else? > > My last question is: What role, if any, the Heisenberg's > uncertainty principle plays in the 4 Ts puzzle? > Think about it: if they are aligned exactly on the line, >then > we can not assure their constant speeds, and under >these > conditions they will miss each other, granted not by a >mile, but > still... Yes, if you try to make this into a "quantum physics" puzzle instead of a "math" puzzle, things are quite different. If you are certain that A, B and C all meet each other, then they cannot say for sure that they stay on their original roads, and the same for A, B and D. Of course in quantum mechanics, they can meet each other even if the roads they are on do not meet, because they are only probably on the roads in the first place. The original solution to the puzzle is only probably true, but the probability increases as the travelers are heavier, because the uncertainty in their momentum does not affect their velocity so much. If the travelers were elephants instead of men, then A and D would be more likely to meet. If they were atoms instead of men, the uncertainty would be so great that whether they met or not would be essentially just random luck. For objects the size of humans, the probability that A and D meet is very close to 100% already. This is all very strange, but as a famous physicist once said, "If you don't find quantum mechanics shocking, you don't really understand it." Stuart Anderson 

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Cpt. Pestich
guest

Dec1605, 02:08 PM (EST) 

6. "RE: 4 Travellers problem"
In response to message #5

Hello fellow travelers, When I think about this problem it is usually along these lines:  it is an old problem (although when I did a search on Google, CTK was the only site that popedup);  for some reason it was composed around number 4, not 5 or 17. it was originated well past the time when 4 was the last number known to mankind, still it got to present times with number 4 in it;  in the old times there were quite a lot of very smart men/women to solve this problem one way or another (look what they accomplished without Sketchpad, armed only with plumes d'oie, compass and ruler);  if a straight line was the only solution it'seems that it wouldn't survived this long as a problem, this solution is too obvious; or it would have reached us as 'as many as you want travelers' problem;  assuming that wayway in the past people like Apollonious decided out of pure modesty to say nothing about adding more fellow travelers and let S. Anderson realize that it is a straight line solution, then the French bunch of Monge, Lemoine, Brocard and Co. would have been only happy to do so. These guys appropriated and put their names on everything that was laying around and gave impression of not being watched. Well, they had an impressive example on European arena to be followed on Euclidean plane, where they fought their battles. After their exploits in that theatre there is practically nothing left for grabs. So it seems to me that there must be something more to this problem. Alex, what is the source of it? Was there a proposed solution? Number 4 is a 'red herring'? I do not think so. I salute you, Cpt. Pestich. 

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Lt. Pestich
guest

Dec1605, 06:34 PM (EST) 

8. "RE: 4 Travelers problem"
In response to message #7

Dear Alex, Was the last paragraph of the solution
https://www.cuttheknot.org/gsolution.shtml in the book? It'seems that it was not. Looks like it was added later, the 'ink' is slightly different. But if it were in the book, ( I do not have this book ) then S. Anderson's recent realization of collinearity with subsequent lemmas and other related statements for it's support would seem to be a bit tardy. What an unexpected low punch it would a been from one of the 'classics' of math! With salute, Lt. Pestich 

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alexb
Charter Member
2631 posts 
Dec1605, 06:56 PM (EST) 

9. "RE: 4 Travelers problem"
In response to message #8

Dear lieutenant, > Was the last paragraph of the solution > in the book? I think I wrote this page early in 1996. For one, I wrote it from memory. It was the problem I loved from my early adolescent years. I had no recollection where I learned about it. Later, when I came across Bolobas' edition and found the problem there, I realized that I saw it in the Russian translation of Mathematical Miscellany by Littlewood himself. The original problem deals with near collisions of 4 ships going on a straight course each. If #1 almost ran into the other three and so did #2, then #3 and #4 had been bound to almost collide as well. > It'seems that it was not. Looks like it was added later, > the 'ink' is slightly different. I received some early responses to the problem. There is a Follow Up section at https://www.cuttheknot.org/gproblems.shtml Somebody made this remark. I added an explanation. > ( I do not have this book ) then S. Anderson's recent > realization of collinearity with subsequent lemmas and > other related statements for it's support would seem to > be a bit tardy. Do not know about that. Stuart approached the problem from a completely different and unexpected angle. His dichotomy is 100% original and an eye opener in some respects. His discovery through the problem of the connection between the Ceva and Menelaus theorems is outright brilliant. Best, etc.


