## Re: [EMHL] Three Triangle Circles

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• Dear Andreas, Do you have a construction of circles (Kabb), (Kacc)? If A1 is the mid point of (Kab), (Kac) then the triangles ABC, A1B1C1 are perspective at
Message 1 of 3 , Feb 23, 2013
Dear Andreas,

Do you have a construction of circles
(Kabb), (Kacc)?

If A1 is the mid point of (Kab), (Kac)
then the triangles ABC, A1B1C1 are perspective
at X(1123) isotomic conjugate of (bc+S:ca+S:ab+S)

If the circles
(Kabb) is tangent externally to (Kab) and to BC, and (Kacc)to
(Kac) and to BC then the triangles ABC, K1K2K3
are perspective at the isotomic conjugate of
(SA + 2S + 2bc : SB + 2S +2ca : . . .)

But your problem seems to be difficult.

Best regards

> Let ABC be a triangle and (Kab),
> (Kac) the two congruent
> and tangent circles such that (Kab) is tangent to the
> sides of the internal angle B and (Kac) to the sides of the
>
> internal angle C.
> Let (Kabb), (Kacc) be the two congruent tangent circles such
> that
> (Kabb) is tangent externally to (Kab) and to AB, and (Kacc)
> to
> (Kac) and to AC.
>
> The quadrilateral KabKabbKaccKac is inscriptible. Let (K1)
> be its incircle. Similarly define (K2), (K3).
>
> (For external angles we get another three circles.)
>
> K1K2K3 is perspective with triangles????
> Is it with ABC?
>
> APH
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
• Dear Nikos, A pure geometric construction of (Kabb), (Kacc) would be quite difficult, since even the construction for lines instead of circles (Kab), (Kac) is
Message 2 of 3 , Feb 23, 2013
Dear Nikos,

A pure geometric construction of (Kabb), (Kacc) would be quite difficult,
since even the construction for lines instead of circles (Kab), (Kac)
is difficult !!
(it is the problem of constucting two mutually tangent congruent
circlles, each one of whose is tangent to the sides of opposite angles

But algebraically/trigonometrically we can compute the radius of the first
circles (Kab), (Kac), and after that the radius of (Kabb), (Kacc), more or less
easily I guess!!

APH

On Sat, Feb 23, 2013 at 4:18 PM, Nikolaos Dergiades <ndergiades@...> wrote:
> Dear Andreas,
>
> Do you have a construction of circles
> (Kabb), (Kacc)?
>
> If A1 is the mid point of (Kab), (Kac)
> then the triangles ABC, A1B1C1 are perspective
> at X(1123) isotomic conjugate of (bc+S:ca+S:ab+S)
>
> If the circles
> (Kabb) is tangent externally to (Kab) and to BC, and (Kacc)to
> (Kac) and to BC then the triangles ABC, K1K2K3
> are perspective at the isotomic conjugate of
> (SA + 2S + 2bc : SB + 2S +2ca : . . .)
>
> But your problem seems to be difficult.
>
> Best regards
>
>
>> Let ABC be a triangle and (Kab),
>> (Kac) the two congruent
>> and tangent circles such that (Kab) is tangent to the
>> sides of the internal angle B and (Kac) to the sides of the
>>
>> internal angle C.
>> Let (Kabb), (Kacc) be the two congruent tangent circles such
>> that
>> (Kabb) is tangent externally to (Kab) and to AB, and (Kacc)
>> to
>> (Kac) and to AC.
>>
>> The quadrilateral KabKabbKaccKac is inscriptible. Let (K1)
>> be its incircle. Similarly define (K2), (K3).
>>
>> (For external angles we get another three circles.)
>>
>> K1K2K3 is perspective with triangles????
>> Is it with ABC?
>>
>> APH
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