## Re: [EMHL] Re: Three Bicentric Quadrangles

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• Dear All My Friends, After a lot of effort I have found only one perspector but I think not so nice: As in my construction, we have three circles (Oa), (Ob),
Message 1 of 17 , Jul 3, 2006
Dear All My Friends,
After a lot of effort I have found only one perspector but I think not so nice:
As in my construction, we have three circles (Oa), (Ob), (Oc) respective with A1, B1, C1. Let's circle (Oa) touches A1B at Ab, Ac. Similarly we have Bc, Ba, Ca, Cb. Three lines AbAc, BcBa, CaCb bound one triangle A2B2C2. This triangle and reference ABC are perspective at one point, say P.
Three sidelines of ABC cut three sidelines of A2B2C2 at six points on one ellipse. Details:
BC cuts A2B2, A2C2 at Xb, Xc respectively
CA cuts B2C2, B2A2 at Yc, Ya respectively
AB cuts C2A2, C2B2 at Za, Zb respectively
Six points Xb, Xc, Yc, Ya, Za, Zb are on one ellipse.
Three lines YaZa, ZbXb, XcYc bound one triangle A3B3C3. Triangle A3B3C3 is perspective with ABC and A2B2C2 with perspector at the same point P.
At the first sight, I think not all inscribed hexagon can have this properties and P may be covers some essential features of this configuration.
Point P may be not in current ETC.
I am working to find some properties of P.
Best regards,
Bui Quang Tuan

Quang Tuan Bui <bqtuan1962@...> wrote: My construction more complicated as following:
Let's I is incenter, Ia is touch point of incircle with BC. Perpendicular bisector of BC cuts circumcircle (O) at A' (the same side with A wrt BC). The circle centered at A' and passing through B, C cuts line IaI at Oa. Draw a circle (Oa) centered at Oa passing through Ia. Two tangent lines from B, C to circle (Oa) cut each other at A1 on circumcircle (O). The quadrangle A1ABC is bicentric. Similarly we can construct B1, C1.

There are three points on circumcircle each bound with reference

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• Dear Nikos. Thanks also to you, for your another simple construction, give us by the same way, all the bicentric quadrilaterals ( with incircles or excircles
Message 2 of 17 , Jul 4, 2006
Dear Nikos.

Thanks also to you, for your another simple construction, give
us by the same way, all the bicentric quadrilaterals ( with incircles
or excircles ).

Best regards.
Kostas Vittas.

>
> The perpendicular bisector of BC meets the circumcircle of ABC
> at the points A' , A".
> If AB=AC then the required point D is A'.
> If AC > AB then the circles (A', A' B), (A'', A'' B),
> meet CA and CD at two points C', C" equidistant from C
> such that CC' = CC" = AC - AB = DC - DB
> Constructing the circle (C, CC') is easy the construction
> of D on the arc BC that contains A'.
> Similarly the construction of the other two points on the
> arcs CA, AB.
>
> Best regards
>
• Dear All My Friends, Let s A1 is a point on circumcircle of triangle ABC that ABCA1 is bicentric quadrangle (A1 respective with A. It means A1 is on arc BC,
Message 3 of 17 , Jul 6, 2006
Dear All My Friends,
Let's A1 is a point on circumcircle of triangle ABC that ABCA1 is bicentric quadrangle (A1 respective with A. It means A1 is on arc BC, opposite to A). Let's Ar is reflection of A1 wrt BC). Line BAr cuts AC at Ac, line CBr cuts AB at Ab. The quadrangle AArAbAc is also bicentric quadrangle. Similarly we can construct two other bicentric quadrangles BBrBcBa, CCrCaCb.
Best regards,
Bui Quang Tuan

Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
After a lot of effort I have found only one perspector but I think not so nice:
As in my construction, we have three circles (Oa), (Ob), (Oc) respective with A1, B1, C1. Let's circle (Oa) touches A1B at Ab, Ac. Similarly we have Bc, Ba, Ca, Cb. Three lines AbAc, BcBa, CaCb bound one triangle A2B2C2. This triangle and reference ABC are perspective at one point, say P.
Three sidelines of ABC cut three sidelines of A2B2C2 at six points on one ellipse. Details:
BC cuts A2B2, A2C2 at Xb, Xc respectively
CA cuts B2C2, B2A2 at Yc, Ya respectively
AB cuts C2A2, C2B2 at Za, Zb respectively
Six points Xb, Xc, Yc, Ya, Za, Zb are on one ellipse.
Three lines YaZa, ZbXb, XcYc bound one triangle A3B3C3. Triangle A3B3C3 is perspective with ABC and A2B2C2 with perspector at the same point P.
At the first sight, I think not all inscribed hexagon can have this properties and P may be covers some essential features of this configuration.
Point P may be not in current ETC.
I am working to find some properties of P.
Best regards,
Bui Quang Tuan

Quang Tuan Bui wrote: My construction more complicated as following:
Let's I is incenter, Ia is touch point of incircle with BC. Perpendicular bisector of BC cuts circumcircle (O) at A' (the same side with A wrt BC). The circle centered at A' and passing through B, C cuts line IaI at Oa. Draw a circle (Oa) centered at Oa passing through Ia. Two tangent lines from B, C to circle (Oa) cut each other at A1 on circumcircle (O). The quadrangle A1ABC is bicentric. Similarly we can construct B1, C1.

There are three points on circumcircle each bound with reference

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• Dear All My Friends, I am sorry for typo mistake. It should be read as following: ... Line BAr cuts AC at Ac, line CAr cuts AB at Ab. ... Best regards, Bui
Message 4 of 17 , Jul 6, 2006
Dear All My Friends,
I am sorry for typo mistake. It should be read as following:
...
Line BAr cuts AC at Ac, line CAr cuts AB at Ab.
...
Best regards,
Bui Quang Tuan

Quang Tuan Bui <bqtuan1962@...> wrote:
Dear All My Friends,
Let's A1 is a point on circumcircle of triangle ABC that ABCA1 is bicentric quadrangle (A1 respective with A. It means A1 is on arc BC, opposite to A). Let's Ar is reflection of A1 wrt BC). Line BAr cuts AC at Ac, line CBr cuts AB at Ab. The quadrangle AArAbAc is also bicentric quadrangle. Similarly we can construct two other bicentric quadrangles BBrBcBa, CCrCaCb.
Best regards,
Bui Quang Tuan

Quang Tuan Bui wrote: Dear All My Friends,
After a lot of effort I have found only one perspector but I think not so nice:
As in my construction, we have three circles (Oa), (Ob), (Oc) respective with A1, B1, C1. Let's circle (Oa) touches A1B at Ab, Ac. Similarly we have Bc, Ba, Ca, Cb. Three lines AbAc, BcBa, CaCb bound one triangle A2B2C2. This triangle and reference ABC are perspective at one point, say P.
Three sidelines of ABC cut three sidelines of A2B2C2 at six points on one ellipse. Details:
BC cuts A2B2, A2C2 at Xb, Xc respectively
CA cuts B2C2, B2A2 at Yc, Ya respectively
AB cuts C2A2, C2B2 at Za, Zb respectively
Six points Xb, Xc, Yc, Ya, Za, Zb are on one ellipse.
Three lines YaZa, ZbXb, XcYc bound one triangle A3B3C3. Triangle A3B3C3 is perspective with ABC and A2B2C2 with perspector at the same point P.
At the first sight, I think not all inscribed hexagon can have this properties and P may be covers some essential features of this configuration.
Point P may be not in current ETC.
I am working to find some properties of P.
Best regards,
Bui Quang Tuan

Quang Tuan Bui wrote: My construction more complicated as following:
Let's I is incenter, Ia is touch point of incircle with BC. Perpendicular bisector of BC cuts circumcircle (O) at A' (the same side with A wrt BC). The circle centered at A' and passing through B, C cuts line IaI at Oa. Draw a circle (Oa) centered at Oa passing through Ia. Two tangent lines from B, C to circle (Oa) cut each other at A1 on circumcircle (O). The quadrangle A1ABC is bicentric. Similarly we can construct B1, C1.

There are three points on circumcircle each bound with reference