ADGEOM 1022 - ADGEOM 1044

#1022

 

 

Happy New Year and a fruitful 2014 to everybody

Here is an easy puzzle with a new triangle center:

X(__) is the only point whose reflections in the sidelines of ABC are collinear and perspective with ABC. The perspector is a triangle center with homogeneous barycentric coordinates


__________________________

and (6,9,13)-search number 0.123280682081... The line containing these reflections is _________________, and the  rectangular circum-hyperbola through this perspector has asymptotes parallel and perpendicular to _____________________.
 It intersects the circumcircle at X(___), which is __________________  (give a simple geometric description).

Best regards
Sincerely
Paul Yiu

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#1024

hints : E574, K060, K037...

 

Could we replace the reflection triangle of P by the homothetic of the pedal triangle of P, center P, ratio k ?

 

With my best wishes for 2014

 

Best regards

 

Bernard Gibert 

 

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#1025

 

Dear Paul

Happy New Year to you and to the ADGeometers

Your conditions mean that your point lies on the circumcircle (collinearity) and on the Neuberg cubic (perspectivity). Thus your point is the isogonal conjugate of the infinite point of the Euler line and the line containing the three reflections, being its Steiner line, is the perpendicular at H to the Euler line.

I fear that we need some computation for your other questions.

Kind regards

Sincerely

Jean-Pierre Ehrmann

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#1026

 

Dear Paul

in fact, if E=X(74) is the isogonal conjugate of the infinite point of the Euler line, your new point is the isogonal conjugate of the homothetic of E in (O,-1/2)

Kind regards

Sincerely

Jean-Pierre Ehrmann

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#1027

 

Dear Bernard

"Could we replace the reflection triangle of P by the homothetic of the pedal triangle of P, center P, ratio k ?"

Yes, but we have not necessarily unicity.

If L is the point of the Euler line such as HL/OL=k-1, the homothetic in (P,k) of the pedal triangle of P is degenerated and perspective with ABC if and only if the Simson line of the antipode of P on the circumcircle goes through L. (Of course, P must lie on the circumcircle)

In any case, I suspect that you knew already the result.

Kind regards

Sincerely Jean - Pierre Ehrmann

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#1029

 

Dear Bernard

I wrote : "If L is the point of the Euler line such as HL/OL=k-1, the homothetic in (P,k) of the pedal triangle of P is degenerated and perspective with ABC if and only if the Simson line of the antipode of P on the circumcircle goes through L. (Of course, P must lie on the circumcircle)"

In this case the perspector is the point Q of the line OP such as PO/OQ=k

Kind regards

Sincerely Jean - Pierre Ehrmann

 

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#1031

 

Dear Jean-Pierre,

 

 

[JPE] I wrote : "If L is the point of the Euler line such as HL/OL=k-1, the homothetic in (P,k) of the pedal triangle of P is degenerated and perspective with ABC if and only if the Simson line of the antipode of P on the circumcircle goes through L. (Of course, P must lie on the circumcircle)"

In this case the perspector is the point Q of the line OP such as PO/OQ=k

Simson lines again ! very nice and elegant.

 

a natural question now : what is locus of Q when k varies ?

 

a nice cubic curve ?

 

Best regards

 

Bernard Gibert

 

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#1032

 

A correction

I wrote "In this case the perspector is the point Q of the line OP such as PO/OQ=k"; the correct statement is

In this case the perspector is the isogonal conjugate of the point Q of the line OP such as PO/OQ=k

Kind regards

Sincerely Jean - Pierre Ehrmann

 

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#1033

 

Dear Bernard

"a natural question now : what is locus of Q when k varies ? a nice cubic curve ?"

 

I've tried unsuccessfully to delete #1029 (and #1032) : I've forgotten "isogonal conjugate of"

The isogonal conjugate Q of the perspector (OP=k.QO) moves on a circular quintic and the perspector on a bicircular septic.

 

Best regards Jean - Pierre Ehrmann

 

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#1034

 

Dear Jean-Pierre,

 

The isogonal conjugate Q of the perspector (OP=k.QO) moves on a circular quintic and the perspector on a bicircular septic.

I find that your quintic is actually Q011.

 

http://bernard.gibert.pagesperso-orange.fr/curves/q011.html

 

Best regards

 

Bernard Gibert

 

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#1036

 

Dear Bernard

Many thanks.

So, we can try to summarize :

suppose that the Simson line of a point M of the circumcircle intersects the Euler line at L.

Then, if P is the antipode of M and k=1+LH/LO, the homothetic of the pedal triangle of P in (P,k) is degenerated and perspective with ABC. The isogonal conjugate Q of the perspector is the common point of the diameter O,M,P and of the Simson line of M (thus Q lies on Q011). More over OM/OQ=k.

If k is given, this is the only way to get the points P (at least 1, at most 3) such as the homothetic of the pedal triangle of P in (P,k) is degenerated and perspective with ABC.

Best regards. Jean - Pierre Ehrmann

 

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#1037

 

Dear Jean-Pierre,

 

If k is given, this is the only way to get the points P (at least 1, at most 3) such as the homothetic of the pedal triangle of P in (P,k) is degenerated and perspective with ABC.

 

if S is such that (k-2)OS - k OH  = 0 (vectors) then these points P lie on the rectangular hyperbola through X3, X110, X2574, X2575 and S.

 

they are the intersections of (O) and pK(X6, S).

 

Best regards
Bernard Gibert

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#1038

 

Dear Bernard

 

I expressed myself poorly (as usual in English)

Of course, there are many ways to construct our points P as the common points (or the antipodes of the common points) of the circumcircle and a rectangular hyperbola.

I was meaning that my summary gives all the solutions.

More over, it's very difficult for me to visualize the points in ETC with a large number; I have to realize that X2574-2575 are the infinite points of the Jerabek hyperbola to follow your statement.

 

Best regards. Jean - Pierre Ehrmann

 

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#1039

 

Dear Bernard

if Q* (isogonal conjugate of Q) is the perspector, then PQ* is perpendicular to the Simson line of P.

More precisely, il P' is the projection of P upon its Simson line, then Vect(PQ*) = 2k/(k+1)Vect(PP')

Kind regards Jean - Pierre Ehrmann

 

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#1040

 

Dear Paul and friends,

(6,9,13)-search number 0.123280682081 is equivalently:

cevapoint of X(13) and X(14)
isogonal conjugate of X(1511)
trilinear pole of line X(1637)X(1989)
the perspector of ABC and the reflection of the Euler triangle in the Euler line

Also, let A'B'C' be the tangential triangle of the Kiepert hyperbola.
Let A" be the intersection, other than X(3258), of the nine-point circle and line A'X(3258); define B", C" cyclically.
The lines AA", BB", CC" concur in this point.

Trilinears: sin A csc 3A/(cos A - 2 cos B cos C) : :
Barycentrics: 1/[3a^6(b^2 + c^2) - 6a^4(b^4 + c^4) + 3a^2(b^6 + c^6) - 2b^8 - 2c^8 +3b^6c^2 + 3b^2c^6 - 6b^4c^4] : :

This point lies on lines 5,1117 30,74 125,477 (at least).

Best regards,
Randy Hutson

 

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#1044

 

Dear Bernard, Peter, Jean-Pierre, Francisco Javier, and Randy,

Thank you very much for your perfect answers and generalizations to the new year puzzle. I particularly like
the description of the perspector

Y = (1/(a^2S_A-2S_{BC})(S^2-3S_{AA})) : ... : ...

as the Cevapoint of the Fermat points.

Best regards
Sincerely
Paul Yiu