In a triangle ABC we inscribe a square BaCaC'aB'a, whose the side BaCa lies on BC. If r_a is the circumradius of the triangle AB'aC'a, show that: Rh_1...
723
Lambrou Michael
lambrou@...
Apr 2, 2000 12:42 pm
... That, of course is what I implied. I know Paris long enough... ... Maybe I will, time permiting... Michael...
724
Lambrou Michael
lambrou@...
Apr 2, 2000 1:04 pm
Dear Antreas, ... Panakis never ceases to impress me. Unfortunatelly I don't have his Geometry nor his Trigonometry, but a friend of mine does, so I will ...
725
xpolakis@...
Apr 2, 2000 4:44 pm
Hallar: La formula que da la suma de los cosenos de los tres angulos de un triangulo en funcion de la suma de los radios de los circulos exinscritos y del...
726
Steve Sigur
ssigur@...
Apr 3, 2000 6:49 am
Hello Hyacynthians, Here I continue to attempt to understand via inversive techniques. In Conway's Global View of inversive triangle geometry it is often easy...
727
Clark Kimberling
ck6@...
Apr 3, 2000 2:28 pm
Philippe Deleham recently sent me a letter citing an early appearance of X(384) and X(385): E. Lemoine, rompte rendus de l'Association Francaise pour...
728
Jean-Pierre.EHRMANN
Jean-Pierre.EHRMANN@...
Apr 3, 2000 4:07 pm
... De : "Clark Kimberling" <ck6@...> À : <Hyacinthos@egroups.com> Envoyé : lundi 3 avril 2000 16:31 Objet : [EMHL] names needed X(i), i=384, 385 ...
729
Jean-Pierre.EHRMANN
Jean-Pierre.EHRMANN@...
Apr 3, 2000 4:13 pm
... De : "Clark Kimberling" <ck6@...> À : <Hyacinthos@egroups.com> Envoyé : lundi 3 avril 2000 16:31 Objet : [EMHL] names needed X(i), i=384, 385 ...
730
xpolakis@...
Apr 3, 2000 8:05 pm
Let ABC be a triangle, and A*,B*,C* the points where the angle (int.) bisectors AA', BB', CC' intersect the circumcircle. Let Iab, Iac; Ibc, Iba; Ica, Icb be...
731
Paul Yiu
yiu@...
Apr 3, 2000 8:12 pm
Dear Antreas, The three Lucas circles are tangent internally to the circumcircle, and externally to each other !! The first of these is obvious. The second can...
732
xpolakis@...
Apr 4, 2000 11:54 am
... Let's prove OX + OY + OZ = R + r, but not trigonometrically: Denote OX := x, OY := y, OZ := z. By applying Ptolemy Theorem in the cyclic 4laterals OYAZ,...
733
Lambrou Michael
lambrou@...
Apr 4, 2000 2:03 pm
... For the benefit of those not versed in Greek, let me elaborate on Antreas' shorthand taken from Euclid's Elements (or any other ancient Greek text, for...
734
John Conway
conway@...
Apr 4, 2000 5:29 pm
... The equation OX + OY + OZ = R + ro has interesting consequences by extraversion. Since O is a strong point, the squares of its distances to the edges...
735
John Conway
conway@...
Apr 4, 2000 5:39 pm
... Yes, this does sound as though it's the reason Lucas was interested in them. Let me check up on their definition, and ask a few questions. These circles...
736
John Conway
conway@...
Apr 4, 2000 7:40 pm
I'm glad to hear that the name "Conway points" has now been outmoded. But I believe quite strongly that people shouldn't start calling these points by either...
737
Ignacio Larrosa Ca...
ilarrosa@...
Apr 4, 2000 8:03 pm
... From: <xpolakis@...> To: <hyacinthos@onelist.com> Sent: Tuesday, April 04, 2000 1:54 PM Subject: [EMHL] Re: "Spanish formula" ... un ... exinscritos...
738
xpolakis@...
Apr 4, 2000 9:25 pm
... Note that we can easily prove it by applying two trigonometrical formulae, namely: 1. r = 4Rsin(A/2)sin(B/2)sin(C/2) 2. cosA + cosB + cosC = 1 +...
739
xpolakis@...
Apr 4, 2000 11:23 pm
... ^^^^^^^^^^^^^^^^^^^^^^ See below a synthetical proof of this. ... Theorem (by Bobillier, in: A. de Gergonne, t.XIX, 1827-8, p. 85 and 90). In every...
740
Steve Sigur
ssigur@...
Apr 5, 2000 5:42 am
Hello all, The other day I computed the vertices of a Tucker hexagon and wanted to share these results. A Tucker hexagon is created by starting on a side of...
741
xpolakis@...
Apr 5, 2000 6:54 am
... Acrually, AA', BB', CC' can be any concurrent cevians; not just the angle bisectors. To prove that these circles meet at I of ABC, it is enough to prove ...
742
Paul A. BLAGA
pablaga@...
Apr 5, 2000 2:18 pm
Dear Andreas, Although I don't have yet a proof, it is almost certain that the answer at the problem 2 is YES. I don't know if you are aware, but there is a...
743
Paul Yiu
yiu@...
Apr 5, 2000 3:05 pm
Dear John and Antreas, Denote the circle BXY (diagram below) by Ob(Rb). Since Rb : R = S : bb+S, [S = 2.area of triangle ABC], the center Ob is given by ...
744
John Conway
conway@...
Apr 5, 2000 3:13 pm
... Then came the proof, for which thanks. However, it quoted a fair bit of geometry, and was still quite long. This set me wondering about the old fight...
745
John Conway
conway@...
Apr 5, 2000 3:43 pm
... Then sums Oa = ( aa(2S+SA) : bbSB : ccSC ) 2S(aa+S) Ob = ( aaSA : bb(2S+SB) : ccSC ) 2S(bb+S) Oc = ( aaSA :...
746
John Conway
conway@...
Apr 5, 2000 4:13 pm
I'm now going to extract a bit more from Paul's message. ... This was a mistake - the second term should be divided by 2SS. ... This prompts a general remark,...
747
Paul Yiu
yiu@...
Apr 5, 2000 4:28 pm
John, I think the centers Ox and the desmic mates Qx are different points. Paul ... From: John Conway[SMTP:conway@...] Reply To:...
748
xpolakis@...
Apr 5, 2000 8:45 pm
Is there a triangle ABC in which the three circles with diameters BC, CA, AB are touching [internally] its incircle? If the answer is yes: Is the triangle...
749
Paul Yiu
yiu@...
Apr 5, 2000 10:19 pm
Dear Antreas, I think there is no such triangle. Let D be the midpoint of BC. The power of D with respect to the incircle is (a/2-(s-b))^2 = (a-(c+a-b)^2/4 =...
750
Steve Sigur
ssigur@...
Apr 5, 2000 10:57 pm
... This has two mistakes which affect little of my conclusions about Tucker circles. John's spiffy circle formula is really S0 - (Px +Qy + Rz)(x+y+z) = 0 ...
751
Steve Sigur
ssigur@...
Apr 5, 2000 11:01 pm
If a circle is written in this form w S0 -(Px+Qy+Rz)(x+y+z) = 0 where S0 is the equation of the circumcircle, its center can be written ... Since I had never...
