- Dear friends:

The incircle and Steiner inellipse intersect in 4 points, one of which, say P, is a triangle center, and the other 3, say A', B', C', form a central triangle (similar to the intersections of NPC and Steiner inellipse = X(115) + medial triangle). P is a non-ETC center, search=2.307963780976356, and A'B'C' is perspective to ABC at non-ETC center, say X, search=1.626790309962831.

Questions:

1.) What are the coordinates of P, A', B', C', and X?

2.) For the ETC reference triangle, A' is the farthest of the 4 intersections from A, and likewise for B' and C'. Does this hold in general?

3.) For the ETC reference triangle, P is the closest of the 4 intersections to X(11). Does this hold in general?

4.) Any other interesting properties?

5.) Generalizations for other inconics?

Thanks in advance,

Randy Hutson > From: rhutson2 <rhutson2@...>

Let t1=sqrt((c+a-b)(a+b-c)), t2=sqrt((a+b-c)(b+c-a)), t3=sqrt((b+c-a)(c+a-b)),

>

> Dear friends:

>

> The incircle and Steiner inellipse intersect in 4 points, one of which, say P,

> is a triangle center, and the other 3, say A', B', C', form a

> central triangle (similar to the intersections of NPC and Steiner inellipse =

> X(115) + medial triangle). P is a non-ETC center, search=2.307963780976356, and

> A'B'C' is perspective to ABC at non-ETC center, say X,

> search=1.626790309962831.

>

> Questions:

>

> 1.) What are the coordinates of P, A', B', C', and X?

with t1,t2,t3 all > 0. Then P=(a-t1, b-t2, c-t3),

A'=(a-t1, b+t2, c+t3), B'=(a+t1, b-t2, c+t3),

C'=(a+t1, b+t2, c-t3) and X=(a+t1, b+t2, c+t3).

>

--

> 2.) For the ETC reference triangle, A' is the farthest of the 4

> intersections from A, and likewise for B' and C'. Does this hold in

> general?

>

> 3.) For the ETC reference triangle, P is the closest of the 4 intersections to

> X(11). Does this hold in general?

>

> 4.) Any other interesting properties?

>

> 5.) Generalizations for other inconics?

>

> Thanks in advance,

> Randy Hutson

Barry Wolk- Dear friends,

[Randy Hutson]> > The incircle and Steiner inellipse intersect in 4

[Barry Wolk]

> points, one of which, say P,

> > is a triangle center, and the other 3, say A', B', C',

> > Questions:

> > 1.) What are the coordinates of P, A', B', C', and X?

> Let t1=sqrt((c+a-b)(a+b-c)), t2=sqrt((a+b-c)(b+c-a)),

[Randy Hutson]

> t3=sqrt((b+c-a)(c+a-b)),

> with t1,t2,t3 all > 0. Then P=(a-t1, b-t2, c-t3),

> A'=(a-t1, b+t2, c+t3), B'=(a+t1, b-t2, c+t3),

> C'=(a+t1, b+t2, c-t3) and

> X=(a+t1, b+t2, c+t3).

> Generalizations for other inconics?

If the isotomic conjugates of the persectors of two

inconics are interior points of ABC with

barycentric coordinates (pp : qq : rr), (PP : QQ : RR)

then their intersections are

S = ( (Qr-qR)^2 : (Rp-rP)^2 : (Pq-pQ)^2 )

A'= ( (Qr-qR)^2 : (Rp+rP)^2 : (Pq+pQ)^2 )

B'= ( (Qr+qR)^2 : (Rp-rP)^2 : (Pq+pQ)^2 )

C'= ( (Qr+qR)^2 : (Rp+rP)^2 : (Pq-pQ)^2 )

and the triangles ABC, A'B'C' are perspective at

X = ( (Qr+qR)^2 : (Rp+rP)^2 : (Pq+pQ)^2 )

Best regards

Nikos Dergiades - Thanks Barry and Nikos. I love it when these problems have such elegant solutions.

Best regards,

Randy Hutson

>________________________________

[Non-text portions of this message have been removed]

> From: Nikolaos Dergiades <ndergiades@...>

>To: Hyacinthos@yahoogroups.com

>Sent: Friday, June 1, 2012 5:33 PM

>Subject: Re: [EMHL] intersections of incircle and Steiner inellipse

>

>Dear friends,

>

>[Randy Hutson]

>> > The incircle and Steiner inellipse intersect in 4

>> points, one of which, say P,

>> > is a triangle center, and the other 3, say A', B', C',

>> > Questions:

>> > 1.) What are the coordinates of P, A', B', C', and X?

>

>[Barry Wolk]

>> Let t1=sqrt((c+a-b)(a+b-c)), t2=sqrt((a+b-c)(b+c-a)),

>> t3=sqrt((b+c-a)(c+a-b)),

>> with t1,t2,t3 all > 0. Then P=(a-t1, b-t2, c-t3),

>> A'=(a-t1, b+t2, c+t3), B'=(a+t1, b-t2, c+t3),

>> C'=(a+t1, b+t2, c-t3) and

>> X=(a+t1, b+t2, c+t3).

>

>[Randy Hutson]

>> Generalizations for other inconics?

>

>If the isotomic conjugates of the persectors of two

>inconics are interior points of ABC with

>barycentric coordinates (pp : qq : rr), (PP : QQ : RR)

>then their intersections are

>S = ( (Qr-qR)^2 : (Rp-rP)^2 : (Pq-pQ)^2 )

>A'= ( (Qr-qR)^2 : (Rp+rP)^2 : (Pq+pQ)^2 )

>B'= ( (Qr+qR)^2 : (Rp-rP)^2 : (Pq+pQ)^2 )

>C'= ( (Qr+qR)^2 : (Rp+rP)^2 : (Pq-pQ)^2 )

>and the triangles ABC, A'B'C' are perspective at

>X = ( (Qr+qR)^2 : (Rp+rP)^2 : (Pq+pQ)^2 )

>

>Best regards

>Nikos Dergiades

>

>

>

>------------------------------------

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