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mr_homm
Member since May2205

Dec1705, 06:33 PM (EST) 

10. "RE: 4 Travelers problem"
In response to message #8

> But if it were in the book, > ( I do not have this book ) then S. Anderson's recent > realization of collinearity with subsequent lemmas and other > related statements for it's support would seem to be a bit > tardy. > What an unexpected low punch it would a been from one > of the 'classics' of math! > > With salute, > > Lt. Pestich >Hello, Lt. Pestich, Let me try to make the situation more clear. I do not and did not claim that the idea of the 4 travelers being on a line is new with me. I fact, in the thread where this was discussed, others mentioned this wellknown fact before I entered the discussion. At first, I did not believe that it could be true, because I understood a different thing by the phrase "general position" than what the problem meant. Therefore, I set out to investigate whether or not it was true that the 4 travelers must be on a straight line. Because my assumption about the roads was different, I found that there were two possibilities: either the 4 travelers were collinear or the 4 roads were concurrent. At that time I realized that the original problem meant by "general position" the assumption that the roads were NOT concurrent at a point. This eliminated the other possibility, leaving only the possibility that the travelers were collinear. So then I was convinced that they were collinear. Then I wrote up my thoughts on the matter, just in case anyone found my analysis interesting, not because I was proving anything new. However, because I had started thinking of the problem a different way, where I made NO assumptions about the roads, instead of assuming (as the original problem did) that they did not all meet at one point, I decided to investigate the other case, and was able to prove that if the travelers were not collinear, then the roads were concurrent. This reminded me of Ceva's and Menelaus' Theorems, which were known and proven long ago, and it showed a new way to prove these old theorems. This is the only thing that I claim was not already known: the application of this puzzle to the joint proof of these theorems. As to your comments about great mathematicians of the past, I am sure that they all could have solved this puzzle as well as or better than I did. Perhaps many of them did. This is a PUZZLE, after all, not an unsolved problem. The solution to it has been known for a long time, and we here are just working on it for fun. There would be no fun in looking up the answer. So we solve for ourselves problems that were solved already long ago. It is also true that this puzzle may not be as ancient as you think. It involves motion, which was not usually considered in classical geometry. The classical Greek idea of Platonic ideal forms has no motion in it, because these ideal forms were supposed to be changeless and permanent. Change was believed to be less perfect than permanence, and so there was not much consideration of motion in geometry. So it is possible that no ancient mathematicians considered this problem at all, since the idea of motion in it would have made it seem to them to be not a "real" geometry problem. Another point about puzzles is that no one tries to make them as general as possible. Theorems should be made general, of course, but puzzles are often stated in very special cases, because they are just for fun and we are not trying to get the greatest generality. Sometimes if the puzzle is stated in the most general form, it is too easy, because the challenge is to find the central idea among the details of the special case. Looking for a needle in a haystack is much more challenging than looking for a needle on the clean surface of an empty desk. As a matter of fact, I did present a general solution to the puzzle for any number of travelers and roads, with some roads meeting concurrently, others not, and some travelers collinear, others not. It is in the thread I referenced before. But if someone tried to make a puzzle out of the general case for many travelers, then just stating what you were supposed to prove would give away too much information, and the puzzle would be so easy that it would not be a puzzle any more. So the short summary is, I did not discover that the travelers are on a straight line, but I did use this puzzle to prove Ceva's and Menelaus' Theorem. This puzzle may not be ancient, but it was solved long ago, and the puzzle does work for more than 4 travelers. I hope this makes the situation more clear. Thank you, Stuart Anderson 

