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Re: [EMHL] Feuerbach
 Dear Randy and Barry!
Yes this is well known. The synthetic proof is in our with Akopjan book "Geometry of conics".
Sincerely Alexey
One generalization:
Let P be any point on the rectangular hyperbola ABCHQ.
Then the cevian circle of P goes through the center of the hyperbola (on the ninepoint circle).
Randy
>________________________________
[Nontext portions of this message have been removed]
> From: Barry Wolk <wolkbarry@...>
>To: "Hyacinthos@yahoogroups.com" <Hyacinthos@yahoogroups.com>
>Sent: Monday, April 30, 2012 5:08 PM
>Subject: [EMHL] Feuerbach
>
>
>
>I have a strange result. Let P be any point on the conic through ABCHI.
>Then the cevian circle of P goes through the Feuerbach point.
>
>Generalizations?
>
>Barry Wolk
>
>
>
>
[Nontext portions of this message have been removed]  APH[ 1 = reflection of I in F, 2 = reflection of F in I, 3 = reflection of 1 in 4]are concurrent.4. A",B",C", resp. (where A", B", C' = the orth. proj. of Ia,Ib,Ic on BC,CA,AB,resp.)3. Ia, Ib, Ic, resp. (where IaIbIc = excentral triangle)2. A', B', C' ,resp. (where A'B'C' = pedal triangle of I)1. A,B,C, resp.The parallels to L1, L2, L3 through:Now, denote: L1, L2, L3 = the Euler lines of AAbAc, BBcBa, CCaCb, resp.Let ABC be a triangle and A'B'C' the cevian triangle of I.APH
Ab := the Orthogonal projection of A on BB'
Ac := the Orthogonal projection of A on CC'
Similarly Bc,Ba; Ca,Cb
The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F,
the Feuerbach point.
https://groups.yahoo.com/neo/ groups/hyacinthos/ conversations/topics/10485  APH[ 1 = reflection of I in F, 2 = reflection of F in I, 3 = reflection of 1 in 4]are concurrent.4. A",B",C", resp. (where A", B", C' = the orth. proj. of Ia,Ib,Ic on BC,CA,AB,resp.)3. Ia, Ib, Ic, resp. (where IaIbIc = excentral triangle)2. A', B', C' ,resp. (where A'B'C' = pedal triangle of I)1. A,B,C, resp.The parallels to L1, L2, L3 through:Now, denote: L1, L2, L3 = the Euler lines of AAbAc, BBcBa, CCaCb, resp.Let ABC be a triangle and A'B'C' the cevian triangle of I.APH
Ab := the Orthogonal projection of A on BB'
Ac := the Orthogonal projection of A on CC'
Similarly Bc,Ba; Ca,Cb
The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F,
the Feuerbach point.
https://groups.yahoo.com/neo/g roups/hyacinthos/conversations /topics/10485Furthermore:Denote:A'b, A'c = the orthogonal projections of A on the external bisectors of B, C (ie on BIa, CIa, resp.)L'1 = the Euler line of AA'bA'c. Similarly L'2, L'3L'1, L'2, L'3 are concurrent at a point F'(they pass through the exFeuerbach points Fa, Fb, Fc, resp.)
The parallels to L'1, L'2, L'3 through:
1. A,B,C, resp.
2. A', B', C' ,resp. (where A'B'C' = pedal triangle of I)
3. Ia, Ib, Ic, resp. (where IaIbIc = excentral triangle)
4. A",B",C", resp. (where A", B", C' = the orth. proj. of Ia,Ib,Ic on BC,CA,AB,resp.)
are concurrent.
[ 1 = reflection of I in 2, 4 = reflection of F' in 3]APH
  Thanks to Angel for his responses !!Now, let's continue.....Let ABC be a triangle.Denote:A'b, A'c = the orthogonal projections of A on the external bisectors of B, C (ie on BIa, CIa, resp.)L'1 = the Euler line of AA'bA'c. Similarly L'2, L'3L'1, L'2, L'3 are concurrent at a point F'(they pass through the exFeuerbach points Fa, Fb, Fc, resp.)
The parallels to L'1, L'2, L'3 through:
[...]5. A1, B1, C1 (where A1, B1, C1 are the orthogonal projections of the excenters Ia, Ib, Ic, resp. on the sides B'C', C'A', A'B' of the pedal triangle of I)6. A2, B2, C2 (where A2, B2, C2 are the orthogonal projections of the vertices A', B', C', resp. of the pedal triangle of I on the sides IbIc, IcIa, IaIb of the excentral triangle)are concurrent (I think !).APH  [APH]:
Let ABC be a triangle.Denote:A'b, A'c = the orthogonal projections of A on the external bisectors of B, C (ie on BIa, CIa, resp.)L'1 = the Euler line of AA'bA'c. Similarly L'2, L'3L'1, L'2, L'3 are concurrent at a point F'(they pass through the exFeuerbach points Fa, Fb, Fc, resp.)
