>

drawing "by

> Consider also to invest into a geometry software instead of

> eye".

Lineal.)

> (there are many free software, one of the best is IMHO Zirkel und

>

THX for the tip. DLed.

Can someone check the following? It doesn't work with

actual test values, and I must have made a stupid error

somewhere.

Call the triangle abc. Draw isosceles triangles

buc,cva and awb and call cot(angle(buc)/2)=u for short.

Assume AU,BV,CW meet in O, make this the origin and

use vector algebra. (A is the vector to a etc.)

Then U=A*j, and since (U-(B+C)/2)(B-C)=0, 2j=(BB-CC)/(AB-AC).

||U-(B+C)/2||/||(B-C)/2||=cos(buc/2)/sin(buc/2)=u.

Thus (2j)^2*AA+BB+CC+2*BC-2*(2j)*(AB+AC)=a^2*u.

(plus 2 "rotated" equations" with v,w)

Also (A-B)^2=AA+BB-2*AB=c^2 (+2 rotated)

Finally (AxB).C=0. Eliminate the scalar products to get

the master equation in u,v,w.

Which is a complete mess, thus it's better to use the

u,v,w you want to test and NSolve[] the mess instead :-)

(So far checking with the circumcenter, where the angles

are pi-@,... failed. I try again.)

Hauke- Dear Hauke, Dear Nikolaos,

In addition to the solutions (and notation) of Nikolaos here are some

extra solutions though not based on isosceles triangles but on

similar triangles.

if s = D/2 then P = X(1)

if s = D then P = X(6)

(D = opposite angle of side on triangle)

if s = E then P = X(4)

if s = 2pi/3 - E then P = X(15)

if s = -2pi/3 + E then P = X(16)

if s = pi - 2E or pi/2 - E then P = X(3)

(E = adjacent angle of side on triangle)

Further, in general can be stated that:

if s = F then P = Q

(F = corresponding angle in pedal triagle of isogonal conjugate of Q)

This means that there is a solution for any point (just a center) by

taking the isogonal conjugate, then its pedal triangle, then by

rising the angles of the pedal triangle on relevant sides.

I hope this is helpfull.

Regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades

<ndergiades@...> wrote:>

(spam); Î¤Î¿ Yahoo! Mail

> Dear Hauke,

> If I understood your problem we have the following:

> Let 2p, 2q, 2r be the peak angles of the isosceles

> triangles A'BC, B'CA, C'AB outside ABC and

> x = cot(p), y = cot(q), z = cot(r)

> X = cot(A), Y = cot(B), Z = cot(C)

> If AA' meets BC at A" then

> BA"/A"C = (ABA')/(AA'C)

> = AB.BA'.sin(B+90-p)/AC.CA'.sin(C+90-p)

> =sinC.cos(p-B)/sinB.cos(p-C)

> =(cotp.cotB+1)/(cotp.cotC+1)

> =(xY+1)/(xZ+1)

> From Ceva's theorem AA", BB", CC" are concurrent

> iff [(x.Y+1)/(x.Z+1)].[(y.Z+1)/(y.X+1)].[(zX+1)/(zY+1)]=1

>

> An obvious solution is x = y = z or p = q = r = s

> and the perspector P is on the Kiepert hyperbola.

> s = pi/2 then P = X(2)

> s = 0 then P = X(4)

> s = pi/6 then P = X(13)

> s = -pi/6 then P = X(14)

> s = pi/3 then P = X(17)

> s = -pi/3 then P = X(18)

> s = pi/4 then P = X(485)

> s = -pi/4 then P = X(486)

>

> Another partial solution is

> p = arccot(cot(A)/n)

> q = arccot(cot(B)/n)

> r = arccot(cot(C)/n)

> where arccot is the inverse function of cot

> If n = 1 then p = A, q = B, r = C and P = X(5)

> If n = -1 then p = -A, q = -B, r = -C and P = X(3).

>

> Another partial solution is p = (pi-A)/2, q = (pi-B)/2, r = (pi-C)/2

> and P = X(1)

>

> Best regards

> Nikos Dergiades

>

>

>

> > >

> > > Yes. And then the locus of X is Kiepert Hyperbola, by

> > definition.

> > >

> > > For instance.

> > > http://faculty.evansville.edu/ck6/bstud/kiepert.html

> > > http://mathworld.wolfram.com/KiepertHyperbola.html

> > >

> > THX, I would have asked that eventually anyway

> > (my memory - x,x,x works - *didn't* fail me this time

> > :-)

> >

> > This leaves (for ending this thread :-) the question

> > which are the points corresponding to isosceles

> > triangles with peak angles 2@,2Ã,2y resp. pi-2@

> > Would do it myself but my PC is currently in repair.

> > I expect some classic points.

> >

> > Hauke

> >

>

>

>

>

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