- Hi Treegoner,

Nice to see you at Hyacinthos!

In Hyacinthos message #10379, you wrote:

>> If I 'm not mistaken, X, Y, Z are the touch

Actually, they are not. X, Y, Z are defined as

>> points of (I_a) and BC, (I_b) and CA, (I_c)

>> and AB.

the touch points of (I_a), (I_b), (I_c) with

the nine-point circle of triangle ABC.

>> Then I think P can't lie on the Euler line

In fact, what you are speaking about are nice

>> of ABC.

>>

>> Because I did prove some properties :

>>

>> 1) UVW and I_aI_bI_c are homothetic. Then

>> their two Euler lines must be parallel

>>

>> 2) UVW and ABC are perspective at a point

>> H which is the circumcenter of UVW.

>>

>> 3) UVW and XYZ are perspective at P which

>> is the orthocenter of UVW.

facts, however not directly related to Juan

Carlos' perspector, since your points X, Y, Z

are not his points X, Y, Z. You can find some

other properties of your configuration in

Hyacinthos message #6729.

Sincerely,

Darij Grinberg - Dear friends,

You 're right Darij. I misunderstood the problem because Carlos wrote NPC which disconcerted me.

But anyway, I have an idea.

1) If you call A', B' be respectively the projections of I_a, and I_b on AB, then U, V, A', B', X, Y lie on the same circle.

2) The orthocenter of ABC is the circumcenter of UVW (my previous post)

From 1) you 'll obtain XYZ and UVW are perspective at P which lies on the line joining the two circumcenters of ABC and UVW which is the Euler line of ABC (from 2)).

In fact, I get stuck with the first one. I spent 3 hours proving that but I didn't make it out. You can just only prove UVXY is cyclic, but I just elaborate two more points A', B'. I 'm quite sure of it (90%) although I do not have a solution to it. I always believe in "my sixth sense of geometry" though sometimes they are wrong. From this proof (if it 's right) you 'll get the equality I think interesting :

Call A_1, A_2 be the projections of I_a on AB, AC, B_1, B_2 be the projections of I_b on BC, AB, C_1, C_2 be the projections of I_a on AC, BC. Then

XA_1 . YB_1 . ZC_1 = XA_2 . YB_2 . ZC_2.

I 'm tired now. I 'll try it again tomorrow. I hope someone will find it out.

Treegoner

PS Hey Darij, how can you remember the number of the point X(?) in ETC ? See you.

Darij Grinberg <darij_grinberg@...> wrote: Hi Treegoner,

Nice to see you at Hyacinthos!

In Hyacinthos message #10379, you wrote:

>> If I 'm not mistaken, X, Y, Z are the touch

Actually, they are not. X, Y, Z are defined as

>> points of (I_a) and BC, (I_b) and CA, (I_c)

>> and AB.

the touch points of (I_a), (I_b), (I_c) with

the nine-point circle of triangle ABC.

>> Then I think P can't lie on the Euler line

In fact, what you are speaking about are nice

>> of ABC.

>>

>> Because I did prove some properties :

>>

>> 1) UVW and I_aI_bI_c are homothetic. Then

>> their two Euler lines must be parallel

>>

>> 2) UVW and ABC are perspective at a point

>> H which is the circumcenter of UVW.

>>

>> 3) UVW and XYZ are perspective at P which

>> is the orthocenter of UVW.

facts, however not directly related to Juan

Carlos' perspector, since your points X, Y, Z

are not his points X, Y, Z. You can find some

other properties of your configuration in

Hyacinthos message #6729.

Sincerely,

Darij Grinberg

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

In Hyacinthos message #10384, you wrote:

>> 1) If you call A', B' be respectively the

That's right. Juan Carlos has communicated

>> projections of I_a, and I_b on AB, then U,

>> V, A', B', X, Y lie on the same circle.

this result to me. However, I have no

synthetic proof yet.

>> 2) The orthocenter of ABC is the

Indeed, this follows from a simple angle

>> circumcenter of UVW

chase once it is shown that the orthocenter

of triangle ABC lies on the lines AU, BV and

CW.

>> From 1) you 'll obtain XYZ and UVW are

Oh, that's nice! In fact, I see what you do:

>> perspective at P which lies on the line

>> joining the two circumcenters of ABC and

>> UVW which is the Euler line of ABC

>> (from 2)).

From 1), the points V, W, Y, Z lie on one

circle; similarly, the points W, U, Z, X lie

on one circle, and similarly, the points U,

V, X, Y lie on one circle. The pairwise

radical axes of these circles are the lines

XU, YV and ZW, and hence these lines concur

at one point, the radical center P of these

three circles. And thus, XP * PU = YP * PV

= ZP * PW is the common power of this point

P with respect to every of these three

circles. But the equation XP * PU = YP * PV

= ZP * PW can be also rewritten as

PX * PU = PY * PV = PZ * PW, and this

yields that there exists an inversion (with

real radius or with imaginary radius) which

maps the points X, Y, Z to the points U, V,

W. And now one readily sees that the

circumcenters of triangles XYZ and UVW lie

on one line with the point P. But the

circumcenter of triangle XYZ is the

nine-point center of triangle ABC, and the

circumcenter of triangle UVW is the

orthocenter of triangle ABC. Hence, the

point P lies on the Euler line of triangle

ABC.

>> Call A_1, A_2 be the projections of I_a

Thanks, a nice formula about the

>> on AB, AC, B_1, B_2 be the projections of

>> I_b on BC, AB, C_1, C_2 be the

>> projections of I_a on AC, BC. Then

>>

>> XA_1 . YB_1 . ZC_1 = XA_2 . YB_2 . ZC_2.

ubiquituous Feuerbach points...

>> PS Hey Darij, how can you remember the

Well, I don't remember anything :-)

>> number of the point X(?) in ETC ?

I just draw a triangle ABC with sides

6, 9, 13, then draw the point of this

triangle whose number (in the ETC) I want

to find, then I measure its distance to

the side BC of this triangle, and then I

look up in the "Search" list of

Kimberling's ETC:

http://faculty.evansville.edu/ck6/encyclopedia/search.html

For more details, see the MathLinks thread

http://www.mathlinks.ro/Forum/viewtopic.php?t=4437

where I explained the method to Grobber.

Friendly,

Darij Grinberg