12119Thank you, Nikos!
- Feb 13, 2006Dear Nikos, dear all,
First, thank you for your calculation for the hyperbola!!....I'm still trying to work it out this case as the following:
As I told you in my first mail, the envelope of the Lemoine's axis is, again, a conic. Now if we have a triangle ABC and let one of the vertex (say A) move along the circumcircle (with cener in O), we get Lemoine´s ellipse E. This can be seen as a point-conic. The ellipse part is due to Lemoine. This ellipse is tangent to the tangential triangle and to the circumcircle in B and C.
Now, the three polars of the symmedians (having the Lemoine's point in common) are collieneal (in fact they all are in Lemoine's axis l). If one takes the the intersection of the symmedian (by A) with the l (the point T) and takes the polar of this point, one gets the original ellipse E as a line-conic. By the way, this polar passes through the point of intersection of BC and l
Using the same arguments, we can get the envelope of the Lemoine's axis (i.e. a line-conic) as a point conic as the locus of T while A moves in the circumcircle, This line-conic is also tangent to the tengencial triangle and to the circumcircle in B and C.
This conic is an ellipse if the angle between OL and OT is acute, a parabola if Ol and OT are perpendicular, and an hyperbola in the obtuse case. (I still wonder why there is nothing literature about it!)
As for the hyperbola that results of taking the intersection of l with the symmedian by B (or by C), and letting A move. This hyperbola is....(I still have to work out the details about polars and quadrangles) as I said a couple of days ago, this hyerpbolas (and some more of all this conics) tend to grow in number with this approach... harmonics, croos-ratio etc are having fun with me.
Tomorrow at my office (where I have a better connection to intenert) I will try to post this figures
a) just as figures (paint brush) and
b) as cabri II plus files (which will have movement and you can replay the consructions). if you have the program.
Again thanks to Nikos, Francois and to all of you for your replies. As soon as the hyperbola is ready, I'll mail the result and post (if I understood how, of couse) the figures.
And I'm still curiuous about the fact that the circumcenter of the tangential triangle lies in the Euler line of the original triangle....I hope to have this worked out also soon.
Ma. de la Paz
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