## Re: Two New Theorems For Complete Quadrilateral

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• Dear Tuan and all, a little correction please. ...
Message 1 of 47 , Apr 30, 2006
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Dear Tuan and all, a little correction please.

--- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...> wrote:
>
>
> Also we have that <HIG = 180 - <MbMaMc (3) ...

<HIG = <MbMaMc (3) ...

Best regards
Kostas vittas
• Dear Jeff Brooks, I m sorry that reply you late because today I am busy with my old bookshelf. I m now is not interesting with this configuration more. If you
Message 47 of 47 , May 14, 2006
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Dear Jeff Brooks,
I'm sorry that reply you late because today I am busy with my old bookshelf. I'm now is not interesting with this configuration more. If you read two Clawson papers, Francois and Jean Pierre messages carefully so you can see that the proof for this configuration is not too hard and we can draw not only 1, 13, 35 concurrent circles. In fact, we can draw INFINITE concurrent circles. If you still want to know so I explain as following:
If we limited with only seven lines as I draw so until now I am very sure that the total number of the concurrent circles is 35. I also have drawn all these 35 circles already some times for checking. When drawing, you could note that: 7 lines cut each other at 7*6/2 = 21 cutting points. In each point pass exactly 5 circumcircles, so the number of concurrent circumcircles can be calculated as 5*21/3=35 circles. (We should divide by 3 because each circle is counted 3 times for one triangle). You can see that if in one cutting point pass not enough 5 circles so you can draw more circles pass it. If in one point already pass 5 circles so you can not draw any more.
Off cause, if you have any new interesting idea so please kindly inform me. I am not always true.
Best regards,
Bui Quang Tuan

Jeff Brooks <trigeom@...> wrote: Dear Tuan,

It's already 2:50 o'clock in the morning here. Are you sure only 35=C
(3,7)=C(4,7) has this property for complete quadrilateral? I have no
reason to dispute this, but I am a bit disappointed if true.

Friendly, Jeff

>
> Dear All My Friends,
> Now in Hanoi already 2 o'clock morning. I have already sleep some
time but in the dream I saw very simple method to count the number of
concurrent circles. So I walk up and post it right now because I
don't want that Jeff Brooks or any one have headache and I also.
> Given one complete quadrilateral. It means we have four lines. By
connecting midpoints of its diagonal triangle sides we have three
lines more. Totally we have seven lines. Choose any four lines from
these seven, we can bond one complete quadrilateral and all 35 these
complete quadrilaterals share one common Miquel point.
> Now we can see that ANY TRIANGLE (formed by any three lines
chosen from these seven lines) have circumcirle as one Miquel circles
and off cause, this circumcirle passes through the common Miquel
point. (This fact is very easy because we can choose one another line
bond with three sidelines of the triangle to create one complete
quadrilateral). Because all these triangles are different each other
so the number of concurrent circles is the number of triangles.
> Now we should calculate the number of the triangles bond by three
lines chosen from seven these lines. This is C(3,7)=7!/(3!*4!)=35. So
we have totally 35 concurrent circles.
> Remark: in this configuration, the number of triangles, the
number of complete quadrilaterals and the number of Miquel circles
which concur at one common Miquel point all are equal and are 35=C
(3,7)=C(4,7). Only 7 can has this property. (Dear Jeff Brooks, you
can see something related Pascal Array here:-)).
> Good night and best regards,
> Bui Quang Tuan
>
> Jeff Brooks <trigeom@...> wrote: Dear Tuan,
>
> You are giving me a headache (as Francois might say :-)
>
> I don't know where the beginning and end meets now!
>
> Fun problem nevertheless!
>
> Sincerely, Jeff

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