The parallels to L'1, L'2, L'3 through:
[...]5. A1, B1, C1 (where A1, B1, C1 are the orthogonal projections of the excenters Ia, Ib, Ic, resp. on the sides B'C', C'A', A'B' of the pedal triangle of I)6. A2, B2, C2 (where A2, B2, C2 are the orthogonal projections of the vertices A', B', C', resp. of the pedal triangle of I on the sides IbIc, IcIa, IaIb of the excentral triangle)are concurrent (I think !).[Angel Montesdeoca]:5. The parallels to L'1, L'2, L'3 through A1, B1, C1 (where A1, B1, C1 are the orthogonal projections of the excenters Ia, Ib, Ic, resp. on the sides B'C', C'A', A'B' of the pedal triangle of I) are concurrent in
W5 = (r+2R) X(1)  r X(21) = (r+4R) X(7)  (r+2R) X(79).
W5 = (a (a^4 (bc)^2a^5 (b+c)+(b^2c^2)^2 (b^2b c+c^2)a (bc)^2 (b^3+4 b^2 c+4 b c^2+c^3)+a^3 (2 b^3+3 b^2 c+3 b c^2+2 c^3)+a^2 (2 b^4+3 b^3 c+6 b^2 c^2+3 b c^32 c^4)):...:...)
with (6913)search number (1.52086596235072, 1.62091170244161, 1.81655670528601).
W5 is the midpoint of X(i) and X(j) for these {i,j}: {3647,3874}.
W5 lies on lines X(i)X(j) for {i,j}: {1,21}, {7,79}, {20,5441}, {27,1844}, {30,553}, {46,7675}, {57,3651}, {65,4304}, {72,5325}, {142,442}, {226,6841}, {354,946}, {377,5883}, {938,2475}, {1012,5884}, {1100,3284}, {1387,2771}, {1697,8000}, {1699,9960}, {1729,2280}, {2646,5427}, {3085,5686}, {3336,7411}, {3555,5837}, {3584,4015}, {3648,9965}, {3833,4197}, {4313,5903}, {5044,6675}, {5249,6701}, {5273,5904}, {5570,6744}, {5719,10021}, {5735,7671}, {5836,8261}.
6. The parallels to L'1, L'2, L'3 through A2, B2, C2 (where A2, B2, C2 are the orthogonal projections of the vertices A', B', C', resp. of the pedal triangle of I on the sides IbIc, IcIa, IaIb of the excentral triangle) are concurrent in
W6 = (r+3R)X(21)  (r+4R)X(142) = (2r+R) X(35)  (2r+5R) X(79).
W6 = (2 a^7a^6 (b+c)(bc)^4 (b+c)^3+5 a b c (b^2c^2)^2+a^3 (b+c)^2 (2 b^23 b c+2 c^2)a^5 (4 b^2+6 b c+4 c^2)+a^4 (b^34 b^2 c4 b c^2+c^3)+a^2 (bc)^2 (b^3+6 b^2 c+6 b c^2+c^3):...:...)
with (6913)search number (1.13150673357715, 1.02446396734287, 4.87214264402659).W6 is the midpoint of X(i) and X(j) for these {i,j}: {79,1770}.
