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Bui Quang Tuan
Member since Jun2307

Jun2307, 08:07 AM (EST) 

"Generalization Of Interesting Lemma"

Dear Alex, Thank you very much for very interesting remark about "Generalization Of Van Schooten's Theorem". It is not easy to identify and I should take some times to recognize it. In this time I often use very simple and interesting lemma to prove some results (such as "Cyclic Hexagon" or "Relations in a Cyclic Polygon"). This lemma can be formulated: In the triangle ABC the altitude AH can be calculated by formula: AH = AB*AC/(2*R) where R is circumradius If we take A' = reflection of A in circumcenter O then the formula can be see as: AH = AB*AC/AA' or AH*AA' = AB*AC (1) Please note that H is on sideline BC and A' is on circumcircle (O) of ABC. Moreover angle(HAA') = angle(CBA)  angle(ACB) (2) (all angles are directed) Based on these analyses we can generalize the lemma as following: If X is any point on sideline BC, Y is a point on circumcircle of triangle ABC such that angle(XAY) = angle(CBA)  angle(ACB) (3) (all angles are directed) then AX*AY = AB*AC Proof: Because all angle are directed, we can calculated: angle(XAH) = angle(XAY)  angle(HAY) angle(YAA') = angle(HAA')  angle(HAY) From (2) and (3) we have angle(XAH) = angle(YAA') moreover angle(XHA) = angle(AYA') = 90, therefore two triangle AHX and AYA' are similar and: AH/AX = AY/AA' or AX*AY = AH*AA' = AB*AC (from the lemma (1)) This is what should be proved. Given a point X on sideline BC, there are a lot of ways to construct Y on circumcircle following condition (3) but I very love this way:  construct circumcircle (Ob) of triangle ABX  construct circumcircle (Oc) of triangle ACX  construct circumcircle (Oa) of triangle AObOc  Y is intersection (other than A) of line AOa and circumcircle (O) of ABC (This configuration is "Two Circumcircles in Triangle" configuration in CTK Geometry) Best regards, Bui Quang Tuan


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alexb
Charter Member
2037 posts 
Jun2307, 01:30 PM (EST) 

2. "RE: Generalization Of Interesting Lemma"
In response to message #0

Dear Bui Quang Tuan: >Thank you very much for very interesting remark about >"Generalization Of Van Schooten's Theorem". It is not easy >to identify and I should take some times to recognize it. No, this is quite easy. You got a/d_{a} = a/(PB·PC)·(2R) = a·PA·(2R)/(PA·PB·PC), b/d_{b} = b/(PC·PA)·(2R) = b·PB·(2R)/(PA·PB·PC), c/d_{c} = c/(PA·PB)·(2R) = c·PC·(2R)/(PA·PB·PC). Assuming a = b = c, a/d_{a} = b/d_{b} + c/d_{c} is equivalent to PA = PB + PC. But then this shows that your extension is wrong. (I've been slow to realize this. I am sorry on both accounts; for my sake and yours.) For, otherwise, we would have, for a regular ngon, that one of the distances PX_{i} equals the sum of the remaining ones. But this is clearly impossible. To see that. Take P opposite a vertex X_{1} and keep both of them fixed while letting n to increase. Then PX_{1} remain fixed but the sum in the right hand side increases without bound simply because the number of summands grows and an increasing number of them become close to PX_{1}. The mistake is subtle and occurs on the inductive step. I began preparing a page when I realized that something is wrong. So the page is practically ready and may serve a good example of an honest mistake. But I shall not make it public without your permission. If I do not mention your name, I do not know why you should object. The whole thing is quite edifying. Please let me know. 

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Bui Quang Tuan
Member since Jun2307

Jun2307, 03:04 PM (EST) 

4. "RE: Generalization Of Interesting Lemma"
In response to message #2

Dear Alex, >But then this shows that your extension is wrong. Thank you very much for your remark about the error. I think we can learn also from our errors. If you think that this error can be interesting so please feel free to make one page to analyse these errors. Tomorrow I should pay some time to check the proof again. Now after checking by calculations and drawings, I see that the fact is still true. Please help me to check it again. I hope that our proof may be false but the fact is still true because it is very interesting fact. Best regards, Bui Quang Tuan


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alexb
Charter Member
2037 posts 
Jun2307, 03:34 PM (EST) 

5. "RE: Generalization Of Interesting Lemma"
In response to message #4

>I hope that our proof may be false but the fact is still >true because it is very interesting fact. I am awfully glad to report that you are indeed right. The circumstance that confused me is twofoild: 1. Juxtaposing with van Schooten's Theorem which singles out the longest distance from a point to the vertices. For an ngon this selection of the special vertex is, for one, untenable (as I argued before), but also can't be used on the inductive step. 2. You make a remark which ratio comes in the lefthand side post factum, after the proof whereas I think it must be a part of the theorem statement, see https://www.cuttheknot.org/pythagoras/PointOnCircumcircle2.shtml For, otherwise, you can't be sure that, for the (k1)gon, it is R_{13} that appears on the left. What do you say? 

