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## [EMHL] Re: Just a center - as always

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• ... drawing by ... Lineal.) ... 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
Message 1 of 11 , Dec 1, 2008
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>
> Consider also to invest into a geometry software instead of
drawing "by
> eye".
> (there are many free software, one of the best is IMHO Zirkel und
Lineal.)
>
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
• ... UNCLE OF EDIT: Flunked a factor of 2. Now it works with the circumcenter (pi-@,pi-ß,pi-y). (@,ß,y) does NOT work. Double drat. (2@,2ß,2y) as well as
Message 2 of 11 , Dec 1, 2008
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> (So far checking with the circumcenter, where the angles
> are pi-@,... failed. I try again.)
>
UNCLE OF EDIT:
Flunked a factor of 2. Now it works with the circumcenter
(pi-@,pi-ß,pi-y). (@,ß,y) does NOT work. Double drat.
(2@,2ß,2y) as well as (pi-2@,pi-2ß,pi-2y) and any (x,x,x)
seem to work.

Hauke
• Hi Hyacinthians, ... [ That is draw on ABC the isosceles triangles BUC, CVA, AWB, with angles U,VW equal to x (similar isosceles triangles) Then AU, BV, CW
Message 3 of 11 , Dec 1, 2008
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Hi Hyacinthians,

Hauke Reddmann wrote:
>> ... and any (x,x,x) seem to work.
>

[ That is draw on ABC the isosceles triangles BUC, CVA, AWB,
with angles U,VW equal to x (similar isosceles triangles)
Then AU, BV, CW concur in some point X ]

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

etc.

There are many known points on this hyperbola.
Have fun with finding your own point ;-)

Regards.
Philippe.
• ... 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
Message 4 of 11 , Dec 4, 2008
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--- In Hyacinthos@yahoogroups.com, "Philippe" <chephip@...> wrote:
>
> 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@,2ß,2y resp. pi-2@...
Would do it myself but my PC is currently in repair.
I expect some classic points.

Hauke
• 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
Message 5 of 11 , Dec 5, 2008
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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

> >
> > 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@,2ß,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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• Dear Alexey, thank you very much. You sent your message only to me and not to hyacinthos. So I tried to answer privately without success. My message returned.
Message 6 of 11 , Dec 6, 2008
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Dear Alexey,
thank you very much.
You sent your message only to me
and not to hyacinthos. So I tried to answer
privately without success. My message returned.
You are right.
I restate your property as following:
If P, Q, R are three points on Kiepert hyperbola
and p, q, r are the corresponding base angles of
the isosceles triangles giving these points
and x = cot(p), y = cot(q), z = cot(r)
then in barycentrics we have
P = (1/(xS+S_A) : 1/(xS+S_B) : (1/(xS+S_C)
Q = (1/(yS+S_A) : 1/(yS+S_B) : (1/(yS+S_C)
R = (1/(zS+S_A) : 1/(zS+S_B) : (1/(zS+S_C)
The isogonal conjugate of any vertex of triangle PQR
lies on the opposite side
(e.g. the isogonal conjugate P' of P lies on line QR)
iff xy + yz + zx = 1
or x = (1-yz)/(y+z) or cot(p) = cot(-q-r)
or p + q + r = k.pi where we can have k = 0, +-1, +-2
For example if
p = pi/2 then P = X(4) and P' = X(3)
q = pi/3 then Q = X(13) and Q' = X(15)
r = pi/6 then R = X(17) and R' = X(61)
since pi/2 + pi/3 + pi/6 = 1.pi
then by the previous property
X(3) lies on the line X(13)X(17)
X(15) lies on the line X(4)X(17)
X(61) lies on the line X(4)X(13)

Best regards

> Dear colleagues!
> I proved interesting property of Kiepert hyperbola.
> Let $ABC'$, $BCA'$, $CAB'$ be similar isosceles
> triangles with $> \angle C'AB=\phi$. We regard $\phi>0$
> when these triangles are
> outside $ABC$ and $\phi<0$ in the opposite case, so
> $-\pi/2\leq\phi > \leq\pi/2$. Note as $X(\phi)$,
> $X'(\phi)$ the perspective center of
> $ABC$ and $A'B'C'$ and the point of line $OL$
> isogonally conjugated
> to it. Then
> $X(\alpha)$, $X(\beta)$, $X'(\gamma)$ are
> collinear iff $\alpha+\beta+ > \gamma=0$.
> Another interesting property. Triangles $ABC$ and
> $A'B'C'$ are
> orthologic. One of orthology centers is $O$, and stcond is
> $X(\pi/2- > \phi)$, so it also lies on Kiepert hyperdola.
>
> Sincerely Alexey

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• ... Yup, I also dimly remembered there was a classic solution with half-angles. Your arccot formula explains many things, although the cases 2A,2B,2C and
Message 7 of 11 , Dec 7, 2008
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--- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades
>
> 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)
>
Yup, I also dimly remembered there was a classic solution
with half-angles. Your arccot formula explains many things,
although the cases 2A,2B,2C and -2A,-2B,-2C (modulo 2pi)
are still not covered by it.
("Just" computing the coordinates of O for the Kimberling
triangle and looking it up smells like problems - the
double of an obtuse angle... :-)

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
Message 8 of 11 , Dec 8, 2008
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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
>
> 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
>
>
>
> > >
> > > 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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