>"X is the intersection of BC with parallel to AI through O".

point of concurs is not in ETC.

> Francisco Javier Garcia Capitan has verified my conjecture and the

> This problem has a relation with a difficult one. I think it is now

solved.

> I must now write a little article on it.

Dear Jean-Louis, It's always thrilling when three lines (constructed

> Cela me fait beaucoup de plaisir d'avoir de tes nouvelles.

> Sincerely

> Jean-Louis

in the same way) coincide. I checked also if this point occurs in ETC

and indeed it is not present. I also made a limited scan of lines on

which it occurs. Here are the lines I found: * X1.X159 *

X3.X960.X1158 * X25.X65 * X31.X56.X154.X221 * X40.X197 *

X64.X71.X198.X220 Besides that it occurs on the

ABC.X154.X220-hyperbola. After all no bad result. Best regards,

Chris van Tienhoven

[Non-text portions of this message have been removed]- Hi Chris & Nikos,

Yes, I have all the ETC barcentric coordinates in a Mathematica file.

In this particular case, the way X(188) of the tangential triangle was found is that once we know the barycentric coordinates of the point P wrt ABC, we can then ask what is that P of the intouch triangle, which ends up at X(188), i.e. P wrt ABC is X(188) wrt tangential. This is done firstly numerically against Clark's {6,9,13} triangle, and then using the same formula but this time symbolically. Sometimes this involves square roots, and needs careful pruning is needed to eek out the answer. The formula is a general one that can in principle find some point wrt some triangle. The answer can be somewhat formidable though. Here is the M code I made up ...

XnBaryTriangle[{b1_,b2_,b3_},{{p1_,p2_,p3_},{q1_,q2_,q3_},{r1_,r2_,r3_}}]:=Module[{siderules},siderules={a->Sqrt[(p1+p2+p3)^2 (((-q2-q3) r1+q1 (r2+r3))^2 SA+((-q1-q3) r2+q2 (r1+r3))^2 SB+(q3 (r1+r2)-(q1+q2) r3)^2 SC)],b->Sqrt[(q1+q2+q3)^2 (((p2+p3) r1-p1 (r2+r3))^2 SA+((p1+p3) r2-p2 (r1+r3))^2 SB+(-p3 (r1+r2)+(p1+p2) r3)^2 SC)],c->Sqrt[(r1+r2+r3)^2 (((-p2-p3) q1+p1 (q2+q3))^2 SA+((-p1-p3) q2+p2 (q1+q3))^2 SB+(p3 (q1+q2)-(p1+p2) q3)^2 SC)]};{{p1,q1,r1},{p2,q2,r2},{p3,q3,r3}}.{(b1/.siderules)/(p1+p2+p3),(b2/.siderules)/(q1+q2+q3),(b3/.siderules)/(r1+r2+r3)}]

So if we want say X(9) of the intouch triangle we would do something like ....

XnBaryTriangle[{a (-a+b+c),b (a-b+c),(a+b-c) c},{{0,1/(a-b+c),1/(a+b-c)},{1/(-a+b+c),0,1/(a+b-c)},{1/(-a+b+c),1/(a-b+c),0}}];

This gives a wretched answer .. but we can ask for the coordinates of test triangle like the {6,9,13}. In this case we get {5.0108402139564704446,1.5503054676667462861}. The y-coordinate (First Trilinear), 1.5503..., is not in the ETC SEARCH page, so we are probably out of luck there, i.e. X(9) of the intouch is not X(something in ETC) of ABC. If we now feel brave, we could contemplate carefully whittling down the symbolic answer, for example to

(a+b-c) (Cos[A/2]-Cos[B/2]+Cos[C/2]) Csc[B/2]+(a-b+c) (Cos[A/2]+Cos[B/2]-Cos[C/2]) Csc[C/2] :: .. on lines {1, 166} & {7, 177}

There are times to be careful .. for example suppose we want X(1) of the orthic triangle, we might be tempted by

XnBaryTriangle[{a,b,c},{{0,a^2+b^2-c^2,a^2-b^2+c^2},{a^2+b^2-c^2,0,-a^2+b^2+c^2},{a^2-b^2+c^2,-a^2+b^2+c^2,0}}]

This would lead us to the B vertex. In this case we need to ask for

XnBaryTriangle[{a,b,-c},{{0,a^2+b^2-c^2,a^2-b^2+c^2},{a^2+b^2-c^2,0,-a^2+b^2+c^2},{a^2-b^2+c^2,-a^2+b^2+c^2,0}}] to find X(4).

Clark's TCCT mentions coordinate transfomation on pages 205 - 209, which might also be of interest.

Best regards,

Peter.

----- Original Message -----

From: chris.vantienhoven

To: Hyacinthos@yahoogroups.com

Sent: Friday, February 13, 2009 10:04 AM

Subject: [EMHL] Re: Three concurrent lines

Hi Peter,

> The relationship was found by initially suspecting that it might

be "something" due to the feel of the configuration, then trying it

out numerically and finally confirmed algebraically. X(188) was not

the "something" I expected though. Then I test it on a drawing in

Sketchpad. The algebraic bit is by no means always straight forward.

> The lines are found by a dramatically unsophisticated brute force

search with Mathematica against the whole list of ETC points.

Very interesting!

I am not quite familiar with Mathematica.

Do I understand it right that you have all formulas of all ETC-points

in Mathematica?

Then I can understand that it is possible to calculate the whole

bunch of ETC-points in for example the tangential triangle

with the outcome if they coincide with the point to be examined or

not. Is that right?

I suppose it takes quite of lot of computertime as well for

calculating in the tangential tringle as well as calculating the

lines.

Can you tell me also (if the subject is not too extensive) how you

algebraically confirm your expectation?

I find it unbelievable that you calculate the trilinear-coordinate-

formulas of some point so quick.

Hope to hear soon.

Best regards,

Chris van Tienhoven

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