W6 lies on lines X(i)X(j) for {i,j}: {4, 1768}, {9, 3648}, {21, 142}, {30, 553}, {35, 79}, {63, 2475}, {191, 3474}, {442, 1155}, {516, 3649}, {1708, 7701}, {1836, 5248}, {3874, 7354}, {3911, 6841}, {5122, 10021}, {5325, 6175}, {5441, 5557}
Angel Montesdeoca
  The L1, L2, L3 concur at Feuerbach point of ABC.L1 = the Euler line of A1A2A3. Similarly L2, L3 = the Euler lines of B1B2B3 and C1C2C3 resp.A1, A2, A3 = the orthogonal projections of A on La, Lb, Lc, resp.La = the Euler line of A'AbAc. Similarly Lb, Lc.Ab, Ac = the reflections of A' in BB', CC', resp.Denote:Let ABC be a triangle and A'B'C' the cevian triangle of I.Note:Let A'b, A'c be the orthogonal projections of A on BB', CC' and L'a the Euler line of AA'bA'c.Similarly L'b, L'c.The L'a, L'b, L'c concur at Feuerbach point (*)Now, the triangles AA'bA'c and A1A2A3 share the same Euler line (the one is the reflection of the other in its Euler line). So L'a = L1 and similarly L'b = L2, L'c = L3.Therefore L1, L2, L3 concur at Feuerbach point.APH
 APH(the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')If so we have four concurrent circlesI think the circumcircle of A'B'C' passes through the Feuerbach point.C' = the intersection of AB and the Ibisector of the triangle AIB.B' = the intersection of CA and the Ibisector of the triangle CIAA' = the intersection of BC and the Ibisector of the triangle BICLet ABC be a triangle.Denote:
 [APH]:[Peter Moses]:(the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')If so we have four concurrent circlesI think the circumcircle of A'B'C' passes through the Feuerbach point.C' = the intersection of AB and the Ibisector of the triangle AIB.B' = the intersection of CA and the Ibisector of the triangle CIAA' = the intersection of BC and the Ibisector of the triangle BICLet ABC be a triangle.Denote:Hi Antreas,>I think the circumcircle of A'B'C' passes through the Feuerbach point.
It does !Best regards,Peter.  [APH]:*****************************************************[Peter Moses]:(the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')If so we have four concurrent circlesI think the circumcircle of A'B'C' passes through the Feuerbach point.C' = the intersection of AB and the Ibisector of the triangle AIB.B' = the intersection of CA and the Ibisector of the triangle CIAA' = the intersection of BC and the Ibisector of the triangle BICLet ABC be a triangle.Denote:Hi Antreas,>I think the circumcircle of A'B'C' passes through the Feuerbach point.
It does !Best regards,Peter.Thanks, Peter!Which point is the center of the circle?I guess it is complicated and not listed in ETC.APH
  [APH]:[APH]:[Peter Moses]:(the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')If so we have four concurrent circlesI think the circumcircle of A'B'C' passes through the Feuerbach point.C' = the intersection of AB and the Ibisector of the triangle AIB.B' = the intersection of CA and the Ibisector of the triangle CIAA' = the intersection of BC and the Ibisector of the triangle BICLet ABC be a triangle.Denote:Hi Antreas,>I think the circumcircle of A'B'C' passes through the Feuerbach point.
It does !Best regards,Peter.Thanks, Peter!Which point is the center of the circle?I guess it is complicated and not listed in ETC.[Peter Moses]:Hi Antreas,The triangle A’B’C’ is the cevian of X(174).
Yes, it appears that way. things quicky became very long and I decided not to bother wrestling with 20 pages of square roots!Best regards,Peter.

  [APH]:(the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')If so we have four concurrent circlesI think the circumcircle of A'B'C' passes through the Feuerbach point.C' = the intersection of AB and the Ibisector of the triangle AIB.B' = the intersection of CA and the Ibisector of the triangle CIAA' = the intersection of BC and the Ibisector of the triangle BICLet ABC be a triangle.Denote:[Peter Moses]:The triangle A’B’C’ is the cevian of X(174).
***************
Whose, other than I, X(174), points the cevian circle passes through the Feuerbach point?APH
 [APH]:
Let ABC be a triangle.
Denote:
A' = the intersection of BC and the Ibisector of the triangle BIC
B' = the intersection of CA and the Ibisector of the triangle CIA
C' = the intersection of AB and the Ibisector of the triangle AIB.
I think the circumcircle of A'B'C' passes through the Feuerbach point.
If so we have four concurrent circles
(the NPC, the cevian and pedal circles of I and the circumcircle of A'B'C')
[Peter Moses]:
The triangle A'B'C' is the cevian of X(174).
[APH]:
Whose, other than I, X(174), points the cevian circle passes through the Feuerbach point?
[Angel Montesdeoca]:
The cevian circle of P passes through the Feuerbach point
If P lies on the Feuerbach Hyperbola or on the quartic (Q): a(bc)y^2z^2+b(ca)z^2x^2+c(a b)x^2y^2 =0.