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Lt. pestich
guest

Dec2005, 12:18 PM (EST) 

11. "RE: 4 Travelers problem"
In response to message #10

Hello fellow travelers! It'seems that my controversial speculation did not find wide support. In fact, there was no support at all, which means one thing only  it has to be dropped. I'm doing it right at this moment. It is a done deal.
Yet, it would be a better puzzle if there would exist a nonlinear solution. As for number 4, it looks like it is a smallest number that would make this puzzle a reasonable one. It is interesting to know if Littlewood was the composer of it, and what was he trying to demonstrate with it. It would seem that it had to be something from projective geometry, and this was what made me to speculate about nonlinear solution. I thought that it was a play on double ratios, that one can projectively transform one quadrilateral into any other. About Ceva and Menelaus: one is the polar transformation of the other: take a point as a circle center in the plane of triangle ABC, its sides cut by a transversal in K,L,M. Poles of the triangle ABC sides make another triangle XYZ and pole of transversal is P. Then polars of K,L,M become cevians of XYZ concurrent in P.
Salute and Season Greatings Lt. Pestich.


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alexb
Charter Member
2631 posts 
Dec2005, 06:31 PM (EST) 

12. "RE: 4 Travelers problem"
In response to message #11

>About Ceva and Menelaus: one is the polar transformation >of the other: take a point as a circle center in the plane >of triangle ABC, its sides cut by a transversal in K,L,M. >Poles of the triangle ABC sides make another triangle XYZ >and pole of transversal is P. Then polars of K,L,M become >cevians of XYZ concurrent in P. Thank you. You deserve to be a colonel. Implicit in the remark is an explanation why the same expressions in Ceva and Menelaus have different signs. This is because the configuration involves 3 harmonic pairs of points. I am eager to add this remark to either https://www.cuttheknot.org/pythagoras/HarmonicRatio.shtml or https://www.cuttheknot.org/Curriculum/Geometry/PolePolarTriangle.shtml No doubt working geometers are cognizant of the fact, but I, like Stuart, have not seen it mentioned anywhere. 

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alexb
Charter Member
2631 posts 
Dec2105, 01:51 PM (EST) 

16. "RE: 4 Travelers problem"
In response to message #14

> Let me stay Lt. If it were only up to me. It's all in the hands of your superiors now. > All credit goes to real General Yaglom. > Take a look at Vol 2 page 100 prob 165 (russian > edition). Yeap. It's volume III (Geometric Transformations, problem 65) in the English edition. However, my proof is more transparent. Thank you again. 

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Lt. Pestich
guest

Dec2105, 09:28 PM (EST) 

18. "RE: 4 Travelers problem"
In response to message #17

Let's assume that 4 travelers problem now set in cyclic environment, 4 circles appropriately intersect so that 1st meets 2nd etc. Will there be a solution? Will it be a concyclic initial position? Will they stay concyclic as they circulate at constant speeds? Can we painlessly add another circulator? Do they have to move all clockwise, or one or more can circulate in counter direction? Salute, Lt. Pestich. 

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Capt. Pestich
guest

Jan2506, 05:32 PM (EST) 

19. "RE: 4 Travelers problem"
In response to message #18

Personal.  Hello Alex!
Recently I reread some material about trilinear polars and it hit me that it might be the FUNdaMENTAL relationship played out in the 4 Ts problem. For any 4th line ( a polar) it is possible to find it's pole wrt triangle made by the other 3 lines so that all the ratios in the picture are harmonic. It makes it possible to assign speeds. This is also what makes Ceva and Menelaos 'meet on these roads' time and time again. Stuart sensed it all right! I still do not like the collinear position of the travelers.
Come to think of it there is too much harmony in the picture. That is why it was so attractive as a possible target to a military mindset and thus caused me to try to bomb it right away at the end of 2005. Salut,
Capt. Pestich 

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