The locus of P, so that the triangles ABC and triangle bounded by the radical axes ((Oa),(O1)),((Ob),(O2)), ((Oc),(O3)) are perspective is the Lucas cubic and...
21595
Antreas
xpolakis
Feb 20, 2013 11:29 am
I am not sure If I have already posted this: Let ABC be a triangle, (N1),(N2),(N3) the reflections of (N)[=NPC] in OA,OB,OC, resp. and A'B'C' the triangle...
21596
Antreas Hatzipolakis
xpolakis
Feb 20, 2013 11:35 am
One more.... HaHbHc = the orthic triangle. (N1) = the circle (Ha, HaN) ie the circle with center Ha and radius HaN =R/2 Similarly (N2),(N3) A'B'C' = the...
21597
Alexey Zaslavsky
zasl@...
Feb 20, 2013 12:04 pm
Dear colleagues! Let a triangle ABC be given. Points A_1, A_2 lie on BC, B_1, B_2 lie on AB, C_1, C_2 lie on AB. Segments A_1B_2, B_1C_2, C_1A_2 are equal,...
21599
Antreas
xpolakis
Feb 20, 2013 9:44 pm
Let A1A2A3A4A5A6 be a regular hexagon and P a point. Name the triangles PA1A2, PA2A3, etc as 1,2,3,4,5,6. The Euler lines of the even triangles (ie 2,4,6) are...
21600
César Lozada
cesar_e_lozada
Feb 21, 2013 4:22 am
1) Let A' be the center of the inverse of the A-excircle w/r to the NPC. Build B', C' cyclically. The lines AA', BB', CC' meet in X=(f(a,b,c) : f(b,c,a):...
21601
Antreas
xpolakis
Feb 21, 2013 9:29 am
One more: Let ABC be an equilateral triangle, P a point, A;B'C' the cevian triangle of P, and A",B",C" the reflections of A,B,C in A',B',C', resp. The Euler...
21602
Alexey Zaslavsky
zasl@...
Feb 21, 2013 9:33 am
Dear colleagues! My previous problem can be reformulated. Consider all pairs of homothetic regular triangles one of which is inscribed in the given triangle...
21603
Nikolaos Dergiades
ndergiades
Feb 21, 2013 5:36 pm
Dear Alexey, I think that the locus is a pair of lines Best regards Nikos Dergiades...
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Vladimir Dubrovsky
vladubr
Feb 21, 2013 8:23 pm
Dear Nikos and Alexei, in fact, there are 6 possible circumscribed triangles for a given inscribed one. For each of them, as my sketch shows, the locus is a ...
21605
rhutson2
Feb 22, 2013 4:41 am
Dear César, Very nice! Some alternate trilinears for your point X: 1 + 2cos(B - C) : : 1 - 4cos^2(B/2 - C/2) : : 3 - 4sin^2(B/2 - C/2) : : Best regards, ...
21606
Alexey Zaslavsky
zasl@...
Feb 22, 2013 5:59 am
Dear Nikos and Vladimir! Of course there are several families of pairs of triangles but I consider the simplest of them. Vertex A_1 of inscribed triangle...
21607
rhutson2
Feb 22, 2013 6:00 am
And some related points: The center of the inverse of the incircle w/r to the NPC is the {X(1),X(5)}-harmonic conjugate of X, with trilinears: 1 - 2cos(B - C)...
21608
rhutson2
Feb 22, 2013 6:07 am
Correction: the 3rd trilinear below should be: 5 - 4sin^2(B/2 - C/2) : : Randy...
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Paul Yiu
yiuatfauedu
Feb 22, 2013 6:41 am
Dear friends, (1) The A-excircle is inverted in the incircle. The center of the inversive image is the point (a(a^2+3(b-c)^2) : b(3a^2+(b-c)^2) :...
21611
Antreas
xpolakis
Feb 22, 2013 11:50 am
Let ABC be a triangle and HaHbHc the orthic triangle. Denote: (Na) = the reflection of the circle (N,NA) in AHa (Nb) = the reflection of the circle (N,NB) in...
21612
Moses, Peter J. C.
peter_mows
Feb 22, 2013 12:23 pm
Hi Antreas, 1) P = X(265) 2) Q = X(52) Best regards, Peter. ... From: Antreas To: Hyacinthos@yahoogroups.com Sent: Friday, February 22, 2013 11:50 AM Subject:...
21613
Antreas
xpolakis
Feb 22, 2013 1:02 pm
Dear Peter Thanks! I am leaving now, so can't check if the following are true and interesting: Let (Oa), (Ob), (Oc) be the reflections of the circles (N,NA),...
21614
Moses, Peter J. C.
peter_mows
Feb 22, 2013 3:27 pm
Hi Antreas, a) X(3) b) No. c) X(52) d) X(52) also OaObOc is perspective to the circumorthic and Carnot triangles at X(110) and to the anticomplementary...
21615
Antreas
xpolakis
Feb 22, 2013 9:19 pm
Let ABC be a triangle and A'B'C' the circumcevian triangle of I. The circle (I,IA') intersects again the circumcircle at A", the (I,IB') at B" and the (I,IC')...
21616
Angel
amontes1949
Feb 23, 2013 12:52 am
Dear Antreas, 1. ABC, A"B"C" are perspective on OI line. Perspector: X(36) 2. Antipedal triangle of I (excentral triangle),A"B"C" are perspective. The first...
21617
Antreas
xpolakis
Feb 23, 2013 1:49 am
Let ABC be a triangle and (Kab), (Kac) the two congruent and tangent circles such that (Kab) is tangent to the sides of the internal angle B and (Kac) to the...
21618
Nikolaos Dergiades
ndergiades
Feb 23, 2013 2:20 pm
Dear Andreas, Do you have a construction of circles (Kabb), (Kacc)? If A1 is the mid point of (Kab), (Kac) then the triangles ABC, A1B1C1 are perspective at...
21619
Antreas Hatzipolakis
xpolakis
Feb 23, 2013 6:15 pm
Dear Nikos, A pure geometric construction of (Kabb), (Kacc) would be quite difficult, since even the construction for lines instead of circles (Kab), (Kac) is...
21621
Antreas Hatzipolakis
xpolakis
Feb 23, 2013 11:13 pm
Let ABC be a triangle and 12, 13 the tangents to the NPC, perpendicular to BC, near to B,C, resp. Similarly the tangents 23,21 and 31,32. The 6 real...
21622
rhutson2
Feb 24, 2013 2:26 am
Let A'B'C' be the medial triangle and A"B"C" be the orthic triangle. (So A', A" are the intersections of the NPC with BC, etc.) Let Ab be the {A',A"}-Harmonic...
21623
César Lozada
cesar_e_lozada
Feb 24, 2013 2:56 am
Answers: X(1) and X(10) Regards César Lozada _____ De: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] En nombre de Antreas Hatzipolakis ...
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Francisco Javier
garciacapitan
Feb 24, 2013 8:14 am
Dear Randy, you can take A'B'C' the medial triangle and A"B"C" the cevian triangle of any point (p:q:r) and this gives a conic with center {(p + q) (p + r) (10...
21625
Jean-Louis Ayme
jeanlouisayme
Feb 24, 2013 1:48 pm
Dear Jean-Pierre and Mathlinkers, working on a generalization of a Nagel's result, I found in the archive of Hyacinthos the message # 19640 Dear Chris [JP] ...
21626
Barry Wolk
wolkbarry
Feb 24, 2013 10:11 pm
... There were several replies, including asking for cyclically symmetric coordinates for those intersection points. The points on the line Px+Qy+Rz=0 can...