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Bui Quang Tuan
Member since Jun2307

Jun2307, 05:13 PM (EST) 

6. "RE: Generalization Of Interesting Lemma"
In response to message #5

Dear Alex, >For, otherwise, you can't be sure that, for the (k1)gon, >it is R_{13} that appears on the left. > >What do you say? I can't sleep because thinking about my proof so I wake up and just read your message and also web site page "An Extension of van Schooten's Theorem". I am very very happy. Thank you very much for your analyses and reply. As you, I also confuse sometimes so I wrote "it is not easy to identify". For the "R13 that appears on the left" matter I think so: 1. Our induction assumptions that the theorem is true for any (k1) gon 2. If P is on arc X1X2 so we can choose (k1) gon as X1X3...Xk (in this case the triangle for proof is X1X2X3) and R13 appears on the left or X2X3...Xk (in this case the triangle for proof is XkX1X2) and Rk2 appears on the left we can use any one of them to prove and the result is same. There is only two these cases can be used. For "beauty" I choose X1X2X3 as the triangle for proof. Now we can sure that our result is true and I can sleep well now. I am very happy to work with you. I would like to wish you nice weekend! Thank you again and best regards, Bui Quang Tuan


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alexb
Charter Member
2037 posts 
Jun2307, 07:36 PM (EST) 

7. "RE: Generalization Of Interesting Lemma"
In response to message #6

>I can't sleep because thinking about my proof so I wake up >and just read your message and also web site page "An >Extension of van Schooten's Theorem". I am very very happy. I am happy too. Luckily the mistake was mine. >For the "R13 that appears on the left" matter I think so: >1. Our induction assumptions that the theorem is true for >any (k1) gon >2. If P is on arc X1X2 so we can choose (k1) gon as >X1X3...Xk (in this case the triangle for proof is X1X2X3) >and R13 appears on the left How can you be sure unless you stipulate so? The theorem simply says that one of the ratios is the sum of the others. It's only after the proof a remark is made how to identify the proper index as if this was following from the theorem. However, to avoid ambiguity, this stipulation must be made in the body of the theorem, not as a postfactum remark. Other than that the proof is quite fine. The new result (with X and Y and signed angles) is quite unexpected and I am grateful to you for posting it here. I am leaving tomorrow for a 2 weeks trip where I am not sure I'll have an internet access. In all likelihood I'll be offline for the next two weeks. I'll take care of that result of yours on my return. >Now we can sure that our result is true and I can sleep well >now. Have good dreams, too. >I am very happy to work with you. Thank you for the kind words, although at best you work through me not with me. All the credit is yours. >I would like to wish you nice weekend! Thank you. Same to you. Mine is going to be relatively long. All the best, Alex 

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Bui Quang Tuan
Member since Jun2307

Jun2407, 00:24 AM (EST) 

8. "RE: Generalization Of Interesting Lemma"
In response to message #7

Dear Alex, I just have found one interesting direct proof which can avoid any matter with induction method. First we still must prove for triangle, n=3 with notation about the biggest ratio is respective with the arc where P is. Suppose now n > 3 and P is on the arc X1X2. Because X1X2...Xn is convex so we can devide it into following triangles by the segments with one end at X1 X1X2X3 X1X3X4 .... X1XkX(k+1) .... X1X(n1)Xn With all these triangles P is always on X1Xk arc (k=1..n1) so we can apply triangle case lemma for each of them: R12=R23+R13 R13=R34+R14R1k=Rk(k+1)+R1(k+1) ... R1(n2)=R(n2)(n1)+ R1(n1) R1(n1)=R(n1)n+R1n Sum all these equations and simplify what common we have the result: R12 = R23 + R34 + ... + R(n1)n + Rn1 I think this proof is more understanding. I don't know if you can receive this message or not. In any case I wish you good travel. Best regards, Bui Quang Tuan


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