Re: [EMHL] Re: Perspector
- 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.
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 touchActually, 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 lineIn 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.
Yahoo! Groups SponsorADVERTISEMENT
Yahoo! Groups Links
To visit your group on the web, go to:
To unsubscribe from this group, send an email to:
Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.
Do you Yahoo!?
Yahoo! Mail Address AutoComplete - You start. We finish.
[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 theThat'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 theIndeed, 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
>> From 1) you 'll obtain XYZ and UVW areOh, 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
>> Call A_1, A_2 be the projections of I_aThanks, 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 theWell, 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
For more details, see the MathLinks thread
where I explained the method to Grobber.