((Q) passes through the vertices (double) of ABC, the vertices of the anticomplementary triangle, X(2), X(7), X(174), X(189), X(366), X(1029), X(5451), X(8046), X(8051))
Triangle center X(i) such that the cevian circle passes through the Feuerbach point, i:
{1, 2, 4, 7, 8, 9, 21, 79, 80, 84, 90, 104, 174, 177, 189, 256, 294, 314, 366, 885, 941, 943, 981, 983, 987, 989, 1000, 1029, 1039, 1041, 1061, 1063, 1156, 1172, 1251, 1320, 1389, 1392, 1476, 1896, 1937, 2298, 2320, 2335, 2344, 2346, 2481, 2648, 2997, 3062, 3065, 3254, 3255, 3296, 3307, 3308, 3427, 3467, 3495, 3551, 3577, 3680, 4180, 4866, 4876, 4900, 5377, 5424, 5451, 5551, 5553, 5555, 5556, 5557, 5558, 5559, 5560, 5561, 5665, 6595, 6596, 6597, 6598, 6599, 6601,
7003, 7049, 7091, 7126, 7133, 7149, 7155, 7160, 7161, 7162, 7261, 7284, 7285, 7317, 7319, 7320, 7595, 7707, 8046, 8051, 8372, 8759, 8809, 9365, 9372, 9442}
Angel Montesdeoca
 APHWe know that the Euler lines of AAbAc, BBcBa, CCaCb concur at the Feuerbach point.Similarly Bc,Ba and Ca, CbIf Ab, Ac are the orthogonal projections of A on IB, IC, then Euler line of AaaAAbAAac = Euler line of AAbAc.L1, L2, L3 concur at the Feuerbach point of ABC.L1 = the Eueler line of AaaAabAac. Similarly L2, L3.Aaa, Aab, Aac = the orthogonal projections of A on INa,INb,INc, resp.Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.Denote:
Let ABC be a triangle.  Ooa, Oob, Ooc = the circumcenters of IObOc, IOcOa, IOaOb, resp.Oa, Ob, Oc = the circumcenters of IB'C', IC'A', IA'B', resp.Let ABC be a triangle, P a point and A'B'C' the pedal triangle of I.Denote:The reflections of A'Ooa, B'Oob, C'Ooc in BC,CA, AB (or equivalently in IA', IB', IC'), resp. are concurrent at the antipode in the incircle of the Feuerbach point.APH
 Another construction of the Feuerbach point:
Let ABC be a triangle and A'B'C' the pedal triangle of I.
Denote:
T = the excentral triangle of A'B'C'
T1, T2, T3 = the pedal triangles of A, B, C wrt triangle T, resp.
The circumcircles of T1, T2, T3 concur at Feuerbach point of ABC.
APH  Another construction of the Feuerbach point:
Let ABC be a triangle and A'B'C' the pedal triangle of I.
Denote:
T = the excentral triangle of A'B'C'
T1, T2, T3 = the pedal triangles of A, B, C wrt triangle T, resp.
The circumcircles of T1, T2, T3 concur at Feuerbach point of ABC.Variation:Let IaIbIc be the excentral triangle of ABC and A', B', C' the orthogonal projections of Ia, Ib, Ic on BC, CA, AB, resp.
T = the excentral triangle of A'B'C'
Let Ta, Tb, Tc be the pedal triangles of A, B, C wrt triangle T.The circumcircles of Ta, Tb, Tc concur at the Feuerbach point of ABC. ABC.
APH  [APH]:Another construction of the Feuerbach point:
Let ABC be a triangle and A'B'C' the pedal triangle of I.
Denote:
T = the excentral triangle of A'B'C'
T1, T2, T3 = the pedal triangles of A, B, C wrt triangle T, resp.
The circumcircles of T1, T2, T3 concur at Feuerbach point of ABC.
*****************************Dear Mr Antreas,
I see a general problem
Let P is a point on Feuerbach Hyperbola of a triangle ABC and A'B'C' is the cevian triangle of P wrt ABC.
T = the excentral triangle of A'B'C'
Let Ta, Tb, Tc be the pedal triangles of A, B, C wrt triangle T.
The circumcircles of Ta, Tb, Tc concur at the Feuerbach point of ABC.And more generalLet ABC be a triangle and A'B'C' is the cevian triangle of any P wrt ABC.H is the orthocenter of ABC.T = the excentral triangle of A'B'C'.Let Ta, Tb, Tc be the pedal triangles of A, B, C wrt triangle T.The circumcircles of Ta, Tb, Tc concur at the center of the Hyperbola through A,B,C,H,P.
Best regards,
Tran Quang Hung.  Let ABC be a triangle and A'B'C' the pedal triangles of I.
Denote:A*, B*, C* = the reflections of A', B', C' in AI, BI, CI resp.Ba = B*C' Intersection BI
Ca = C*B' Intersection CISimilarly Cb, Ab, Ac, Bc
The Euler lines of ABaCa, BCbAb, CAcBc concur at Feuerbach point.Note:
It is equivalent to Hyacinthos #10485(for Ba, Ca are the orthogonal projections of A on BI, CI, resp.)
APH