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darij
Member since Jan-9-04
Jan-11-04, 08:15 PM (EST)

"Synthetic proof of CB's Conjecture"

Subject     Author     Message Date     ID
Synthetic proof of CB's Conjecture darij Jan-11-04 TOP
Autopolar ? Bractals Feb-09-04 1
RE: Autopolar ? darij Feb-10-04 2
RE: Autopolar ? Bractals Feb-10-04 3
Three more cases Bractals Feb-13-04 4
RE: Three more cases Bractals Mar-19-04 5
RE: Three more cases darij Apr-06-04 6

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Bractals
Member since Jun-9-03
Feb-09-04, 01:28 PM (EST)

1. "Autopolar ?"
In response to message #0

 Hi darij,Do you know a place on the internet where I can find a proof (in english) that the diagonal triangle of a cyclic quadrilateral is autopolar?Thanks,Bractals

darij
Member since Jan-9-04
Feb-10-04, 08:40 AM (EST)

2. "RE: Autopolar ?"
In response to message #1

 Dear Bractals,You wrote:>> Do you know a place on the internet where I can find a>> proof (in english) that the diagonal triangle of a>> cyclic quadrilateral is autopolar?Well, it is just necessary to show that if we have fourpoints A, B, C, D on a circle k, and K is theintersection of AB and CD, L is the intersection of BCand DA, and M is the intersection of AC and BD, thentriangle KLM is autopolar with respect to k.Theorem 1 of Vladimir's "Quadrilateral inversions" postin this forum (College Math), applied to the cyclicquadrilateral ABCD, yields that L lies on the polar ofK with respect to k. The same fact, applied to thecyclic quadrilateral ABDC, shows that M lies on thepolar of K. Hence, the line LM is the polar of K withrespect to k. Similarly, MK is the polar of L, and KLis the polar of M, so that the triangle KLM is indeedautopolar. Sincerely, Darij Grinberg

Bractals
Member since Jun-9-03
Feb-10-04, 08:39 PM (EST)

3. "RE: Autopolar ?"
In response to message #2

 Hi Darij,Thanks for the pointer to Vladimir's Theorem #1. Bractals

Bractals
Member since Jun-9-03
Feb-13-04, 01:34 PM (EST)

4. "Three more cases"
In response to message #0

 Hi Darij,I have followed your proof through paragraph 5 for Case I (see below) the configuration of Christopher Bradley. Using the following notation: ` i(XYZ) denotes the incenter of triangle XYZ e(XYZ) denotes the excenter of triangle XYZ on the bisector of angle XYZ e(YZX) denotes the excenter of triangle XYZ on the bisector of angle YZX e(ZXY) denotes the excenter of triangle XYZ on the bisector of angle ZXY`The proof also holds for cases II - IV (replacing 'in' with 'ex' where required).` CASE I II III IV A' i(DAB) e(DAB) i(DAB) e(DAB) B' i(ABC) e(ABC) e(ABC) i(ABC) C' i(BCD) e(BCD) i(BCD) e(BCD) D' i(CDA) e(CDA) e(CDA) i(CDA) X i(PAB) e(BPA) e(ABP) e(PAB) Y i(PBC) e(CPB) e(PBC) e(BCP) Z i(PCD) e(DPC) e(CDP) e(PCD) W i(PDA) e(APD) e(PDA) e(DAP) `Note: The four cases exhaust the sixteen "centers" of the triangles PAB, PBC, PCD, and PDA.Bractals

Bractals
Member since Jun-9-03
Mar-19-04, 03:25 PM (EST)

5. "RE: Three more cases"
In response to message #4

 Hi Darij,Here is a proof for Cases I - IV in the previous post. If you have a chance to look it over, you might tell me why I did not need Monge's theorem. Is it only because the results of Monge's theorem are needed in your proof from paragraph 6 and beyond?Thanks,Bractals

darij
Member since Jan-9-04
Apr-06-04, 03:32 PM (EST)

6. "RE: Three more cases"
In response to message #5

 Dear Bractals,You wrote:>> Here is a proof for Cases I - IV in the previous>> post. If you have a chance to look it over, you>> might tell me why I did not need Monge's theorem.Thanks - your proof is indeed a bit shorter thanmine, since I use Monge and Desargues, while you usedDesargues only. Once we know that the lines AC, A'C',XY (your A*B*) and ZW (your C*D*) concur, it istrivial that their point of concurrence is theexsimilicenter of the circles (A') and (C'), the oneof the circles (X) and (Y), and the one of (Z) and(W), since AC is a common tangent of these pairs ofcircles and A'C', XY and ZW are their respectivecentral lines. Hence, we don't need Monge at all! Sincerely, Darij Grinberg